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Dunagan's Commentary
Dunagan's Commentary on the Bible
Wells of Living Water
McGarvey'S Commentaries
Hampton's Commentary
THE BOOK OF ACTS | CHAPTER 4
OUTLINE AND COMMENTARY - MARK DUNAGAN
I. OUTLINE OF CHAPTER FOUR:
I. Peter And John Arrested: 4:1-5
II. Trial Before The Sanhedrin: 4:6-22
III. Prayer For Boldness: 4:23-31
IV. Caring For Christians In Need: 4:32-37
II. INTRODUCTORY COMMENTS:
In this chapter the Church experiences its first run-in with the Jewish religious authorities. Chapter 4 continues the events of chapter 3, for the temple guard intervenes as Peter and John are still preaching (4:1). We should note that it was the "content" of the apostles' preaching that upset the Jewish rulers (4:2), more than the healing of the lame man. We also should be reminded that the gospel message will always offend someone ( 1Co_1:21 ). 'Such a crowd gathered round Peter and John while they spoke thus in Solomon's colonnade that the temple authorities intervened.' (Bruce p. 95)
III. COMMENTARY ON CHAPTER FOUR:
Act_4:1 And as they spake unto the people, the priests and the captain of the temple and the Sadducees came upon them,
'AND AS THEY'-Apparently both Peter and John were preaching.
'THE PRIESTS'-'The priests were divided into twenty-four courses, each of which served a week in the temple; the priests mentioned here are probably those on duty for that week.' (Boles p. 63)
'CAPTAIN OF THE TEMPLE'-'This is the official referred to in rabbinical literature as the chief "sagan"..He belonged to one of the chief-priestly families, and in the temple he ranked next to the high priest. The temple guard which he commanded was a picked body of Levites...was responsible for maintaining order in the temple courts..' (Bruce p. 95) ( Act_5:24-25 ) 'These Levitical soldiers acted as sentinels at night in the temple area, and preserved peace during the day..in New Testament times numbered 400 men..One of the duties of the temple guard was to preserve the orderly, worshipful atmosphere in the temple..' (Reese pp. 171-172)
'SADDUCEES'-This is the "party" to which the chief-priestly families belonged. In contrast to the Pharisees, who were often from the middle-class, the Sadducees derived its members from the Jewish aristocracy, and was virtually a closed society. While the Pharisees laid great stress on the traditions of the elders, the Sadducees claimed to limit their beliefs to the doctrines they could find in the first five books of the O.T. They denied the existence of the soul, angels, spirits and the resurrection ( Act_23:8 ). Stott notes, 'They were the ruling class of wealthy aristocrats. Politically, they ingratiated themselves with the Romans, and followed a policy of collaboration, so that they feared the subversive implications of the apostles' teaching. Theologically, they believed that the Messianic age had begun in the Maccabean period; so they were not looking for the Messiah.' (p. 95)
'a small but powerful party of the priestly nobles who were supported by the temple dues, and had come to regard religion as a matter of profitable living rather than a service to God. They disliked any popular movement which might disturb the steady accumulation of temple revenues..' (Boles p. 64) See Joh_11:47-50 .
'CAME UPON THEM'-'Burst upon them suddenly or stood by them in a hostile attitude.' (Robertson p. 49). While the Pharisees controlled the synagogues, the Temple constituted the Sadducees "turf".
Act_4:2 being sore troubled because they taught the people, and proclaimed in Jesus the resurrection from the dead.
'SORE TROUBLED'-'vexed through and through' (Vincent p. 464). 'much annoyed' (TCNT); 'exasperated' (NEB).
'BECAUSE..PROCLAIMED IN JESUS THE RESURRECTION FROM THE DEAD'-They opposed not only the resurrection of Jesus, but the very idea of anyone being raised. 'They thus saw the apostles as...both disturbers of the peace and enemies of the truth.' (Stott p. 95)
Act_4:3 And they laid hands on them, and put them in ward unto the morrow: for it was now eventide.
'PUT THEM'-Peter, John and possibly the lame man. For when the trial begins the next morning, the lame man is present (4:14).
'IN WARD UNTO THE MORROW: FOR IT WAS NOW EVENTIDE'-The lame man had been healed around 3:00 p.m. (3:3). The Jewish "evening" started around 6:00 p.m. Apparently, Peter and John had preached for two or three hours after the man was healed. 'and there was no time to hold an inquiry into the apostle's conduct before sundown. They were locked up for the night.' (Bruce p. 96) 'It is one of the great examples of how a party of men, in order to retain their vested interests, would not themselves listen to the truth or give anyone else a chance to hear it.' (Barclay p. 34)
Point to Note:
Erdman reminds us, 'It should be observed that skeptics like the Sadducees have always been the most bitter enemies of Christ..' (p. 48) Remember, the Sadducees constituted the "liberal" or left-wing of First Century Judaism. Far from being "tolerant", they are very intolerant.
The following comment should remind us that those who claim to be tolerant in our own generation, are often the best example of intolerance: 'We must never lose sight of the fact that liberals claim a right for themselves which they never wish to grant to others. Once liberals have taken over a school, or a congregation, have you ever heard them inviting someone to teach or to speak from a conservative view...How many pleas would you expect to hear then, saying, please, please, elders...and editors, let the conservative voices be heard?...After the liberals are in control, the conservatives are then regarded as troublemakers and fanatics...These pleas for tolerance are made only in behalf of those who want to change the church..' 1
Act_4:4 But many of them that heard the word believed; and the number of the men came to be about five thousand.
'NUMBER OF MEN'-Not including the women. 'The increase since the day of Pentecost must have been very rapid, for doubtless many of those baptized then must have departed to their distant homes, and still the increase had been more than two thousand, without counting women.' (McGarvey p. 68)
'Luke assures his readers immediately that the opposition of men did not hinder the Word of God. The Sadducees could arrest the apostles, but not the gospel.' (Stott p. 96) ( 2Ti_2:9 'but the word of God is not imprisoned.')
Act_4:5 And it came to pass on the morrow, that their rulers and elders and scribes were gathered together in Jerusalem;
'ON THE MORROW'-'The meetings of the Sanhedrin customarily began at 10 a.m.' (Reese p. 174)
'RULERS'-'These three groups of officers made up the body called the Sanhedrin. "Sanhedrin" comes from the Greek sun and hedra, which means "to sit together". The Sanhedrin was the high court of the Jewish people. It was composed of 72 members (authorities vary from 70 to 72..). It would have its counterpart in our American society in the Supreme Court. So, these men were before the supreme court of the Jewish people...Rulers would be the chief priests, the heads of the 24 courses. The elders were the 24 elderly men picked by the people to sit on the court. These would be men of age, influence, and position. The scribes...were the learned men, the men skilled and familiar with the Law because they worked with it constantly, making copies of it. They kept the records of the courts of justice and the family registers in the synagogues, wrote articles of contract and sale, and bills of divorce...They were not, however, a religious sect, but might be either Pharisee or Sadducee by belief.' (Reese pp. 174-175)
We should remember that during Roman times the authority of the Sanhedrin was limited, in that it did not have the authority to execute the death penalty. The occupying Roman government retained this right.
Act_4:6 and Annas the high priest was there, and Caiaphas, and John, and Alexander, and as many as were of the kindred of the high priest.
'ANNAS THE HIGH PRIEST'-That is, the legal high priest according to family succession. Although the Romans had replaced him in A.D. 15, he still retained among the Jewish people the prestige, influence and title ( Luk_3:2 ). 'Though he was no longer recognized by the Romans as high priest and not allowed to officiate in the temple, he still kept the presidency of the Sanhedrin.' (Reese p. 175)
Barclay notes, 'In the great days the High Priesthood had been hereditary and it had been for life; but in the Roman times the office of High Priest was the subject of intrigue, bribery and corruption and High Priests rose and fell so that between 37 B.C. and 67 A.D. there were no fewer than 28 High Priests. But even after a High Priest had been deposed he often remained the power behind the throne. Second, although the High Priesthood has ceased to be hereditary it was still the prerogative of a very few families. Of the 28 High Priests already mentioned all but 6 came from 4 priestly families..' (pp. 35-36)
'CAIAPHAS'-The son-in-law of Annas ( Joh_18:13 ), who had advised the Jewish rulers to put an end to Jesus (18:14; 11:49-51). He was appointed to office in A.D. 18 and will be removed during the passover season in A.D. 36. Thus we know that Acts chapter 4 happens prior to A.D. 36.
'JOHN AND ALEXANDER'-apparently two prominent members of the chief-priestly families. Since Luke mentions these two individuals, they must have been prominent individuals in Jewish circles. Some suggest, 'that these were the members of the family of the high priest whose ancestors had lately enjoyed the high priesthood.' (Boles p. 5) Bruce notes, that John may refer to the Jonathan, son of Annas who became High Priest in A.D. 36.
'AND AS MANY AS WERE OF THE KINDRED OF THE HIGH PRIEST'-'and all the High Priest's relations' (TCNT); 'other members of the High Priest's family' (Wey). 'This might have included some ex-high priests (there were three in office between Annas' high priesthood, and that of Caiaphas).' (Reese p. 176)
'the chief-priestly, Sadducean element in its membership was specially well represented..Only a few weeks (?) had passed since they had both taken a hand in the condemnation of Jesus ( Joh_18:13-24 ). Their hope that they had got rid of Him was but short-lived: it looked as if they were going to have even more trouble on His account than they had had before His crucifixion.' (Bruce p. 98)
'memories of the trial of Jesus must have flooded the apostles' minds. Was history to repeat itself? They could hardly have expected justice from that court, which had listened to false witnesses and unjustly condemned their Lord. Were they to suffer the same fate? Would they too be handed over to the Romans and crucified?' (Stott p. 96)
Act_4:7 And when they had set them in the midst, they inquired, By what power, or in what name, have ye done this?
'SET THEM IN THE MIDST'-The Sanhedrin sat customarily in a semi-circle.
'BY WHAT POWER, OR IN WHAT NAME'-'By what authority, in whose name' (Mof); 'what power on whose authority' (Rieu). They had asked Jesus the same type of question ( Mat_21:23 ). 'They ask as if they would accuse them of referring to some magical name or formula for the performance of the miracles.' (Gr. Ex. N.T. p. 125) 'the chief priests cunningly framed an indefinite question in hope that the defendants, in their confusion, would furnish a ground of accusation...It was You, you apostles who did it. Where did the power come from? It was demonic, wasn't it?' (Reese p. 176)
'YE'-'There is a scornful emphasis in the position of the pronoun.."People like you"' (Bruce p. 99)
PETER'S SERMON:
Act_4:8 Then Peter, filled with the Holy Spirit, said unto them, Ye rulers of the people, and elders,
'FILLED WITH THE HOLY SPIRIT'-Jesus had promised the apostles inspiration in the face of their enemies ( Mat_10:17-20 ; Mar_13:11 ; Luk_12:12 ; Luk_21:14-15 .)
'YE RULERS OF THE PEOPLE, AND ELDERS'-'Christianity demands that we be respectful even to our enemies..' (Boles p. 66) ( 1Pe_2:17 )
'Now, when we read the speech of Peter, we must remember...It was spoken to an audience of the wealthiest, the most intellectual and the most powerful in the land, and yet Peter, the Galilean fisherman, stands before them rather as their judge than as their victim.' (Barclay p. 36)
The question posed to Peter and John was intentionally vague, for they hadn't done anything wrong. This vagueness now starts working against the Sanhedrin, for it gives Peter an open forum to preach the Gospel to these men.
Act_4:9 if we this day are examined concerning a good deed done to an impotent man, by what means this man is made whole;
'ARE EXAMINED CONCERNING A GOOD DEED DONE TO A IMPOTENT MAN'-Notice the touch of sarcasm. Right off the back, God turns the tables and puts the Jewish rulers on the defensive. What could be objectionable about healing a lame man? What kind of law jails people for doing good deeds? 'Surely these honorable judges will not object to the apostles doing a good deed to help a sick man!...This is a tacit condemnation of the unrighteousness of the course the Sanhedrin was following.' (Reese p. 177)
'BY WHAT MEANS THIS MAN IS MADE WHOLE'-Nobody could deny the miracle, for the man was "standing" (4:14) in the courtroom and was clearly "well". 'as to how this man has been cured'.
Act_4:10 be it known unto you all, and to all the people of Israel, that in the name of Jesus Christ of Nazareth, whom ye crucified, whom God raised from the dead, even in him doth this man stand here before you whole.
'BE IT KNOW UNTO YOU ALL, AND TO ALL THE PEOPLE OF ISRAEL'-'Peter wants everyone to know the source of the miracle...' (Reese p. 178)
'JESUS CHRIST OF NAZARETH'-the very same Jesus they had crucified.
'WHOM YE CRUCIFIED'-'Instead of the Sanhedrin placing Peter and John on trial, these apostles now put the Sanhedrin on trial.' (Boles p. 67) (2:23-24; 3:15)
'DOTH THIS MAN STAND HERE BEFORE YOU WHOLE'-'in perfect health' (Wey). And how could the Sanhedrin contradict such an assertion? The man standing before them was cured, that was a fact! 'To deny the power was divine, would have been absurd in the estimation of all the people; and to have rejected the explanation given by those through whom the power was exerted, would have been not less so.' (McGarvey p. 71)
Act_4:11 He is the stone which was set at nought of you the builders, which was made the head of the corner.
'HE IS THE STONE'-This is a quotation from Psa_118:22 . Jesus Himself had used this Scripture against the Jewish religious leaders ( Mar_12:10 f; Mat_21:42-46 ). Thus some in the Sanhedrin that day had probably heard this Scripture used against them before.
'SET AT NOUGHT OF YOU THE BUILDERS, WHICH WAS MADE THE HEAD OF THE CORNER'-'he puts his judges and accusers in the ridiculous attitude of builders laying the foundation of a house, but rejecting the stone which was cut out for the corner.' (McGarvey p. 72)
'SET AT NOUGHT'-1848. exoutheneo ex-oo-then-eh'-o; a variation of 1847 and meaning the same: -contemptible, despise, least esteemed, set at nought.
-'scorned' (TCNT); 'treated with contempt' (Wey).
Points to Note:
'He had just appealed to their own senses; he now adds the witness of their own prophets'. (P.P. Comm. p. 124)
The stone that was the 'Head of the Corner', was either the capstone, or the cornerstone, which united two walls and on which they rested and were made firm. This stone was of such importance, that nothing else could be built until this stone was laid, 'The entire building must await the cornerstone.' (Reese p. 179)
This verse infers that what the Jewish religious leaders were building--wasn't the same thing that God was building. For Jesus was the cornerstone in God's building ( Eph_2:20 ; 1Pe_2:4-6 ).
Act_4:12 And in none other is there salvation: for neither is there any other name under heaven, that is given among men, wherein we must be saved.
'AND'-Seeing that God only has one "household" in which Jesus is the chief cornerstone.
'IN NONE OTHER'-A very exclusive statement. There isn't even a "second name" by which one can be saved. Jesus taught the same truth ( Joh_8:24 ; Joh_14:6 ). 'By Peter's inspired statement every other major (and minor) world religion, including the Jewish, stands impoverished when it comes to saving a man from his sins..None are valid at all! None put a man right in God's sight.' (Reese p. 180)
'ANY OTHER NAME UNDER HEAVEN'-Inferring that Jesus is the only Savior for all time, all cultures, all races and all nations ( Mar_16:15-16 ).
'WE MUST BE SAVED'-'In the Greek, the "we" is emphatic, since it is the last word in the sentence. It means WE--whether we are priests, elders, scribes, fishermen, or ex-beggars.' (Reese p. 180) This even applied to the members of the Sanhedrin.
'MUST'-If you are going to be saved, then you "must" have Christ.
Unfortunately, a number of modern denominations insist that one can be saved outside of Jesus Christ. They contend that our religious neighbors (Jews, Buddhists, Moslems, etc..) are simply serving the same God. For all practical purposes, such a claim puts one in the same position as the religious leaders who rejected Jesus in the first place. If you can be saved outside of Jesus Christ, then Jesus isn't the chief cornerstone! That is, then you really don't need Him in building a relationship with God.
Act_4:13 Now when they beheld the boldness of Peter and John, and had perceived that they were unlearned and ignorant men, they marvelled; and they took knowledge of them, that they had been with Jesus.
'BOLDNESS'-3954. parrhesia par-rhay-see'-ah; from 3956 and a derivative of 4483; all out-spokenness, i.e. frankness, bluntness, publicity; by implication, assurance: -bold (X -ly, -ness, -ness of speech), confidence, X freely, X openly, X plainly(-ness).
-'telling it all' (Robertson p. 51); 'free or outspokenness' (P.P. Comm. p. 124) 'Instead of answering evasively, or timidly, as was expected of men in their social position..' (McGarvey p. 72)
'HAD PERCEIVED THAT THEY WERE UNLEARNED'-'untrained laymen' (NEB); 'uneducated, common men' (RSV); 'uneducated persons with no advantages' (Gspd). 'Unlettered men without technical training in the professional rabbinical schools.' (Robertson p. 52)
'IGNORANT'-2399. idiotes id-ee-o'-tace; from 2398; a private person, i.e. (by implication) an ignoramus (compare "idiot"): -ignorant, rude, unlearned.
-'Originally, one in a private station, as opposed to one in office or in public affairs. Therefore one without professional knowledge, a layman..common.' (Vincent p. 465) 'laymen with no professional qualifications. The Sanhedrin, as it were, regarded them as men without a college education and with no professional status.' (Barclay p. 37)
'THEY TOOK KNOWLEDGE OF THEM, THAT THEY HAD BEEN WITH JESUS'-'Imperfect...they began to recognize them as men that they had seen with Jesus.' (Robertson p. 52) We know that John was acquainted with some in the High Priestly family ( Joh_18:15 ). This verse may also mean, that what surprised the Sanhedrin, was that these followers of Jesus, who had been so timid, were now so bold.
'He (Jesus) too had sat at the feet of no eminent rabbi, and yet He taught with an authority which they could well remember. People expressed the same surprise about Him..( Joh_7:15 ). None could equal Him in His sure handling of OT scripture...He had supported His teaching by the mighty works which He performed; and now Peter and John had done the same..' (Bruce p. 102)
Notice how truth prevails. All the training, skills, wisdom and knowledge of the Sanhedrin (a collection of the best minds in Judaism, a religious think-tank), stood helpless before God's truth ( Psa_119:97-100 ).
Act_4:14 And seeing the man that was healed standing with them, they could say nothing against it.
'STANDING WITH THEM'-'That the cripple had been cured was evident; he stood before them as a witness to the fact.' (Bruce p. 102) 'We are not told whether the man was a prisoner with the disciples' (Gr. Ex. N.T. p. 129)
