Solution to Partial Differential Equations: Methods and Applications (Robert McOwen) Section 4.4

1. Consider (74) with $\Omega=(0,a)\times(0,b)$, the rectangle of Example 1.
   1. Use the eigenvalues and eigenfunctions computed in the text to express the solution (77).
   2. Find the frequencies of the special solutions (78).
   3. Assume $a=b/2$. Find the two smallest frequencies and their corresponding nodal curves.
Solution
1. We firstly recall that the eigenvalues and eigenfunctions for $-\Delta$ on $\Omega=(0,a)\times(0,b)$:

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

   This means $-\Delta\phi_{mn}=\lambda_{mn}\phi_{mn}$ for $m,n\in\mathbb N$ and $(x,y)\in\Omega$. In addition, we recall the wave equation with initial/boundary value problem (74) can be denoed as

   $\left\{\begin{aligned} &u_{tt}=u_{xx}+u_{yy}\quad\text{for}~(x,y)\in(0,a)\times(0,b)~\text{and}~t>0,\\&u(x,y,0)=g(x,y),\quad u_t(x,y,0)=h(x,y,0)\quad\text{for}~(x,y)\in(0,a)\times(0,b),\\&u(x,y,0)=0\quad\text{for}~(x,y)\in\partial\Omega~\text{and}~t>0.\end{aligned}\right.$

   Using the eigenfunction expansions, the initial data can be represented as

   $\begin{aligned} &g(x,y)=\sum_{m,n=1}^\infty G_{mn}\phi_{mn}(x,y)=\sum_{m,n=1}^\infty G_{mn}\sin\frac{m\pi x}a\sin\frac{n\pi y}b,\\&h(x,y)=\sum_{m,n=1}^\infty H_{mn}\phi_{mn}(x,y)=\sum_{m,n=1}^\infty H_{mn}\sin\frac{m\pi x}a\sin\frac{n\pi y}b,\end{aligned}$

   where

   $\begin{aligned} &G_{mn}=\frac4{ab}\int_0^a\!\int_0^b\!g(x,y)\phi_{mn}(x,y)\,\mathrm dy\,\mathrm dx=\frac4{ab}\int_0^a\!\int_0^b\!g(x,y)\sin\frac{m\pi x}a\sin\frac{n\pi y}b\,\mathrm dy\,\mathrm dx,\\&H_{mn}=\frac4{ab}\int_0^a\!\int_0^b\!h(x,y)\phi_{mn}(x,y)\,\mathrm dy\,\mathrm dx=\frac4{ab}\int_0^a\!\int_0^b\!h(x,y)\sin\frac{m\pi x}a\sin\frac{n\pi y}b\,\mathrm dy\,\mathrm dx.\end{aligned}$

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

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

   Assuming sufficient convergence to pass the derivatives inside the summation, this gives $v_{mn}''(t)+\lambda_{mn}v_{mn}(t)=0$ for $t>0$ and $m,n\in\mathbb N$. Since $\lambda_{mn}>0$ for all $m,n\in\mathbb N$, this ordinary differential equation has general solution

   $v_{mn}(t)=A_{mn}\cos(t\sqrt{\lambda_{mn}})+B_{mn}\sin(t\sqrt{\lambda_{mn}})$.

   At $t=0$, we get

   $\begin{aligned} &\sum_{m,n=1}^\infty G_{mn}\phi_{mn}(x,y)=g(x,y)=u(x,y,0)=\sum_{m,n=1}^\infty v_{mn}(0)\phi_{mn}(x,y)=\sum_{m,n=1}^\infty A_{mn}\phi_{mn}(x,y),\\&\sum_{m,n=1}^\infty H_{mn}\phi_{mn}(x,y)=h(x,y)=u_t(x,y,0)=\sum_{m,n=1}^\infty v_{mn}'(0)\phi_{mn}(x,y)=\sum_{m,n=1}^\infty B_{mn}\sqrt{\lambda_{mn}}\phi_{mn}(x,y).\end{aligned}$

   This gives $A_{mn}=G_{mn}$ and $B_{mn}=H_{mn}/\sqrt{\lambda_{mn}}$, which implies

   $\begin{aligned} &A_{mn}=\frac4{ab}\int_0^a\!\int_0^b\!g(x,y)\sin\frac{m\pi x}a\sin\frac{n\pi y}b\,\mathrm dy\,\mathrm dx,\\&B_{mn}=\frac4{ab\sqrt{\lambda_{mn}}}\int_0^a\!\int_0^b\!h(x,y)\sin\frac{m\pi x}a\sin\frac{n\pi y}b\,\mathrm dy\,\mathrm dx.\end{aligned}$

