PHYSICS 312
Solution to problem set 5 1999:

Problem 1:
a:
If $x=L\theta$ represents distances along the perimeter of the ring the differential equation is

$$\frac{\partial ^2 T}{\partial x^2}=\frac{1}{k}\frac{\partial T}{\partial t}$$

In terms of the angle $\theta$ the differential equation becomes

$$\frac{\partial ^2 T}{\partial \theta^2}=\frac{L^2}{k}\frac{\partial T}{\partial t}$$

Separating the variables

$$T(\theta,t)=\phi(\theta)\tau(t)$$

we find

$$\tau^\prime=-\frac{k^2\lambda^2}{L^2}\tau$$

$$\phi^{\prime\prime}+\lambda^2\phi=0$$

The general solution to the equation for $\phi$ is

$$\phi=A\cos(\lambda\theta)+B\sin(\lambda\theta)$$

If the angle is incremented by $2\pi$ we are back were we started, hence $\lambda=n=$ integer or zero. The solution for $\tau$ is

$$\tau=const\exp(-\frac{kn^2t}{L^2})$$

The solution to the problem can thus be expressed on the Fourier series form

$$T(\theta,t)=A_0+\sum_{n=0}^\infty \exp(-\frac{kn^2t}{L^2})(A_n\cos(n\theta)+B_n\sin(n\theta))$$

where

$$A_0=\frac{1}{2\pi}\int_0^{2\pi}d\theta f(\theta)$$

$$A_n=\frac{1}{\pi}\int_0^{2\pi}d\theta f(\theta)\cos(n\theta)$$

$$B_n=\frac{1}{\pi}\int_0^{2\pi}d\theta f(\theta)\sin(n\theta)$$

b
In this case $\cos\theta$ is an eigenfunction with eigenvalue $\lambda=n=1$ Hence we put $A_n=0$ for $n\neq 1$ and $B_n=0$ for all $n$

$$T(\theta,t)=T_0\exp(-\frac{kt}{L^2})\cos\theta$$

The time it takes for the temperature difference between the hottest and coldest spot to halve is thus

$$t_{1/2}=\frac{L^2}{k}\ln 2$$

Problem 2:
The general solution to the 2-dimensional Laplace equation

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

in polar coordinates is

$$u=a+b\ln r +\sum_{n=1}^\infty(a_n\cos(n\theta)+b_n\sin(n\theta)) (\frac{\alpha_n}{r^n}+\beta r^n)$$

Because of the boundary conditions

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

only the cosine term with $n=1$ will contribute to the solution so we can write

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

Substituting for the conditions at $r=a$ and $r=2a$ we find

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

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

we find

$$\alpha=2a/3, \beta=1/(3a)$$

and

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

Problem 3:
The differential equation

$$\nabla^2u=1$$

becomes in spherical coordinates

$$\frac{1}{r^2}\frac{\partial }{\partial r}(r^2\frac{\partial u... ...ac{1}{r^2\sin^2\theta}\frac{\partial ^2 u}{\partial \phi^2}= 1$$

The boundary conditions are such that $u$ does not depend on either $\phi$ nor $\theta$. The differential equation then simplifies to

$$\frac{1}{r^2}\frac{d}{d r}(r^2\frac{d u}{d r})=1$$

The general solution to the homogeneous equation

$$\frac{d}{d r}(r^2\frac{d u}{d r})=0$$

is

$$u_h=\frac{c}{r}+b$$

where $c$ and $b$ are constants. We require that $u$ is bounded at $r=0$ hence $c=0$. We must add to the homogeneous solution a particular solution to the inhomogeneous equation. It was suggested to try a solution on the form $u=ar^2$. Substituting into the differential equation gives

$$\frac{1}{r^2}\frac{d}{\d r}(r^22br)=6a=1$$

Giving $a=1/6)$ The boundary condition

$$u(1,\theta,\phi)=0$$

gives $b=1/6$ Hence

$$u(r)=(r^2-1)/6$$

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