PHYSICS 312
Sessional Examination, April 2002, Solution

Problem 1:
Substituting

$$y(x)=\sum_{n=0}^\infty a_nx^n$$

into the differential equation

$$\frac{d^2y}{dx^2}-2x\frac{dy}{dx}-2y=0$$

gives

$$\sum_{n=2}^\infty a_n(n-1)(n-2)x^{n-2}-\sum_{n=0}^\infty 2a_n(n+1)x^n$$

Substituting $m=n-2$ in the first term and $n=m$ in the second gives

$$\sum_{m=0}^\infty x^m[(m+2)(m+1)a_{m+2}-2(m+1)a_m]=0$$

$$a_{m+2}=\frac{2a_m}{m+2}$$

a: If

$$y(0)=1,\;\;\frac{dy}{dx}\vert _{x=0}=0$$

the power series will only contain even terms. With

$$k=2m$$

We find

$$a_{2k+2}=\frac{a_{2k}}{k+1}$$

$$y(x)=\sum_{k=0}^\infty \frac{x^{2k}}{k!}$$

b: We recognize this power series as that of

$$y(x)=e^{x^2}$$

We have

$$\frac{dy}{dx}=2xe^{x^2};\;\;2x\frac{dy}{dx}=4x^2e^{x^2}$$

$$\frac{d^2y}{dx^2}=4x^2e^{x^2}+2e^{x^2}$$

We find by substitution that the differential equation is satisfied.
c:
>From a: we have

$$b_{m+2}=\frac{2b_m}{m+2}$$

The boundary condition require $b_0=0$, $b_1=1$ hence

$$b_3=\frac{2}{3};\;b_5=\frac{2}{5}\cdot{2}{3}\cdots$$

$$y(x)=x+\frac{2x^3}{3}+\frac{2^2x^5}{3\cdot 5}+\frac{2^3x^7}{3\cdot 5\cdot 7}\cdots$$

Problem 2:
a: Substituting

$$u=e^{\lambda x}$$

into the fourth order differential equation

$$a^4\frac{d^4u}{d\; x^4}-\omega^2 u=0$$

gives the characteristic equation

$$a^4\lambda^4-\omega^2=0$$

with roots

$$\lambda_1=i\frac{\omega^{1/2}}{a};\;\;\lambda_2=-i\frac{\omeg... ...rac{\omega^{1/2}}{a};\;\;\lambda_4=-\frac{\omega^{1/2}}{a};\;\;$$

or

$$u(x)=b_1\exp(i\frac{\omega^{1/2}}{a})+b_2\exp(-i\frac{\omega^... ..._3\exp(\frac{\omega^{1/2}}{a})+b_4\exp(-\frac{\omega^{1/2}}{a})$$

an alternative form which is a bit more convenient for the remainder of the problem is

$$u(x)=\alpha\cos(\frac{\omega^{1/2}}{a})+\beta\sin(\frac{\omeg... ...sh(\frac{\omega^{1/2}}{a})+\delta\sinh(-\frac{\omega^{1/2}}{a})$$

b:With

$$u(x,t)=f(x)g(t)$$

we find

$$\frac{a^4f^{(4)}}{f}=\frac{g^{\prime\prime}}{g}=-\omega^2$$

$$\frac{g^{\prime\prime}}{g}=-\omega^2$$

$$g(t)=A\cos(\omega t)+B\sin(\omega t)$$

We identify $\omega$ as the frequency of vibration and obtain

$$a^4\frac{d^4u}{d\; x^4}-\omega^2 u=0$$

giving the modes

$$u(x)=[\alpha\cos(\frac{\omega^{1/2}x}{a})+\beta\sin(\frac{\om... ...\frac{\omega^{1/2}x}{a})+\delta\sinh(-\frac{\omega^{1/2}x}{a})]$$

$$\times[A\cos(\omega t)+B\sin(\omega t)]$$

c: If the end at $x=0$ is clamped down we must have

$$\alpha+\gamma=0;\;\;\beta+\delta=0$$

The conditions at the other end then yield

$$0=(\cos(\frac{\omega^{1/2}L}{a})-\cosh(\frac{\omega^{1/2}L}{a... ...n(\frac{\omega^{1/2}L}{a})-\sinh(\frac{\omega^{1/2}L}{a}))\beta$$

$$0=(-\sin(\frac{\omega^{1/2}L}{a})-\sinh(\frac{\omega^{1/2}L}{... ...s(\frac{\omega^{1/2}L}{a})-\cosh(\frac{\omega^{1/2}L}{a}))\beta$$

