# Toronto Math Forum

## APM346-2015S => APM346--Tests => Final Exam => Topic started by: Victor Ivrii on April 14, 2015, 07:39:04 PM

Title: FE Problem 1
Post by: Victor Ivrii on April 14, 2015, 07:39:04 PM
Solve by the method of characteristics the BVP for a wave equation
\begin{equation*}
\left\{\begin{aligned}
&u_{tt}-  4u_{xx}=0,\qquad 0<x<\infty , \; t>0\\[2pt]
& u(x,0)=f(x),\\[2pt]
& u_t(x,0)=g(x),\\[2pt]
& u_x (0,t)= h(t)
\end{aligned}\right.
\end{equation*}
with $f(x)=0$,     g(x)=\left\{\begin{aligned} &1\quad && 0<x<1,\\ &0 &&1<x<\infty,\end{aligned}\right.   and   h(t)=\left\{\begin{aligned} &1\quad && 0<t<1,\\ &0 &&1<t<\infty. \end{aligned}\right.

Title: Re: FE Problem 1
Post by: Jessica Chen on April 14, 2015, 11:40:50 PM
:P
Title: Re: FE Problem 1
Post by: Victor Ivrii on April 17, 2015, 08:09:46 AM
It looks like I made it rather tedious. So I decided to grade it as is but leave exam mark out of 105 points (without scaling down to 100)

From
\begin{equation}
u(x,t)=\varphi(x+2t)+\psi (x-2t)
\label{p1-5}
\end{equation}
and initial conditions  we see that for $x>0$
\begin{equation*}
&0 &&1<x<\infty,\end{aligned}\right.
\end{equation*} and then
\begin{equation}
\varphi(x)=\left\{\begin{aligned}
&\frac{1}{4}x && x<1,\\
&\frac{1}{4} && x>1
\end{aligned}\right.
\label{p1-6}
\end{equation}
(we set $\varphi(0)=0$ as it can be set additionally without affecting solution), $\psi (x)=-\varphi(x)$  and then
\begin{equation}
u(x,t)=\left\{\begin{aligned}
& 0 && x>2t+1,\\
& \frac{1}{4} (1+2t-x) && 2t<x <2t+1.
\end{aligned}\right.
\label{p1-7}
\end{equation}
which could be obtained by D'Alembert formula as well.

From boundary condition we see
\varphi'(2t)+\psi'(-2t)= \left\{\begin{aligned} &1\quad && 0<t<1,\\ &0 &&1<t<\infty \end{aligned}\right. \\
\iff \varphi'(-x)+\psi'(x)= \left\{\begin{aligned} &1\quad && -2<x<0,\\ &0 &&x<-2 \end{aligned}\right.\\
\iff \psi(x)=\varphi(-x)+ \left\{\begin{aligned} &x\quad && -2<x<0,\\ &-2 &&x<-2 \end{aligned}\right.
\begin{equation*}
\iff  \psi(x)=\left\{\begin{aligned}
&\frac{3}{4}x && -1<x<0,\\
&\frac{1}{4}+x &&-2<x<-1,\\
&-\frac{7}{4} && x<-2
\end{aligned}\right.
\end{equation*}
and finally as $0<x<2t$
\begin{equation*}
u(x,t)=\left\{\begin{aligned}
&\frac{3}{4}(x-2t) && 2t-1<x<2t,\\
&\frac{1}{4}+x-2t &&2t-2<x<2t-1,\\
&-\frac{7}{4} && x<2t-2
\left\{\begin{aligned}
&\frac{1}{4}x + \frac{1}{2}t && x<1-2t,\\
&\frac{1}{4} && x>1-2t
\end{aligned}\right.
\end{equation*}
Title: Re: FE Problem 1
Post by: Victor Ivrii on April 17, 2015, 08:13:18 AM
Alternative solution

One can solve first problem with $h(t)=0$:
\begin{equation*}
v(x,t)=\frac{1}{2}\Bigl[ F(x+2t)+F(x-2t)\Bigr]+ \frac{1}{4}\int _{x-2t}^{x+2t} G(x')\,dx' \qquad x>0
\end{equation*}
where $F(x')=0$,
G(x')= \left\{\begin{aligned}&1 &&|{x'}|<1,\\ &0 &&|{x'}|>1\end{aligned}\right. and we applied method of continuation; then
integral is taken over $\max(x-2t,-1)< x'<\min (x+2t,1)$ which is empty as $x>2t+1$ and otherwise it is $\min (x+2t,1)-\max(x-2t,-1)= \min (x+2t,1)+\min(-x+2t,1)=\min (4t, -x+2t+1, 2)$. So
\begin{align*}
v(x,t)=&\left\{\begin{aligned}
&0 && x>2t+1\\
&\frac{1}{4}\min (4t, -x+2t+1, 2) && 0<x<2t+1
\end{aligned}\right.=\\
&\left\{\begin{aligned}
&0 && x>2t+1\\
&\frac{1}{4}(2t+1-x)&& \max(1-2t,0)<x<1+2t,\\
& t && 0<x<1-2t.\
\end{aligned}\right.
\end{align*}
Next solve problem with $g(t)=0$. It is $w(x,t)=\psi (x-2t)$ with $\psi(0)=0$ and $\psi'(-2t)=h(t)$; then
\psi'(x)=\left\{\begin{aligned} & 1 && 0>x>-2,\\ & 0 && x<-2\end{aligned}\right. and \\
\psi(x)=\left\{\begin{aligned} & x && 0>x>-2,\\ & -2 && x<-2\end{aligned}\right. and
\begin{equation*}
w(x,t)=\left\{\begin{aligned}
&0 && x>2t,\\
& x-2t && 2t-2< x <2t,\\
&-2 && x<2t-2.
\end{aligned}\right.
\end{equation*}
Finally,
\begin{align*}
u(x,t)=v(x,t)+w(x,t)=&\left\{\begin{aligned}
&0 && x>2t+1\\
&\frac{1}{4}(2t+1-x)&& \max(1-2t,0)<x<1+2t,\\
& t && 0<x<1-2t
\end{aligned}\right.
\\ + &\left\{\begin{aligned}
&0 && x>2t,\\
& x-2t && 2t-2< x <2t,\\
&-2 && x<2t-2.
\end{aligned}\right.
\end{align*}