PHYSICS 312. Midterm Examination. February 27, 2000
SOLUTION

Problem 1 :

$$f(x)=\sin(x)+\cos(x);\;0<x<\pi$$

The function g(x) in the range $-\pi<x<0$ that constitutes the
a: periodic extension of f(x) is

$$-\sin(x)-\cos(x)$$

b: the odd periodic extension of f(x) is

$$\sin(x)-\cos(x)$$

c: the even periodic extension of f(x) is

$$-\sin(x)+\cos(x)$$

Problem 2:
Let us define $k=\sqrt{\lambda}$ The general solution to

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

is

$$\phi=A\sin(kx)+B\cos(kx)$$

and

$$\phi^\prime=Ak\cos(kx)-Bk\sin(kx)$$

$$\phi(0)+\frac{d\phi}{dx}\vert _{x=0}=B+Ak$$

Hence B=-Ak and

$$\phi(a)+\frac{d\phi}{dx}\vert _{x=a}=(A-Bk)\sin(ka)+(B+Ak)\cos(ka)=0$$

The cosine term above vansihes and we get

$$A(1+k^2)\sin(ka)=0$$

or

$$k=\frac{n\pi}{a};\;\;n=1,3\cdots$$

The eigenfunctions can then be written

$$\phi_n=A(\sin(\frac{n\pi x}{a})-\frac{n\pi}{a}\cos(\frac{n\pi x}{a}))$$

and the eigenvalues

$$\lambda_n=(\frac{n\pi}{a})^2$$

Problem 3:
The steady state temperature differential equation is

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

The ends x=0 and x=L were kept at the temperature T=0
a: We integrate this equation to get

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

Integrating once more gives

$$\kappa_0T+\frac{\alpha}{2}T^2=\frac{Q}{2}x^2+bx+c$$

The boundary condition T=0 at x=0 gives c=0 while T=0 for x=L gives b=-QL/2. This gives us the quadratic equation

$$\kappa_0T+\frac{\alpha}{2}T^2=\frac{Q}{2}(x^2-Lx)$$

which can be solved for T to give for $a\neq 0$

$$T=-\frac{\kappa_0}{\alpha}\pm\sqrt{\frac{\kappa_0^2}{\alpha^2}+\frac{Q}{\alpha}(x^2-Lx)}$$

b: For $\alpha>0$ we must choose the positive root to get T=0 for x=0 and x=L. Hence

$$T=-\frac{\kappa_0}{\alpha}+\sqrt{\frac{\kappa_0^2}{\alpha^2}+\frac{Q}{\alpha}(x^2-Lx)}$$

c: $\alpha=0$ The division by $\alpha$ in the quadratic equation doesn't work and we get instead

$$T=\frac{Q}{2\kappa_0}(x^2-Lx)$$

d: $\alpha<0$. We must now pick the negative root in the solution to the quadratic

$$T=-\frac{\kappa_0}{\alpha}-\sqrt{\frac{\kappa_0^2}{\alpha^2}+\frac{Q}{\alpha}(x^2-Lx)}$$

Bonus question: If $\vert\alpha\vert$ is too large and $\alpha<0$ there no longer is a real solution in the middle of the bar. The maximum value of x²-Lx occurs for x=L/2 and is L²/4. Hence when

$$Q>\frac{\-4\kappa_0^2}{L^2\alpha}$$

the calculation breaks down. What happens is that it gets so hot in the middle of the bar that the thermal conductivity turns negative and we get a runaway solution with more and more heat pouring in.