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2.5 HW 5

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2.5.1 Section 34, Problem 3

Figure 2.57:Problem statement

Solution

The boundary conditions are

Figure 2.58:Boundary conditions

Let $$u\left ( x,y\right ) =X\left ( x\right ) Y\left ( y\right )$$ Substitution in the PDE $u_{xx}+y_{yy}=0$ leads to $$\begin{align*} X^{\prime \prime }Y+Y^{\prime \prime }Y & =0\\ \frac{X^{\prime \prime }}{X} & =-\frac{Y^{\prime \prime }}{Y}=-\lambda \end{align*}$$

Where $\lambda$ is the separation constant. We obtain two ODE’s $$\begin{align} X^{\prime \prime }+\lambda X & =0\tag{1}\\ Y^{\prime \prime }-\lambda Y & =0\tag{2} \end{align}$$

We use the $X\left ( x\right )$ ODE (1) to determine the eigenvalues, since that ODE has both boundary conditions specified: $$\begin{align*} X^{\prime \prime }+\lambda X & =0\\ X^{\prime }\left ( 0\right ) & =0\\ X^{\prime }\left ( \pi \right ) & =0 \end{align*}$$

Case $\lambda <0$

Solution is $$\begin{align*} X\left ( x\right ) & =A\cosh \left ( \sqrt{-\lambda }x\right ) +B\sinh \left ( \sqrt{-\lambda }x\right ) \\ X^{\prime }\left ( x\right ) & =A\sqrt{-\lambda }\sinh \left ( \sqrt{-\lambda }x\right ) +B\sqrt{-\lambda }\cosh \left ( \sqrt{-\lambda }x\right ) \end{align*}$$

At $x=0$ the above gives $$\begin{align*} 0 & =B\sqrt{-\lambda }\cosh \left ( 0\right ) \\ & =B\sqrt{-\lambda } \end{align*}$$

Hence $B=0$ and the solution (3) reduces to $$\begin{align*} X\left ( x\right ) & =A\cosh \left ( \sqrt{-\lambda }x\right ) \\ X^{\prime }\left ( x\right ) & =A\sqrt{-\lambda }\sinh \left ( \sqrt{-\lambda }x\right ) \end{align*}$$

At $x=\pi$ the above becomes $$0=A\sqrt{-\lambda }\sinh \left ( \sqrt{-\lambda }\pi \right )$$ For non-trivial solution we want $\sinh \left ( \sqrt{-\lambda }\pi \right ) =0$, but $\sinh$ is only zero when its argument is zero, which is not possible here, since $\lambda \neq 0$. Therefore $\lambda <0$ is not possible.

Case $\lambda =0$

Solution becomes $X=Ax+B$. Hence $X^{\prime }=A$. At $x=0$ this leads to $A=0$. Therefore the solution now becomes $X=B$. Hence $X^{\prime }=0$. Therefore the second boundary conditions at $x=\pi$ is automatically satisfied. Hence the solution is $X\left ( x\right ) =B$, a constant. We pick $X\left ( x\right ) =1$. Therefore $\lambda =0$ is eigenvalue with associated eigenfunction $X_{0}\left ( x\right ) =1$.

Case $\lambda >0$

The solution becomes $$\begin{align*} X\left ( x\right ) & =A\cos \left ( \sqrt{\lambda }x\right ) +B\sin \left ( \sqrt{\lambda }x\right ) \\ X^{\prime }\left ( x\right ) & =-A\sqrt{\lambda }\sin \left ( \sqrt{\lambda }x\right ) +B\sqrt{\lambda }\cos \left ( \sqrt{\lambda }x\right ) \end{align*}$$

At $x=0$ the above becomes $$0=B\sqrt{\lambda }$$ Hence $B=0$ and the solution reduces to $$\begin{align*} X\left ( x\right ) & =A\cos \left ( \sqrt{\lambda }x\right ) \\ X^{\prime }\left ( x\right ) & =-A\sqrt{\lambda }\sin \left ( \sqrt{\lambda }x\right ) \end{align*}$$

At $x=\pi$ the above gives $$\begin{align*} 0 & =-A\sqrt{\lambda }\sin \left ( \sqrt{\lambda }\pi \right ) \\ \sin \left ( \sqrt{\lambda }\pi \right ) & =0 \end{align*}$$

