Uniqueness Theorem for Laplace Equation
This is the final tutorial of this chapter and we will conclude by showing the uniqueness of solution for the three major types of second-order linear PDEs introduced before. We will first start with the Laplace equation with a homogeneous Dirichlet B.C. We first need Green’s first identity:
\begin{equation}
\int_{\partial\Omega} \phi\nabla\psi \cdot d\vec{\sigma} = \int_{\Omega} (\phi\nabla^2\psi + \nabla\phi \cdot \nabla\psi) dV \tag{1}
\end{equation}
where \(\partial \Omega\) is the boundary of the domain \(\Omega\), and \(d\vec{\sigma} = \hat{n} d\sigma\) is the normal component of surface element. This is briefly established by the Divergence Theorem:
\begin{align}
\int_{\partial\Omega} \phi\nabla\psi \cdot d\vec{\sigma} &= \int_{\Omega} \nabla\cdot(\phi\nabla\psi) dV \\
&= \int_{\Omega} (\phi\nabla^2\psi + \nabla\phi \cdot \nabla\psi) dV \tag{2}
\end{align}
where we have used the vector identity \( \nabla\cdot(A\nabla B) = A\nabla^2B + \nabla A \cdot \nabla B\). Now, assume that there exist two solutions \( u_1,u_2 \) to the Laplace equation problem and denote their difference by \( w = u_1 -u_2 \). Then, it is clear that \( w \) also satisfies the Laplace equation:
\begin{align}
\nabla^2 w &= \nabla^2 (u_1 -u_2) \\
&= \nabla^2 u_1 -\nabla^2 u_2 = 0-0 = 0 \tag{3}
\end{align}
Plugging \(w\) into (1) to obtain
\begin{align}
\int_{\partial\Omega} w \frac{\partial w}{\partial n}d\sigma = \int_{\partial\Omega} w \nabla w \cdot d\vec{\sigma} = \int_{\Omega} (w \nabla^2 w + \nabla w \cdot \nabla w) dV \tag{4}
\end{align}
where \( \partial w/\partial n\) is the derivative along the normal to the boundary. From (3), we know that \( \nabla^2 w = 0 \) so the first term on the R.H.S. of (4) vanishes, and by the homogeneous Dirichlet B.C., \( w = 0 \) along the boundary, hence the L.H.S. also vanishes. What remains shows that
\begin{equation}
\int_{\Omega} \nabla w \cdot \nabla w dV = \int_{\Omega} |\nabla w|^2 dV = 0 \tag{5}
\end{equation}
This implies that \( |\nabla w| = 0 \) and hence \( w \) is a constant everywhere in the domain. Since \( w = u_1-u_2 = 0 \) on the boundary as they have to obey the same Dirichlet B.C., the only possibility is that \( w = 0 \) and \( u_1 = u_2 \) everywhere, proving the uniqueness theorem of the Laplace equation with a Dirichlet B.C.
For a homogeneous Neumann B.C., (5) and its implication that \( w \) is a constant throughout the domain are still valid. However, \( w \) is not necessarily zero and can be equal to any fixed constant \( c \). This \(c\) will not have any effect on the Neumann B.C. because the (normal) derivative of a constant is just \(0\). Therefore, if any \(u_1\) is a solution to the Laplace equation with a Neumann B.C., \( u_2 = u_1 + c \) will also be another valid solution. We hence say that the uniqueness of solution to a Neumann problem in this case is only up to an additive constant. Another caveat is that, if the Neumann B.C. is not homogeneous, so that
\begin{align}
\frac{\partial u}{\partial n} = g(x,y,\ldots) \tag{6}
\end{align}
along \( \partial\Omega \). Then it is required that the surface integral, by Divergence Theorem:
\begin{align}
\int_{\partial\Omega} g(x,y,\ldots) d\sigma &= \int_{\partial\Omega} \frac{\partial u}{\partial n} d\sigma \\
&= \int_{\partial\Omega} \nabla u \cdot d\vec{\sigma} \\
&= \int_{\Omega} (\nabla\cdot\nabla u) dV \\
&= \int_{\Omega} (\nabla^2u) dV = 0 \tag{7}
\end{align}
Physically, it means that the inflow and outflow of the material across the boundary must balance.
