Web Bonus Problem to Week 7 (#1)

[Victor Ivrii]

Problem 4.2.6 http://www.math.toronto.edu/courses/apm346h1/20159/PDE-textbook/Chapter4/S4.2.P.html#problem-4.2.P.6

[Emily Deibert]

We start by assuming a solution: \begin{equation}

U(x,t) = X(x)T(x) \end{equation}

Then plugging this into the wave equation: \begin{equation}

X(x)T''(t) - c^2T(t)X''(x) = 0 \end{equation}

Dividing through by $U(x,t)$ gives: \begin{equation}

\frac{T''(t)}{T(t)} = c^2\frac{X''(x)}{X(x)} = \lambda \end{equation}

In this case we will first consider $\lambda = \omega{}^2 > 0$. Then this gives us two ODEs. Solving for $T(t)$: \begin{equation}

\frac{T''(t)}{c^2T(t)} = \omega{}^2 \rightarrow T''(t) - \omega{}^2c^2T(t) = 0 \end{equation}

The characteristic equation is: \begin{equation}

r^2 = \omega{}^2 = 0 \rightarrow r^2 = \omega{}^2 \rightarrow r = \pm \omega \end{equation}

We solve this to get the solution for $T(t)$: \begin{equation}

T(t) = Ae^{c\omega{}t} + Be^{-c\omega{}t} \end{equation}

Likewise, we will have the following equation for $X(x)$: \begin{equation}

\frac{X''(x)}{X(x)} = \omega{}^2 \end{equation}

We can solve this to show that: \begin{equation}

X(x) = Ce^{\omega{}x} + De^{-\omega{}x} \end{equation}

I will continue the rest of the problem soon, or anyone else can contribute if they want to.

[Victor Ivrii]

Hint: your $\lambda<0$ so $\lambda=-\omega^2$.

Hint: Look at boundary condition $(u_x+\alpha u_t)|_{x=0}=0$ which would imply that $T(t)$ is "some" exponent so $T=e^{\pm i\omega t}$. Investigate each sign.

[Xi Yue Wang]

So, let $$\frac{X''(x)}{X(x)} = -\lambda,\ \frac{T''(t)}{T(t)} = -c^2\lambda$$
Then we have,$$ X''(x) + \lambda X(x) = 0,\ T''(t) + c^2\lambda T(t) = 0$$
And let $-\lambda = \omega^2$, We get $$X(x) = A\cos(\omega x) +B\sin(\omega x),\ T(t) = C\cos(c\omega t) + D\sin(c\omega t)$$
Given conditions that, $u_x(0,t) = 0,\ (u_x + i\alpha u_t)(l,t) = 0$
We have $$X'(0)T(t) = 0\\X'(0) = 0 = \omega B \\B = 0$$
Hence, $$X(x) = A\cos(\omega x)$$
$$X'(l)T(t) + i\alpha X(l)T'(t) = 0\\ -\omega A\sin(\omega l ) T(t) + i\alpha A\cos(\omega l)T'(t) = 0\\T'(t) = \frac{\omega\sin(\omega l)}{i\alpha\cos(\omega l)}T(t)\\ T'(t) = \frac{\omega}{i\alpha}\tan(\omega l)T(t)\\T'(t) = \frac{-i\omega}{\alpha}\tan(\omega l)T(t)$$
Solve this ODE, we get $$ T(t) =Ke^{\frac{-i\omega t}{\alpha}\tan(\omega l)}$$