Solution to Partial Differential Equations: Methods and Applications (Robert McOwen) Section 5.1

1. A square two-dimensional plate of side length $a$ is heated to a uniform temperature $U_0>0$. Then at $t=0$ all sides are reduced to zero temperature. (Assume that the top and bottom of the plate are insulated so that the heat flow can be restricted to two dimensions.) Describe the heat diffusion $u(x,y,t)$.
Solution

The initial/boundary value problem for heat equation can be formulated as

$\left\{\begin{aligned} &u_t(x,y,t)=k\Delta u(x,y,t)=k(u_{xx}(x,y,t)+u_{yy}(x,y,t))\quad\text{for}~(x,y,t)\in(0,a)\times(0,a)\times(0,\infty),\\&u(x,y,0)=U_0>0\quad\text{for}~(x,y)\in(0,a)\times(0,a),\\&u(0,y,t)=u(a,y,t)=0\quad\text{for}~y\in[0,a]~\text{and}~t>0,\\&u(x,0,t)=u(x,a,t)=0\quad\text{for}~x\in[0,a]~\text{and}~t>0.\end{aligned}\right.$

Recall that the eigenvalues and associated eigenfunctions of $-\Delta$ with zero Dirichlet boundary condition over the two-dimensional square are

$\displaystyle\lambda_{mn}=\frac{\pi^2}{a^2}(m^2+n^2),\quad\phi_{mn}(x,y)=\sin\frac{m\pi x}a\sin\frac{n\pi y}a\quad\text{for}~m,n\in\mathbb N$.

The initial condition can be represented as

$\displaystyle U_0=\sum_{m,n=1}^\infty A_{mn}\phi_{mn}(x,y)=A_{mn}\sin\frac{m\pi x}a\sin\frac{n\pi y}a,$

where $A_{mn}$ is given by

$\begin{aligned}A_{mn}&=\frac4{a^2}\int_0^a\!\int_0^a\!u(x,y,0)\sin\frac{m\pi x}a\sin\frac{n\pi y}a\,\mathrm dy\,\mathrm dx\\&=\frac4{a^2}\cdot U_0\cdot\frac{a^2}{mn\pi^2}\cdot\left.\cos\frac{m\pi x}a\right|_0^a\cdot\left.\cos\frac{n\pi y}a\right|_0^a\\&=\frac{4(1-(-1)^m)(1-(-1)^n)}{mn\pi^2}U_0=\begin{cases}\displaystyle\frac{16}{mn\pi^2}U_0&\text{if}~m~\text{and}~n~\text{are odd},\\0&\text{otherwise}.\end{cases}\end{aligned}$

Let us assume that the solution $u$ can be expanded in the eigenfunctions with coefficients depending on $t$:

$\displaystyle u(x,y,t)=\sum_{m,n=1}^\infty v_{mn}(t)\phi_{mn}(x,y)=\sum_{m,n=1}^\infty v_{mn}(t)\sin\frac{m\pi x}a\sin\frac{n\pi y}a.$

Assuming sufficient convergence to pass the derivatives inside the summation, this gives

$v_{mn}'(t)+k\lambda_{mn}v_{mn}(t)=0\quad\text{for}~t>0~\text{and}~m,n\in\mathbb N$.

This ordinary differential equation has general solution

$v_{mn}(t)=v_{mn}(0)\exp(-k\lambda_{mn}t)\quad\text{for}~t\geq0$.