   Therefore, the solution of (74) can be given by

   $\begin{aligned}u(x,y,t)&=\sum_{m,n=1}^\infty(A_{mn}\cos(t\sqrt{\lambda_{mn}})+B_{mn}\sin(t\sqrt{\lambda_{mn}}))\sin\frac{m\pi x}a\sin\frac{n\pi y}b\\&=\frac4{ab}\sum_{m,n=1}^\infty\sin\frac{m\pi x}a\sin\frac{n\pi y}b\cos(t\sqrt{\lambda_{mn}})\int_0^a\!\int_0^b\!g(s,t)\sin\frac{m\pi s}a\sin\frac{n\pi t}b\,\mathrm dt\,\mathrm ds\\&\quad+\frac4{ab}\sum_{m,n=1}^\infty\sin\frac{m\pi x}a\sin\frac{n\pi y}b\frac{\sin(t\sqrt{\lambda_{mn}})}{\sqrt{\lambda_{mn}}}\int_0^a\!\int_0^b\!h(s,t)\sin\frac{m\pi s}a\sin\frac{n\pi t}b\,\mathrm dt\,\mathrm ds.\end{aligned}$

2. By a phase shift, we may assume $B_{mn}=0$ and then

   $v_{mn}(t)\phi_{mn}=A_{mn}\cos(t\sqrt{\lambda_{mn}})\phi_{mn}(x,y)$.

   It is clear that the frequency $\displaystyle f_{mn}=\frac{\sqrt{\lambda_{mn}}}{2\pi}=\frac12\sqrt{\frac{m^2}{a^2}+\frac{n^2}{b^2}}$ for $m,n\in\mathbb N$.
   
   The original exercise says that "show that thery are not integral multiples of each other", which is not true. In fact, it is possible that some of them are integral multiple of each other. For example, one may find that

   $\displaystyle f_{nn}=\frac1{\frac{n^2}{a^2}+\frac{n^2}{b^2}}=\frac n2\sqrt{\frac1{a^2}+\frac1{b^2}}=nf_{11}\quad\text{for}~n\in\mathbb N$.

   More generally, we have

   $\displaystyle f_{(mk)(nk)}=\frac12\sqrt{\frac{(mk)^2}{a^2}+\frac{(nk)^2}{b^2}}=\frac k2\sqrt{\frac{m^2}{a^2}+\frac{n^2}{b^2}}=kf_{mn}\quad\text{for}~m,n,k\in\mathbb N$.

3. When $a=b/2$, then the frequencies $f_{mn}$ can be denoted as

   $\displaystyle f_{mn}=\frac12\sqrt{\frac{m^2}{a^2}+\frac{n^2}{4a^2}}=\frac{\sqrt{4m^2+n^2}}{4a}$.

   Clearly, the two smallest frequencies are $\displaystyle f_{11}=\frac{\sqrt5}{4a}=\frac{\sqrt5}{2b}$ and $\displaystyle f_{12}=\frac{\sqrt2}{2a}=\frac{\sqrt2}b$.
   • For $m=n=1$, the nodal curve is the set of zeros of $\phi_{11}(x,y)=0$, i.e.,

     $\displaystyle\sin\frac{\pi x}a\sin\frac{\pi y}{2a}=0$,

     which is impossible because $x\in(0,a)$ and $y\in(0,b)=(0,2a)$. This means there is no any nodal curves.
   • For $m=1$ and $n=2$, the nodal curve is the set of zeros of $\phi_{12}(x,y)=0$, i.e.,

     $\displaystyle\sin\frac{\pi x}a\sin\frac{2\pi y}{2a}=0$,

     which gives the horizontal line $y=a=b/2$ as the nodal curve.


2. Let $\Omega=(0,a)\times(0,b)\times(0,c)\subseteq\mathbb R^3$. Find the Dirichlet eigenvalues and eigenfunctions for $\Delta$ in $\Omega$.
Solution

Consider the following Dirichelt eigenvalue problem:

$\left\{\begin{aligned} &\Delta u+\lambda u=u_{xx}+u_{yy}+u_{zz}+\lambda u=0\quad\text{in}~\Omega=(0,a)\times(0,b)\times(0,c),\\&u(0,y,z)=u(a,y,z)=0\quad\text{for}~(y,z)\in[0,b]\times[0,c],\\&u(x,0,z)=u(x,b,z)=0\quad\text{for}~(x,z)\in[0,a]\times[0,c],\\&u(x,y,0)=u(x,y,c)=0\quad\text{for}~(x,y)\in[0,a]\times[0,b].\end{aligned}\right.$

Let $u(x,y,z)=X(x)Y(y)Z(z)$. Then from the partial differential equation, we have

$X''(x)Y(y)Z(z)+X(x)Y''(y)Z(z)+X(x)Y(y)Z''(z)+\lambda X(x)Y(y)Z(z)=0\quad\text{for}~(x,y,z)\in(0,a)\times(0,b)\times(0,c)$,

which implies

$\displaystyle\frac{X''(x)}{X(x)}+\frac{Y''(y)}{Y(y)}+\frac{Z''(z)}{Z(z)}=-\lambda\quad\text{for}~(x,y,z)\in(0,a)\times(0,b)\times(0,c).$

Since $X''/X$, $Y''/Y$ and $Z''/Z$ are single-variable functions, we get the ordinary differential equations