'THEY COULD SAY NOTHING AGAINST IT'-'Not one was ready to contradict anything he had said, or to rebuke him for saying it. Their embarrassment was painful.' (McGarvey p. 73)
Act_4:15 But when they had commanded them to go aside out of the council, they conferred among themselves,
'THEY CONFERRED AMONG THEMSELVES'-'The silence was broken by a proposal that the prisoners be withdrawn.' (McGarvey p. 73) Here they are trying to buy time and 'come up with some way of rationalizing the situation and cutting their loses to the minimum.' (Reese p. 183)
'Liberal critics have enjoyed themselves in asking how Luke could have known what went on in the Sanhedrin's confidential discussion...But Paul may have been there..' (Stott p. 98) In addition, God was there! Remember, Luke wrote by inspiration. Others note, 'remember that "a great company of the priests" afterward became obedient to the faith (6:7); and they would not hesitate to confess the villainy of their previous beliefs and actions.' (Reese p. 184)
Act_4:16 saying, What shall we do to these men? for that indeed a notable miracle hath been wrought through them, is manifest to all that dwell in Jerusalem; and we cannot deny it.
'NOTABLE'-1110. gnostos gnoce-tos'; from 1097; well-known: -acquaintance, (which may be) known, notable.
-'unmistakable' (Wms). The same was true concerning the miracles that Jesus performed ( Joh_11:47 ).
'IS MANIFEST TO ALL THAT DWELL IN JERUSALEM'-Thus denying it, would only weaken their credibility in the eyes of the people.
Act_4:17 But that it spread no further among the people, let us threaten them, that they speak henceforth to no man in this name.
'LET US THREATEN THEM'-'It expresses intensity..a most solemn threat' (Reese p. 183)
Act_4:18 And they called them, and charged them not to speak at all nor teach in the name of Jesus.
'It is particularly striking that neither on this nor on any subsequent occasion...did the Sanhedrin take any serious action to disprove the apostles' central affirmation--the resurrection of Jesus. Had it seemed possible to refute them on this point, how readily would the Sanhedrin have seized the opportunity! Had they succeeded, how quickly and completely the new movement would have collapsed! It is plain that the apostles meant a physical resurrection when they said that Jesus was risen; it is equally plain that the rulers understood them in this sense. The body of Jesus had vanished so completely that all the authority they had at their command could not produce it..' (Bruce p. 103)
The above statement should really make us think. For the best Jewish minds, the total religious and intellectual resources of the Jewish nation could not contradict the claims of two former fishermen. Where is the great Jewish argument against the resurrection? Why didn't they even argue the point? Why didn't they present their Scriptures to refute such an idea?
The best argument that this body of intelligent men came up with against the apostles' claim was to threaten them.
Act_4:19 But Peter and John answered and said unto them, Whether it is right in the sight of God to hearken unto you rather than unto God, judge ye:
'WHETHER IT IS RIGHT IN THE SIGHT OF GOD TO HEARKEN UNTO YOU RATHER THAN UNTO GOD, JUDGE YE'-'you must decide' (TCNT). 'If they thought that any heed would be paid to this ban, they were quickly disillusioned' (Bruce p. 103) Notice how God immediately takes all the wind out of their threat. (1) God had obviously healed the lame man. (2) God had worked through the apostles, who healed him in the name of Jesus. (3) They were told to stop speaking in that name. (4) You figure it out! Notice how God goes after the consciences of these judges. "Would you keep silent, if God had healed through you?"
Act_4:20 for we cannot but speak the things which we saw and heard.
'WHICH WE SAW AND HEARD'-'Their message was no carried tale. They knew at first-hand that it was true; and they were so sure of it that they were willing to stake their life upon it.' (Barclay p. 38)
These verses infer: (a) We are obligated to obey the laws of the land and human authorities up to the point that such laws don't violate God's laws ( Act_5:29 ). (b) No human authority is absolute. No human authority has the right to impose laws which violate God's laws. (c) The Sanhedrin was in opposition to the will of God.
Act_4:21 And they, when they had further threatened them, let them go, finding nothing how they might punish them, because of the people; for all men glorified God for that which was done.
'FURTHER THREATENED THEM'-Repeated the same threats, but were completely at a loss for a good argument.
'LET THEM GO'-Not because reconciliation happened, but because the Sanhedrin couldn't find any way to punish these men, especially in view of the fact that they had become very popular with the people.
'BECAUSE OF THE PEOPLE'-These rulers were also concerned about their "popularity". Their "position" was more important to some of them, than "truth" ( Joh_12:42-43 ).
'FOR ALL MEN GLORIFIED GOD FOR THAT WHICH WAS DONE'-'kept on glorifying' (Robertson p. 53)
Act_4:22 For the man was more than forty years old, on whom this miracle of healing was wrought.
'MORE THAN FORTY YEARS OLD'-'made him well known and an object of universal sympathy.' (McGarvey p. 75) 'he had reached an age when such cures simply do not occur.' (Bruce p. 104)
Some claim that the key to church growth in the first century was the fact that they had miraculous powers. But this section of Scripture reminds us that the miracles didn't convince everyone! See also ( Luk_16:30-31 )
BACK WITH THE OTHER APOSTLES:
Act_4:23 And being let go, they came to their own company, and reported all that the chief priests and the elders had said unto them.
'TO THEIR OWN COMPANY'-The fact that the Holy Spirit comes upon this company (4:31), and yet it is still only the apostles that are mentioned as being the workers of the miracles (4:33; 5:12), proves that "their own company" in this verse refers to the other 10 apostles.
Act_4:24 And they, when they heard it, lifted up their voice to God with one accord, and said, O Lord, thou that didst make the heaven and the earth and the sea, and all that in them is:
'WITH ONE ACCORD'-(1:14; 2:46; 5:12; 7:57; 15:25). It could be that one prayed and all others mentally joined in or said Amen. Or, all 12 apostles by inspiration uttered these words at the same time.
'O LORD' despotes des-pot'-ace; perhaps from 1210 and posis (a husband); an absolute ruler ("despot"): -Lord, master.
-'a term used of a slave owner and of ruler of unchallengeable power. The Sanhedrin might utter warnings, threats and prohibitions, and try to silence the church, but their authority was subject to a higher authority still, and the edicts of men cannot overturn the decrees of God.' (Stott p. 99)
-We should also note that the "Lord" in these verse refers to the Father (4:27).
'IT IS THOU WHO DIDST MAKE THE HEAVEN AND THE EARTH'-'First, he is the God of creation' (Stott p. 99). The God revealed in the O.T. is the same as the God revealed in the N.T. ( Exo_20:11 ; Neh_9:6 ; Psa_146:6 ).
Act_4:25 who by the Holy Spirit, by the mouth of our father David thy servant, didst say, Why did the Gentiles rage, And the peoples imagine vain things?
'WHO BY THE HOLY SPIRIT...DIDST SAY'-God is also the God of revelation. Again, we find another claim that the writings of the O.T. were inspired (2:30; 3:18; 2Pe_1:20-21 )
'BY THE MOUTH OF OUR FATHER DAVID'-What will follow is a quotation from Psa_2:1-2 . 'already in the first century B.C. recognized as Messianic (by the Jewish people).' (Stott p. 100) 'This psalm...had been interpreted of the coming deliverer of David's line at least as early as the middle of the first century B.C...' (Bruce p. 106)
'WHY DID THE GENTILES'-'The Gentiles raged against Jesus in the person of the Romans who sentenced Him to the cross.' (Bruce p. 106)
'RAGE'-5433. phruasso froo-as'-so; akin to 1032, 1031; to snort (as a spirited horse), i.e. (figuratively) to make a tumult: -rage.
-'Originally, to neigh or snort like a horse. Of men, to give one's self haughty airs, and to act and speak insolently.' (Vincent p. 466)
'THE PEOPLES'-i.e. the Jewish people.
'IMAGINE VAIN THINGS'-'form futile plans' (Mon); 'cherish vain dreams' (Knox); 'lay their plans in vain' (NEB).
'IMAGINE'-3191. meletao mel-et-ah'-o; from a presumed derivative of 3199; to take care of, i.e. (by implication) revolve in the mind: -imagine, (pre-)meditate.
-'means to ponder meditatively, and to practise like orators in order to perfect their style..The Jewish people used thought, plan, purpose, in opposing the Messiah.' (Reese p. 188) ( Joh_11:53 ).
Act_4:26 The kings of the earth set themselves in array, And the rulers were gathered together, Against the Lord, and against his Anointed:
'AGAINST HIS ANOINTED'-i.e. Jesus
'ANOINTED'-5548. chrio khree'-o; probably akin to 5530 through the idea of contact; to smear or rub with oil, i.e. (by implication) to consecrate to an office or religious service: -anoint.
- Act_10:38 .
Act_4:27 for of a truth in this city against thy holy Servant Jesus, whom thou didst anoint, both Herod and Pontius Pilate, with the Gentiles and the peoples of Israel, were gathered together,
'FOR OF A TRUTH'-The apostles now declare that what had happened at Jerusalem was the fulfillment of Psa_2:1-12 .
'WHO THOU DIDST ANOINT'- Act_10:38 ; Luk_4:18 ; Isa_61:1 .
'BOTH HEROD AND PONTIUS PILATE'-The "kings and rulers" mentioned in the Psalm ( Luk_23:7 ff). 'Luke is the only Gospel writer who records the part taken by Herod (Antipas) in conjunction with Pilate in the condemnation of Jesus.' (Reese p. 189)
'WITH THE GENTILES AND THE PEOPLES OF ISRAEL'-the other groups mentioned in the Psalm, including the soldiers that mocked and crucified Him, and the Jewish council and mob that cried out for Him to be put to death.
Act_4:28 to do whatsoever thy hand and thy council foreordained to come to pass.
'They were gathered together for the purpose of executing their own will, as they thought, but really to fulfill the purpose of God.' (P.P. Comm. p. 125)
Again, the apostles declare that the death of Jesus was the original plan, what the prophets had spoken (2:23; 3:18).
Notice how God's foreordained plan didn't violate the freewill of anyone. Herod, Pilate, the Roman solders, the Jewish leaders and the Jewish mob were all freewilled individuals. God is so wise that He can even bring about His purposes, without forcing anyone to do anything. He can even use the hate, jealousy and stubbornness of men to achieve His plans.
Notice, 'God made', 'God spoke', and 'God decided'.
Thus how foolish and shortsighted to argue that the rebellion of mankind could stop God's purposes.
Act_4:29 And now, Lord, look upon their threatenings: and grant unto thy servants to speak thy word with all boldness,
'LOOK UPON'-Could mean, simply take note of. Or it could involve of ideas of frustrating and hindering the plans of wicked men. 'It is beautiful to notice how, in the heat of the unjust persecutions, the Church hands over her quarrel to her Lord, and is only careful that she be not stopped in her work by the threatenings of her enemies.' (P.P. Comm. p. 125)
'THEIR THREATENINGS'-i.e. the threats just issued by the Sanhedrin.
'GRANT..TO SPEAK THY WORK WITH ALL BOLDNESS'-'with all fearlessness' (TCNT); 'with fearless courage' (Wey); 'confidently' (Knox).
They had already been very bold (4:19-20), but they realized that this was just the beginning of persecution and not the end.
Notice how the apostles didn't pray for obstacles to be removed. But rather for the courage to face them. It is unreasonable for a Christian to expect that God will protect them from opposition ( Act_14:22 ; 2Ti_3:12 ).
The Gospel message deserves to be presented in a confident manner, without apology or embarrassment ( Rom_1:16 ).
The apostles were remembering two very important things: (a) The God they served was all-powerful. (b) All human rebellion is futile. In the end, God always wins, and in the end, His purposes are always fulfilled.
'The Sanhedrin might threaten, but their threats called not for fear and silence but for increased boldness of speech.' (Bruce pp. 106-107)
Act_4:30 while thy stretchest forth thy hand to heal; and that signs and wonders may be done through the name of thy holy Servant Jesus.
'TO HEAL'-'As Alexander pointed out, "their demand is not now for miracles of vengeance or destruction, such as fire from heaven ( Luk_9:54 ), but for miracles of mercy.' (Stott p. 100)
'SIGNS AND WONDERS'-Again we should note that the miracles were for the purpose of confirming the message being spoken through the apostles ( Heb_2:3-4 ).
Act_4:31 And when they had prayed, the place was shaken wherein they were gathered together; and they were all filled with the Holy Spirit, and they spake the word of God with boldness.
God granted them a direct and immediate answer to their prayer. 'as Chrysostom commented, "that made them the more unshaken".' (Stott p. 100)
"Shaken"-see Act_16:25-26 .
'filled with the Holy Spirit'-like that on the Day of Pentecost ( Act_2:4 ).
Notice how this text doesn't fit the theology of the Pentecostal movement. This movement claims that speaking in tongues (devotional, private "prayer tongues") is the evidence of being baptized or filled with the Spirit. But the apostles aren't praying in tongues, rather the result of this filling is that they spoke the Word of God with boldness. In addition, carefully note that the apostles didn't spend hours or days in prayer, before the Spirit filled them.
As to why we have a second "filling", 'Some suggest that the presence of the baptismal power of the Holy Spirit was not an abiding thing--but each man was empowered from time to time as the occasion demanded.' (Reese p. 191) That may or may not be the case.
But this section of Scripture does tell us: Even though these men already had the Holy Spirit, the Spirit didn't overrule their freewill, and neither did He completely control all their thoughts ( 1Co_14:32 ), i.e. the Spirit didn't make them immune from fear or concern.
UNITY AND GENEROSITY OF THE EARLY CHURCH:
Act_4:32 And the multitude of them that believed were of one heart and soul: and not one of them said that aught of the things which he possessed was his own; but they had all things common.
'MULTITUDE OF THEM THAT BELIEVED'-The last count was well over 5000 (4:4). The word multitude also suggests that the "boldness" of the apostles preaching was continuing to attract converts.
'WERE OF ONE HEART AND SOUL'-Again, we are given another glimpse at the internal life of the early church (2:42ff). 'One heart-they felt alike, were interested in the same things. The word represents the intellectual side of their Christian life...One soul--indicates the close and loving union of the early believers. This word speaks of the emotional side of their Christian life.' (Reese p. 192)
Again, we see that unity in a congregation composed of different personalities, is a real possibility. They had unity because they all recognized the same standard-i.e. the apostles teaching (2:42).
'BUT THEY HAD ALL THINGS COMMON'-see comments on Act_2:44-45 . 'The account of the community of goods is repeated here in order to introduce the contrasted episodes of Barnabas and Ananias' (Bruce p. 108)
Act_4:33 And with great power gave the apostles their witness of the resurrection of the Lord Jesus: and great grace was upon them all.
'WITH GREAT POWER'-Indicating that God also granted their request that He would continue to heal through them (4:30). The fact that the apostles are specifically mentioned confirms our belief that those praying in 4:23-31 are only the apostles.
'THEIR WITNESS OF THE RESURRECTION'-The apostles kept right on doing what the Sanhedrin had forbidden them to do.
'GREAT GRACE WAS UPON THEM ALL'-'great favour' (Rhm). Which may indicate that the people held them in favor (2:47; 5:13); or that God's favor rested upon them, or both. God's favor was upon them because they were doing His will. And the people couldn't help but admire the generosity, self-sacrifice and unity which was found among these believers.
Act_4:34 For neither was there among them any that lacked: for as many as were possessors of lands or houses sold them, and brought the prices of the things that were sold,
'LACKED'-1729. endees en-deh-ace'; from a compound of 1722 and 1210 (in the sense of lacking); deficient in: -lacking.
-'there was no poverty among them' (TCNT); 'none of them was destitute' (Knox). 'Some of them were in need, not because they had been idle ( 2Th_3:10 ), neither because they had squandered their possessions, nor yet because they were shiftless...' (Boles p. 76)
'POSSESSORS OF LANDS OR HOUSES'-Carefully note, that they didn't sell all their possessions: According to the tense of the verse, the "selling" happened from time to time, as there was occasion by reason of need. In addition, members still had their own houses ( Act_12:12 ). Plus, remember that this has been going on since chapter 2. If everyone had sold every possession that they had owned in chapter 2, then nothing would be left to sell in this chapter. Reese reminds us, 'Communism says, "What is yours is mine, and I'll take it". Fellowship says, "What is mine is yours; I'll share it!"' (p. 193)
Act_4:35 and laid them at the apostles' feet: and distribution was made unto each, according as any one had need.
'LAID THEM AT THE APOSTLES' FEET'-Who were viewed as those possessing authority in the early church.
'ACCORDING AS ANYONE HAD NEED'-Funds were only given when a need arose. The money wasn't just divided up among all the members, and neither was it the exclusive possession of the apostles. Carefully note, that while poor non-Christians existed, this money collected wasn't used to relieve the needs of those outside the Church. Thus the Church under the guidance of the Apostles never viewed itself as a welfare agency for the community.
Act_4:36 And Joseph, who by the apostles was surnamed Barnabas (which is, being interpreted, Son of exhortation), a Levite, a man of Cyprus by race,
'WHO BY THE APOSTLES WAS SURNAMED'-'who had received from the Apostles the additional name of' (TCNT).
'BARNABAS (WHICH IS, BEING INTERPRETED, SON OF EXHORTATION'-'Son of encouragement' (RSV), 'the man of encouragement' (Knox) ( Act_11:23 )
'A LEVITE'-from the priestly tribe. This man may have been converted when he came to Jerusalem to serve in the Temple.
'OF CYPRUS BY RACE'-Probably born on the island of Cyprus to Jewish parents. 'We know that a great many Jews were settled in Cyprus.' (P.P. Comm. p. 126).
-The Island of Cyprus is located in the northeastern corner of the Mediterranean Sea, about 60 miles off the coast of Syria, and about 41 miles off the coast of Turkey.
Act_4:37 having a field, sold it, and brought the money and laid it at the apostles' feet.
'HAVING A FIELD'-We aren't told whether this field existed in Palestine or on the island of Cyprus. McGarvey notes, 'Inasmuch as the law of Moses made no appropriation of lands for the tribe of Levi, but provided that it should be supported by the tithes from the other tribes, some surprise has been expressed that this Levite was the owner of real estate. But it should be remembered that the original allotment of certain lands to certain tribes, and certain cities to the Levites, had been completely broken up by the Assyrian and Babylonian captivities..This state of things left the Levites to shift for themselves to a great extent, and there was no law, nor had there ever been, to prevent them from acquiring individual landed possessions.' (pp. 81-82)
Dunagan, Mark. "Commentary on Acts 4". "Dunagan's Commentaries on the Bible". https://www.studylight.org/commentaries/eng/dun/acts-4.html. 1999-2014.


