The uniqueness theorem for Laplace’s equation proves that a solution to the equation within a given volume is unique once the boundary conditions are specified. The most common proof, often called the “energy method,” uses a proof by contradiction and Green’s first identity (a form of the divergence theorem).
The Proof
Let’s assume there are two different solutions, V1 and V2, that both satisfy Laplace’s equation, \(∇^2V = 0\), within a volume v and have the same specified value on the boundary surface S.
$$\nabla^2V_1 = 0 \quad \text {and} \quad \nabla^2V_2 = 0 $$
$$V_1 = V_2$$
Now, let’s define a new function, V3, as the difference between the two solutions.
$$ V_3=V_1−V_2 $$
Since Laplace’s equation is linear, the difference between two solutions is also a solution.
$$\nabla^2V_3 = \nabla^2(V_1 – V_2) = \nabla^2V_1 – \nabla^2V_2 = 0 – 0 = 0$$
So, V3 also satisfies Laplace’s equation.
Since V1 and V2 are equal on the boundary surface S, the difference V3 must be zero on the boundary.
$$V_3 = V_1 −V_2=0 $$
Green’s first identity is a fundamental vector calculus identity that relates a volume integral to a surface integral. It is given by:
$$\int_v (\phi \nabla^2 \psi + \nabla\phi \cdot \nabla\psi) \, dv = \oint_S (\phi \nabla\psi) \cdot d\mathbf{a}$$
On applying this identity by setting both ϕ and ψ to be the difference function, V3, we get:
$$\int_v (V_3 \nabla^2 V_3 + \nabla V_3 \cdot \nabla V_3) \, dv = \oint_S (V_3 \nabla V_3) \cdot d\mathbf{a}$$
Since, \(∇^2 V_3=0\) and also that V3=0 on the boundary surface S, substituting these into the identity.
$$\int_v (V_3(0) + |\nabla V_3|^2) \, dv = \oint_S (0 \cdot \nabla V_3) \cdot d\mathbf{a}$$
This simplifies to:
$$\int_v |\nabla V_3|^2 \, dv = 0$$
The term ∣∇V3∣2 is the square of the magnitude of the gradient, which is always a non-negative value. The only way for the integral of a non-negative function over a volume to be zero is if the function itself is zero everywhere within that volume.
$$|\nabla V_3|^2 = 0$$
This implies that the gradient of V3 is zero everywhere inside the volume.
$$\nabla V_3 = 0$$
A function with a zero gradient is a constant. Therefore, V3 must be a constant everywhere in the volume.
$$ V_3 = \text{constant} $$
It is already known that V3 is zero on the boundary surface S and since V3 is a constant, and it is zero on the boundary, it must be zero everywhere within the volume.
V3=0 (everywhere in volume v)
Substituting back the original definition, we get:
V1−V2=0⟹V1=V2
This proves that the two solutions must be identical, thus proving the uniqueness theorem for Dirichlet boundary conditions.