$X''(x)=\lambda_1X(x),\quad Y''(y)=\lambda_2Y(y),\quad Z''(z)=\lambda_3Z(z).$

On the other hand, since $X$, $Y$ and $Z$ are nonzero, the boundary conditions give $X(0)=X(a)=0$, $Y(0)=Y(b)=0$, $Z(0)=Z(c)=0$. If $\lambda_i\geq0$ for some $i\in\{1,2,3\}$, then we can use the boundary condition to obtain $X\equiv0$, $Y\equiv0$ or $Z\equiv0$. Thus, $\lambda_i$ are negative for $i=1,2,3$ so we write $\lambda_i=-\mu_i^2$ and the ordinary differential equations become

$X''(x)=-\mu_1^2X(x),\quad Y''(y)=-\mu_2^2Y(y),\quad Z''(z)=-\mu_3^2Z(z)$

with $\lambda=\mu_1^2+\mu_2^2+\mu_3^2$. It is easy to check that $\displaystyle\mu_1=\frac{m\pi}a$, $\displaystyle\mu_2=\frac{n\pi}b$ and $\displaystyle\mu_3=\frac{k\pi}c$ for $m,n,k\in\mathbb N$. Therefore, the eigenvalues are

$\displaystyle\lambda_{mnk}=\pi^2\left(\frac{m^2}{a^2}+\frac{n^2}{b^2}+\frac{k^2}{c^2}\right)$

associated with the eigenfunctions

$\displaystyle u_{mnk}(x,y,z)=\sin\frac{m\pi x}a\sin\frac{n\pi y}b\sin\frac{k\pi z}c\quad\text{for}~m,n,k\in\mathbb N.$

Remark. In most books, we usually regard $\lambda$ as the eigenvalues for $-\Delta$ rather than for $\Delta$.



3. Consider the initial/boundary value problem with forcing term

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

   Use Duhamel's principle and an expansion of $f$ in eigenfunctions to obtain a (formal) solution.
Solution 1

Based on Duhamel's principle, we firstly consider the following equations

$\left\{\begin{aligned} &U_{tt}(x,t,s)-\Delta U(x,t,s)=0\quad\text{for}~x\in\Omega~\text{and}~t,s>0,\\&U(x,0,s)=0\quad\text{for}~x\in\Omega~\text{and}~s>0,\\&U_t(x,0,s)=f(x,s)\quad\text{for}~x\in\Omega~\text{and}~s>0,\\&U(x,t,s)=0\quad\text{for}~x\in\partial\Omega~\text{and}~t,s>0.\end{aligned}\right.$

Let $\lambda_n$ and $\phi_n$ be eigenvalues and associated eigenfunctions of $-\Delta$ with the zero Diricihlet boundary condition. Then we may write

$\displaystyle f(x,s)=\sum_{n=1}^\infty b_n(s)\phi_n(x)$.

Suppose the solution $U$ has the following form:

$\displaystyle U(x,t,s)=\sum_{n=1}^\infty V_n(t,s)\phi_n(x).$

Since $\lambda_n$ are positive, it is easy to obtain

$V_n(t,s)=A_n(s)\cos(t\sqrt{\lambda_n})+B_n(s)\sin(t\sqrt{\lambda_n}),$

and hence

$\displaystyle U(x,t,s)=\sum_{n=1}^\infty[A(s)\cos(t\sqrt{\lambda_n})+B_n(s)\sin(t\sqrt{\lambda_n})]\phi_n(x)$.

Now we use the initial condition, we get $A_n(s)\equiv0$ and $B_n(s)=b_n(s)/\sqrt{\lambda_n}$ so the solution $U$ becomes

$\displaystyle U(x,t,s)=\sum_{n=1}^\infty\frac{b_n(s)}{\sqrt{\lambda_n}}\sin(t\sqrt{\lambda_n})\phi_n(x)$.

Finally, we apply Duhamel's principle to get the solution $u$ given by

$\displaystyle u(x,t)=\int_0^t\!U(x,t-s,s)\,\mathrm ds=\sum_{n=1}^\infty\frac{\phi_n(x)}{\sqrt{\lambda_n}}\int_0^t\!b_n(s)\sin[(t-s)\sqrt{\lambda_n}]\,\mathrm ds$.

Solution 2

Let $\lambda_n$ and $\phi_n$ be eigenvalues and associated eigenfunctions of $-\Delta$ with the zero Diricihlet boundary condition. Moreover, we suppose the solution $u$ and the forcing term $f$ can be expanded in eigenfunctions:

$\displaystyle u(x,t)=\sum_{n=1}^\infty a_n(t)\phi_n(t),\quad f(x,t)=\sum_{n=1}^\infty b_n(t)\phi_n(x),$

where

$\displaystyle a_n(t)=\int_\Omega\!u(x,t)\phi_n(x)\,\mathrm dx,\quad b_n(t)=\int_\Omega\!f(x,t)\phi_n(x)\,\mathrm dx$.