I have an obsession with all things Winter. But in Southern California where I'm from, to really enjoy a snowy Winter means a 2-3 hour drive up the mountains, and I would have to get a 4WD. Or worse, chains! So, rather than going through the trouble, I was inspired to organize a Winter themed wedding shoot to satisfy my Winter fantasies.
I've envisioned a Winter wonderland in the woods, filled with the things I love about the season that are lacking where I live – snow, frost, pines, pine cones, and berries. I incorporated these elements into the details, from the bridal hair wreath, bouquet, and boutonniere (all put together myself), to invitation designs and table settings. And what better place to do it in than under a romantic and magical oak tree branching over us?
I am most proud of the Wintery bouquet that I put together. I didn't want a typical or traditional flower bouquet that you see at most weddings or styled shoots. I wanted something unique, adding a bit of winter sparkle. Keeping the elements in mind, I opted for several frosty branches, long berry stems, and a pair of icy blue and silver poinsettias to complete the look.
With the help of my amazing team of event planner, models, and vendors, every detail came together beautifully, bringing my vision to life. This shoot was a dream come true for me and I am beyond excited for an opportunity to share it with you.
Previous Post MDP Epic Wedding Featured on Woman Getting Married!


text:
News > Idaho
New tax deals cost Idaho millions
Lawyer says formal decision on plane sets precedent
Wed., Sept. 15, 2010
By Betsy Z. Russell betsyr@spokesman.com(208) 336-2854
A 2008 report revealed secret deals had become so frequent that corporations routinely protest their state tax bills to get their "Idaho tax break."
Read previous coverage and find the Tax Commission decision at spokesman.com/tags /secret-tax-deals.
BOISE – Idaho stands to lose more than $75 million from tax compromises that are now in the works, according to a group of whistleblowers who are current and former employees of the state Tax Commission.
A $203,000 tax break is about to be handed to a taxpayer on his $7 million private plane.
"This is the worst scandal I've seen in Idaho since I first came here in 1950," declared Robert Huntley, the former Idaho Supreme Court justice who's representing state Rep. Shirley Ringo, D-Moscow, for free in her lawsuit against the state over the controversial deals.
Ringo has offered to put the lawsuit on hold in favor of an immediate state investigation, along with job protection for the current Tax Commission auditors and managers who came forward with sworn statements about the deals. But so far, the state hasn't taken her up on the offer.
"It's our office policy not to comment on pending litigation," said Bob Cooper, spokesman for the Idaho attorney general's office. "What we have to say to the court will be said to the court."
A hearing is scheduled Nov. 4 in Ada County District Court on the state's motion to dismiss the lawsuit, charging that Ringo doesn't have standing to sue as a state legislator, a contention Huntley disputes.
"I think we have just hit the tip of the iceberg on what's gone wrong here," Huntley said. "Most of the employees are terribly afraid to speak."
In 2008, longtime senior state tax auditor Stan Howland sent lawmakers, the governor and the attorney general a 17-page whistleblower report charging that tax commissioners routinely excuse large sums in taxes owed by large, multistate corporations, and use confidentiality laws to prevent anyone from finding out about it. He said the deals have become so frequent that corporations routinely protest their state tax bills to get their "Idaho tax break."
Since then, two state investigations have concluded no laws were broken, but seven more longtime Tax Commission employees have come forward with sworn statements about what they consider other questionable tax deals, including three current commission employees who came forward in the past two weeks. All said reform legislation enacted in 2009 after Howland's revelations didn't fix the problem, and may even have made it worse.
Among the deals they've described in affidavits:
• Paul Chugg, a tax auditor for the commission for the past 28 years and a certified public accountant, said he was threatened with disciplinary action when he objected to waiving $400,000 in penalties for a taxpayer who falsely claimed more than $5 million in Idaho tax credits. The penalties were waived in full. "A few years later, I was subpoenaed to testify by the U.S. Department of Justice at the criminal trial of the taxpayer's former employee who was responsible for the fraudulent claims," Chugg said in his affidavit.
• Three auditors said a large, multistate taxpayer with sales affiliates in Idaho wrongly claimed the affiliates weren't taxable in the state, and the commission decided to settle the case for $1.2 million less than the amount owed, though the auditors said there was "absolutely no legal support for this extremely troubling compromise."
• Former longtime auditor Gary Mattox said a golf club owed more than $700,000 in sales taxes on high-priced membership fees, but the commission forgave the full amount and didn't pursue it.
• Barbara Nichols, manager of multistate income tax audits for the commission and an employee since 1983, said a taxpayer wrongly claimed nonbusiness income, and the commission decided to compromise by allowing 50 percent of that income to be treated that way. "This compromise was made in direct violation of Idaho law and cost the state approximately $680,000," according to her affidavit.
Most of the compromises described in the affidavits are secret deals; until 2009, a single tax commissioner could strike a deal with a taxpayer. The 2009 legislation added some new requirements to that process, though the deals remain secret.
But the deal involving the private plane is even more troubling, Huntley said, because it is a pending formal, published decision and will set precedent. In that case, the owner of the $7 million plane claimed an Idaho investment tax credit on the full value of the plane, though by law, the credit is available only for use of the plane in Idaho. The portion of a plane's value eligible for the credit is determined by the ratio between Idaho departures and out-of-state departures; according to its flight log, the plane had 45 Idaho departures among its 107 total departures.
But the taxpayer argued he should be able to get investment tax credits for the full value of the plane, because many of the out-of-state departures occurred when he'd leased the plane out for use by others, and he received that leasing income in Idaho.
In the proposed decision, the commission agrees, allowing the additional investment tax credit and waiving all penalties, excusing the taxpayer from $202,252 in taxes, penalties and interest.
"The statute and rule involved has been to the Idaho Supreme Court two times, approved both times," Huntley said. "That's slam-dunk."
Published: Sept. 15, 2010, midnight
Tags: Barbara Nichols, Bob Cooper, Gary Mattox, Idaho attorney general's office, idaho supreme court, Idaho Tax Commission, Paul Chugg, robert huntley, secret tax deals, Shirley Ringo, stan howland, tax breaks, tax deals, U.S. Department of Justice
Top stories in Idaho
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Moscow property taxes expected to rise 25 percent next year …
Coeur d'Alene police tight-lipped about 18-year-old's medical condition after officer-involved shooting …
Reports of gunfire prompt SWAT standoff at Coeur d'Alene apartment; no one injured …
Eye on Boise: North Idaho lawmaker continues pushing federal appeal to have guns …
Labrador narrowly elected chairman of Idaho Republican Party …