Therefore $\sqrt{\lambda }\pi =n\pi$ for $n=1,2,3,\cdots$. Hence $$\lambda _{n}=n^{2}\qquad n=1,2,3,\cdots$$ And the solution (corresponding eigenfunctions) is $$\begin{align*} X_{n}\left ( x\right ) & =\cos \left ( \sqrt{\lambda _{n}}x\right ) \\ & =\cos \left ( nx\right ) \end{align*}$$

In summary, the solution to the $X$ ODE resulted in $$\begin{align} X_{0}\left ( x\right ) & =1\qquad n=0\tag{3}\\ X_{n}\left ( x\right ) & =\cos \left ( nx\right ) \qquad n=1,2,3,\cdots \nonumber \end{align}$$

Now we solve for the $Y$ ODE $$\begin{align*} Y^{\prime \prime }-\lambda Y & =0\\ Y\left ( 0\right ) & =0 \end{align*}$$

We are only given boundary conditions on bottom edge.

case $\lambda =0$ $$Y=Ay+B$$ When $y=0$ the above leads to $0=B$. Hence the corresponding eigenfunction is $Y_{0}\left ( y\right ) =y$.

case $\lambda >0$

The solution becomes $$Y\left ( y\right ) =A\cosh \left ( \sqrt{\lambda }y\right ) +B\sinh \left ( \sqrt{\lambda }y\right )$$ At $y=0$ the above gives $$\begin{align*} 0 & =A\cosh \left ( 0\right ) \\ & =A \end{align*}$$

Hence the solution reduces to $$Y\left ( y\right ) =B\sinh \left ( \sqrt{\lambda }y\right )$$ Therefore the eigenfunctions for $n=1,2,3,\cdots$ are $Y_{n}\left ( y\right ) =\sinh \left ( ny\right )$ since $\lambda _{n}=n^{2}$ for $n=1,2,3,\cdots$.

In summary, the solution to the $Y$ ODE resulted in $$\begin{align} Y_{0}\left ( y\right ) & =y\qquad n=0\tag{4}\\ Y_{n}\left ( x\right ) & =\sinh \left ( ny\right ) \qquad n=1,2,3,\cdots \nonumber \end{align}$$

From (3,4) we see that $$u_{n}\left ( x,y\right ) =X_{n}\left ( x\right ) Y_{n}\left ( y\right )$$ For $n=0$ the above becomes $$\begin{align*} u_{0}\left ( x,y\right ) & =\left ( 1\right ) \left ( y\right ) \\ & =y \end{align*}$$

And for $n=1,2,3,\cdots$ $$\begin{align*} u_{n}\left ( x,y\right ) & =\sinh \left ( ny\right ) \\ & =\cos \left ( nx\right ) \sinh \left ( ny\right ) \end{align*}$$

Using superposition, then $$\begin{align*} u\left ( x,y\right ) & =X\left ( x\right ) Y\left ( y\right ) \\ & =A_{0}u_{0}+\sum _{n=1}^{\infty }A_{n}u_{n}\\ & =A_{0}y+\sum _{n=1}^{\infty }A_{n}\cos \left ( nx\right ) \sinh \left ( ny\right ) \end{align*}$$

QED.

2.5.2 Section 37, Problem 1

Figure 2.59:Problem statement

Solution

The boundary conditions now become as follows

Figure 2.60:Boundary conditions

From the above problem we know the general solution is $$\begin{equation} u\left ( x,y\right ) =A_{0}y+\sum _{n=1}^{\infty }A_{n}\cos \left ( nx\right ) \sinh \left ( ny\right ) \tag{1} \end{equation}$$ Now we impose the remaining boundary condition $u\left ( x,2\right ) =f\left ( x\right )$. Therefore the above becomes $$f\left ( x\right ) =2A_{0}+\sum _{n=1}^{\infty }A_{n}\cos \left ( nx\right ) \sinh \left ( 2n\right )$$ Multiplying both sides by $\cos \left ( mx\right )$ integrating w.r.t. $x$ from $x=0$ to $x=\pi$ results in $$\begin{align*} \int _{0}^{\pi }f\left ( x\right ) \cos \left ( mx\right ) dx & =\int _{0}^{\pi }2A_{0}\cos \left ( mx\right ) dx+\left [ \int _{0}^{\pi }\sum _{n=1}^{\infty }A_{n}\cos \left ( nx\right ) \cos \left ( mx\right ) \sinh \left ( 2n\right ) dx\right ] \\ \int _{0}^{\pi }f\left ( x\right ) \cos \left ( mx\right ) dx & =\int _{0}^{\pi }2A_{0}\cos \left ( mx\right ) dx+\left [ \sum _{n=1}^{\infty }A_{n}\sinh \left ( 2n\right ) \left ( \int _{0}^{\pi }\cos \left ( nx\right ) \cos \left ( mx\right ) dx\right ) \right ] \end{align*}$$