Hence we get

$\displaystyle u(x,y,t)=\sum_{m,n=1}^\infty v_{mn}(0)\exp(-k\lambda_{mn}t)\sin\frac{m\pi x}a\sin\frac{n\pi y}a.$

By the initial condition, we get $v_{mn}(0)=A_{mn}$ for all $m,n\in\mathbb N$. Therefore, we obtain

$\begin{aligned}u(x,y,t)&=\sum_{m,n=1}^\infty A_{mn}\exp(-k\lambda_{mn}t)\sin\frac{m\pi x}a\sin\frac{n\pi y}a\\&=\frac{16U_0}{\pi^2}\sum_{m',n'=1}^\infty\frac1{(2m'-1)(2n'-1)}\exp\left(-\frac{k\pi^2[(2m-1)^2+(2n-1)^2]}{a^2}t\right)\sin\frac{(2m-1)\pi x}a\sin\frac{(2n-1)\pi y}a.\end{aligned}$



2. Describe how you would solve the initial/boundary value problem (3) with the homogeneous Dirichlet condition replaced by $u(x,t)=h(x)$ for $x\in\partial\Omega$ and $t>0$. What happens to $u(x,t)$ as $t\to\infty$?
Solution

Consider the following Laplace equation with the Dirichlet boundary condition

$\left\{\begin{aligned} &\Delta w(x)=0\quad\text{for}~x\in\Omega,\\&w(x)=h(x)\quad\text{for}~x\in\partial\Omega.\end{aligned}\right.$

Then we consider the following initial value with homogeneous Dirichlet condition for heat equation:

$\displaystyle\left\{\begin{aligned} &v_t(x,t)=\Delta v(x,t)\quad\text{for}~x\in\Omega~\text{and}~t>0,\\&v(x,0)=g(x)-w(x)\quad\text{for}~x\in\bar\Omega,\\&v(x,t)=0\quad\text{for}~x\in\partial\Omega~\text{and}~t>0.\end{aligned}\right.$

By solving the above two problems, we can find that $u(x,t)=v(x,t)+w(x)$ solves the following initial/boundary value problem

$\left\{\begin{aligned} &u_t(x,t)=\Delta u(x,t)\quad\text{for}~x\in\Omega~\text{and}~t>0,\\&u(x,0)=g(x)\quad\text{for}~x\in\bar\Omega,\\&u(x,t)=h(x)\quad\text{for}~x\in\partial\Omega~\text{and}~t>0.\end{aligned}\right.$

Now we study the asymptotic behaviour of $u$ as $t$ tends to infinity. Clearly, $w$ is independent of $t$. Suppose that $\lambda_n$ and $\phi_n$ are eigenvalues and eigenfunctions for $-\Delta$ over $\Omega$ with the Dirichlet boundary condition, i.e., $-\Delta\phi_n=\lambda_n\phi_n$ in $\Omega$ with $\phi_n=0$ on $\partial\Omega$. Note that $\lambda_n>0$ for all $n\in\mathbb N$ and $\lambda_n\to\infty$ as $n\to\infty$. Then $v$ can be represented as

$\displaystyle v(x,t)=\sum_{n=1}^\infty e^{-\lambda_nt}\phi_n(x)\int_\Omega\!\phi_n(y)(g-w)(y)\,\mathrm dy$.

Since $\displaystyle\lim_{t\to\infty}e^{-\lambda_nt}=0$, we obtain $\displaystyle\lim_{t\to\infty}v(x,t)=0$, which implies

$\displaystyle\lim_{t\to\infty}u(x,t)=w(x)+\lim_{t\to\infty}v(x,t)=w(x)\quad\text{for}~x\in\bar\Omega$.



3. Suppose $\displaystyle f(x,t)=\sum_{n=1}^\infty a_n(t)\phi_n(x)$ converges absolutely and uniformly for $(x,t)\in\bar U_T$ for every $T>0$. Let $g(x)\equiv0$ and $h(x,t)\equiv0$, and describe the solution of (9).
Solution

Since $g\equiv0$ and $h\equiv0$, equations (9) become

$\left\{\begin{aligned} &u_t(x,t)=\Delta u(x,t)+f(x,t)\quad\text{for}~x\in\Omega~\text{and}~t>0,\\&u(x,0)=0\quad\text{for}~x\in\bar\Omega,\\&u(x,t)=0\quad\text{for}~x\in\partial\Omega~\text{and}~t>0.\end{aligned}\right.$

Let us assume that the soluion $u$ can be expanded in the eigenfunctions with coefficients depending on $t$:

$\displaystyle u(x,t)=\sum_{n=1}^\infty v_n(t)\phi_n(x)$.