redpajama_set_name: RedPajamaCommonCrawl


text:
\section{Introduction} \label{section1}
\setcounter{equation}{0}

Let $\Omega$ be the multiply connected domain in the extended complex plane $\overline{{\mathbb C}}={\mathbb C}\cup\{\infty\}$ obtained by removing $m$ non-overlapping rectilinear segments from the infinite strip $S=\{z\,:\, \left|\Im z\right|<\pi/2\}$.
Solving boundary value problems in such a domain with complicated boundaries is not as easy as it is for domains with smooth boundaries.
A possible remedy is to find a conformal mapping from $\Omega$ onto a multiply connected domain $G$ bordered by smooth Jordan curves. However, up to our knowledge, there is no analytical or numerical method in the literature to compute such a conformal mapping.

The above domain $\Omega$ is one of the canonical domains for conformal mapping of multiply connected domains~\cite[p.~128]{wen}.
An efficient method for numerical computation of the conformal mapping $\Phi$ from domains with smooth boundaries $G$ onto the canonical domain $\Omega$ is presented in~\cite{NF13}. Still, in this method, the domain $G$ is supposed given while $\Omega$ is unknown and should be computed alongside the conformal mapping $\Phi$ from $G$ onto $\Omega$.
On the contrary, in this paper, we take up the case when $\Omega$ is known. Our objective will be then to find an unknown preimage domain $G$ bordered by smooth Jordan curves, and to determine a conformal mapping $\Phi$ from $G$ onto $\Omega$. This means that the method presented in~\cite{NF13} is not directly applicable in our context, but this method is still useful to develop an iterative scheme for finding the unknown preimage domain $G$ as well as the conformal mapping $\Phi$.
In this way, the inverse mapping $\Phi^{-1}$ would be the desired conformal mapping from the domain $\Omega$ onto the domain $G$.
It is known that Laplace equation in the plane is invariant under conformal mappings. Thus, with the help of the inverse conformal mapping $\Phi^{-1}$, boundary value problems for the Laplace equation in the domain $\Omega$ will be solved easily in the domain $G$.

The idea of our iterative method is similar in spirit to the approach employed in~\cite{AST13,N18,NG18} but for other canonical domains. The proposed numerical method will be useful in solving several problems in strip with rectilinear slit domains. Applications of the proposed method to two types of such problems will be considered in this paper.

As a main illustration of how useful is the developed numerical conformal mapping method of this paper, we study a domain functional, the conformal capacity of a condenser.
Recall that a \emph{condenser} is a pair $(D,E)$ where $ D \subset
{\mathbb C}$ is a domain and $E \subset D$ is a non-empty compact set, and the
\emph{conformal capacity} is defined as follows
\begin{align}\label{def_condensercap}
{\rm cap}(D,E)=\inf_{u\in \mathcal{A}}\int_{D}|\nabla u|^2 \, dm,
\end{align}
where $dm$ stands for the 2-dimensional Lebesgue measure and $\mathcal{A}$ is the family of all harmonic functions in $D$ with values in $[1,\infty)$ on $E$ and approaching $0$ on $\partial D$.
If $D\setminus E$ is a multiply connected domain such that all of its boundary components are smooth Jordan curves, then the infimum in \eqref{def_condensercap} is known to be attained by a harmonic function \cite[p. 65]{ah}. In fact,
this extremal function is the solution of the
Laplace equation with boundary values equal to $1$ on $E$ and vanishing on $\partial D$. The conformal capacity, which is a mathematical model
of the capacity of a physical condenser, has numerous applications also
to potential theory and to geometric function theory \cite{du, hkv, sol}.

As is generally known, the set function $E \mapsto {\rm cap}(D,E)$
looks similar to an outer measure \cite[Lemma 7.1, Theorem 9.6]{hkv}.
For instance, the subadditivity property of the condenser capacity
for compact sets $E_j \subset D, j=1,...,m,$ says that
\begin{equation} \label{subadd}
{\rm cap} (D, \cup_{j=1}^m E_j)\le \sum_{j=1}^m {\rm cap} (D,E_j).
\end{equation}
This is reminiscent of the subadditivity
of  Lebesgue measurable sets $E_j \subset {\mathbb R},
j=1,2,...,$
\[
{\rm m}( \cup_{j=1}^{\infty}{E_j}) \le \sum_{j=1}^{\infty}{\rm m}({E_j})
\]
with equality for separate sets. For the capacity case \eqref{subadd} the
situation is different, no equality statement is known for separate sets. Moreover, even in the
case $m=2$ it is not easy to give non-trivial examples of sets with strict
inequality in \eqref{subadd}.

Our numerical computation leads
to two novel observations. First, we
will show here that an asymptotic equality holds,
the lower bound in \eqref{subadd} will be arbitrarily close to the upper
bound for some sets, far away from each other. This asymptotic equality is
a  manifestation of some kind of  ``\emph{weak additivity}'' for such sets. Second,
our computational experiments have revealed an inequality for the capacity of condensers in the strip domain $S.$ More precisely, if the condenser is of the form
$(S,E)$ and $E= E_1 \cup E_2 \subset {\mathbb R}$, $E_1$ and $E_2$ are segments, then
\begin{equation}\label{eq:ineq}
 {\rm cap}(S,E)\ge {\rm cap}(S,H)
\end{equation}
where $H \subset {\mathbb R}$ is a segment with diameter equal to the sum
of diameters of $E_1$ and $E_2.$ Due to the conformal invariance of the capacity,
the inequality \eqref{eq:ineq} admits various extensions to simply connected plane
domains, for instance to the case when the strip domain $S$ is replaced by the
unit disk. In that case we have to use the hyperbolic metric. The case of multiply
connected domains seems to offer problems for further research.

In the second application, we consider computing the complex potential for a uniform inviscid and incompressible flow past multiple disjoint segment obstacles in the strip $S$ in the case when the circulations around the segments are zeros.
A numerical method for approximating such complex potentials when the obstacles have smooth Jordan curves boundaries has been presented in~\cite{sak12} where the domain $\Omega$ was called a channel domain.
The idea used in~\cite{sak12} is based on constructing a conformal mapping $w=F(z)$ from $\Omega$ onto the domain obtained by removing horizontal slits from the strip $S$. Then, $W(z)=F(z)$ represents a complex potential for the uniform flow in $\Omega$. The streamlines for a uniform flow in $\Omega$ are then the contour plots of the imaginary part of the complex potential $W(z)$.

In this paper, the geometry of the domain $\Omega$ is more complicated as the obstacles are slits. However, the method used here is similar to the method presented in~\cite{sak12} and hinges on the numerical computation of the conformal mapping $w=F(z)$ from $\Omega$ onto a domain $H$ obtained by removing horizontal slits from the strip $S$. To compute such a conformal mapping, we first apply the proposed iterative method to compute a conformal mapping from $\Omega$ to a domain $G$ bordered by smooth Jordan curves, and then use the method of~\cite{NF13} to conformally map $G$ onto the domain $H$ obtained by removing horizontal slits from the strip $S$.

Further applications of the proposed iterative method are possible. For example, with the help of the iterative method, one can extend the method presented in~\cite{NasVla} for simulating local fields in carbon nanotube reinforced composites for infinite strip with circular voids and elliptic CNTs to the case of slit CNTs. Another example is to extend the method in~\cite{NG18} to compute ideal fluid flow in channel domains with multiple slit stirrers.

\section{Conformally mapping  a strip with rectilinear slits onto a domain with smooth Jordan boundaries}
\label{sec:strip}

Suppose that $\Omega$ is the canonical multiply connected domain obtained by removing $m$ non-overlapping rectilinear slits $[a_j,b_j]$ from the infinite strip $S=\{z\,:\, \left|\Im z\right|<\pi/2\}$ where $a_j,b_j\in S$ are complex numbers, $j=1,\ldots,m$, i.e. $\Omega=S \setminus \cup_{j=1}^{m}[a_j,b_j]$; see Figure~\ref{fig1} for an example of $\Omega $ with $m=4$.

\subsection{Boundary integral equation for the conformal mapping}

In this subsection, we briefly review the method presented in~\cite{NF13} to compute a conformal mapping from  a multiply connected domain $G$ with smooth Jordan curves as boundaries onto the above domain $\Omega$. The domain $G$ is of the form $\mathbb{D}\setminus \cup_{j=1}^m \overline{E_j}$ where the $\{E_j : j-1,\ldots m\}$ is a collection of disjoint simply connected  domains that are bounded by smooth Jordan curves $\Gamma_1,...,\Gamma_m$.
The external boundary of $G$ is the unit circle $\Gamma_0 := \partial \mathbb{D}.$
In this setting, we have the existence of a unique conformal mapping $\Phi$ from $G$ onto $\Omega$ such that~\cite[p.~128]{wen}
\begin{equation}\label{eq:strip-cond}
\Phi(\pm 1)=\pm\infty+\i 0 \quad\mbox{and}\quad \Phi(\i)=\frac{\pi}{2}\i.
\end{equation}
For $j=1,\ldots,m$, let $\ell_j=|b_j-a_j|$ be the length of the segment $L_j=[a_j,b_j]$,  $c_j=(a_j+b_j)/2$ be the center of $L_j$, and  $\theta_j$ be the angle between $L_j$ and the positive real axis. The values of the real constants $\ell_1,\ldots,\ell_m$ and the complex constants $c_1,\ldots,c_m$ are undetermined and should be computed alongside the conformal mapping $\Phi$. These constants are uniquely determined by the domain $G$. On the other hand, the values of the angles $\theta_1,\ldots,\theta_m$ can be fixed in advance.