Then it suffices to determine the unknown coefficient $a_n$. For this purpose, we observe that

$\begin{aligned}a_n''(t)&=\int_\Omega\!u_{tt}(x,t)\phi_n(x)\,\mathrm dx=\int_\Omega\!(\Delta u(x,t)+f(x,t))\phi_n(x)\,\mathrm dx\\&=\int_\Omega\!\Delta u(x,t)\phi_n(x)\,\mathrm dx+\int_\Omega\!f(x,t)\phi_n(x)\,\mathrm dx\\&=\int_\Omega\!u(x,t)\Delta\phi_n(x)\,\mathrm dx+\int_\Omega\!f(x,t)\phi_n(x)\,\mathrm dx\quad\text{(by integration by parts twice)}\\&=-\lambda_n\int_\Omega\!u(x,t)\phi_n(x)\,\mathrm dx+\int_\Omega\!f(x,t)\phi_n(x)\,\mathrm dx\\&=-\lambda_na_n(t)+b_n(t).\end{aligned}$

In addition, the initial value data of $a_n$ are given by

$\displaystyle a_n(0)=\int_\Omega\!u(x,0)\phi_n(x)\,\mathrm dx=0,\quad a_n'(0)=\int_\Omega\!u_t(x,0)\phi_n(x)\,\mathrm dx=0.$

Hence $a_n$ satisfies

$\left\{\begin{aligned} &a_n''(t)+\lambda_na_n(t)=b_n(t)\quad\text{for}~t>0,\\&a_n(0)=a_n'(0)=0.\end{aligned}\right.$

Based on Duhamel's principle, we consider the following equations

$\left\{\begin{aligned} &A_n''(t,s)+\lambda_nA(t,s)=0\quad\text{for}~t,s>0,\\&A_n(0,s)=0\quad\text{for}~s>0,\\&A_n'(0,s)=b_n(s)\quad\text{for}~s>0.\end{aligned}\right.$

It is easy to solve

$A_n(t,s)=C_n(s)\cos(t\sqrt{\lambda_n})+D_n(s)\sin(t\sqrt{\lambda_n})$.

Then by initial condition, we get $C_1\equiv0$ and $D_n(s)=b_n(s)/\sqrt{\lambda_n}$. Thus, we get

$\displaystyle A_n(t,s)=\frac{b_n(s)}{\sqrt{\lambda_n}}\sin(t\sqrt{\lambda_n})$.

By Duhamel's principle, we obtain

$\displaystyle a_n(t)=\int_0^t\!A_n(t-s,s)\,\mathrm ds=\frac1{\sqrt{\lambda_n}}\int_0^t\!b_n(s)\sin[(t-s)\sqrt{\lambda_n}]\,\mathrm ds$.

Therefore, the solution $u$ can be represented as

$\displaystyle u(x,t)=\sum_{n=1}^\infty a_n(t)\phi_n(x)=\sum_{n=1}^\infty\frac{\phi_n(x)}{\sqrt{\lambda_n}}\int_0^t\!b_n(s)\sin[(t-s)\sqrt{\lambda_n}]\,\mathrm ds$.



4. Suppose the forcing term in the previous exercise is $f(x,t)=A(x)\sin\omega t$. Find the (formal) solution when (a) $\omega^2\neq\lambda_n$ for any of the eigenvalues $\lambda_n$, and (b) $\omega^2=\lambda_k$ for some $k$ (resonance).
Solution
We firstly expand $f$ as
$\displaystyle f(x,t)=\sum_{n=1}^\infty\phi_n(x)\int_\Omega\![A(y)\sin(\omega t)\phi_n(y)]\,\mathrm dy=\sum_{n=1}^\infty\!a_n\sin(\omega t)\phi_n(x):=\sum_{n=1}^\infty\!b_n(t)\phi_n(x),$
where $\displaystyle a_n=\int_\Omega\!A(y)\phi_n(y)\,\mathrm dx$ and $b_n(t)=a_n\sin(\omega t)$.
1. If $\omega^2\neq\lambda_n$ for any $n\in\mathbb N$, then we note the following trigometric integral

   \begin{align}\int_0^t\!\sin(\omega s)\sin[(t-s)\sqrt{\lambda_n}]\,\mathrm ds&=\frac12\int_0^t\!\{\cos[(\omega+\sqrt{\lambda_n})s-t\sqrt{\lambda_n}]-\cos[(\omega-\sqrt{\lambda_n})s+t\sqrt{\lambda_n}]\}\,\mathrm ds\notag\\&\label{eq:3}=\left.\frac12\left\{\frac{\sin[(\omega+\sqrt{\lambda_n})s-t\sqrt{\lambda_n}]}{\omega+\sqrt{\lambda_n}}-\frac{\sin[(\omega-\sqrt{\lambda_n})s+t\sqrt{\lambda_n}]}{\omega-\sqrt{\lambda_n}}\right\}\right|_0^t\\&=\frac12\left(\frac{\sin(\omega t)+\sin(t\sqrt{\lambda_n})}{\omega+\sqrt{\lambda_n}}-\frac{\sin(\omega t)-\sin(t\sqrt{\lambda_n})}{\omega-\sqrt{\lambda_n}}\right)\notag\\&=\frac{\omega\sin(t\sqrt{\lambda_n})-\sqrt{\lambda_n}\sin(\omega t)}{\omega^2-\lambda_n}\notag.\end{align}