Assuming sufficient convergence to pass the derivatives inside the summation, this gives

$v_n'(t)+\lambda_nv_n(t)=a_n(t)\quad\text{for}~t>0~\text{and}~n\in\mathbb N\quad\text{with}~v_n(0)=0$,

where $\lambda_n$ is the eigenvalue of $-\Delta$. Note that $\lambda_n>0$ for all $n\in\mathbb N$. Multiplying it by $e^{\lambda_nt}$, we get

$(e^{\lambda_nt}v_n(t))'=e^{\lambda_nt}v_n'(t)+\lambda_ne^{\lambda_nt}v_n(t)=e^{\lambda_nt}a_n(t)\quad\text{for}~t>0~\text{and}~n\in\mathbb N$,

which implies

$\displaystyle v_n(t)=\int_0^t\!e^{\lambda_n(s-t)}a_n(s)\,\mathrm ds$.

Therefore, the solution $u$ is given by

$\displaystyle u(x,t)=\sum_{n=1}^\infty\phi_n(x)\int_0^t\!e^{\lambda_n(s-t)}a_n(s)\,\mathrm ds$.



4. More generally than in Exercise 2, describe how you would solve the initial boundary value problem (3) with the homogeneous Dirichlet condition replaced by $u(x,t)=h(x,t)$ for $x\in\partial\Omega$ and $t>0$.
Solution

Suppose that $h(x,t)$ can be extends to $x\in\bar\Omega$ such that $h$ has the eigenfunction expansion:

$\displaystyle h(x,t)=\sum_{n=1}^\infty a_n(t)\phi_n(x)$,

which converges absolutely and uniformly on $\bar U_T$ for every $T>0$. Let $f(x,t)=h_t(x,t)-\Delta h(x,t)$ for $x\in\Omega$ and $t>0$. Clearly, $f$ admits the eigenfunction expansion:

$\displaystyle f(x,t)=\sum_{n=1}^\infty(a_n'(t)+\lambda_na_n(t))\phi_n(x)$,

which also converges absolutely and uniformly on $\bar U_T$ for every $T>0$. Then we consider the following initial/boundary value problem with homogeneous Dirichlet condition and forcing term:

$\left\{\begin{aligned} &v_t(x,t)=\Delta v(x,t)+f(x,t)\quad\text{for}~x\in\Omega~\text{and}~t>0,\\&v(x,0)=g(x)-h(x,0)\quad\text{for}~x\in\bar\Omega,\\&v(x,t)=0\quad\text{for}~x\in\bar\Omega~\text{and}~t>0.\end{aligned}\right.$

Using the solution of $v$, we may find that $u(x,t)=v(x,t)+h(x,t)$ solves

$\left\{\begin{aligned} &u_t(x,t)=\Delta u(x,t)\quad\text{for}~x\in\Omega~\text{and}~t>0,\\&u(x,0)=g(x)\quad\text{for}~x\in\bar\Omega,\\&u(x,t)=h(x,t)\quad\text{for}~x\in\partial\Omega~\text{and}~t>0,\end{aligned}\right.$

which is desired.

Now we derive the expansion solution of $u$ as follows. By the results of Exercise 2 and Exercise 3, we can get

$\displaystyle v(x,t)=\sum_{n=1}^\infty e^{-\lambda_nt}\phi_n(x)\left[\int_\Omega\!\phi_n(y)[g(y)-h(y,0)]\,\mathrm dy+\int_0^t\!e^{\lambda_ns}[a_n'(s)+\lambda_na_n(s)]\,\mathrm ds\right]$.

