6.6. Multidimensional equations


## 6.6. Multidimensional equations

In this section we consider separation of variables for multidimentional equations in the simplest cases. More complicated cases will be considered in Chapter 7.

### Non-stationary equations

We consider only wave equation $$u_{tt}-c^2\Delta u =0 \label{eq-6.6.1}$$ in the rectangular box $\Omega=\{\mathbf{x}=(x_1,x_2,\ldots,x_n)\colon 0< x_1 <a_1,\ldots, 0<x_n< a_n\}$.

For $n=2$ we have a rectangle $\Omega=\{(x,y)\colon 0< x <a,0<y< b\}$.

Then, plugging $u(x,y,t)=X(x)Y(y)T(t)$ we get \begin{equation*} T''(t)X(x)Y(y)- c^2 T(t)X''(x)Y(y)- c^2 T(t)X(x)Y''(y), \end{equation*} which can be rewritten as $$\underbracket{\frac{T''(t)}{T(t)}}- c^2 \underbracket{\frac{X''(x)}{X(x)}}- c^2 \underbracket{\frac{Y''(y)}{Y(y}}=0. \label{eq-6.6.2}$$ Observe that each of selected terms depends only on it's own variable, and repeating arguments of 1D-case we conclude that it must be a constant: \begin{equation*} \frac{X''}{X}=-\lambda, \frac{Y''}{Y}=-\mu, \quad \frac{T''}{T}=-\nu = -c^2(\lambda+\mu). \end{equation*} In other words $$X''+\lambda X=0,\qquad Y''+\mu Y=0,\qquad T''+\nu T=0.\label{eq-6.6.3}$$

Next, we assume that $u$ must satisfy Dirichlet boundary conditions: $$u|_{\Gamma}=0, \label{eq-6.6.4}$$ where $\Gamma=\partial\Omega$ is the boundary of $\Omega$, which in our case consists of four segments $\{(x,y)\colon x=0,\ 0<y<b\}$, $\{(x,y)\colon x=a,\ 0<y<b\}$, $\{(x,y)\colon y=0,\ 0<x<a\}$, $\{(x,y)\colon y=b,\ 0<x<a\}$.

(In $n$-dimensional case $\Gamma$ consists of $2n$ rectangular boxes of dimension $(n-1)$ each).

Then, epeating arguments of 1D-case we conclude that $$X(0)=X(a)=0,\qquad Y(0)=Y(b)=0. \label{eq-6.6.5}$$ Thus we got the same problems for $X$ and for $Y$ as in 1D-case, and therefore \begin{align} &X_m(x)=\sin (\frac{\pi m x}{a}),
&&Y_n (y)=\sin (\frac{\pi n y}{b}), \label{eq-6.6.6}\\ &\lambda_m= (\frac{\pi m }{a})^2, &&\mu_n =(\frac{\pi n }{b})^2 \label{eq-6.6.7} \end{align} with $m,n=1,2,\ldots$, and therefore \begin{gather} T_{mn}(t)= A_{mn}\cos (\omega_{mn}t)+ B_{mn}\sin (\omega_{mn}t),\label{eq-6.6.8}\\ \omega_{mn}= c \pi \bigl( \frac{m^2}{a^2}+\frac{n^2}{b^2}\bigr)^{\frac{1}{2}}. \label{eq-6.6.9} \end{gather} Finally, \begin{multline} u_{mn}(x,y,t)=\\ \Bigl(A_{mn}\cos (\omega_{mn}t)+ B_{mn}\sin (\omega_{mn}t)\Bigr) \sin (\frac{\pi m x}{a}) \sin (\frac{\pi n y}{b}) \label{eq-6.6.10} \end{multline} and the general solution is derived in the form of multidimensional Fourier series \begin{multline} u(x,y,t)=\\ \sum_{m=1}^\infty \sum_{n=1}^\infty \Bigl(A_{mn}\cos (\omega_{mn}t)+ B_{mn}\sin (\omega_{mn}t)\Bigr) \sin (\frac{\pi m x}{a}) \sin (\frac{\pi n y}{b}). \label{eq-6.6.11} \end{multline} One can find coefficients from initial conditions $u|_{t=0}=g(x,y)$ and $u_t|_{t=0}=h(x,y)$,

Visual examples (animation)

Remark 1. While solution (\ref{eq-6.6.10}) is periodic with respect to $t$, solution (\ref{eq-6.6.11}) is not (in the generic case).

Remark 2.

1. The same arguments work in higher dimensions.

2. The same arguments work for Neumann and Robin boundary conditions and on each face of $\Gamma$ could be its own boundary condition.

3. The same arguments work for many other uquations, like heat equation, or Schrödinger equation.

### Stationary equations

Consider Laplace equation in the rectangular box $\Omega$, with the Dirichlet boundary conditions on $\Gamma=\partial\Omega$: \begin{align} &\Delta u=0, \label{eq-6.6.12}\\ &u|_\gamma =g. \label{eq-6.6.13} \end{align} We consider $n=3$, so $\Omega=\{(x,y,z)\colon 0<x< a, \ 0<y< b, 0<z< c\}$. Without any loss of the generality we can assume that $g=0$ on all faces, except two opposite $z=0$ and $z=c$.

Again we are looking for solution $u(x,y,z)=X(x)Y(y)Z(z)$ and separating variables we get $$\underbracket{\frac{X''(x)}{X(x)}} + \underbracket{\frac{Y''(y)}{Y(y)}}+ \underbracket{\frac{Z''(z)}{Z(z}}=0\label{eq-6.6.14}$$ and therefore $$X''+\lambda X=0,\qquad Y''+\mu Y=0,\qquad Z''-\nu Z=0\label{eq-6.6.15}$$ with $\nu =\lambda+\mu$.

We also get (\ref{eq-6.6.5}) and then (\ref{eq-6.6.6})-(\ref{eq-6.6.7}). Therefore \begin{gather} Z_{mn}(z)= A_{mn}\cosh (\omega_{mn}z)+ B_{mn}\sinh (\omega_{mn}z),\label{eq-6.6.16}\\ \omega_{mn}= \pi \bigl( \frac{m^2}{a^2}+\frac{n^2}{b^2}\bigr)^{\frac{1}{2}}. \label{eq-6.6.17} \end{gather} Finally, \begin{multline} u_{mn}(x,y,t)=\\ \Bigl(A_{mn}\cosh (\omega_{mn}z)+ B_{mn}\sinh (\omega_{mn}z)\Bigr) \sin (\frac{\pi m x}{a}) \sin (\frac{\pi n y}{b}) \label{eq-6.6.18} \end{multline} and the general solution is derived in the form of multidimensional Fourier series \begin{multline} u(x,y,t)=\\ \sum_{m=1}^\infty \sum_{n=1}^\infty \Bigl(A_{mn}\cosh (\omega_{mn}z)+ B_{mn}\sinh (\omega_{mn}z)\Bigr) \sin (\frac{\pi m x}{a}) \sin (\frac{\pi n y}{b}). \label{eq-6.6.19} \end{multline} One can find coefficients from initial conditions $u|_{z=0}=g(x,y)$ and $u|_{z=c}=h(x,y)$.

Remark 3.

1. The same arguments work in higher dimensions.

2. The same arguments work for Neumann and Robin boundary conditions and on each face of $\Gamma$ could be its own boundary condition.