   By the result of Exercise 3, the solution $u$ can be represented as

   $\begin{aligned}u(x,t)&=\sum_{n=1}^\infty\frac{a_n}{\sqrt{\lambda_n}}\frac{\omega\sin(t\sqrt{\lambda_n})-\sqrt{\lambda_n}\sin(\omega t)}{\omega^2-\lambda_n}\phi_n(x)\\&=\sum_{n=1}^\infty\frac{\phi_n(x)}{\sqrt{\lambda_n}}\frac{\omega\sin(t\sqrt{\lambda_n})-\sqrt{\lambda_n}\sin(\omega t)}{\omega^2-\lambda_n}\int_\Omega\!A(y)\phi_n(y)\,\mathrm dy.\end{aligned}$

2. Suppose $\omega^2=\lambda_k$ for some $k\in\mathbb N$. Then for $n\neq k$, the same calculation in \eqref{eq:3} gives

   $\displaystyle\int_0^t\sin(s\sqrt{\lambda_k})\sin[(t-s)\sqrt{\lambda_n}]\,\mathrm ds=\frac{\sqrt{\lambda_k}\sin(t\sqrt{\lambda_n})-\sqrt{\lambda_n}\sin(t\sqrt{\lambda_k})}{\lambda_k-\lambda_n}.$

   But for $n=k$, we have

   $\begin{aligned}\int_0^t\!\sin(s\sqrt{\lambda_k})\sin[(t-s)\sqrt{\lambda_k}]\,\mathrm ds&=\frac12\int_0^t\!\{\cos[2s\sqrt{\lambda_k}-t\sqrt{\lambda_k}]-\cos(t\sqrt{\lambda_k})\}\,\mathrm ds\\&=\left.\frac12\left(\frac{\sin[2s\sqrt{\lambda_k}-t\sqrt{\lambda_k}]}{2\sqrt{\lambda_k}}-s\cos(t\sqrt{\lambda_k})\right)\right|_0^t\\&=\frac{\sin(t\sqrt{\lambda_k})-t\sqrt{\lambda_k}\cos(t\sqrt{\lambda_k})}{2\sqrt{\lambda_k}}.\end{aligned}$

   Therefore, by the result of Exercise 3, the solution $u$ can be represented as

   $\displaystyle u(x,t)=\frac{a_k[\sin(t\sqrt{\lambda_k})-t\sqrt{\lambda_k}\cos(t\sqrt{\lambda_k})]}{2\lambda_k}\phi_k(x)+\sum_{n\neq k}\frac{a_n}{\sqrt{\lambda_n}}\frac{\sqrt{\lambda_k}\sin(t\sqrt{\lambda_n})-\sqrt{\lambda_n}\sin(t\sqrt{\lambda_k})}{\lambda_k-\lambda_n}\phi_n(x).$

   Here we recall that $\displaystyle a_m=\int_\Omega\!A(y)\phi_m(y)\,\mathrm dy$ for all $m\in\mathbb N$.


5. Let $\Omega=(0,a)\times(0,b)$ and consider the initial/boundary value problem

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

   1. Find the eigenvalues and eigenfunctions for Laplaican operator associated Neumann problem on $\Omega$.
   2. Find the solution as an expansion similar to (77).
Solution
1. By separation of variables, we let $u(x,y)=X(x)Y(y)$. Then the partial differential equation $\Delta u+\lambda u=0$ in $\Omega$ becomes

   $X''(x)Y(y)+X(x)Y''(y)+\lambda X(x)Y(y)=0\quad\text{for}~(x,y)\in(0,a)\times(0,b)$,

   which implies

   $\displaystyle\frac{X''(x)}{X(x)}+\frac{Y''(y)}{Y(y)}=-\lambda\quad\text{for}~(x,y)\in(0,a)\times(0,b).$

   Since $X''/X$ and $Y''/Y$ are single-variable functions, we get the ordinary differential equations

   $X''(x)=\lambda_1X(x),\quad Y''(y)=\lambda_2Y(y)$.

   On the other hand, since $X$ and $Y$ are nonzero, the Neuamann boundary condition give $X'(0)=X'(a)=0$ and $Y'(0)=Y'(b)=0$. If $\lambda_i>0$, then we can use the boundary condition to obtain $X\equiv0$ or $Y\equiv0$. Thus, $\lambda_i$ are nonpositive for $i=1,2,3$ so we write $\lambda_i=-\mu_i^2$ and the ordinary differential equations become

   $X''(x)=-\mu_1^2X(x),\quad Y''(y)=-\mu_2^2Y(y)$

   with $\lambda=\mu_1^2+\mu_2^2$. It is easy to check that $\displaystyle\mu_1=\frac{m\pi}a$ and $\displaystyle\mu_2=\frac{n\pi}b$ for $m,n\in\mathbb N\cup\{0\}$. Therefore, the eigenvalues are

