PHYSICS 312. Midterm. February 10 2003.Solution.
Problem 1:
a:

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

We have

$$f+g=e^{-x^2}$$

$$cf^\prime-cg^\prime=-2xce^{-x^2}$$

Integrating the last equation we find

$$f(x)-g(x)=e^{-x^2}+a$$

where $a$ is a constant. Solving for $f$ and $g$ gives

$$f(x+ct)=\exp(-(x+ct)^2)+\frac{a}{2}$$

$$g(x-ct)=-\frac{a}{2}$$

Adding the terms gives

$$u(x,t)=\exp(-(x+ct)^2)$$

b: This is a bell shaped pulse moving left, without changing its shape from its initial form.
Problem 2:
a:We have

$$\phi=a\sin(\lambda x)+b\cos(\lambda x)$$

The boundary condition for $x=0$ gives $a=0$. The boundary condition at $x=a$ gives

$$\lambda=\frac{(2n+1)\pi}{2a};\;\;n=0,2,3,\dots$$

b: The three lowest eigenfunctions can be written

$$f_1=b\cos(\frac{\pi}{2a});\;\;f_2=b\cos(\frac{3\pi}{2a});\;\;f_3=b\cos(\frac{5\pi}{2a});\;\;$$

drawn below. The constant $b$ is arbitrary and we choose it to be unity.

Problem 3:
a: We have

$$(\kappa_0+\alpha T)\frac{dT}{dx}=const$$

Choosing the constant to be $c$ and integrating once more gives

$$\kappa T+\frac{\alpha}{2}T^2=cx+d$$

where $d$ is another constant.
b: The solution to the equation above is

$$T=-\frac{\kappa_0}{\alpha}\pm \sqrt{\frac{\kappa_0^2}{\alpha^2}-\frac{2(cx+d)}{\alpha}}$$

The boundary condition at $x=0$ forces us to choose the positive root and put $d=0$. The boundary condition at $x=L$ gives

$$(T_0+\frac{\kappa_0}{\alpha})^2=\frac{\kappa_0^2}{\alpha^2}-\frac{2(cx+d)}{\alpha}$$

or

$$c=\frac{T_0^2\alpha+2\kappa_0T_0}{2L}$$

c: We must use the positive root.