case $m=0$ $$\begin{align} \int _{0}^{\pi }f\left ( x\right ) dx & =\int _{0}^{\pi }2A_{0}dx\nonumber \\ & =2A_{0}\pi \nonumber \\ A_{0} & =\frac{1}{2\pi }\int _{0}^{\pi }f\left ( x\right ) dx \tag{2} \end{align}$$

case $m=1,2,\cdots$ $$\int _{0}^{\pi }f\left ( x\right ) \cos \left ( mx\right ) dx=\sum _{n=1}^{\infty }A_{n}\sinh \left ( 2n\right ) \left ( \int _{0}^{\pi }\cos \left ( nx\right ) \cos \left ( mx\right ) dx\right )$$ But $\int _{0}^{\pi }\cos \left ( nx\right ) \cos \left ( mx\right ) dx=0$ for all $m\neq n$ and $\frac{\pi }{2}$ when $m=n$. Hence the above simplifies to $$\begin{align*} \int _{0}^{\pi }f\left ( x\right ) \cos \left ( mx\right ) dx & =\frac{\pi }{2}A_{m}\sinh \left ( 2m\right ) \\ A_{m} & =\frac{2}{\pi \sinh \left ( 2m\right ) }\int _{0}^{\pi }f\left ( x\right ) \cos \left ( mx\right ) dx \end{align*}$$

Since $m$ is summation index, we can rename it to $n$ and the above becomes $$\begin{equation} A_{n}=\frac{2}{\pi \sinh \left ( 2n\right ) }\int _{0}^{\pi }f\left ( x\right ) \cos \left ( nx\right ) dx \tag{3} \end{equation}$$ Using (2,3) in (1) gives the final solution $$u\left ( x,y\right ) =\left ( \frac{1}{2\pi }\int _{0}^{\pi }f\left ( x\right ) dx\right ) y+\sum _{n=1}^{\infty }\left ( \frac{2}{\pi \sinh \left ( 2n\right ) }\int _{0}^{\pi }f\left ( x\right ) \cos \left ( nx\right ) dx\right ) \cos \left ( nx\right ) \sinh \left ( ny\right )$$

2.5.3 Section 37, Problem 3

Figure 2.61:Problem statement

Part (a)

$$u_{xx}-xtu_{tt}=0$$ Let $u=X\left ( x\right ) T\left ( t\right )$. Substituting this into the above PDE gives $$X^{\prime \prime }T-xtT^{\prime \prime }X=0$$ Dividing by $XT\neq 0$ gives $$\frac{X^{\prime \prime }}{X}-xt\frac{T^{\prime \prime }}{T}=0$$ Diving by $x$ gives $$\begin{align*} \frac{1}{x}\frac{X^{\prime \prime }}{X}-t\frac{T^{\prime \prime }}{T} & =0\\ \frac{1}{x}\frac{X^{\prime \prime }}{X} & =t\frac{T^{\prime \prime }}{T}=-\lambda \end{align*}$$

Hence it possible to separate them. The generated ODE’s are $$\begin{align*} X^{\prime \prime }+\lambda xX & =0\\ T^{\prime \prime }+\lambda \frac{T}{t} & =0 \end{align*}$$

Part (b)

$$\left ( x+t\right ) u_{xx}-u_{t}=0$$ Let $u=X\left ( x\right ) T\left ( t\right )$. Substituting this into the above PDE gives $$\left ( x+t\right ) X^{\prime \prime }T-T^{\prime }X=0$$ Dividing by $XT\neq 0$ gives $$x\frac{X^{\prime \prime }}{X}+t\frac{X^{\prime \prime }}{X}-\frac{T^{\prime }}{T}=0$$ It is not possible to separate them.

Part (c)

$$xu_{xx}+u_{xt}+tu_{tt}=0$$ Let $u=X\left ( x\right ) T\left ( t\right )$. Substituting this into the above PDE gives $$\begin{align*} xX^{\prime \prime }T-\frac{\partial }{\partial t}\left ( X^{\prime }T\right ) +tT^{\prime \prime }X & =0\\ xX^{\prime \prime }T-X^{\prime }T^{\prime }X+tT^{\prime \prime }X & =0 \end{align*}$$

Dividing by $XT\neq 0$ gives $$x\frac{X^{\prime \prime }}{X}-X^{\prime }T^{\prime }+t\frac{T^{\prime \prime }}{T}=0$$ It is not possible to separate them.