From $u(x,t)=v(x,t)+h(x,t)$, we obtain

$\displaystyle u(x,t)=\sum_{n=1}^\infty\phi_n(x)\left[a_n(t)+e^{-\lambda_nt}\int_\Omega\!\phi_n(y)[g(y)-h(y,0)]\,\mathrm dy+\int_0^t\!e^{-\lambda_n(t-s)}[a_n'(s)+\lambda_na_n(s)]\,\mathrm ds\right]$.



5. Suppose the square plate of Exercise 1 is given an initial temperature distribution $u(x,y,0)=g(x,y)$ and then insulated so that heat cannot flow across $\partial\Omega$ (i.e., $\partial u/\partial\nu=0$). Use an expansion in appropriate eigenfunctions to describe the heat diffusion $u(x,y,t)$.
Solution

The initial/boundary value problem for heat equation can be formulated as

$\left\{\begin{aligned} &u_t(x,y,t)=k\Delta u(x,y,t)=k(u_{xx}(x,y,t)+u_{yy}(x,y,t))\quad\text{for}~(x,y,t)\in(0,a)\times(0,a)\times(0,\infty),\\&u(x,y,0)=g(x,y)\quad\text{for}~(x,y)\in(0,a)\times(0,a),\\&u_x(0,y,t)=u_x(a,y,t)=0\quad\text{for}~y\in[0,a]~\text{and}~t>0,\\&u_y(x,0,t)=u_y(x,a,t)=0\quad\text{for}~x\in[0,a]~\text{and}~t>0.\end{aligned}\right.$

Recall that the eigenvalues and associated eigenfunctions of $-\Delta$ with zero Neumann boundary condition over the two-dimensional square are

$\displaystyle\lambda_{mn}=\frac{\pi^2}{a^2}(m^2+n^2),\quad\phi_{mn}(x,y)=\cos\frac{m\pi x}a\cos\frac{n\pi y}a\quad\text{for}~m,n\in\mathbb N\cup\{0\}.$

Using the eigenfunction expansion, the initial data can be represetned as

$\displaystyle g(x,y)=\sum_{m,n=0}^\infty G_{mn}\phi_{mn}(x,y)=\sum_{m,n=0}^\infty G_{mn}\cos\frac{m\pi x}a\cos\frac{n\pi y}a,$

where

$G_{mn}=\begin{cases}\displaystyle\frac1{a^2}\int_0^a\!\int_0^a\!g(x,y)\,\mathrm dy\,\mathrm dx&\text{if}~m=n=0,\\\displaystyle\frac2{a^2}\int_0^a\!\int_0^a\!g(x,y)\cos\frac{m\pi x}a\,\mathrm dy\,\mathrm dx&\text{if}~m\neq0,~n=0,\\\displaystyle\frac2{a^2}\int_0^a\!\int_0^a\!g(x,y)\cos\frac{n\pi y}a\,\mathrm dy\,\mathrm dx&\text{if}~m=0,~n\neq0,\\\displaystyle\frac4{a^2}\int_0^a\!\int_0^a\!g(x,y)\cos\frac{m\pi x}a\cos\frac{n\pi y}a\,\mathrm dy\,\mathrm dx&\text{if}~m\neq0,~n\neq0.\end{cases}$

Let us assume that the solution $u$ can be expanded in the eigenfunctions with coefficients depending on $t$:

$\displaystyle u(x,y,t)=\sum_{m,n=0}^\infty v_{mn}(t)\phi_{mn}(x,y)=\sum_{m,n=0}^\infty v_{mn}(t)\sin\frac{m\pi x}a\sin\frac{n\pi y}a.$

Assuming sufficient convergence to pass the derivatives inside the summation, this gives

$v_{mn}'(t)+k\lambda_{mn}v_{mn}(t)=0\quad\text{for}~t>0~\text{and}~m,n\in\mathbb N\cup\{0\}$.

