Content deleted Content added
No edit summary |
m →Proof |
||
Line 32:
Where <math>S_i</math> are boundary surfaces specified by boundary conditions.
Since <math>\epsilon > 0</math> and <math>(\mathbf{\nabla}\phi)^2 \ge 0</math>, then <math>\mathbf{\nabla}\phi</math> must be zero everywhere (and so <math>\mathbf{\nabla}\varphi_{1} = \mathbf{\nabla}\varphi_{2}</math>) when the surface integral vanishes.
This means that the gradient of the solution is unique when
|