Part (d)

$$u_{xx}-u_{tt}-u_{t}=0$$ Let $u=X\left ( x\right ) T\left ( t\right )$. Substituting this into the above PDE gives $$X^{\prime \prime }T-T^{\prime \prime }X-T^{\prime }X=0$$ Dividing by $XT\neq 0$ gives $$\begin{align*} \frac{X^{\prime \prime }}{X}-\frac{T^{\prime \prime }}{T}-\frac{T^{\prime }}{T} & =0\\ \frac{X^{\prime \prime }}{X} & =\frac{T^{\prime \prime }}{T}+\frac{T^{\prime }}{T}=-\lambda \end{align*}$$

It is possible to separate them. The ODE’s are $$\begin{align*} X^{\prime \prime }+\lambda X & =0\\ T^{\prime \prime }+T^{\prime }+\lambda T & =0 \end{align*}$$

2.5.4 Section 37, Problem 5

Figure 2.62:Problem statement

Case $\lambda <0$

Solution is $$X\left ( x\right ) =A\cosh \left ( \sqrt{-\lambda }x\right ) +B\sinh \left ( \sqrt{-\lambda }x\right )$$ At $x=0$ the above gives $$0=A$$ Hence the solution becomes $$X\left ( x\right ) =B\sinh \left ( \sqrt{-\lambda }x\right )$$ At $x=c$ the above becomes $$0=B\sinh \left ( \sqrt{-\lambda }c\right )$$ For non-trivial solution we want $\sinh \left ( \sqrt{-\lambda }c\right ) =0$. But $\sinh$ is zero only when its argument is zero. Which means $\sqrt{-\lambda }c=0$ which is not possible. Hence $\lambda <0$ is not possible.

Case $\lambda =0$

Solution is $$X\left ( x\right ) =Ax+B$$ At $x=0$ the above gives $$0=B$$ Hence the solution becomes $$X\left ( x\right ) =B$$ At $x=c$ the above becomes $$0=B$$ Which gives trivial solution. Hence $\lambda =0$ is not possible.

Case $\lambda >0$

Solution is $$X\left ( x\right ) =A\cos \left ( \sqrt{\lambda }x\right ) +B\sin \left ( \sqrt{\lambda }x\right )$$ At $x=0$ the above gives $$0=A$$ Hence the solution becomes $$X\left ( x\right ) =B\sin \left ( \sqrt{\lambda }x\right )$$ At $x=c$ the above becomes $$0=B\sin \left ( \sqrt{\lambda }c\right )$$ For non trivial solution we want $\sin \left ( \sqrt{\lambda }c\right ) =0$ which implies $$\begin{align*} \sqrt{\lambda }c & =n\pi \qquad n=1,2,3,\cdots \\ \lambda _{n} & =\left ( \frac{n\pi }{c}\right ) ^{2} \end{align*}$$

Therefore the eigenvalues are $\lambda _{n}=\left ( \frac{n\pi }{c}\right ) ^{2}$ for $n=1,2,3,\cdots$ and the eigenfunctions are $X_{n}\left ( x\right ) =\sin \left ( \frac{n\pi }{c}x\right )$ for $n=1,2,3,\cdots$.

2.5.5 Section 39, Problem 2

Figure 2.63:Problem statement

Solution

Example 1 is: Solve $u_{t}=ku_{xx}$ with $u\left ( 0,t\right ) =0$ and $u\left ( \pi ,t\right ) =0$. We now use initial conditions $u\left ( x,0\right ) =\sin \left ( x\right )$. The eigenvalues are $\lambda _{n}=n^{2}$ for $n=1,2,3,\cdots$ and eigenfunctions are $\sin \left ( nx\right )$. The general solution for this example is given in the book as $$u\left ( x,t\right ) =\sum _{n=1}^{\infty }B_{n}e^{-kn^{2}t}\sin \left ( nx\right )$$ At $t=0$ the above becomes $$\begin{equation} \sin x=\sum _{n=1}^{\infty }B_{n}\sin \left ( nx\right ) \tag{1} \end{equation}$$ By comparing sides, we see that only $n=1$ term exist. Hence $B_{1}=1$ and all other terms are zero. Hence the solution is, for $n=1$ $$u\left ( x,t\right ) =e^{-kt}\sin \left ( x\right )$$ To verify this, we start with (1) and multiply both sides by $\sin \left ( mx\right )$ and integrate which gives $$\begin{align*} \int _{0}^{\pi }\sin x\sin \left ( mx\right ) dx & =\int _{0}^{\pi }\sum _{n=1}^{\infty }B_{n}\sin \left ( nx\right ) \sin \left ( mx\right ) dx\\ & =\sum _{n=1}^{\infty }B_{n}\left ( \int _{0}^{\pi }\sin \left ( nx\right ) \sin \left ( mx\right ) dx\right ) \end{align*}$$

