text: # Diff Eq Translated Matrix?
  
  • Mar 6th 2014, 05:36 AM
  stealth4933
  Diff Eq Translated Matrix?
  This is probably a trivial problem, considering I have done most of the work already:
  
  -2x-y+y^2=x'
  -4x+y=y'
  
  This system is almost linear with Crit Points at 0,0 and 3/8,3/2
  
  My problem is that I cannot figure out for the life of me what the "translated matrix" of the point is 3/8,3/2 is in order to classify it. Does anyone know how to do this?
  
  First do the translation: you want to "move" x and y so that x= 3/8, y= 3/2 is moved to 0, 0. Do that by taking u= x- 3/8 and v= y- 3/2. Then x= u+ 3/8, y= v+ 3/2, x'= u, y'= v', and $\displaystyle y^2= (v+ 3/2)^2= v^2+ 3v+ 9/4$ so the equations become $\displaystyle x'= u'= -2(u+ 3/8)- (v+ 3/2)+ v^2+ 3v+ 9/2= -2u+ 2v+ v^2$ and [tex]y'= v'= -4(u+ 3/8)+ (v+ 3/2)= -4u+ v.
  Now there are two, equivalent and very similar, ways to continue. The first is to recognize that the first equation is non-linear because of the "$\displaystyle v^2$". The only way to "linearize" that is to drop the $\displaystyle y^2$: u'= -2u+ 2v and v'= -4u+ v. The "translated matrix" is $\displaystyle \begin{bmatrix}-2 & 2 \\ -4 & 1\end{bmatrix}$.
  The other way is to form the "Jacobian". If $\displaystyle x'= f(x,y)$ and $\displaystyle y'= g(x,y)$ the Jacobian is $\displaystyle \begin{bmatrix}\frac{\partial f}{\partial x} & \frac{\partial g}{\partial x} \\ \frac{\partial f}{\partial y} & \frac{\partial g}{\partial y}\end{bmatrix}$
  In this case that would be [tex]\begin{bmatrix}$\displaystyle \begin{bmatrix}-2 & 2+ 2v \\ -4 & 1\end{bmatrix}$. Now let u and v be 0 which gives the same matrix as before. This second method is more "formal" and helps determine how to "linearize" a formula.
explanation: The document contains a math question related to differential equations and matrix translation. The user asks how to find the "translated matrix" for a given critical point in a system of differential equations. The response provides a detailed explanation of how to perform the translation and linearization, leading to the desired matrix.
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distilabel_metadata.role: system
distilabel_metadata.content: # Diff Eq Translated Matrix?
  
  • Mar 6th 2014, 05:36 AM
  stealth4933
  Diff Eq Translated Matrix?
  This is probably a trivial problem, considering I have done most of the work already:
  
  -2x-y+y^2=x'
  -4x+y=y'
  
  This system is almost linear with Crit Points at 0,0 and 3/8,3/2
  
  My problem is that I cannot figure out for the life of me what the "translated matrix" of the point is 3/8,3/2 is in order to classify it. Does anyone know how to do this?
  
  First do the translation: you want to "move" x and y so that x= 3/8, y= 3/2 is moved to 0, 0. Do that by taking u= x- 3/8 and v= y- 3/2. Then x= u+ 3/8, y= v+ 3/2, x'= u, y'= v', and $\displaystyle y^2= (v+ 3/2)^2= v^2+ 3v+ 9/4$ so the equations become $\displaystyle x'= u'= -2(u+ 3/8)- (v+ 3/2)+ v^2+ 3v+ 9/2= -2u+ 2v+ v^2$ and [tex]y'= v'= -4(u+ 3/8)+ (v+ 3/2)= -4u+ v.
  Now there are two, equivalent and very similar, ways to continue. The first is to recognize that the first equation is non-linear because of the "$\displaystyle v^2$". The only way to "linearize" that is to drop the $\displaystyle y^2$: u'= -2u+ 2v and v'= -4u+ v. The "translated matrix" is $\displaystyle \begin{bmatrix}-2 & 2 \\ -4 & 1\end{bmatrix}$.
  The other way is to form the "Jacobian". If $\displaystyle x'= f(x,y)$ and $\displaystyle y'= g(x,y)$ the Jacobian is $\displaystyle \begin{bmatrix}\frac{\partial f}{\partial x} & \frac{\partial g}{\partial x} \\ \frac{\partial f}{\partial y} & \frac{\partial g}{\partial y}\end{bmatrix}$
  In this case that would be [tex]\begin{bmatrix}$\displaystyle \begin{bmatrix}-2 & 2+ 2v \\ -4 & 1\end{bmatrix}$. Now let u and v be 0 which gives the same matrix as before. This second method is more "formal" and helps determine how to "linearize" a formula.
distilabel_metadata.role: user
distilabel_metadata.raw_output_q_a_detector_0: The document contains a math question related to differential equations and matrix translation. The user asks how to find the "translated matrix" for a given critical point in a system of differential equations. The response provides a detailed explanation of how to perform the translation and linearization, leading to the desired matrix.
  
  Label: question-and-answer