This ordinary differential equation has general solution

$v_{mn}(t)=v_{mn}(0)\exp(-k\lambda_{mn}t)\quad\text{for}~t\geq0.$

Hence we get

$\displaystyle u(x,y,t)=\sum_{m,n=0}^\infty v_{mn}(0)\exp(-k\lambda_{mn}t)\cos\frac{m\pi x}a\cos\frac{n\pi y}a$.

By the initial condition, we get $v_{mn}(0)=G_{mn}$ for $m,n\in\mathbb N\cup\{0\}$. Therefore, we get

$\displaystyle u(x,y,t)=\sum_{m,n=0}^\infty G_{mn}\exp(-k\lambda_{mn}t)\cos\frac{m\pi x}a\cos\frac{n\pi y}a$.



6. Suppose $U=\Omega\times(0,T)$ and $u\in C^{2;1}(U)\cap C(\bar U)$ satisfies $u_t\leq\Delta u+cu$ in $U$, where $c\leq0$ is a constant. If $u\geq0$, show that (11) holds for $u$. Give a counterexample when $c<0$ and $u<0$ in $\bar U$.
Solution

If the condition "$u\geq0$" means $u\geq0$ in $\bar U$, then it is easy to find that $cu\leq0$ in $U$, and hence $v_t\leq \Delta u+cu\leq\Delta u$. Clearly, (11) follows from the weak maximum principle.

Now we consider the case that $u(x_0,t_0)\geq0$ for some $(x_0,t_0)\in\bar U$. Without loss generality, we may assume that $\displaystyle u(x_0,t_0)=\max_{\bar U}u\geq0$. Let us firstly assume that $u_t<\Delta u+cu$ in $U$. If $(x_0,t_0)\in U\cup(\partial\Omega\times\{T\})$, then $\nabla u(x_0,t_0)=0$ and $v_t(x_0,t_0)\geq0$ and $\Delta v(x_0,t_0)\leq0$, which implies

$0\leq u_t(x_0,t_0)<\Delta u(x_0,t_0)+cu(x_0,t_0)\leq0$.

This leads a contradiction, and hence $(x_0,t_0)\in\Gamma$, i.e., (11) holds true. For the case that $u_t\leq\Delta u+cu$ in $U$, we define $u^\epsilon(x,t)=u(x,t)-\epsilon t$ for $(x,t)\in\bar U$ and $\epsilon>0$. Then $u^\epsilon$ satisfies

$u^\epsilon_t=u_t-\epsilon\leq\Delta u+cu-\epsilon=\Delta u^\epsilon+cu^\epsilon+\epsilon(ct-1)<\Delta u^\epsilon+cu^\epsilon$.

Thus, by the previous case, we have

$\displaystyle\max_{(x,t)\in\bar U}[u(x,t)-\epsilon t]=\max_{(x,t)\in\bar U}u^\epsilon(x,t)=\max_{(x,t)\in\Gamma}u^\epsilon(x,t)=\max_{(x,t)\in\Gamma}[u(x,t)-\epsilon t]$.

Note that

$\begin{aligned} &\max_{(x,t)\in\bar U}u(x,t)-\epsilon T\leq\max_{(x,y)\in\bar U}[u(x,t)-\epsilon t]\leq\max_{(x,t)\in\bar U}u(x,t),\\&\max_{(x,t)\in\Gamma}u(x,t)-\epsilon T\leq\max_{(x,y)\in\Gamma}[u(x,t)-\epsilon t]\leq\max_{(x,t)\in\Gamma}u(x,t)\quad\text{for all}~\epsilon>0.\end{aligned}$

Letting $\epsilon\to0^+$, we obtain (11). The proof is complete.

Suppose that $c<0$ and $u<0$ in $\bar U$. I thinkt the simplest example is that $v(x,t)=-e^{ct}$ for $(x,t)\in\bar U$. Then it is easy to verify that

$v_t(x,t)=-ce^{-ct}=cv=\Delta v+cu\quad\text{in}~U$.

Since $c<0$, function $v$ is increasing in $t$ and the maximum value of $v$ occurs at $(x,t)\in\bar\Omega\times\{T\}$. Clearly, (11) does not hold.

