Technique used to prove lemmas in structural graph theory
The discharging method is a technique used to prove lemmas in structural graph theory. Discharging is most well known for its central role in the proof of the four color theorem. The discharging method is used to prove that every graph in a certain class contains some subgraph from a specified list. The presence of the desired subgraph is then often used to prove a coloring result.
Most commonly, discharging is applied to planar graphs.
Initially, a "charge" is assigned to each face and each vertex of the graph.
The charges are assigned so that they sum to a small positive number. During the "Discharging Phase" the charge at each face or vertex may be redistributed to nearby faces and vertices, as required by a set of discharging rules. However, each discharging rule maintains the sum of the charges. The rules are designed so that after the discharging phase each face or vertex with positive charge lies in one of the desired subgraphs. Since the sum of the charges is positive, some face or vertex must have a positive charge. Many discharging arguments use one of a few standard initial charge functions (these are listed below). Successful application of the discharging method requires creative design of discharging rules.
An example.
In 1904, Wernicke introduced the discharging method to prove the following theorem, which was part of an attempt to prove the four color theorem.
Theorem: If a planar graph has minimum degree 5, then it either has an edge
with endpoints both of degree 5 or one with endpoints of degrees 5 and 6.
Proof:
We use formula_1, formula_2, and formula_3 to denote the sets of vertices, faces, and edges, respectively.
We call an edge "light" if its endpoints are both of degree 5 or are of degrees 5 and 6.
Embed the graph in the plane. To prove the theorem, it is sufficient to only consider planar triangulations (because, if it holds on a triangulation, when removing nodes to return to the original graph, neither node on either side of the desired edge can be removed without reducing the minimum degree of the graph below 5). We arbitrarily add edges to the graph until it is a triangulation. 
Since the original graph had minimum degree 5, each endpoint of a new edge has degree at least 6. 
So, none of the new edges are light.
Thus, if the triangulation contains a light edge, then that edge must have been in the original graph.
We give the charge formula_4 to each vertex formula_5 and the charge formula_6 to each face formula_7, where formula_8 denotes the degree of a vertex and the length of a face. (Since the graph is a triangulation, the charge on each face is 0.) Recall that the sum of all the degrees in the graph is equal to twice the number of edges; similarly, the sum of all the face lengths equals twice the number of edges. Using Euler's Formula, it's easy to see that the sum of all the charges is 12:
formula_9
We use only a single discharging rule:
We consider which vertices could have positive final charge.
The only vertices with positive initial charge are vertices of degree 5. 
Each degree 5 vertex gives a charge of 1/5 to each neighbor. 
So, each vertex is given a total charge of at most formula_10. 
The initial charge of each vertex v is formula_4. 
So, the final charge of each vertex is at most formula_12. Hence, a vertex can only have positive final charge if it has degree at most 7. Now we show that each vertex with positive final charge is adjacent to an endpoint of a light edge.
If a vertex formula_5 has degree 5 or 6 and has positive final charge, then formula_5 received charge from an adjacent degree 5 vertex formula_15, so edge formula_16 is light. If a vertex formula_5 has degree 7 and has positive final charge, then formula_5 received charge from at least 6 adjacent degree 5 vertices. Since the graph is a triangulation, the vertices adjacent to formula_5 must form a cycle, and since it has only degree 7, the degree 5 neighbors cannot be all separated by vertices of higher degree; at least two of the degree 5 neighbors of formula_5 must be adjacent to each other on this cycle. This yields the light edge.
References.
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