Solutions to 1992 114B exam

1.

(a)Following hint, try y = x^p. Plugging into ODE:

$$p(p-1)x^{p-2} + \omega^2 x^{p-2} = 0$$

which implies $p(p-1) + \omega^2 = 0$. Solving quadratic equation yields

$$p_{\pm} = {1 \pm \sqrt{1-4\omega^2} \over 2}$$

which we write (since $\omega > 1/2$) as $p_{\pm} = {1 \pm i \sqrt{4\omega^2 -1 } \over 2}$.

(b) The general solution is then y = C₊x^p₊ + C_-x^p_-, and at x=1, y=0. Using this we obtain C₊ + C_- =0. Therefore $y = C_+(x^{p_+} - -x^{p_-}) = C\sqrt{x} \sin (\sqrt{\omega^2 -{1\over 4}}\ln x)$.

2. Write y(x,t)  =  X(x)T(t) and plugging into

$${1\over x^2} {\partial^2 y\over \partial t^2} ~=~ {\partial^2 y\over \partial x^2}$$

gives

$${T''\over T} = {x^2X\over X}$$

which then must equal a constant, call it $- \omega^2$. With the boundary condition from above, i.e. y(1) =0, we have

$$T(t) = e^{\pm i\omega t}$$

and

$$X(x) = \sqrt{x} \sin (\sqrt{\omega^2 -{1\over 4}}\ln x).$$

(b) At x=2, y=0, so $\sin(\sqrt{\omega^2 -{1\over 4}}\ln 2) = 0$, which implies $\sqrt{\omega^2 -{1\over 4}}\ln 2 ~=~ n \pi$.

(c) Solving for $\omega$, $\omega_n = \sqrt{({n\pi\over \ln 2})^2 + {1\over 4}}$for $n ~=~ 1,2,3,\dots$.

(d) Summing over all solutions:

$$y = \sum_{n=1}^\infty (A_n \cos (\omega_n t) + B_n \sin (\omega_n t))\sqrt{x} \sin ({n\pi \over \ln 2} \ln x) .$$

(e) Graph!

3. $f(x,t) \equiv e^{-i\omega t} f(x)$ and

$${1\over x^2} {\partial^2 y\over \partial t^2} ~=~ {\partial^2 y\over \partial x^2} + F(x,t)$$

Try $y(x,t) = e^{-i\omega t} X(x)$. This implies

$$X''(x) + {\omega^2\over x^2} X = f(x).$$

4. (a) From 1(b) $y = A(x') \sqrt{x} \sin (\sqrt{\omega^2 -{1\over 4}}\ln x)$ for x < x'.

(b) Rescale x: $y = C_+({x\over 2})^{p_+} + C_-({x\over 2})^{p_-}$, with the boundary condition y(x/2) = 0. Therefore C₊ + C_- =0, and so $y = B(x')(({x\over 2})^{p_+} - -({x\over 2})^{p_-}) = B(x')\sqrt{x} \sin (\sqrt{\omega^2 -{1\over 4}}\ln (x/2))$. for x> x'.

(c)

$$\int_{x'-\epsilon}^{x'+\epsilon} ~~ (G'' + {\omega^2 \over x^2} G) dx ~=~ \int_{x'-\epsilon}^{x'+\epsilon} \delta(x-x') dx$$

and taking $\lim_{\epsilon\rightarrow 0}$ gives

$$G'(x+\epsilon,x') - G'(x-\epsilon,x') = 1$$

(d)Define $D \equiv \sqrt{\omega^2 -{1\over 4}}$. From continuity $G'(x+\epsilon,x') = G'(x-\epsilon,x')$, so

$$A(x')\sin (D \ln x') = B(x')\sin (D \ln (x'/2)).$$

and from 4(c)

$$B(x')\cos (D \ln (x'/2)) - A(x')\cos (D \ln (x')) ~=~ \sqrt{x'}/D.$$

Solving for A and B gives

$$A(x') = {\sqrt{x'} \over D} {\sin (D \ln (x'/2)) \over \sin (D \ln (2))}$$

and

$$B(x') = {\sqrt{x'} \over D} {\sin (D \ln (x')) \over \sin (D \ln (2))}$$

(e) $y = \int_1^2 G(x,x') f(x') dx'$.