We parametrize each boundary component $\Gamma_j$ by a $2\pi$-periodic complex-valued function $\eta_j(t)$, $t\in J_j:=[0,2\pi]$, $j=0,\ldots,m$.
Henceforth, we define the total parameter domain $J$ as the disjoint union of the $m+1$ intervals $J_j=[0,2\pi]$, $j=0,\ldots,m$. In this way,
the whole boundary $\Gamma$ is parametrized by
\begin{equation}\label{eq:eta}
\eta(t)= \left\{ \begin{array}{l@{\hspace{0.5cm}}l}
\eta_0(t),&t\in J_0,\\
\vdots & \\
\eta_m(t),&t\in J_m.
\end{array}
\right.
\end{equation}
See~\cite{Nas-ETNA} for more details.

We take $\theta_0=0$, assume that $\theta_1,\ldots,\theta_m$ are given real constants, and consider the complex function $A$ defined by
\begin{equation}\label{eq:A}
A(t) = e^{\i(\frac{\pi}{2}-\theta(t))}(\eta(t)-\alpha),
\end{equation}
where $\alpha\in G$ is given and $\theta$ is defined on $J$ by
\[
\theta(t)=\theta_j \quad {\rm for}\quad t\in J_j, \quad j=0,\ldots,m,
\]
i.e., the function $\theta$ is constant on each interval $J_j$.
Then, the kernel $N(s,t)$ defined on $J\times J$ by
\begin{equation}\label{eq:N}
N(s,t) :=
\frac{1}{\pi}\Im\left(\frac{A(s)}{A(t)}\frac{\dot\eta(t)}{\eta(t)-\eta(s)}\right) \quad {\rm for}\quad (s,t)\in J\times J,
\end{equation}
is known as the \emph{generalized Neumann kernel}.
We also define the kernel $M(s,t)$ on $J\times J$ by
\begin{equation}\label{eq:M}
M(s,t) :=
\frac{1}{\pi}\Re\left(\frac{A(s)}{A(t)}\frac{\dot\eta(t)}{\eta(t)-\eta(s)}\right)  \quad {\rm for}\quad (s,t)\in J\times J.
\end{equation}
Note that $N(s,t)$ is continuous while $M(s,t)$ is singular with its singular part involving the cotangent function. Hence, the integral operator ${\bf N}$ with the kernel $N(s,t)$ is compact and the integral operator ${\bf M}$ with the kernel $M(s,t)$ is singular. Further details can be found in~\cite{Weg-Nas}.

The method proposed in~\cite{NF13} for computing the conformal mapping $\Phi$ from the domain $G$ onto the domain $\Omega$ is summarized in the following theorem. For this, let us set
\begin{equation*}\label{eq:gama-strip}
	\gamma(t)= \left\{ \begin{array}{l@{\hspace{0.5cm}}l}
		0,&t\in J_0,\\
		\Im\left[e^{-\i\theta_j}\Psi(\eta(t))\right],&t\in J_j,\quad j=1,\ldots,m, \\
	\end{array}
	\right.
\end{equation*}
where
\begin{equation}\label{eq:Psi}
	\Psi(w) = \log\frac{1+w}{1-w}.
\end{equation}
The function $\zeta=\Psi(w)$ is a conformal mapping from the unit disk ${\mathbb D}$ onto the infinite strip $-\pi/2<\Im\zeta<\pi/2$.

\begin{theorem}[\cite{NF13}]\label{thm:cm-strip}
If $\rho$ is the unique solution of the boundary integral equation
\begin{equation}\label{eq:ie}
({\bf I}-{\bf N})\rho=-{\bf M}\gamma,
\end{equation}
and the piecewise constant function $h$ is given by
\begin{equation}\label{eq:h}
h=[{\bf M}\rho-({\bf I}-{\bf N})\gamma]/2,
\end{equation}
then the conformal mapping $\Phi$ from $G$ onto $\Omega$ is given by
\begin{equation}
\label{eq:Phi-f}
\Phi(w)=-\i f(\i)+w f(w)+\Psi(w), \quad w\in G\cup\Gamma,
\end{equation}
where $f$ is the analytic function in $G$ with the boundary values
\begin{equation}\label{eq:strip-f}
A(t)f(\eta(t))=\gamma(t)+h(t)+\i\rho(t).
\end{equation}
\end{theorem}

\subsection{Computing the preimage domain $G$}

We assume that the strip with rectilinear slits domain $\Omega$ is given. This means that the values of the constants $\ell_j$, $c_j$, and $\theta_j$ are now known for $j=1,\ldots,m$.
The method shown in Theorem~\ref{thm:cm-strip} will be used in this subsection to develop an iterative scheme to find a bounded multiply connected preimage domain $G$ in the exterior of $m$ smooth Jordan curves $\Gamma_1,\ldots,\Gamma_m$ and the interior of the unit circle $\Gamma_0$ as well as the conformal map $z=\Phi(w)$ from $G$ onto $\Omega$ that satisfies the normalization conditions~\eqref{eq:strip-cond}.

\begin{figure}[ht]\label{fig1}
\centerline{
\scalebox{0.38}[0.38]{\includegraphics[trim=0 -1.0cm 0 0,clip]{strip_fig_dom}}
\hfill
\scalebox{0.38}[0.38]{\includegraphics[trim=0 -1.0cm 0 0,clip]{strip_fig_domc}}
\hfill
\scalebox{0.38}[0.38]{\includegraphics[trim=0 0    0 0,clip]{strip_fig_domi}}
}
\caption{On the left, the domain $\Omega$ with four rectilinear slits. In the center, the initial intermediate domain $\hat\Omega^0$ where thin ellipses are chosen on the given slits with $r=0.4$. On the right, the domain interior to the unit circle and exterior to the dot-red curves is the initial domain $G^0$. The computed preimage domain $G$ is the domain interior to the unit circle and exterior to the solid-black curves.}
\label{fig:strip}
\end{figure}

For $k=0,1,2,3,\ldots$, where $k$ is the iteration number, we let $\hat\Omega^k$ to be the multiply connected domain obtained by removing the $m$ ellipses $\hat L_j^k$ parametrized by
\[
\hat\eta_j^k(t)= z_j^k+0.5a_j^ke^{\i\theta_j}(\cos t-\i r\sin t), \quad t\in J_j, \quad j=1,\ldots,m,
\]
from the infinite strip $S=\{\xi\in{\mathbb C}\,:\, \left|\Im \xi\right|<\pi/2\}$. Here $r$, $0<r\le1$, is the ratio of the lengths of the major to the minor axes of the ellipse $\hat L_j^k$.
We denote by $G^k$ the image of the multiply connected domain $\hat\Omega^k$ under the conformal mapping $w=\Psi^{-1}(\xi)$ where
\[
w=\Psi^{-1}(\xi)=\frac{\exp(\xi)-1}{\exp(\xi)+1}=\tanh(\xi/2).
\]
Hence, the preimage domain $G^k$ (see Figure~\ref{fig:strip} (right) for an example) is the bounded multiply connected domain interior to the unit circle parametrized by
\[
\eta^k_0(t)=e^{\i t}, \quad  t\in J_0,
\]
and exterior to the $m$ quasi-ellipses $\Gamma_1,\ldots,\Gamma_m$ parametrized by
\[
\eta^k_j(t)=\Psi^{-1}\left(\hat\eta_j^k(t)\right), \quad t\in J_j, \quad j=1,\ldots,m.
\]
Thus, determining the the preimage domain $G^k$ requires computing the parameters $z_j^k$ and $a_j^k$ of the ellipses $\hat L_j^k$, $j=1,\ldots,m$, which will be accomplished using the following iterative method.
We point out that the initial domain $\hat\Omega^0$ is obtained from the given domain $\Omega$ by replacing each slit $L_j$ by a thin ellipse $\hat L_j^0$ whose major axis is on the slit $L_j$ (see Figure~\ref{fig:strip} (center)).\\

\noindent{\bf Initialization:}\\
Set
\[
z_j^0=c_j, \quad a_j^0=(1-0.5r)\ell_j, \quad j=1,\ldots,m.
\]
\noindent{\bf Iterations:} \\
For $k=1,2,3,\ldots$,
\begin{itemize}
	\item The method in Theorem~\ref{thm:cm-strip} is used to compute the conformal mapping from the domain $G^{k-1}$ onto the domain $\Omega^k$ (the infinite strip $|\Im z|<\pi/2$ with $m$ slits $L_1^k,\ldots,L_m^k$ such that the angle between the slit $L_j^k$ and the positive real axis is $\theta_j$, $j=1,\ldots,m$).
	\item If $c_j^k$ is the center of the slit $L_j^k$ and $\ell_j^k$ is its length, then the parameters $z_j^k$ and $a_j^k$ are updated through
\begin{eqnarray*}
     z_j^{k} &=& z_j^{k-1}-(c_j^{k}-c_j), \\
     a_j^{k} &=& a_j^{k-1}-(1-0.5r)(\ell_j^k -\ell_j),
\end{eqnarray*}
for $j=1,\ldots,m$.
  \item Stop the iterations if
	\begin{equation}\label{eq:Ek}
	E_k=\frac{1}{2m}\sum_{j=1}^{m}\left(|c_j^{k} -c_j|+|\ell_j^k -\ell_j|\right)<\varepsilon \quad{\rm or}\quad k>{\tt Max}
	\end{equation}
	where $\varepsilon$ is a fixed tolerance and ${\tt Max}$ is the maximum number of iterations to be not exceeded. In our numerical experiments, we set $\varepsilon=10^{-14}$ and ${\tt Max}=100$.
\end{itemize}

The iterations above produce a sequence of multiply connected preimage domains $G^0$, $G^1$, $G^2$, $G^3$, $\ldots$, which numerically converges to the required domain $G$. This method also produces a conformal map $z=\Phi(w)$ from the computed preimage domain $G$ onto the given domain $\Omega$. Similar iterative procedures have been experimented for other canonical domains in~\cite{NG18,N18} where the numerical examples demonstrate the fast convergence even for domains with high connectivity.

It is understandable that at every iteration of the above method ones needs to solve the integral equation~\eqref{eq:ie} and to compute the piecewise constant function $h$ in~\eqref{eq:h}. This can be done with the fast method presented in~\cite{Nas-ETNA} by applying the MATLAB function \verb|fbie| in which we discretize~\eqref{eq:ie} by the Nystr\"om method using the trapezoidal rule with $n$ equidistant nodes in each sub-interval $J_j$, $j=0,\ldots,m$. This produces a $(m+1)n\times(m+1)n$ linear system, which in turn is solved by the generalized minimal residual method.
More precisely, we employ the MATLAB function \verb|gmres| without restart where the tolerance and maximum number of iterations are chosen to be $10^{-14}$ and $100$, respectively.
The matrix-vector product in \verb|gmres| is computed through the Fast Multipole Method.
by calling \verb|zfmm2dpart| from the MATLAB toolbox~\cite{Gre-Gim12} with a  tolerance of $0.5\times10^{-15}$.
The complexity of each iteration of the presented iterative method is $O((m+1)n\log n)$.
We refer to ~\cite{Nas-ETNA} for further details.

For the parameter $r$ in the above iterative method, in general, the value of $0<r\le 1$ is chosen such that the inner boundary components are not overlapping, i.e., we need to choose a small value for $r$ if the slits are close to each others. On one hand, the geometry of the preimage domain $G$ will be simpler if we choose $r=1$ or close to $1$ (when it is possible). On the other hand, the rate of convergence of the iterative method depends on the value of $r$ where the method converges faster for small $r$ (see Figure~\ref{fig:ex-m22-err} (left) below for the example considered in Section~\ref{ex:cm}). However, for small values of $r$, the method requires more GMRES iterations (see Figure~\ref{fig:ex-m22-err} (right)). Furthermore, for small values of $r$, the inner boundary components will be thin and hence one needs to consider a larger value of $n$ to achieve a satisfactory accuracy. The preimage domain in Figure~\ref{fig:ex-m22} is computed with $r=0.2$.

\subsection{Computing the conformal mapping from $\Omega$ onto $G$}

The iterative method presented in the preceding subsection allows us to compute the preimage domain $G$ and the conformal mapping from $z=\Phi(w)$ from $G$ onto $\Omega$. More precisely, it provides us with a parametrization $\eta(t)$, $t\in J$, of the boundary $\partial G$. As described in Theorem~\ref{thm:cm-strip}, solving~\eqref{eq:ie} and computing $h$ in~\eqref{eq:h} permit to determine the boundary values of the auxiliary analytic function $f$.
Consequently, the boundary values $\Phi(\eta(t))$ of $\Phi$ can be computed by~\eqref{eq:Phi-f}.
To determine the values $\Phi(w)$ for $w\in G$, we first compute the values of $f(w)$ using the Cauchy integral formula. In our numerical computations, the values $f(w)$
for $w\in G$ are computed accurately using the MATLAB function \verb|fcau| presented in~\cite{Nas-ETNA}.