But $\int _{0}^{\pi }\sin \left ( nx\right ) \sin \left ( mx\right ) dx=0$ for $m\neq n$ and $\frac{\pi }{2}$ for $n=m$. Hence the above gives $$\int _{0}^{\pi }\sin x\sin \left ( mx\right ) dx=B_{m}\frac{\pi }{2}$$ Similarly, $\int _{0}^{\pi }\sin x\sin \left ( mx\right ) dx=0$ for $m\neq 1$ and $\frac{\pi }{2}$ when $m=1$, therefore the above becomes $$\begin{align*} \frac{\pi }{2} & =B_{1}\frac{\pi }{2}\\ B_{1} & =1 \end{align*}$$

And all other $B_{n}=0$. Which gives the same result obtain above, which is $u\left ( x,t\right ) =e^{-kt}\sin \left ( x\right )$

2.5.6 Section 39, Problem 4

Figure 2.64:Problem statement

Solution

We need to solve $$u_{t}=ku_{xx}\qquad t>0,0<x<\pi$$ With boundary conditions $$\begin{align*} u\left ( 0,t\right ) & =u_{0}\\ u\left ( \pi ,t\right ) & =0 \end{align*}$$

And initial conditions $$u\left ( x,0\right ) =0$$ Solution (15) is $$\begin{equation} u\left ( x,t\right ) =\frac{u_{0}}{\pi }\left [ x+2\sum _{n=1}^{\infty }\frac{\left ( -1\right ) ^{n}}{n}e^{-n^{2}kt}\sin \left ( nx\right ) \right ] \tag{15} \end{equation}$$ Replacing $x$ by $\pi -x$ in (15) gives $$\begin{align} u\left ( x,t\right ) & =\frac{u_{0}}{\pi }\left [ \left ( \pi -x\right ) +2\sum _{n=1}^{\infty }\frac{\left ( -1\right ) ^{n}}{n}e^{-n^{2}kt}\sin \left ( n\left ( \pi -x\right ) \right ) \right ] \nonumber \\ & =\frac{u_{0}}{\pi }\left ( \pi -x\right ) +2\frac{u_{0}}{\pi }\sum _{n=1}^{\infty }\frac{\left ( -1\right ) ^{n}}{n}e^{-n^{2}kt}\sin \left ( n\pi -nx\right ) \tag{2} \end{align}$$

Using $\sin \left ( A-B\right ) =\sin A\cos B+\cos A\sin B$, then $$\sin \left ( n\pi -nx\right ) =\sin \left ( n\pi \right ) \cos \left ( nx\right ) +\cos \left ( n\pi \right ) \sin \left ( nx\right )$$ But $\sin \left ( n\pi \right ) =0$ since $n$ is integer and $\cos \left ( n\pi \right ) =\left ( -1\right ) ^{n}$, then $\sin \left ( n\pi -nx\right ) =\left ( -1\right ) ^{n}\sin \left ( nx\right )$. Substituting this in (2) gives $$\begin{align*} u\left ( x,t\right ) & =u_{0}-u_{0}\frac{x}{\pi }+2\frac{u_{0}}{\pi }\sum _{n=1}^{\infty }\frac{\left ( -1\right ) ^{n}}{n}e^{-n^{2}kt}\left ( -1\right ) ^{n}\sin \left ( nx\right ) \\ & =u_{0}\left [ 1-\frac{x}{\pi }+\frac{2}{\pi }\sum _{n=1}^{\infty }\frac{\left ( -1\right ) ^{2n}}{n}e^{-n^{2}kt}\sin \left ( nx\right ) \right ] \\ & =u_{0}\left [ 1-\frac{x}{\pi }+\frac{2}{\pi }\sum _{n=1}^{\infty }\frac{1}{n}e^{-n^{2}kt}\sin \left ( nx\right ) \right ] \end{align*}$$

Which is the result required.