Chapter 10.7, Problem 2E

In Problems 1-5, find a formal solution to the given boundary value problem.

$\begin{array}{rll}\frac{{\partial }^{2}u}{\partial x^2}+\frac{{\partial }^{2}u}{\partial y^2}& =& 0,0<x<\pi ,0<y<\pi ,\\ \frac{\partial u}{\partial x}\left(0,y\right)& =& \frac{\partial u}{\partial x}\left(\pi ,y\right)& =& 0,0\le y\le \pi ,\\ u\left(x,0\right)& =& \cos x-2\cos 4x,0\le x\le \pi ,\\ u\left(x,\pi \right)& =& 0,0\le x\le \pi \end{array}$

To determine

The formal solution to the given initial value problem $u\left(x,0\right)=\cos x-2\cos 4x,0\le x\le \pi ,u\left(x,\pi \right)=0,0\le x\le \pi$.

Answer to Problem 2E

Solution:

The formal solution to the given initial value problem is $u\left(x,y\right)=E_0\left(y-\pi \right)+{\displaystyle \sum _{n=1}^{\infty }{E}_n\cos nx\sinh n\left(y-\pi \right)}$.

Explanation of Solution

It is given that

$\frac{{\partial }^{2}u}{\partial x^2}+\frac{{\partial }^{2}u}{\partial y^2}=0,0<x<\pi ,0<y<\pi \mathrm{...}\left(1\right)$

With boundary conditions,

$\frac{\partial u}{\partial x}\left(0,y\right)=\frac{\partial u}{\partial x}\left(\pi ,y\right)=0,0\le y\le \pi \mathrm{...}\left(2\right)$

and initial conditions,

$u\left(x,0\right)=\cos x-2\cos 4x,0\le x\le \pi ,\mathrm{...}\left(3\right)$

$u\left(x,\pi \right)=0,0\le x\le \pi \mathrm{...}\left(4\right)$

Consider the solution of the equation be $u\left(x,y\right)=X\left(x\right)Y\left(y\right)$.

Above equation satisfies equation (1), so it becomes,

$\begin{array}{rll}{X}^{″}\left(x\right)Y\left(y\right)+X\left(x\right)Y^{″}\left(y\right)& =& 0\\ \frac{X^{″}\left(x\right)}{X\left(x\right)}& =& -\frac{Y^{″}\left(y\right)}{Y\left(y\right)}& =& -\lambda \\ X^{″}\left(x\right)+\lambda X\left(x\right)& =& 0\mathrm{...}\left(5\right)\end{array}$

and

$Y^{″}\left(y\right)-\lambda Y\left(y\right)=0\mathrm{...}\left(6\right)$

Boundary condition in equation (2) becomes,

$X^{\prime }\left(0\right)=X^{\prime }\left(\pi \right)=0$

So, equation (5) becomes, $X^{″}\left(x\right)+\lambda X\left(x\right)=0$ with boundary conditions $X^{\prime }\left(0\right)=X^{\prime }\left(\pi \right)=0$.

The auxiliary equation becomes $r^2+\lambda =0$.

If $\lambda =0$, then

$\begin{array}{rll}{r}^2& =& 0\\ r& =& 0\end{array}$

Thus $r=0$ is the repeated root.

So, the general solution becomes $X\left(x\right)=C_{1}x+C_2$.

By boundary condition,

$\begin{array}{rll}{X}^{\prime }\left(0\right)& =& C_1\\ C_1& =& 0\end{array}$

and

$\begin{array}{rll}{X}^{\prime }\left(\pi \right)& =& C_1\\ C_1& =& 0\end{array}$

As, $C_1=0$ then $C_2$ can be any constant. Then $X\left(x\right)=c_2$.