Since $\Phi$ maps $\partial G$ onto $\partial\Omega$, we can see that
\begin{equation}\label{eq:zet}
\zeta(t) = \Phi(\eta(t)) = -\i f(\i)+\eta(t) f(\eta(t))+\Psi(\eta(t)), \quad t\in J,
\end{equation}
is a parametrization of $\partial\Omega$, and $\Phi^{-1}(\zeta(t))=\eta(t)$. The boundary of the domain $\Omega$ passes through the point at infinity, which means the Cauchy integral formula cannot be employed to directly compute $\Phi^{-1}(z)$ for $z\in\Omega$. Nevertheless, we write $\Phi^{-1}(z)$ as
\begin{equation}\label{eq:Phi-1-g}
\Phi^{-1}(z)=(\Phi^{-1}\circ\Psi\circ\Psi^{-1})(z) = g(\Psi^{-1}(z))
\end{equation}
where $g=\Phi^{-1}\circ\Psi$ and $\Psi$ is given by~\eqref{eq:Psi}. The mapping function $\Psi^{-1}$ maps the unbounded domain $\Omega$ onto the bounded domain $\tilde\Omega=\Psi^{-1}(\Omega)$ (see Figure~\ref{fig:t-omega}). The domain $\tilde\Omega$ is interior to the unit circle and exterior to $m$ slits (which are not rectilinear) and the boundary of $\tilde\Omega$ is parametrized by
\[
\tilde\zeta(t) = \Psi^{-1}(\zeta(t)), \quad t\in J.
\]
The boundary values of $g$ are then given by
\[
g(\tilde\zeta(t))=\Phi^{-1}(\Psi(\tilde\zeta(t)))=\Phi^{-1}(\zeta(t))=\eta(t), \quad t\in J.
\]
For $z\in\Omega$ we have $\tilde z=\Psi^{-1}(z)\in\tilde\Omega$, and the values of $g(\tilde z)$ can now be computed using the Cauchy integral formula.
Consequently, the values of $\Phi^{-1}(z)$ can be computed through~\eqref{eq:Phi-1-g}, that is,
\[
\Phi^{-1}(z)= g(\Psi^{-1}(z))
= \frac{1}{2\pi\i} \int_{\partial\tilde\Omega} \frac{g(\tilde\zeta)}{\tilde\zeta-\Psi^{-1}(z)}d\tilde\zeta
= \frac{1}{2\pi\i} \int_{J} \frac{\eta(t)}{\tilde\zeta(t)-\Psi^{-1}(z)}\tilde\zeta'(t)dt.
\]

\begin{figure}[ht] %
\centerline{
\scalebox{0.4}[0.4]{\includegraphics[trim=0 0    0 0,clip]{strip_fig_domic}}
}
\caption{The bounded domain $\tilde\Omega$ for the domain $\Omega$ shown in Figure~\ref{fig:strip}(left).}
\label{fig:t-omega}
\end{figure}

In the numerical computations presented below, the values of $g(\tilde z)$ are computed by the MATLAB function \verb|fcau|. However, to use the function \verb|fcau|, we need to compute $\tilde\zeta'(t)$ for $t\in J$ in advance. This derivative is computed by first approximating the real and imaginary parts of $\tilde\zeta(t)$ on each sub-interval $J_j$, $j=0,\ldots,m$, with trigonometric interpolating polynomials, and then differentiating these polynomials. Note that the interpolating polynomials can be computed by applying the fast Fourier transform~\cite{Weg05}.

\subsection{A numerical example}\label{ex:cm}

We present a numerical example that illustrates how the iterative method introduced in Section~\ref{sec:strip} can be applied to determine the preimage $G$ of an infinite strip with $m=21$ rectilinear slits as shown in Figure~\ref{fig:ex-m22} (left). The method is applied with $n=2^{11}$ and $r=0.2$ and the obtained preimage domain is displayed in Figure~\ref{fig:ex-m22} (right). We also compute the error $E_k$ defined by~\eqref{eq:Ek} and the number of GMRES iterations with $n=2^{11}$ for $r=0.05$, $r=0.2$, and $r=0.5$. The computed error and the number of GMRES iterations vs. the number of iteration $k$ is given in Figure~\ref{fig:ex-m22-err}. The total CPU time required by the iterative method is $68.91$ sec for $r=0.05$, $70.15$ sec for $r=0.2$, and $127.72$ for $r=0.5$.

\begin{figure}[ht] %
	\centerline{
		\scalebox{0.45}{\includegraphics[trim=0 0 0 0,clip]{fig_dom_O}}
		\hfill
		\scalebox{0.45}{\includegraphics[trim=0 0 0 0,clip]{fig_dom_G}}
	}
	\caption{The domain $\Omega$ with $21$ rectilinear slits (left) and the preimage domain $G$ (right).}
	\label{fig:ex-m22}
\end{figure}

\begin{figure}[ht] %
	\centerline{
		\scalebox{0.45}{\includegraphics[trim=0 0.0cm  0 0,clip]{map_error}}
		\hfill
		\scalebox{0.45}{\includegraphics[trim=0 0.0cm  0 0,clip]{map_gmres}}
	}
	\caption{The error $E_k$ for the domain in Figure~\ref{fig:ex-m22} (left) for $r=0.1$, $r=0.2$, and $r=0.3$.}
	\label{fig:ex-m22-err}
\end{figure}

\section{Capacity of generalized condensers}
\label{sec:strip-g-cap}

\subsection{Numerical computation of capacity of generalized condensers}

We aim to compute the capacity of generalized condensers of the form $C=(S,E,\delta)$ where $S$ is the infinite strip
\[
S=\{z\,:\, \left|\Im z\right|<\pi/2\},
\]
$E=\{E_j\}_{j=1}^m$ is a collection of $m$ nonempty closed pairwise disjoint segments $E_j=[a_j,b_j]$ with complex numbers $a_j,b_j\in S$, and $\delta=\{\delta_j\}_{j=1}^{m}$ is a collection of real numbers. We assume $m>1$ and $\delta$ contains at least two different numbers. The domain $\Omega=S \setminus E =S \setminus \cup_{j=1}^{m}[a_j,b_j]$ is known as the field of the condenser $C$, the sets $E_k$ as the plates, and the numbers $\delta_k$ as the levels of the potential of the plates $E_j$, $j=1,\ldots,m$~\cite[p.~12]{du}  (see Figure~\ref{fig:domaingcS} (left) for $m=4$).

\begin{figure}[ht] %
\centerline{
\scalebox{0.45}[0.45]{\includegraphics[trim=0 0 0 0,clip]{domaingcs}}
\hfill
\scalebox{0.45}[0.45]{\includegraphics[trim=0 0 0 0,clip]{domaingcd}}
}
\caption{The domain $\Omega$ (left) and the domain $G$ (right) for $m=4$.}
\label{fig:domaingcS}
\end{figure}

The conformal capacity of the generalized condenser $C$, $\mathrm{cap}(C)$, is given by the Dirichlet integral
\begin{equation}\label{eq:cap}
\mathrm{cap}(C)=\iint\limits_{\Omega}|\nabla u|^2dxdy
\end{equation}
where $u$ is the potential function of the condenser $C$, i.e.,
the function $u$ is the unique solution of the Dirichlet BVP~\cite[p.~13, p.~305]{du}
\begin{subequations}\label{eq:u-Dir}
\begin{eqnarray}
\label{eq:u-Dir-1}
\Delta u&=& 0, \quad\; {\rm on}\;\;\Omega, \\
\label{eq:u-Dir-2}
u      &=& 0, \quad\; {\rm on}\;\;\partial S, \\
\label{eq:u-Dir-3}
u      &=& \delta_j, \quad {\rm on}\;\;E_j, \quad j=1,\ldots,m.
\end{eqnarray}
\end{subequations}

If all values of $\delta_k$ are equal to $1$ or if $m=1$ (i.e., $E$ consists of only one segment), then $C=(S,E,\delta)$ is a classical condenser and it is shortly denoted by $C=(S,E)$. In this case, the capacity $\mathrm{cap}(C)$ may simply be written as $\mathrm{cap}(S,E)$, see Section~\ref{section1}.

One important property of the capacity of generalized condensers is its invariance  under conformal mappings, which implies that domains having complex geometry like $G$  can be treated with the aid of conformal mappings~\cite{avv,dnv,dt,du,ps,sl,Vas02,vu88}. See also~\cite{bbgghv,dek,et}.

The capacity $\mathrm{cap}(C) = \mathrm{cap}(S,E,\delta)$ will be computed in two steps:\\
\noindent{\bf Step 1.}  For the given domain $\Omega=S \setminus \cup_{j=1}^{m}[a_j,b_j]$, the iterative method proposed in Section~\ref{sec:strip} will be used to find a preimage domain $G$ interior to the unit circle $\Gamma_0$ and exterior to $m$ ellipses $\Gamma_j$, $j=1,\ldots,m$. Let $D_j$ be the simply connected domain interior to $\Gamma_j$ for $j=1,\ldots,m$, let $\hat E=\cup_{j=1}^{m}\overline{D}_j$, and let $\hat C=({\mathbb D},\hat E,\delta)$ (see Figure~\ref{fig:domaingcS} (right)). Then $\mathrm{cap}(C)=\mathrm{cap}(\hat C)$.\\
\noindent{\bf Step 2.}  The capacity $\mathrm{cap}(\hat C)$ will be evaluated by the method that we will describe in the sequel.

We point out that we use the same number $n$ of discretization points for both steps.
Now, let us present the method needed in Step 2 to compute the capacity of the generalized condenser $\hat C=({\mathbb D},\hat E,\delta)$ where ${\mathbb D}$ is the unit disk, $\hat E=\{\hat E_j\}_{j=1}^m$ is a family of $m$ nonempty closed and pairwise disjoint sets $\hat E_j=D_j\cup\Gamma_j\subset {\mathbb D}$, and $\delta=\{\delta_j\}_{j=1}^{m}$ is a collection of real numbers. Here, $\Gamma_j=\partial \hat E_j$ is a smooth Jordan curve for $j=1,\ldots,m$. Hence, $G={\mathbb D} \setminus \hat E$ is a bounded multiply connected domain of connectivity $m+1$, and $\Gamma=\partial G=\cup_{j=0}^{m}\Gamma_j$ where $\Gamma_0$ is the unit circle and $\Gamma_1,\ldots,\Gamma_m$ are ellipses enclosed in $\Gamma_0$ (see Figure~\ref{fig:domaingcS} (right) for $m=4$).

The above generalized condenser $\hat C=({\mathbb D},\hat E,\delta)$ is a special type of the generalized condenser considered in~\cite[p.~12]{du} and~\cite{Nvm}. Hence, the numerical method presented in~\cite{Nvm} can be used to compute the capacity of the above generalized condenser. For the  convenience of the reader, we review this method below.

The conformal capacity of the generalized condenser $\hat C$ is
\begin{equation}\label{eq:capU}
\mathrm{cap}(\hat C)=\iint\limits_{G}|\nabla U|^2dxdy
\end{equation}
where the potential function $U$ is now the unique solution of the Dirichlet BVP
\begin{subequations}\label{eq:U-Dir}
\begin{eqnarray}
\label{eq:U-Dir-1}
\Delta U&=& 0, \quad\; {\rm on}\;\; G, \\
\label{eq:U-Dir-2}
U      &=& 0, \quad\; {\rm on}\;\;\Gamma_0, \\
\label{eq:U-Dir-3}
U      &=& \delta_j, \quad {\rm on}\;\;\Gamma_j, \quad j=1,\ldots,m.
\end{eqnarray}
\end{subequations}
The harmonic function $U$ is the real part of an analytic function $F$ in $G$ which is not necessarily single-valued. If we take a given point $\alpha_k\in G_k$ for each $k=1,\ldots,m$, then $F$ can be written as~\cite{Gak,garmar,Mik64,Mus}
\begin{equation}\label{eq:F-u}
F(z)=g(z)-\sum_{k=1}^{m} a_k\log(z-\alpha_k)
\end{equation}
where $g$ is a single-valued analytic function in $G$ and $a_1,\ldots,a_{m}$ are undetermined real constants satisfying~\cite[\S31]{Mik64}
\begin{equation}\label{eq:ak}
a_k=\frac{1}{2\pi}\int_{\Gamma_k}\frac{\partial U}{\partial{\bf n}}ds, \quad k=1,\ldots,m.
\end{equation}
Using Green's formula~\cite[p.~4]{du}, and in view of~\eqref{eq:U-Dir-2}--\eqref{eq:U-Dir-3} and~\eqref{eq:ak}, equation~\eqref{eq:capU} can be written as
\begin{equation}\label{eq:cap-n}
\mathrm{cap}(\hat C)=\int_{\Gamma}u\frac{\partial u}{\partial{\bf n}}ds
=\sum_{k=1}^{m}\delta_k\int_{\Gamma_k}\frac{\partial u}{\partial{\bf n}}ds=2\pi\sum_{k=1}^{m}\delta_k a_k.
\end{equation}
Consequently, by the conformal invariance of the capacity, we have
\begin{equation}\label{eq:cap-a}
\mathrm{cap}(C)=2\pi\sum_{k=1}^{m}\delta_k a_k.
\end{equation}

The problem~\eqref{eq:U-Dir} above is a particular case of the problem considered in~\cite[Eq.~(4)]{Nvm}. Let $\eta(t)$, $t\in J$, be a parametrization of the boundary $\Gamma=\partial G$ and let $A(t)$ be defined~\eqref{eq:A}. Then, the constants $a_1,\ldots,a_m$ in~\eqref{eq:cap-a} will be computed as in the following theorem from~\cite[Theorem 4]{Nvm}.

\begin{theorem}\label{thm:method}
For each $k=1,\ldots,m$, let the function $\gamma_k$ be defined by
\begin{equation}\label{eq:gam-k}
\gamma_k(t)=\log|\eta(t)-\alpha_k|,
\end{equation}
let $\rho_k$ be the unique solution of the integral equation
\[
({\bf I}-{\bf N})\rho_k=-{\bf M}\gamma_k,
\]
and let the piecewise constant function $h_k=(h_{0,k},h_{1,k},\ldots,h_{m,k})$ be given by
\[
h_k=[{\bf M}\rho_k-({\bf I}-{\bf N})\gamma_k]/2.
\]
Then, the $m+1$ real constants $a_1,\ldots,a_{m},c$ are the unique solution of the linear system
\begin{equation}\label{eq:sys-method}
\left[\begin{array}{ccccc}
h_{0,1}    &h_{0,2}    &\cdots &h_{0,m}      &1       \\
h_{1,1}    &h_{1,2}    &\cdots &h_{1,m}      &1       \\
\vdots     &\vdots     &\ddots &\vdots       &\vdots  \\
h_{m,1}    &h_{m,2}    &\cdots &h_{m,m}      &1       \\
\end{array}\right]
\left[\begin{array}{c}
a_1    \\a_2    \\ \vdots \\ a_{m} \\  c
\end{array}\right]
= \left[\begin{array}{c}
0 \\  1 \\  \vdots \\ 1
\end{array}\right].
\end{equation}
\end{theorem}

The linear system~\eqref{eq:sys-method} is usually small and can be solved using Gaussian elimination. Henceforth, the capacity can be computed by~\eqref{eq:cap-a}. Note that, in our computation in this paper, we do not need the value of the real constant $c$ in~\eqref{eq:sys-method}.

