PHYSICS 312. Midterm Examination, March 2 2001, Solution

Problem 1 :
a:
Let us write

f(x)=f₁(x)+f₂(x)

$$f_1(x)=\sin(x)$$

$$f_2(x)=\cos(x)$$

$$g_1(x)=-\sin(x),\;\; -\pi<x<0$$

is the periodic extension of f₁(x) and also the even periodic extension. Next

$$g_2(x)=\cos(x),\;\; -\pi<x<0$$

is the even periodic extension, but not the periodic extension of f₂(x). So statements 1 and 2 and 4 are false, while 3 is true!
b: Since we are dealing with the even periodic extension the Fourier series is a cosine series. So statement 1 is truewhile 2 and 3 are false!
c: Calculate the coefficient $\alpha_0$.

$$\alpha_0=\frac{1}{2\pi}\int_0^\pi\sin(x)dx-\frac{1}{2\pi}\int... ...dx+ \int_0^{2\pi}\cos(x)dx=\frac{1}{2\pi}(2+2+0)=\frac{2}{\pi}$$

Problem 2:
By solving the characteristic equation

$$\gamma^2+2\gamma+\lambda$$

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

we find that the general solution to

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

is

$$\phi(x)=e^{-x}(\alpha\cos(x\sqrt{\lambda-1})+\beta\sin(x\sqrt{\lambda-1}))$$

if

$$\lambda>1$$

and

$$\phi(x)=(\alpha+\beta x)e^{-x}$$

for $\lambda=1$ and

$$\phi(x)=e^{-x}(\alpha e^{x\sqrt{1-\lambda}}+\beta e^{-x\sqrt{1-\lambda}})$$

for $\lambda<1$. The boundary conditions were

$$\phi(0)=0$$

$$\phi(a)=0$$

a: For $\lambda>1$ we have $\alpha=0$ and

$$\sqrt{\lambda-1}=\frac{n\pi}{a}$$

or

$$\lambda=(\frac{n\pi}{a})^2+1$$

with eigenfunctions

$$\phi(x)=e^{-x}\sin(\frac{n\pi}{a}),\;\;n=1,2,3,...$$

b: For $\lambda<1$ including $\lambda<0$ it is impossible to satisfy the boundary conditions. The same holds for $\lambda=1$, so $\lambda=1$ is not an eigenvalue.


Problem 3:
The steady state solution satisfies

$$\frac{\partial}{\partial x}(\kappa(x)\frac{\partial T_S(x,t)}{\partial x})=0$$

$$\kappa(x)\frac{\partial T_S(x,t)}{\partial x}=c$$

$$\frac{\partial T_S(x,t)}{\partial x}=\frac{c}{\kappa_0}(1+\alpha x)$$

$$T_S=\frac{c}{\kappa_0}(x+\alpha\frac{x^2}{2})+d$$

The boundary condition at x=0 gives d=0, while at the other end

$$T_0=\frac{c}{\kappa_0}(L+\frac{\alpha L^2}{2})$$

so

$$T_S=T_0\frac{x+\frac{\alpha x^2}{2}}{L+\frac{\alpha L^2}{2}}$$

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