HA3-P5

[Victor Ivrii]

Problem 1 here
http://www.math.toronto.edu/courses/apm346h1/20159/PDE-textbook/Chapter2/S2.4.P.html#problem-2.4.P.1

[Zaihao Zhou]

The general solution is
\begin{equation} u(x,t) = \frac{1}{2c}\int_0^t{\int_{x-c(t-t')}^{x+c(t-t')} f(x',t')dx'dt'}   \end{equation}
For problem: \begin{equation}u_{tt} - c^2u_{xx} = f(x,t)  \quad and \quad g(x)=0,\ h(x)=0\end{equation}
To solve problems in P1 (4) - (7), substitute the f(x,t) to corresponding equations.

Below are my results, not sure if they are right.
a)
\begin{equation} u(x,t) = \frac{1}{2\alpha^2c^2}[sin(\alpha x + \alpha ct) - sin(\alpha x - \alpha ct)]\end{equation}

b):  \begin{equation} u(x,t) = \frac{sin(\alpha x)}{\beta^2 - \alpha^2c^2}[sin(\beta t) - sin(\alpha ct)] \end{equation}

c):  \begin{equation} u(x,t) = \frac{1}{2c^2}[F(x+ct)-F(x-ct)] \end{equation}

d): stuck at  \begin{equation} u(x,t) = \frac{1}{2c}\int^t_0[F''(x+ct-ct')-F''(x-ct+ct')]t'dt' \end{equation}

Kinda feel we can still do the integration but don't know how after this.

[Chi Ma]

This is what I got for c:

\begin{align} u(x,t) = \frac{1}{2c^2}[F(x+ct)+F(x-ct)-2F(x)] \end{align}

For d, apply integration by part, which is why we need the third derivative $f(x)=F'''(x)$.

\begin{align}
u(x,t) &= \frac{1}{2c}\int^t_0[t'F''(x+c(t-t'))-t'F''(x-c(t-t'))]dt' \\
&= -\frac{1}{2c^2} \left\{  \int^t_0 t'dF'(x+c(t-t')) + \int^t_0 t'dF'(x-c(t-t')) \right\} \\
&= -\frac{1}{c^2}tF'(x) + \frac{1}{2c^3} [F(x+ct)+F(x-ct)]
\end{align}

[Andrew Lee Chung]

@Zaihao

I got this for
a)
\begin{equation} u(x,t) = \frac{1}{2\alpha^2c^2}[2\sin(\alpha x) - \sin(\alpha (x - ct))- \sin(\alpha (x + ct))] \end{equation}

b)
\begin{equation} u(x,t) = \frac{sin(\alpha x)}{c \alpha(\beta^2 - \alpha^2c^2)}[(\beta) sin(\alpha ct) - (\alpha c) sin(\beta t)] \end{equation}

c), d) Same as Chi Ma

Could someone else check a) and b)?

[Zaihao Zhou]

Sorry my fault. Lost one of negative signs in the process throughout. Thank you for correcting me.
Confirmed all, you guys are correct