Energy Method for Heat and Wave Equations
To show the uniqueness of solution for both the heat and wave equations, we need another approach called the energy method. We will start with the heat equation first and the argument is similar for the wave equation. We will also work with a one-dimensional space but it can be easily generalized to higher-dimensional spaces. Again, let \( u_1,u_2 \) be two solutions and \( w = u_1 -u_2 \) be their difference. Then this \(w\) also satisfies the heat equation (recall (7) of this tutorial):
\begin{align}
w_t -\kappa w_{xx} = 0 \tag{8}
\end{align}
Multiplying by \( w \) and integrating over the domain to obtain
\begin{align}
\int_{D} (w w_t -\kappa w w_{xx}) dx &= 0 \\
\int_{D} (\frac{1}{2}w^2)_t dx -\kappa [w w_{x}]_{\partial D} + \kappa \int_{D} (w_x)^2 dx &= 0 \\
\int_{D} (\frac{1}{2}w^2)_t dx -(0) + \kappa \int_{D} (w_x)^2 dx &= 0 \tag{9}
\end{align}
where we have integrated by parts and the resulting boundary terms vanish due to either Dirichlet/Neumann B.C. Hence
\begin{align}
\frac{d}{dt}\int_{D} \frac{1}{2}w^2 dx = \int_{D} (\frac{1}{2}w^2)_t dx &= -\kappa \int_{D} |w_x|^2 dx \leq 0 \tag{10}
\end{align}
as \( |w_x|^2 \geq 0 \). The quantity \( \int_{D} \frac{1}{2}w^2 dx \) is then the “energy”, denoted as \(E(t)\), and by (10) it is always decreasing in time, hence \( E(t) \leq E(0) \), i.e.
\begin{align}
0 \leq \int_{D} \frac{1}{2}w^2(x,t) dx \leq \int_{D} \frac{1}{2}w^2(x,0) dx = 0 \tag{11}
\end{align}
because \( u_1 \) and \( u_2 \) will obey the same I.C. and therefore \( w(x,0) = 0 \) at initial time. (11) then enforces that \( w(x,t) = 0 \) for all time, and thus \( u_1(x,t) = u_2(x,t) \) and we are done.
For the wave equation, the only difference is that the integration at the first step will require multiplying by \(w_t\) and look like
\begin{align}
\int_{D} (w_t w_{tt} -c^2 w_t w_{xx}) dx &= 0 \\
\int_{D} ((\frac{1}{2}w_t^2)_t + c^2 w_{xt} w_{x}) dx &= 0 \\
\int_{D} ((\frac{1}{2}w_t^2)_t + (\frac{1}{2}c^2 w_{x}^2)_t) dx &= 0 \tag{12} \\
\end{align}
where we used integration by parts, and the boundary term vanishes due to a suitable B.C. again. The energy in this scenario is then
\begin{align}
E(t) = \int_{D} (\frac{1}{2}w_t^2 + \frac{1}{2}c^2 w_{x}^2) dx \tag{13}
\end{align}
The subsequent logic will then be the same as above, leading to \( w(x,t) = 0 \).
Exercise
Prove the weak maximum principle for the Laplace equation, i.e. if \(u(x,y)\) is a solution to the Laplace equation over a domain \( \Omega \), then the maximum (and also minimum) value of \(u\) will always be attained at the boundary and never in the interior. Hint: Let \( w(x,y) = u(x,y) + \epsilon (x^2 + y^2) \) for some \( \epsilon > 0 \) and compute \( \nabla^2 w \).
Answer
The proof is from the book Applied Partial Differential Equations by Logan. Following the hint, we have
\begin{align}
\nabla^2 w = \nabla^2 u + 4 \epsilon = (0) + 4\epsilon > 0
\end{align}
Therefore \( \nabla^2 w > 0 \) in \( \Omega \). If \( w \) has a maximum within the interior, by multivariable Calculus we must have \( w_{xx}, w_{yy} \leq 0 \) at that point, contradicting \( \nabla^2 w > 0 \) just derived. So the maximum of \( w \) will occur at the boundary, let’s say at \( (x_b, y_b) \). Subsequently,
\begin{align}
u(x,y) &= w(x,y) -\epsilon (x^2 + y^2) \\
&< w(x,y) \\
&\leq w(x_b, y_b) \\
&= u(x_b, y_b) + \epsilon(x_b^2 + y_b^2)
\end{align}
Since \( \epsilon \) is arbitrary, we conclude that \( u(x,y) \leq u(x_b, y_b) \) for all \( (x,y) \in \Omega \). This inequality becomes strict if \( u \) is not a constant function.
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