   $\displaystyle\lambda_{mn}=\pi^2\left(\frac{m^2}{a^2}+\frac{n^2}{b^2}\right)$

   associated with the eigenfunctions

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

2. Using the eigenfunction expansion, the initial data can be represented as

   $\begin{aligned} &g(x,y)=\sum_{m,n=0}^\infty G_{mn}\phi_{mn}(x,y)=\sum_{m,n=0}^\infty G_{mn}\cos\frac{m\pi x}a\cos\frac{n\pi y}b,\\&h(x,y)=\sum_{m,n=0}^\infty H_{mn}\phi_{mn}(x,y)=\sum_{m,n=0}^\infty H_{mn}\cos\frac{m\pi x}a\cos\frac{n\pi y}b,\end{aligned}$

   where

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

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

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

   Assuming sufficient convergence to pass the derivatives inside the summation, this gives $v_{mn}''(t)+\lambda_{mn}v_{mn}(t)=0$ for $t>0$ and $m,n\in\mathbb N\cup\{0\}$. Since $\lambda_{mn}\geq0$ for all $m,n\in\mathbb N\cup\{0\}$, this ordinary differential equaiton has general solution

   $v_{mn}(t)=\begin{cases}A_{00}+B_{00}t&\text{if}~m=n=0,\\A_{mn}\cos(t\sqrt{\lambda_{mn}})+B_{mn}\sin(t\sqrt{\lambda_{mn}})&\text{if}~m^2+n^2>0.\end{cases}$

   Note that $v_{mn}(0)=A_{mn}$ for all $m,n\in\mathbb N\cup\{0\}$, $v_{00}'(0)=B_{00}$ and $v_{mn}'(0)=B_{mn}\sqrt{\lambda_{mn}}$ for $(m,n)\in(\mathbb N\cup\{0\})^2-\{(0,0)\}$. At $t=0$, we get

   $\begin{aligned} &\sum_{m,n=0}^\infty G_{mn}\phi_{mn}(x,y)=g(x,y)=u(x,y,0)=\sum_{m,n=0}^\infty v_{mn}(0)\phi_{mn}=\sum_{m,n=0}^\infty A_{mn}\phi_{mn}(x,y),\\&\sum_{m,n=0}^\infty H_{mn}\phi_{mn}(x,y)=h(x,y)=u_t(x,y,0)=\sum_{m,n=0}^\infty v_{mn}'(0)\phi_{mn}(x,y).\end{aligned}$

   This gives $A_{mn}=G_{mn}$ for $m,n\in\mathbb N\cup\{0\}$, and $B_{00}=H_{00}$ and $B_{mn}=H_{mn}/\sqrt{\lambda_{mn}}$ for $(m,n)\in(\mathbb N\cup\{0\})^2-\{(0,0)\}$. Therefore, the solution $u$ can be given by

   $\displaystyle u(x,y,t)=\sum_{m,n=0}^\infty\!(A_{mn}\cos(t\sqrt{\lambda_{mn}})+B_{mn}\sin(t\sqrt{\lambda_{mn}}))\cos\frac{m\pi x}a\cos\frac{n\pi y}b$,

   where the coefficents are given above.


6. The ordinary differential equation $r^2y''+ry'+(r^2-n^2)y=0$ for $y(r)$ in $r>0$ is called Bessel's equation of order $n$, where $n=0,1,2,\dots$. This equation has a solution $y=J_n(r)$, which behaves like $r^n$ as $r\to0$, and has an infinite number of zeros $J_n(r)$ at $r=\rho_{n1},\,\rho_{n2},\dots\to\infty$.
   1. If $\lambda>0$, show that $y=J_n(\sqrt\lambda r)$ solves the ordinary differential equation

      $r^2y''+ry'+(\lambda r^2-n^2)y=0.$

   2. If $\Omega=\{(x_1,x_2)\in\mathbb R^2\,:\,x_1^2+x_2^2<a\}$, then use separation of variables and part (a) to obtain eigenvalues and eigenfunctions for

      $\left\{\begin{aligned} &\Delta u+\lambda u=0\quad\text{in}~\Omega,\\&u=0\quad\text{on}~\partial\Omega.\end{aligned}\right.$

   3. Assuming the functions $g(r,\theta)$ and $h(r,\theta)$ may be expanded in the eigenfunctions of (b), find the solution of the initial/boundary value problem

      $\left\{\begin{aligned} &u_{tt}=\Delta u\quad\text{in}~\Omega,\\&u(r,\theta,0)=g(r,\theta)\quad\text{for}~0\leq r<a~\text{and}~0\leq\theta<2\pi,\\&u_t(r,\theta,0)=h(r,\theta)\quad\text{for}~0\leq r<a~\text{and}~0\leq\theta<2\pi,\\&u(a,\theta,t)=0\quad\text{for}~0\leq\theta<2\pi\quad\text{and}~t>0.\end{aligned}\right.$

Solution

1. We can compute the derivatives of $y$ directly to get

   $y'=\sqrt\lambda J_n'(\sqrt\lambda r),\quad y''=\lambda J_n''(\sqrt\lambda r)$.