This gives rise to a determinantal equation for the frequencies $\omega$

$$(\cos(\frac{\omega^{1/2}L}{a})-\cosh(\frac{\omega^{1/2}L}{a}))(\cos(\frac{\omega^{1/2}L}{a})-\cosh(\frac{\omega^{1/2}L}{a}))$$

$$-(-\sin(\frac{\omega^{1/2}L}{a})-\sinh(\frac{\omega^{1/2}L}{a... ...\sin(\frac{\omega^{1/2}L}{a})-\sinh(\frac{\omega^{1/2}L}{a}))=0$$

You were not asked to go beyond this point, but if we multiply out we find using

$$\cos^2x+\sin^2x=1;\;\;\cosh^2x-\sinh^2x=1$$

$$1=\cos(\frac{\omega^{1/2}L}{a})\cosh(\frac{\omega^{1/2}L}{a}))$$

This is a transcendental equation which can be solved numerically.
Problem 3:
a: Substituting

$$u(x,t)=e^{-\lambda t}v(x)$$

into

$$\frac{\partial ^2u}{\partial x^2}-\frac{\partial u}{\partial t}=\exp(-\lambda t)$$

gives the differential equation for $v(x)$

$$\frac{d^2v}{dx^2}+\lambda v=1$$

b: The general solution to this equation can be written

$$v(x)=\frac{1}{\lambda}+a\cos(x\sqrt{\lambda})+b\sin(x\sqrt{\lambda})$$

In order to satisfy the boundary condition at $x=0$ we must put

$$a=-\frac{1}{\lambda}$$

The boundary condition at the other end then gives

$$0=\frac{1}{\lambda}-\frac{1}{\lambda}\cos(L\sqrt{L\lambda})+b\sin(x\sqrt{L\lambda})$$

$$b=-\frac{\frac{1}{\lambda}-\frac{1}{\lambda}\cos(L\sqrt{\lambda})} {\sin(L\sqrt{\lambda})}$$

This will work for almost all values of $\lambda$. If by accident

$$\sin(L\sqrt{\lambda})=0$$

some special attention has to be paid.
c:
We can solve the boundary value problem

$$\frac{\partial ^2u}{\partial x^2}-\frac{\partial u}{\partial t}=\exp(-\lambda t)$$

$$u(0,t)=u(L,t)=0,\;u(x,0)=f(x)$$

by writing

$$u(x,t)=v(x)e^{-\lambda t}+w(x,t)$$

$w(x,t)$ will then have to satisfy

$$\frac{\partial ^2w}{\partial x^2}-\frac{\partial w}{\partial t}=0$$

with boundary condition

$$w(0,t)=w(L,t)=0;\;\;w(x,0)=f(x)-v(x)$$

where $v(x)$ was found above. This problem can be solved by separation of variables and expanding the initial condition in a Fourier sine series in the usual way.
Problem 4:
Substituting a:

$$u(x,t)=h(x)\tau (t)$$

into

$$\frac{\partial ^2 u}{\partial t^2}+2f\frac{\partial u}{\partial t}-c^2\frac{\partial ^2u} {\partial x^2}=0$$

gives

$$\frac{\tau^{\prime\prime}+2f\tau^\prime}{c^2\tau}=\frac{h^{\prime\prime}}{h} =-\lambda^2=const$$