Now, if $\lambda >0$, then the roots becomes,

$\begin{array}{rll}{r}^2+\lambda & =& 0\\ r^2& =& -\lambda \\ r& =& \pm i\sqrt{\lambda }\end{array}$

The general solution becomes,

$X\left(x\right)=c_1\cos \sqrt{\lambda }x+c_2\sin \sqrt{\lambda }x$

By boundary condition,

$\begin{array}{rll}{X}^{\prime }\left(0\right)& =& -\sqrt{\lambda }{c}_1\sin 0+\sqrt{\lambda }{c}_2\cos 0\\ 0& =& c_2\end{array}$

and

$\begin{array}{rll}{X}^{\prime }\left(\pi \right)& =& -\sqrt{\lambda }{c}_1\sin \sqrt{\lambda }\pi +\sqrt{\lambda }{c}_2\cos \sqrt{\lambda }\pi \\ 0& =& \sqrt{\lambda }{c}_1\sin \sqrt{\lambda }\pi \end{array}$

Here $\sqrt{\lambda }\ne 0,c_1\ne 0$, then

$\begin{array}{rll}\sin \sqrt{\lambda }\pi & =& 0\\ \sin \sqrt{\lambda }\pi & =& \sin n\pi \\ \sqrt{\lambda }& =& n\\ \lambda & =& n^2\end{array}$

which is given by

$\begin{array}{rll}X\left(x\right)& =& c_1\cos nx\\ X_n\left(x\right)& =& c_n\cos nx\end{array}$

Now, put the value of $\lambda$ in (6),

$Y^{″}\left(y\right)-n^{2}Y\left(y\right)=0$

The auxiliary equation will be,

$\begin{array}{rll}{r}^2-n^2& =& 0\\ r& =& \pm n\end{array}$

The solution becomes,

$Y_n\left(y\right)=c_1{e}^{ny}+c_2{e}^{-ny}$

Apply $\cosh z=\frac{e^z+e^{-z}}{2}$ and $\sinh z=\frac{e^z-e^{-z}}{2}$ in above equation,

$Y_n\left(y\right)=A_n\cosh \left(ny\right)+B_n\sinh \left(ny\right)\mathrm{...}\left(7\right)$

Now, the boundary conditions in (4) will be satisfied if $Y\left(\pi \right)=0$, so let $y=\pi$ in equation (7).

$\begin{array}{rll}{Y}_n\left(\pi \right)& =& A_n\cosh \left(n\pi \right)+B_n\sinh \left(n\pi \right)\\ 0& =& A_n\cosh \left(n\pi \right)+B_n\sinh \left(n\pi \right)\\ A_n\cosh \left(n\pi \right)& =& -B_n\sinh \left(n\pi \right)\\ A_n& =& -B_n\tanh \left(n\pi \right)\end{array}$

Put in equation (7),

$\begin{array}{rll}{Y}_n\left(y\right)& =& \left(-B_n\tanh \left(n\pi \right)\right)\cosh \left(ny\right)+B_n\sinh \left(ny\right)\\ Y_n\left(y\right)& =& B_n\left[\left(-\frac{\sinh \left(n\pi \right)}{\cosh \left(n\pi \right)}\right)\cosh \left(ny\right)+\sinh \left(ny\right)\right]\\ Y_n\left(y\right)& =& \frac{B_n}{\cosh \left(n\pi \right)}\left[-\sinh \left(n\pi \right)\cosh \left(ny\right)+\cosh \left(n\pi \right)\sinh \left(ny\right)\right]\\ Y_n\left(y\right)& =& C_n\sinh n\left(y-\pi \right)\end{array}$

Put the value $X_n$ and $Y_n$ in $u_n\left(x,y\right)=X_n\left(x\right)Y_n\left(y\right)$,

$\begin{array}{rll}{u}_n\left(x,y\right)& =& \left(c_n\cos nx\right)\left(C_n\sinh n\left(y-\pi \right)\right)\\ u_n\left(x,y\right)& =& E_n\cos nx\sinh n\left(y-\pi \right)\end{array}$

Then the formal solution obtained is,

$u\left(x,y\right)=E_0\left(y-\pi \right)+{\displaystyle \sum _{n=1}^{\infty }{E}_n\cos nx\sinh n\left(y-\pi \right)}$

where,

$\begin{array}{rll}{E}_0& =& -\frac{1}{{\pi }^2}{\displaystyle \underset{0}{\overset{\pi }{\int }}\left(\cos x-2\cos 4x\right)dx}\\ E_n& =& \frac{2}{\pi \sinh \left(-n\pi \right)}{\displaystyle \underset{0}{\overset{\pi }{\int }}\left(\cos x-2\cos 4x\right)\cos nxdx}\end{array}$

The formal solution to the given initial value problem is $u\left(x,y\right)=E_0\left(y-\pi \right)+{\displaystyle \sum _{n=1}^{\infty }{E}_n\cos nx\sinh n\left(y-\pi \right)}$.