Remark 1. It seems useless that consider $v(x,t)=e^{-ct}u(x,t)$ for $(x,t)\in\bar U$. Surely, one may obtain $$v_t(x,t)=-ce^{-ct}u(x,t)+e^{-ct}u_t(x,t)\leq-cv(x,t)+e^{-ct}\Delta u(x,t)+ce^{-ct}u(x,t)=\Delta v(x,t)\quad\text{in}~U.$$ Note that $e^{-ct}\geq1$ in $\bar U$ when $c\leq0$. Thus, by the weak maximum principle for $v$, we get $$\max_{(x,t)\in\bar U}u(x,t)\leq\max_{(x,t)\in\bar U}[e^{-ct}u(x,t)]=\max_{(x,t)\in\bar U}v(x,t)=\max_{(x,t)\in\Gamma}v(x,t)=\max_{(x,t)\in\Gamma}e^{-ct}u(x,t),$$ where $\Gamma=(\partial\Omega\times[0,T])\cup(\bar\Omega\times\{0\})$. This seems impossbile to derive (11).

Remark 2. The original question is that "If $u\geq0$, show that (11) holds for $u$. Give a counterexample without the condition $u\geq0$." The condition $u\geq0$ is ambiguous, which is already dealt within the solution. However, when $c=0$, the weak maximum principle still holds true so we need to require $c<0$. In addition, we need to make the assumption that $u<0$ in $\bar U$.

Remark 3. When $c\leq0$, we have $\displaystyle\max_{(x,t)\in\bar U}u(x,t)\leq\max_{(x,t)\in\Gamma}u^+(x,t)$, where $u^+=\max\{u,0\}$.



7. If $u$ satisfies (1), define its heat energy by $\displaystyle\mathcal E(t)=\int_\Omega\!u^2(x,t)\,\mathrm dx$.
   1. If $U=\Omega\times(0,\infty)$ and $u\in C^{2;1}(\bar U)$ satisfies (1) and either (i) $u=0$ on $\partial\Omega$, or (ii) $\partial u/\partial\nu=0$ on $\partial\Omega$, then prove that $\mathcal E(t)$ is nonincreasing in $t$.
   2. Use (a) to conclude the uniqueness of a solution $u\in C^{2;1}(\bar U)$ for either the nonhomogeneous Dirichlet problem (9) or the corresponding nonhomogeneous Neumann problem.
Solution
1. Since $u\in C^{2,1}(\bar U)$ satisfies $u_t=k\Delta u$ in $U$, we have

   $\begin{aligned}\mathcal E'(t)&=2\int_\Omega\!u(x,t)u_t(x,t)\,\mathrm dx=2k\int_\Omega\!u(x,t)\Delta u(x,t)\,\mathrm dx\\&=2k\int_{\partial\Omega}\!u(x,t)\frac{\partial u}{\partial\nu}(x,t)\,\mathrm dS_x-2k\int_\Omega|\nabla u(x,t)|^2\,\mathrm dx=-2k\int_\Omega|\nabla u(x,t)|^2\,\mathrm dx\leq0\quad\text{($k>0$)}.\end{aligned}$

   Here we have used the fact that $\displaystyle u\frac{\partial u}{\partial\nu}\equiv0$ on $\partial\Omega$, which follows from either (i) $u\equiv0$ on $\partial\Omega$ or (ii) $\displaystyle\frac{\partial u}{\partial\nu}\equiv0$ on $\partial\Omega$. Threfore, $\mathcal E$ is nonincreasing in $t$. The proof is complete.
2. • Suppose that there are two solutions $\tilde u$ and $u$ of the nonhomogeneous Dirichlet problem (9), i.e.,