   Then one may verify that

   $\begin{aligned}r^2y''+ry'+(\lambda r^2-n^2)y&=r^2\cdot\lambda J_n''(\sqrt\lambda r)+r\cdot\sqrt\lambda J_n'(\sqrt\lambda r)+(\lambda r^2-n^2)J_n(\sqrt\lambda r)\\&=(\sqrt\lambda r)^2J_n''(\sqrt\lambda r)+(\sqrt\lambda r)J_n'(\sqrt\lambda r)+((\sqrt\lambda r)^2-n^2)J_n(\sqrt\lambda r)=0.\end{aligned}$

   Hence function $J_n(\sqrt\lambda r)$ solve the ordinary differential equation $r^2y''+ry'+(\lambda r^2-n^2)y=0$.
2. Let $\tilde u(r,\theta)=u(x_1,x_2)$ with $r=\sqrt{x_1^2+x_2^2}$ and $\theta=\tan^{-1}(x_2/x_1)$, or equivalently, $x_1=r\cos\theta$ and $x_2=r\sin\theta$, where $r\in[0,a)$ and $\theta\in[0,2\pi)$. Then the partial differential equation with boundary condition becomes

   $\left\{\begin{aligned} &\tilde u_{rr}(r,\theta)+\frac1r\tilde u_r(r,\theta)+\frac1{r^2}\tilde u_{\theta\theta}(r,\theta)+\lambda\tilde u(r,\theta)=0\quad\text{for}~(r,\theta)\in[0,a)\times[0,2\pi),\\&\tilde u(a,\theta)=0\quad\text{for}~\theta\in[0,2\pi).\end{aligned}\right.$

   Using the separation of variables, we let $\tilde u(r,\theta)=R(r)\Theta(\theta)$. Plugging into the partial differential equation, we get

   $(r^2R''(r)+rR'(r))\Theta(\theta)+R(r)\Theta''(\theta)+\lambda R(r)\Theta(\theta)=0$,

   which implies

   $\displaystyle\frac{r^2R''(r)+rR'(r)+\lambda r^2R(r)}{R(r)}+\frac{\Theta''(\theta)}{\Theta(\theta)}=0.$

   Since $R$ and $\Theta$ are single-variable functions, there exsits a constant $\mu$ such that

   $\left\{\begin{aligned} &r^2R''(r)+rR'(r)+(\lambda r^2-\mu)R(r)=0,\\&\Theta''(\theta)+\mu\Theta(\theta)=0.\end{aligned}\right.$

   From the boundary condition, we have $R(a)=0$. In addition, we have the perodic condition: $\Theta^{(k)}(0)=\Theta^{(k)}(2\pi)$ for all $k\in\mathbb N\cup\{0\}$.
   • If $\mu<0$, then we write $\mu=-\eta^2$ for $\eta>0$. From $\Theta''(\theta)-\eta^2\Theta(\theta)=0$, we get

     $\Theta(\theta)=C_1e^{\eta\theta}+C_2e^{-\eta\theta}$.

     Since $\Theta(0)=\Theta(2\pi)$ and $\Theta'(0)=\Theta(2\pi)$, it is easy to derive $C_1=C_2=0$, which gives $\Theta\equiv0$. This is a trivial solution.
   • If $\mu=0$, then we have $\Theta''(\theta)=0$, which gives $\Theta(\theta)=C_1+C_2\theta$. By $\Theta(0)=\Theta(2\pi)$, we get $\Theta\equiv C_1$. This gives a nonzero solution, provided $C_1\neq0$.
   • If $\mu>0$, then we write $\mu=\kappa^2$ for $\kappa>0$. From $\Theta''(\theta)+\kappa^2\Theta(\theta)=0$, we get

     $\Theta(\theta)=C_1\cos(\kappa\theta)+C_2\sin(\kappa\theta)$.

     By $\Theta(0)=\Theta(2\pi)$, we find $\kappa=m\in\mathbb N$, i.e., $\mu=m^2$ for $m\in\mathbb N$.
   Thus, $\mu=m^2$ for $m\in\mathbb N\cup\{0\}$. Then $R(r)=J_m(\sqrt\lambda r)$. Then by $R(a)=0$, we $\sqrt\lambda a=\rho_{mn}$, which gives $\lambda_{mn}=\rho_{mn}^2/a^2$ for $m\in\mathbb N\cup\{0\}$ and $n\in\mathbb N$. Therefore, the eigenvalues are $\lambda_{mn}=\rho_{mn}^2/a^2$ and associated eigenfunctions are

   $\phi_{mn}(r,\theta):=J_m\left(\frac{\rho_{mn}r}a\right)(C_1\cos(m\theta)+C_2\sin(m\theta))\quad\text{for}~m\in\mathbb N\cup\{0\}~\text{and}~n\in\mathbb N.$

