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.