     $\left\{\begin{aligned} &\tilde u_t(x,t)=\Delta\tilde u(x,t)+f(x,t)\quad\text{for}~x\in\Omega~\text{and}~t>0,\\&\tilde u(x,0)=g(x)\quad\text{for}~x\in\bar\Omega,\\&\tilde u(x,t)=h(x,t)\quad\text{for}~x\in\partial\Omega~\text{and}~t>0,\end{aligned}\right.\quad\left\{\begin{aligned} &u_t(x,t)=\Delta u(x,t)+f(x,t)\quad\text{for}~x\in\Omega~\text{and}~t>0,\\&u(x,0)=g(x)\quad\text{for}~x\in\bar\Omega,\\&u(x,t)=h(x,t)\quad\text{for}~x\in\partial\Omega~\text{and}~t>0,\end{aligned}\right.$

     Let $w(x,t)=\tilde u(x,t)-u(x,t)$ for $(x,t)\in\bar U$. Then $w$ satisfies

     $\left\{\begin{aligned} &w_t(x,t)=\Delta w(x,t)\quad\text{for}~x\in\Omega~\text{and}~t>0,\\&w(x,0)=0\quad\text{for}~x\in\bar\Omega,\\&w(x,t)=0\quad\text{for}~x\in\partial\Omega~\text{and}~t>0\end{aligned}\right.$

     and its heat energy is given by $\mathcal E_w(t)=\int_\Omega\!w^2(x,t)\,\mathrm dx$. By (a), it is clear that

     $\displaystyle\int_\Omega\!w^2(x,t)\,\mathrm dx=\mathcal E_w(t)\leq\mathcal E_w(0)=\int_\Omega\!w^2(x,0)\,\mathrm dx=0$,

     which implies $w\equiv0$ in $\bar U$ and hence $\tilde u\equiv u$ in $\bar U$. Therefore, we complete the proof of uniquenees of solution of the nonhomogeneous Dirichlet problem.
   • Suppose that there are two solutions $\tilde u$ and $u$ of the nonhomogeneous Neumann problem, i.e.,

     $\left\{\begin{aligned} &\tilde u_t(x,t)=\Delta\tilde u(x,t)+f(x,t)\quad\text{for}~x\in\Omega~\text{and}~t>0,\\&\tilde u(x,0)=g(x)\quad\text{for}~x\in\bar\Omega,\\&\frac{\partial\tilde u}{\partial\nu}(x,t)=h(x,t)\quad\text{for}~x\in\partial\Omega~\text{and}~t>0,\end{aligned}\right.\quad\left\{\begin{aligned} &u_t(x,t)=\Delta u(x,t)+f(x,t)\quad\text{for}~x\in\Omega~\text{and}~t>0,\\&u(x,0)=g(x)\quad\text{for}~x\in\bar\Omega,\\&\frac{\partial u}{\partial\nu}(x,t)=h(x,t)\quad\text{for}~x\in\partial\Omega~\text{and}~t>0,\end{aligned}\right.$

     Let $w(x,t)=\tilde u(x,t)-u(x,t)$ for $(x,t)\in\bar U$. Then $w$ satisfies

     $\left\{\begin{aligned} &w_t(x,t)=\Delta w(x,t)\quad\text{for}~x\in\Omega~\text{and}~t>0,\\&w(x,0)=0\quad\text{for}~x\in\bar\Omega,\\&\frac{\partial w}{\partial\nu}(x,t)=0\quad\text{for}~x\in\partial\Omega~\text{and}~t>0\end{aligned}\right.$

     and its heat energy is given by $\mathcal E_w(t)=\int_\Omega\!w^2(x,t)\,\mathrm dx$. By (a), it is clear that

     $\displaystyle\int_\Omega\!w^2(x,t)\,\mathrm dx=\mathcal E_w(t)\leq\mathcal E_w(0)=\int_\Omega\!w^2(x,0)\,\mathrm dx=0$,

     which implies $w\equiv0$ in $\bar U$ and hence $\tilde u\equiv u$ in $\bar U$. Therefore, we complete the proof of uniquenees of solution of the nonhomogeneous Neumann problem.