3. Based on the assumption that functions $g$ and $h$ can be exapnded in the eigenfunctions of (b), we wrtie

   $\begin{aligned} &g(r,\theta)=\sum_{m=0,n=1}^\infty J_m\left(\frac{\rho_{mn}r}a\right)\left(G_{1,mn}\cos(m\theta)+G_{2,mn}\sin(m\theta)\right),\\&h(r,\theta)=\sum_{m=0,n=1}^\infty J_m\left(\frac{\rho_{mn}r}a\right)\left(H_{1,mn}\cos(m\theta)+H_{2,mn}\sin(m\theta)\right).\end{aligned}$

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

   $\displaystyle u(r,\theta,t)=\sum_{m=0,n=1}^\infty J_m\left(\frac{\rho_{mn}r}a\right)(v_{mn}(t)\cos(m\theta)+w_{mn}(t)\sin(m\theta))$.

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

   $\displaystyle v_{mn}''(t)+\frac{\rho_{mn}^2}{a^2}v_{mn}(t)=0,\quad w_{mn}''(t)+\frac{\rho_{mn}^2}{a^2}w_{mn}(t)=0.$

   Hence we get

   $\displaystyle v_{mn}(t)=A_{mn}\cos\left(\frac{\rho_{mn}t}a\right)+B_{mn}\sin\left(\frac{\rho_{mn}t}a\right),\quad w_{mn}(t)=C_{mn}\cos\left(\frac{\rho_{mn}t}a\right)+D_{mn}\sin\left(\frac{\rho_{mn}t}a\right).$

   At $t=0$, we get

   $\begin{aligned} &\quad\sum_{m=0,n=1}^\infty J_m\left(\frac{\rho_{mn}r}a\right)(G_{1,mn}\cos(m\theta)+G_{2,mn}\sin(m\theta))=g(r,\theta)=u(r,\theta,0)\\&=\sum_{m=0,n=1}^\infty J_m\left(\frac{\rho_{mn}r}a\right)(v_{mn}(0)\cos(m\theta)+w_{mn}(0)\sin(m\theta))\\&=\sum_{m=0,n=1}^\infty J_m\left(\frac{\rho_{mn}r}a\right)(A_{mn}\cos(m\theta)+C_{mn}\sin(m\theta)),\\&\quad\sum_{m=0,n=1}^\infty J_m\left(\frac{\rho_{mn}r}a\right)(H_{1,mn}\cos(m\theta)+H_{2,mn}\sin(m\theta))=h(r,\theta)=u_t(r,\theta,0)\\&=\sum_{m=0,n=1}^\infty J_m\left(\frac{\rho_{mn}r}a\right)(v_{mn}'(0)\cos(m\theta)+w_{mn}'(0)\sin(m\theta))\\&=\sum_{m=0,n=1}^\infty J_m\left(\frac{\rho_{mn}r}a\right)\left[\frac{\rho_{mn}}aB_{mn}\cos(m\theta)+\frac{\rho_{mn}}aD_{mn}\sin(m\theta)\right].\end{aligned}$

   Thus, $A_{mn}=G_{1,mn}$, $B_{mn}=aH_{1,mn}/\rho_{mn}$, $C_{mn}=G_{2,mn}$ and $D_{mn}=aH_{2,mn}/\rho_{mn}$ for $m\in\mathbb N\cup\{0\}$ and $n\in\mathbb N$. Therefore, the solutino $u$ is given by

   $\begin{aligned}u(r,\theta,t)&=\sum_{m=0,n=1}^\infty J_m\left(\frac{\rho_{mn}r}a\right)\left(G_{1,mn}\cos\left(\frac{\rho_{mn}t}a\right)+a\frac{H_{1,mn}}{\rho_{mn}}\sin\left(\frac{\rho_{mn}t}a\right)\right)\cos(m\theta)\\&\quad+\sum_{m=0,n=1}^\infty J_m\left(\frac{\rho_{mn}r}a\right)\left(G_{2,mn}\cos\left(\frac{\rho_{mn}t}a\right)+a\frac{H_{2,mn}}{\rho_{mn}}\sin\left(\frac{\rho_{mn}t}a\right)\right)\sin(m\theta).\end{aligned}$



7. If $\lambda\leq0$ and $\partial\Omega$ is smooth, show that only solution $u\in C^2(\bar\Omega)$ of (62) is the trivial solution $u\equiv0$.
Solution

Multiplying (62) by $u$ and then integrating it over $\Omega$, we get

$\displaystyle0\leq-\int_\Omega\!(|\nabla u|^2-\lambda u^2)\,\mathrm dx=\int_\Omega\!(u\Delta u+\lambda u^2)\,\mathrm dx=0$.

Here we have used the integration by parts with the zero Dirichlet boundary condition and the fact that $|\nabla u|^2-\lambda u^2\geq0$ in $\Omega$ because $\lambda\leq0$. Thus, we get $|\nabla u|^2-\lambda u^2\equiv0$, and hence $u\equiv0$. The proof is complete.

Remark. This means there is no negative eigenvalue for $-\Delta$, i.e., $-\Delta u=\lambda u$.