We now present several examples to illustrate how the above proposed method can be applied to estimate the capacity $\mathrm{cap}(S,E,\delta)$.

\subsection{A strip with one rectilinear slit}

\begin{example}\label{ex:msi-si}
$E=[-s\i,s\i]$ where $0<s<\pi/2$ is a real number.
\end{example}

The exact value of the capacity of the condenser $(S,E)$ can be given in terms of special functions
for this example. For $\Psi$ defined by~\eqref{eq:Psi}, the mapping function
\[
z\mapsto-\i\Psi^{-1}(z)=-\i\tanh(z/2)
\]
maps conformally the domain $S\backslash[-s\i,s\i]$ onto the domain ${\mathbb D}\backslash[-\tan(s/2),\tan(s/2)]$. Then, by the M{\"o}bius transformation
\[
z\mapsto \frac{z+\tan(s/2)}{1+z\,\tan(s/2)}
\]
the domain ${\mathbb D}\setminus[-\tan(s/2),\tan(s/2)]$ is mapped onto the domain ${\mathbb D}\setminus[0,\sin(s)]$.
Owing to the conformal invariance of the capacity, we have
\[
\mathrm{cap}(S,[-s\i,s\i]) = \mathrm{cap}({\mathbb D},[0,\sin(s)]).
\]
This yields (see \cite{LV}, \cite[Thm 8.6(1)]{vu88}),
\begin{equation}\label{eq:strip-exact-i}
\mathrm{cap}(S,[-s\i,s\i])=\frac{2\pi}{\mu(\sin s)}
\end{equation}
where
\begin{equation}\label{eq:mu}
\mu(r)=\frac{\pi}{2}\frac{\mathcal{K}'(r)}{\mathcal{K}(r)},\quad
\mathcal{K}(r)=\int^1_0 \frac{dx}{\sqrt{(1-x^2)(1-r^2x^2)}},
\quad \mathcal{K}'(r)=\mathcal{K}(r'), \quad r'=\sqrt{1-r^2}\,.
\end{equation}
Here, $\mathcal{K}(r)$ and $\mathcal{K}'(r)$ are the elliptic integrals of the first kind, and $\mu:(0,1)\to(0,\infty)$ is a decreasing homeomorphism. For the numerical computation of the values of $\mu(r)$, we use the method described in~\cite{Nv}.

We set $n=2^{10}$ and $r=0.2$ to compute $\mathrm{cap}(S,E)$ for several values of $s$ in $(0,\pi/2)$, and present the numerical results in Table~\ref{tab:strip-seg-err-i}.
The relative error in the approximate values for $s=0.5$ and $s=1.55$ vs. $n$, the number of discretization points on each boundary component, are depicted in Figure~\ref{fig:sing-msi-err} (left). The error is $O(e^{-0.217n})$ for $s=0.5$ and $O(e^{-0.091n})$ for $s=1.55$.

\begin{table}[ht]
\caption{The capacity $\mathrm{cap}(S,[-s\i,s\i])$ for several values of $s$.}
\label{tab:strip-seg-err-i}%
\centerline{
\begin{tabular}{c|c|c|c|c|c}\hline
  $s$  & Estimated value      & Exact value        & Relative Error      & Time (sec) & Iterations \\ \hline
$0.1$  & $1.703662933054205$ & $1.703662933054233$ & $1.7\times10^{-14}$ & 3.0  & 13\\
$0.25$ & $2.270486340549139$ & $2.270486340549105$ & $1.5\times10^{-14}$ & 2.9  & 13  \\
$0.5$  & $3.053400295538014$ & $3.053400295538072$ & $1.9\times10^{-14}$ & 2.7  & 12  \\
$1$    & $4.885188789695857$ & $4.885188789695905$ & $9.8\times10^{-15}$ & 2.5  & 10  \\
$1.5$  & $10.27204980801009$ & $10.27204980801069$ & $8.5\times10^{-14}$ & 6.9  & 32  \\
$1.55$ & $13.39253761345057$ & $13.39253761345013$ & $3.3\times10^{-14}$ & 12.6 & 56  \\
\hline
\end{tabular}}
\end{table}

 \begin{figure}[ht] %
\centerline{
\scalebox{0.6}{\includegraphics[trim=0 0.0cm 0 0cm,clip]{sing_msi_si_err}}
\hfill
\scalebox{0.6}{\includegraphics[trim=0 0.0cm 0 0cm,clip]{sing_ms_s_err}}
}
\caption{On the left, the relative error in the approximated values for Example~\ref{ex:msi-si} for $s=0.5$ and $s=1.55$. On the right, the relative error in the approximated values for Example~\ref{ex:ms-s} for $s=0.5$ and $s=2$.}
\label{fig:sing-msi-err}
\end{figure}

\begin{example}\label{ex:ms-s}
$E=[-s,s]$ where $s>0$ is a real number.
\end{example}

In this case again, the exact value of the capacity of the condenser $(S,E)$ can be determined.
Clearly,
$
z\mapsto\Psi^{-1}(z)=\tanh(z/2)
$
maps conformally the domain $S\setminus[-s,s]$ onto the domain ${\mathbb D}\setminus[-\tanh(s/2),\tanh(s/2)]$. Then, the M{\"o}bius transformation
\[
z\mapsto \frac{z+\tanh(s/2)}{1+z\,\tanh(s/2)}
\]
maps the domain ${\mathbb D}\setminus[-\tanh(s/2),\tanh(s/2)]$ onto the domain ${\mathbb D}\setminus[0,\tanh(s)]$. Hence, as in the previous example, we have
\begin{equation}\label{eq:strip-exact}
\mathrm{cap}(S,[-s,s]) = \mathrm{cap}({\mathbb D},[0,\tanh(s)]) = \frac{2\pi}{\mu(\tanh s)}.
\end{equation}

We now take $n=2^{10}$ and $r=0.2$ to compute $\mathrm{cap}(S,E)$ for several values of $s$.
For $s=0.5$ and $s=2$, the relative error in the approximate values vs. $n$ are given in Figure~\ref{fig:sing-msi-err} (right). The error is $O(e^{-0.217n})$ for $s=0.5$ and $O(e^{-0.215n})$ for $s=2$.

\begin{example}\label{ex:ver}
$E=\i s+[-\i,\i]$ with $-0.55\le s\le0.55$.
\end{example}

In this example, we consider the vertical segment $[-\i,\i]$ and study the effect of vertically shifting this segment on the capacity of the condenser $(S,E)$.
On the left of Figure~\ref{fig:sing-cant-mi-i}, we display the graph of the capacity $\mathrm{cap}(S,\i s+[-\i,\i])$ as a function of $s\in[-0.55, 0.55]$. The computation is performed with $n=2^{11}$ and $r=0.1$.

\begin{figure}[ht] %
\centerline{
\scalebox{0.5}[0.5]{\includegraphics[trim=0 0.0cm 0 0cm,clip]{sing_cont_ver_is}}
\hfill
\scalebox{0.5}[0.5]{\includegraphics[trim=0 0.0cm 0 0cm,clip]{sing_cont_hor_is}}
}
\caption{The graph of the capacity functions $\mathrm{cap}(S,\i s+[-\i,\i])$ in Example~\ref{ex:ver} (left) and $\mathrm{cap}(S,\i s+[-1,1])$ in Example~\ref{ex:hor} (right).}
\label{fig:sing-cant-mi-i}
\end{figure}

\begin{example}\label{ex:hor}
$E=\i s+[-1,1]$ with $-1.55\le s\le 1.55$.
\end{example}

We now consider the horizontal segment $[-1,1]$ and study the effect of vertically shifting this segment on the capacity of the condenser $(S,E)$.
The graph of the capacity as a function of $s\in[-1.55, 1.55]$, computed using the same parameters as in the previous example, is shown in Figure~\ref{fig:sing-cant-mi-i} (right).

By looking at Figure~\ref{fig:sing-cant-mi-i}, we can notice an increase in the capacity as the segment moves vertically close the boundary of $S$.
It is worth noting that horizontal movement of a segment in a strip does not change the capacity value because of translation invariance. In fact, for any segment $[a,b]$ in $S$ and any real number $s$, the linear transformation
\[
z\mapsto z-s
\]
maps the domain $S\backslash(s+[a,b])$ onto the domain $S\backslash[a,b]$, and hence by conformal invariance of the capacity, we have
\begin{equation}\label{eq:shit-s}
\mathrm{cap}(S,s+[a,b])=\mathrm{cap}(S,[a,b]), \quad a,b\in S, \quad s\in{\mathbb R}.
\end{equation}

\begin{example}
Segments with constant capacity.
\end{example}

For a given point $a$ in $S$, consider all points $x+\i y\in S\backslash\{a\}$ such that the capacity $\mathrm{cap}(S,[a,x+\i y])$ is constant.
The contour lines of the function $\mathrm{cap}(S,[a,x+\i y])$ in the sub-domain $x+\i y\in [-3, 3]+\i (-\pi/2, \pi/2)\setminus \{a\}$ are displayed in Figure~\ref{fig:sing-cant-0-yi} on the left for $a=0$ and on the right for $a=\i$. For each contour line, we have the same capacity for all segments with one end at $a$ and the other end on the contour line. These results are again computed with $n=2^{11}$ and $r=0.1$.

\begin{figure}[ht] %
\centerline{
\scalebox{0.45}{\includegraphics[trim=0 0 0 0,clip]{sing_cont_0_x_yi}}
\hfill
\scalebox{0.45}{\includegraphics[trim=0 0 0 0,clip]{sing_cont_i_x_yi}}
}
\caption{The contour lines of the capacity functions $\mathrm{cap}(S,[0,x+\i y])$ (left) and  the function $\mathrm{cap}(S,[\i,x+\i y])$ (right).}
\label{fig:sing-cant-0-yi}
\end{figure}

\subsection{A strip with two slits}

In this subsection, we want to estimate the capacity $\mathrm{cap}(S,E)$ where $E$ is the union of two disjoint segments.

\begin{example}
$E=[a,b]\cup[c,d]$ for several values of $a,b,c,d\, \in S$ as in Table~\ref{tab:cap-abcd}.
\end{example}

The above described method with $n=2^{10}$ and $r=0.2$ is used to compute the capacity $\mathrm{cap}(S,[a,b]\cup[c,d])$ for several values of $a$, $b$, $c$ and $d$. The obtained results are presented in
Table~\ref{tab:cap-abcd}.

\begin{table}[ht]
\caption{The approximate values of the capacity $\mathrm{cap}(S,[a,b]\cup[c,d])$.}
\label{tab:cap-abcd}%
\centerline{
\begin{tabular}{c|c|c|c|c}\hline
$a$     & $b$     & $c$     & $d$      & $\mathrm{cap}(S,[a,b]\cup[c,d])$\\ \hline %
$-1$    & $-1+\i$ & $1$     & $1-\i$   & $6.0697365159628$  \\
$-1$    & $-1+\i$ & $1$     & $1+\i$   & $6.0193744425645$  \\
$-1$    & $-2$    & $1$     & $2$      & $5.6844096460738$  \\
$-1+\i$ & $1+\i$  & $-1-\i$ & $1-\i$   & $11.029565510437$  \\
\hline
\end{tabular}}
\end{table}

\begin{example}\label{ex:x+J}
$E=E_1\cup E_2$ with $E_1=-x+J$ and $E_2=x+J$ where $J=[-\i,\i]$ and $x>0$ is a real number.
\end{example}

In this example, we consider the two vertical segments $E_1$ and $E_2$ centered on the $x$-axis (middle line of the strip) and far from each other by a distance of $2x>0$. We study the effect of this distance on $\mathrm{cap}(S,E_1\cup E_2)$.
We take $n=2^{11}$ and $r=\min\{0.2,x/2\}$, and compute the values of $\mathrm{cap}(S,[-x-\i,-x+\i]\cup[x-\i,x+\i])$ as a function of $x$ for $0.01\le x\le 4$. The graph of this function is shown in Figure~\ref{fig:2v-u}.

