PHYSICS 312, Solution to Midterm Examination, February 26 1999

Problem 1:
The steady state temperature T_S satisfies the differential equation

$$\frac{d^2T_S}{dx^2}=h^2(T_S-T_0-bx)$$

With the boundary conditions

T_S(0)=T_S(a)=0

We introduce the new variable

u=T_S-T₀-bx

and substitute into the differential eqation to find

$$\frac{d^2u}{dx^2}=h^2u$$

The general solution for u is

$$u=A\sinh(hx)+B\cosh(hx)$$

where A and B are constants to be determined. We find

$$T_S=A\sinh(hx)+B\cosh(hx)+T_0+bx$$

The boundary condition at x=0 gives

B=-T₀

The boundary condition at x=a gives

$$0=A\sinh(ha)-T_0\cosh(ha)+T_0+ba$$

$$A=\frac{T_0(\cosh(ha)-1)-ba}{\sinh(ha)}$$

Giving



Problem 2:
In order to solve

$$\frac{d^2\phi}{dx^2}+2\frac{d\phi}{dt}+\lambda^2\phi=0$$

we first solve the characteristic equation

$$\gamma^2+2\gamma+\lambda^2=0$$

$$\gamma=-1\pm i\sqrt{\lambda^2-1}$$

giving the general solution to the differential equation

$$\phi=e^{-x}(A\sin(x\sqrt{\lambda^2-1})+B\cos(x\sqrt{\lambda^2-1})$$

The boundary condition at x=0 gives B=0. The boundary condition at $x=\pi$ gives

$$\sqrt{\lambda^2-1}=n;\;\;n=1,2,3\cdots$$

or

$$\lambda^2=1+n^2$$

The three lowest eigenvalues and eigenfunctions are

$$\lambda_1^2=2;\;\phi_1=e^{-x}\sin x$$

$$\lambda_2^2=5;\;\phi_2=e^{-x}\sin(2x)$$

$$\lambda_2^3=10;\;\phi_3=e^{-x}\sin(3x)$$

You were only asked to give a sketch of the eigenfunctions. A more accurate plot is given on the attached worksheet.

Problem 3:
a:
We attempt a separation of variables solution to

$$\frac{\partial^2u}{\partial x^2}=\frac{1}{k}\frac{\partial u}{\partial t},\; 0<x<\infty,\;0<t$$

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

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

$$\tau(t)=c\;e^{-k\lambda t}$$

where without loss of genrality we can put c=1

$$\rho(x)=F\sin(\lambda x)+G\cos(\lambda x)$$

The boundary condition at x=0 gives G=0. The eigenvalue $\lambda$ is now a continuous variable and the eigenfunctions are on the form

$$\rho(x)=\sin(\lambda x)$$

The coefficients $F(\lambda)$ can be obtained from the Fourier sine transform. We can thus write for the complete solution

$$u(x,t)=\int_0^\infty dx F(\lambda)\sin(\lambda x)e^{-k\lambda^2t}$$

where

$$F(\lambda)=\frac{2}{\pi}\int_0^\infty dx\sin(\lambda x)f(x)$$

b:
In the special case

$$f(x)=T_0\sin(x)$$

where T₀ is a constant, the initial condition is already in the form of an eigenfunction with $\lambda=1$ and we have the solution

$$u(x,t)=T_0\sin(x)e^{-kt}$$

return to title page.