Toronto Math Forum
APM3462015F => APM346Home Assignments => HA3 => Topic started by: Victor Ivrii on September 28, 2015, 01:05:56 PM

Problem 1 here
http://www.math.toronto.edu/courses/apm346h1/20159/PDEtextbook/Chapter2/S2.4.P.html#problem2.4.P.1 (http://www.math.toronto.edu/courses/apm346h1/20159/PDEtextbook/Chapter2/S2.4.P.html#problem2.4.P.1)

The general solution is
\begin{equation} u(x,t) = \frac{1}{2c}\int_0^t{\int_{xc(tt')}^{x+c(tt')} 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(xct)] \end{equation}
d): stuck at \begin{equation} u(x,t) = \frac{1}{2c}\int^t_0[F''(x+ctct')F''(xct+ct')]t'dt' \end{equation}
Kinda feel we can still do the integration but don't know how after this.

This is what I got for c:
\begin{align} u(x,t) = \frac{1}{2c^2}[F(x+ct)+F(xct)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(tt'))t'F''(xc(tt'))]dt' \\
&= \frac{1}{2c^2} \left\{ \int^t_0 t'dF'(x+c(tt')) + \int^t_0 t'dF'(xc(tt')) \right\} \\
&= \frac{1}{c^2}tF'(x) + \frac{1}{2c^3} [F(x+ct)+F(xct)]
\end{align}

@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)?

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