It follows from~\eqref{eq:shit-s} that
\[
\mathrm{cap}(S,-x+J)=\mathrm{cap}(S,x+J)=\mathrm{cap}(S,J).
\]
Hence, in view of~\eqref{eq:strip-exact-i}, we have
\[
\mathrm{cap}(S,E_1)=\mathrm{cap}(S,E_2)=\mathrm{cap}(S,[-\i,\i])=\frac{2\pi}{\mu(\sin(1))}.
\]
Then, by~\cite[Theorem 1.8]{du} and \cite[Lemma 7.1]{hkv},
\[
\frac{2\pi}{\mu(\sin(1))}=\mathrm{cap}(S,E_1)\le \mathrm{cap}(S,E_1\cup E_2) \le \mathrm{cap}(S,E_1)+\mathrm{cap}(S,E_2)=\frac{4\pi}{\mu(\sin(1))},
\]
which is validated numerically by the presented method in Figure~\ref{fig:2v-u}.
Further, Figure~\ref{fig:2v-u} suggests that
\[
\lim_{x\to0^+}\mathrm{cap}(S,E_1\cup E_2)=\mathrm{cap}(S,[-\i,\i])=\frac{2\pi}{\mu(\sin(1))},
\]
and for large $x$,
\[
\mathrm{cap}(S,E_1\cup E_2)\approx \mathrm{cap}(S,E_1)+\mathrm{cap}(S,E_2)=\frac{4\pi}{\mu(\sin(1))}.
\]

\begin{figure}[ht] %
\centerline{\scalebox{0.5}{\includegraphics[trim=0 0 0 0,clip]{2v_cont_u}}}
\caption{Results for Example~\ref{ex:x+J}: The graph of the capacity $\mathrm{cap}(S,E_1\cup E_2)$ where $E_1=[-x-\i,-x+\i]$ and $E_2=[x-\i,x+\i])$.}
\label{fig:2v-u}
\end{figure}

\begin{example}\label{ex:x+I}
$E=E_1\cup E_2$ with $E_1=-x+I$ and $E_2=x+I$ where $I=[-1,1]$ and $x>1$ is a real number.
\end{example}

We consider here the two horizontal segments $E_1$ and $E_2$ located on the $x$-axis where the distance between them is $2x>2$.
Note that $\lim_{x\to1^+}E_1\cup E_2=[-2,2]$. Further, it follows from~\eqref{eq:strip-exact} and~\eqref{eq:shit-s} that
\[
\mathrm{cap}(S,E_1)=\mathrm{cap}(S,E_2)=\mathrm{cap}(S,I)=\frac{2\pi}{\mu(\tanh(1))}.
\]

We use our method with $n=2^{11}$ and $r=0.2$ to compute the values of $\mathrm{cap}(S,E_1\cup E_2)$ for $1.01\le x\le 4$. The graph $\mathrm{cap}(S,E_1\cup E_2)$, as a function of $x$, is shown in Figure~\ref{fig:2h-u}. The obtained numerical results show that
\[
\frac{2\pi}{\mu(\tanh(2))}=\mathrm{cap}(S,[-2,2])\le \mathrm{cap}(S,E_1\cup E_2) \le \mathrm{cap}(S,E_1)+\mathrm{cap}(S,E_2)=\frac{4\pi}{\mu(\tanh(1))}.
\]
Moreover, Figure~\ref{fig:2h-u} reveals that
\[
\lim_{x\to1^+}\mathrm{cap}(S,E_1\cup E_2)=\mathrm{cap}(S,[-2,2])=\frac{2\pi}{\mu(\tanh(2))},
\]
and for large $x$,
\[
\mathrm{cap}(S,E_1\cup E_2)\approx \mathrm{cap}(S,E_1)+\mathrm{cap}(S,E_2)=\frac{4\pi}{\mu(\tanh(1))}.
\]

Figure~\ref{fig:2h-u} also leads to a new conjecture about the capacity of condensers in the strip domain $S$. Indeed, if the condenser is of the form
$(S,E)$ and $E= E_1 \cup E_2 \subset {\mathbb R}$ , $E_1$ and $E_2$ are segments, and $E_2$ is approaching $E_1,$
then
$$ {\rm cap}(S,E)\ge {\rm cap}(S,H)$$
where $H \subset {\mathbb R}$ is a segment with diameter equal to the sum of diameters
of $E_1$ and $E_2.$

\begin{figure}[ht] %
\centerline{\scalebox{0.5}{\includegraphics[trim=0 0 0 0,clip]{2h_cont_u}}}
\caption{Results for Example~\ref{ex:x+I}: The graph of the function $\mathrm{cap}(S,E_1\cup E_2)$ where $E_1=[-x-1,-x+1]$ and $E_2=[x-1,x+1]$.}
\label{fig:2h-u}
\end{figure}

\begin{example}\label{ex:xi+I}
$E=E_1\cup E_2$ where $E_1=-x\i+I$, $E_2=x\i+I$, $I=[-1,1]$, and $x>0$ is a real number.
\end{example}

In the final example concerning the case of two slits, we consider the two parallel horizontal segments $E_1$ and $E_2$, where both of them are centered on the $y$-axis, and study the effect of the distance between them, which is  $2x$, on the capacity $\mathrm{cap}(S,E_1\cup E_2)$.
We use our method with $n=2^{12}$ and $r=\min\{0.2,x/2,1.53-x\}$ to compute the values of $\mathrm{cap}(S,E_1\cup E_2)$ as a function of $x$ for $0.01\le x\le 1.5$ (see Figure~\ref{fig:2g-u} for the graph of this function). For this case, the exact values of the capacities $\mathrm{cap}(S,E_1)$ and $\mathrm{cap}(S,E_1)$ are unknown and hence will be computed numerically using the proposed method.

By looking at Figure~\ref{fig:2g-u}, we can see that
\[
\mathrm{cap}(S,E_1)=\mathrm{cap}(S,E_2) \le \mathrm{cap}(S,E_1\cup E_2) \le \mathrm{cap}(S,E_1)+\mathrm{cap}(S,E_2).
\]
It is also clear that $\mathrm{cap}(S,E_1\cup E_2) \approx \mathrm{cap}(S,E_1)$ for small $x$, and $\mathrm{cap}(S,E_1\cup E_2) \approx \mathrm{cap}(S,E_1)+\mathrm{cap}(S,E_2)$ for large $x$.

\begin{figure}[ht] %
\centerline{\scalebox{0.5}{\includegraphics[trim=0 0 0 0,clip]{2hi_cont_u}}}
\caption{Results for Example~\ref{ex:xi+I}: The graph of the function $\mathrm{cap}(S,E_1\cup E_2)$ where $E_1=-x\i+I$, $E_2=x\i+I$, $I=[-1,1]$, and $x>0$ is a real number.}
\label{fig:2g-u}
\end{figure}

\subsection{A strip with many rectilinear slits}

The last two examples illustrate that our iterative approach can be employed in the case of a multiply connected infinite strip with a high connectivity for which the exact value of the capacity is not known.

\begin{example} We first consider the case of four rectilinear slits
$E_1=[2-\i , 3.5+0.5\i]$, $E_2=[1+\i , -1+\i]$, $E_3=[-\i , -2.5+0.5\i]$ and $E_4=[-3-\i , -3+\i]$ with $\delta_1=1$, $\delta_2=2$, $\delta_3=3$ and $\delta_4=4$.
The approximate value of $\mathrm{cap}(S,E,\delta)$ obtained with $n=2^{11}$ and $r=0.2$ is
\[
\mathrm{cap}(S,E,\delta)=41.8434999283923.
\]
\end{example}

\begin{example}\label{ex:100}
We finally take up a collection of $m$ disjoints horizontal intervals of length $2/m$ with random location on the real axis between $-4$ and $4$; see Figure~\ref{fig:100} (left) for $m=100$.
Clearly, the sum of the diameters of all these intervals is $2$ which is equal to the diameter of $E=[-1,1]$. Note that by~\eqref{eq:strip-exact} the exact capacity involving the single plate $E$ is $\mathrm{cap}(S,E)=2\pi/\mu(\tanh(1))$. We  also consider the case when the $m$ collection of disjoints horizontal intervals is randomly distributed in the strip $[-4,4]\times[-1,1]$, and denote it if so by $(\hat E_j)_{j=1}^m$; see Figure~\ref{fig:100} (right) for $m=100$.
\end{example}

To estimate $\mathrm{cap}\left(S,\cup_{j=1}^{m}E_j\right)$ and $\mathrm{cap}\left(S,\cup_{j=1}^{m}\hat E_j\right)$, we employ our method using $n=2^{10}$ and $r=0.2$. We run the code for $10$ times so that to get $10$ different locations for these slits. The computed values of the capacities for these $10$ locations are shown in Figure~\ref{fig:100-10}. Note that $\cup_{j=1}^{m}E_j$ and $\cup_{j=1}^{m}\hat E_j$ have the same diameter as $E$ where the center of the horizontal slits $E_j$ are real numbers in $[-4,4]$ and the centers of the horizontal slits $\hat E_j$ are complex numbers in $[-4,4]\times[-1,1]$. The curves displayed in Figure~\ref{fig:100-10} indicate that regardless of the slits positions, we have the following inequalities
\[
\mathrm{cap}\left(S,\cup_{j=1}^{m}\hat E_j\right) > \mathrm{cap}\left(S,\cup_{j=1}^{m} E_j\right)
> \mathrm{cap}\left(S,E\right).
\]
This result is somehow expected as the slits $\hat E_j$ are less condensed and closer to the boundary of $S$ when compared to the slits $E_j$.

\begin{figure}[ht] %
	\centerline{
		\scalebox{0.45}{\includegraphics[trim=0 0  0 0,clip]{slits100_re}}
		\hfill
		\scalebox{0.45}{\includegraphics[trim=0 0  0 0,clip]{slits100_x}}
	}
	\caption{The domains $S\setminus \cup_{j=1}^{m} E_j$ (left) and $S\setminus \cup_{j=1}^{m}\hat E_j$ (right) in Example~\ref{ex:100} for $m=100$.}
	\label{fig:100}
\end{figure}

\begin{figure}[ht] %
\centerline{\scalebox{0.5}{\includegraphics[trim=0 0 0 0,clip]{slits100cap}}}
\caption{Results for Example~\ref{ex:100}: The computed values of the capacities for $10$ random locations of the slits.}
\label{fig:100-10}
\end{figure}

\section{Uniform potential flow in multiply connected channel domains}

In this section, we present a fast and accurate numerical method for constructing the complex potential function $W(z)$ for a uniform incompressible, inviscid and irrotational flow past multiple disjoint segment obstacles in the strip $S=\{z\,:\, |\Im z|<-\pi/2\}$ in the case when the circulations around the segments are zeros.
The method is based on using the above iterative method to construct a conformal mapping $w=F(z)$ from a domain $\Omega$ obtained by removing $m$ non-overlapping rectilinear slits from the strip $S$ (see Figure~\ref{fig:channel-flow} (left)) onto a domain $H$ obtained by removing $m$ horizontal slits from the strip $S$ (see Figure~\ref{fig:channel-flow}(right)). Then
\[
W(z)=F(z)
\]
represents the complex potential for a uniform flow in the domain $\Omega$ and the level curves of $\Im[W(z)]$ represent streamlines of the flow.

To construct the conformal mapping $w=F(z)$, we first use the iterative method described in Section~\ref{sec:strip} to obtain a preimage domain $G$ bordered by smooth Jordan curves and the conformal mapping $z=\Psi(\zeta)$ from $G$ onto $\Omega$ (see Figure~\ref{fig:channel-flow}(center)). Then the method presented in Theorem~\ref{thm:cm-strip} is used to compute a conformal mapping $w=\Upsilon(\zeta)$ from the domain $G$ on the domain $H$. Thus, the function
\[
F(z)=(\Upsilon\circ\Psi^{-1})(z)
\]
is the required conformal mapping from the given domain $\Omega$ onto the domain $H$.
We have applied the method with $n=2^{11}$ and $r=0.2$ to compute the streamlines of a uniform flow for two channel domains as shown in Figure~\ref{fig:channel-uflow}.

\begin{figure}[ht] %
\centerline{
\scalebox{0.38}{\includegraphics[trim=0 0 0 0,clip]{flow_domO}}
\hfill
\scalebox{0.38}{\includegraphics[trim=0 0 0 0,clip]{flow_domG}}
\hfill
\scalebox{0.38}{\includegraphics[trim=0 0 0 0,clip]{flow_domH}}
}
\caption{The domain $\Omega$ (left), the domain $G$ (center), and the domain $H$ (right) for $m=4$. The domain $G$ is computed with $n=2^{11}$ and $r=0.5$.}
\label{fig:channel-flow}
\end{figure}

\begin{figure}[ht] %
\centerline{
\scalebox{0.55}{\includegraphics[trim=0 2cm 0 2cm,clip]{flow_uniform4}}
\hfill
\scalebox{0.55}{\includegraphics[trim=0 2cm 0 2cm,clip]{flow_uniform21}}
}
\caption{Streamlines for a uniform flow in the multiply connected channel domain $\Omega$ for $m=4$ (left) and $m=21$ (right).}
\label{fig:channel-uflow}
\end{figure}

\section*{Acknowledgements}
We are indebted to Prof. A. Solynin  and Prof.
D. Betsakos who have independently provided an analytic argument to confirm our
experimental discovery~\eqref{eq:ineq} and Example~\ref{ex:x+J}.

\bibliographystyle{amsplain}

redpajama_set_name: RedPajamaArXiv


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redpajama_set_name: RedPajamaGithub