The solution to the differential equation for $h$ is with the boundary condition that $h$ is zero at $x=0$ and $x=L$

$$h(x)=A\sin(\lambda x);\;\; \lambda=\frac{n\pi}{L};\;\;n=1,2.\cdots$$

The differential equation for $\tau$

$$\tau^{\prime\prime}+2f\tau^\prime+c^2\lambda^2\tau$$

has solutions

$$\tau=e^{-ft}(A \sin(\sqrt{c^2\lambda^2-f^2}) +B \cos(\sqrt{c^2\lambda^2-f^2})$$

giving

$$u(x,t)=\sum_{n=1}^\infty\alpha_n\sin(\frac{n\pi x}{L})e^{-ft} (A_n\cos(t\sqrt{\frac{c^2n^2\pi^2}{L^2}-f^2})$$

$$+ B_n \sin(t\sqrt{\frac{c^2n^2\pi^2}{L^2}-f^2}))$$

We expand the initial condition $u(x,0)=g(x)$ in a sine series and find

$$A_n=\frac{2}{L}\int_0^Ldx\;g(x)(\sin\frac{n\pi x}{L})$$

The condition

$$\frac{du}{dt}u(x,t)\vert _{t=0}=0$$

Gives

$$B_n=\frac{f A_n}{\sqrt{\frac{c^2n^2\pi^2}{L^2}-f^2}}$$

where $g(x)$ is a known function.
b:Since the damping term $e^{-ft}$ is independent of the mode number $n$ all the modes die out at the same rate
c: The maximum value of the damping constant for oscillations to occur at the n'th frequencies is

$$f=\frac{cn\pi}{L}=$$

As the damping is increased the lower frequency modes become over-damped before the high frequency modes.
Problem 5
The general solution to the 2-dimensional Laplace equation in polar coordinates

$$\nabla^2u(r,\theta)=\frac{\partial ^2u}{\partial r^2}+\frac{1... ...rtial r}+\frac{1}{r^2}\frac{\partial ^2 u}{\partial \theta^2}=0$$

is

$$u=\alpha_0+\beta_0 \ln r +\sum_{n=1}^\infty(a_n\cos(n\theta)+b_n\sin(n\theta)) (\frac{\alpha_n}{r^n}+\beta_n r^n)$$

a: The solution to

$$\nabla^2u(r,\theta)=0$$

in the region $a<r<2a$ with

$$u(a,\theta)=1,\;u(2a,\theta)=2$$

will be on the form

$$\alpha+\beta \ln \frac{r}{a}$$

We have

$$\alpha_0=1$$

$$\alpha+\beta\ln 2=2$$

we find

$$u(r,\theta)=1+\frac{\ln\frac{r}{a}}{\ln 2}$$

b:

$$u(a,\theta)=\cos\theta,\;u(2a,\theta)=\cos\theta$$

The solution is now on the form

$$(\frac{\alpha a}{r}+\frac{\beta r}{a})\cos\theta$$

we find

$$\alpha+\beta=1$$

$$\frac{\alpha}{2}+2\beta=1$$

giving

$$\beta=\frac{1}{3},\;\alpha=\frac{2}{3}$$

$$u(r,\theta)=(\frac{2a}{3r}+\frac{r}{3a})cos\theta$$

c:The solution will depend on $r$ only. We rewrite the differential equation as

$$\frac{1}{r}\frac{d}{d r}(r\frac{d u}{d r})=1$$

giving

$$r\frac{d u}{d r}=\frac{r^2}{2}+b$$

Integrating once more gives

$$u(r)=\frac{r^2}{4}+b\ln\frac{r}{a}+c$$

With the boundary conditions $u(a,\theta)=1,\;u(2a,\theta)=2$ we find

$$\frac{a^2}{4}+c=1$$

$$a^2+b\ln 2+c=2$$

with solution

$$b=-\frac{3a^2-4}{4\ln(2)},c=-\frac{a^2}{4}+1$$

and we find

$$u(r)=\frac{r^2}{4}+\frac{-3a^2+4}{4\ln(2)}\ln\frac{r}{a}-\frac{a^2}{4}+1$$