# Toronto Math Forum

## APM346-2015S => APM346--Tests => Test 1 => Topic started by: Victor Ivrii on February 12, 2015, 07:25:41 PM

Title: TT1 Problem 2
Post by: Victor Ivrii on February 12, 2015, 07:25:41 PM
Solve the wave equation with the following initial conditions
\begin{equation*}
\left\{\begin{aligned}
& u_{tt}- 4 u_{xx}= x t e^{-x^2} ,\qquad&& -\infty <x< \infty\\
&u (x,0) = e^{-x^2} ,\\
&u_t(x,0)= x e^{-x^2}\end{aligned}\right.
\end{equation*}
Title: Re: TT1 Problem 2
Post by: Jessica Chen on February 12, 2015, 10:23:59 PM
I don't know how to attach a pdf properly such that we can see its preview without download it.

You cannot. So scan to png or jpg. V.I.
Title: Re: TT1 Problem 2
Post by: Biao Zhang on February 13, 2015, 05:28:50 AM
To solve the question by using D'Alemeber formula:\\
\begin{multline*}
u(x,t)= \underbrace{\frac{1}{2}\bigl( G(x+ct)+G(x-ct) \bigr)}_1
+ \underbrace{\frac{1}{2c}\int_{x-ct} ^{x+ct} H(x')\,dx'}_2 +
\underbrace{\frac{1}{2c} \int_0^t \int _{x-c(t-t')} ^{x+c(t-t')} F(x',t')\, dx' dt'}_3
\end{multline*}
we have
\begin{align*}
c=2\\
G(x)=e^{-x^2}\\
H(x)=xe^{-x^2}\\
F(x,t)=xte^{-x^2}
\end{align*}
\begin{align*}
(1): \frac{1}{2}(e^{-(x+2t)^2}+e^{-(x-2t)^2})\\
(2): \frac{1}{4}\int_{x-2t} ^{x+2t} H(x')\,dx'\\
\hspace{1cm} =\frac{1}{8}(e^{-(x-2t)^2}-e^{-(x+2t)^2})
\end{align*}
combine (1) and (2) we have:
\begin{equation*}
\frac{3}{8}e^{-(x+2t)^2} +\frac{5}{8}e^{-(x-2t)^2}
\end{equation*}
now we need find (3):
\begin{align*}
\frac{1}{4} \int_0^t \int _{x-2(t-t')} ^{x+2(t-t')} x't'e^{-x'^2}\, dx' dt'\\
\frac{1}{8}  \int_0^t t'(e^{-(x+2t-2t')^2}-e^{-(x-2t+2t')^2})\, dt'
\end{align*}
This part i don't know how to integrate this part . but answer should be:
\begin{equation*}
\frac{3}{8}e^{-(x+2t)^2} +\frac{5}{8}e^{-(x-2t)^2}+\frac{1}{8}  \int_0^t t'(e^{-(x+2t-2t')^2}-e^{-(x-2t+2t')^2})\, dt'
\end{equation*}
Title: Re: TT1 Problem 2
Post by: Victor Ivrii on February 14, 2015, 05:57:59 AM
OK.

Remark. The last integral is non-elementary but changing variable we will get something like $\int_X^Y (A+B s) e^{-s^2/2}\,ds$ where one part is integrable and another could be expressed through $\operatorname{erf}$. But this is optional. I have not punished those who did not the last step and marginally punished those who did incorrectly.

Consider the last term obtained by Biao (which is contribution of the RHE after we integrated by $x'$), but Biao erred with sign

\frac{1}{8}\int_0^t  s\Bigl( e^{-(x-2t+2s)^2}  -  e^{-(x+2t-2s)^2} \Bigr)\, ds=
F(x,t,2)-F(x,t,-2)
\label{a}

with

F(x,t,k)= \frac{1}{8}\int_0^t  s e^{-(x-k t+ks)^2}   \, ds=
\frac{1}{8k^2}\int_{x-kt}^{x}  (x-kt+y)  e^{-y^2}\,dy
\label{b}

where we made a substitution $y=x-kt+ks$.Then
\begin{equation*}
F(x,t,k)=
\frac{1}{8k^2}\biggl[(x-kt) \int_{x-kt}^x   e^{-y^2}\,dy - \int_{x-kt}^x  y  e^{-y^2}\,dy\biggr];
\end{equation*}
then (\ref{a}) equals
\begin{gather*}
\frac{1}{32}
\biggl[(x-2t) \int_{x-2t}^x e^{-y^2}\,dy - (x+2t) \int_{x+2t}^x   e^{-y^2}\,dy
-\int_{x-2t}^x  y  e^{-y^2}\,dy + \int_{x+2t}^x  y  e^{-y^2}\,dy\biggr]\\[4pt]
=\frac{1}{32}\biggl[ (x-2t) \bigl( E(x)-E(x-2t)\bigr)- (x+2t) \bigl( E(x)-E(x+2t)\bigr) \biggr]+
\frac{1}{64}\biggl[ e^{-(x+2t)^2}-e^{-(x-2t)^2}\biggr]
\end{gather*}
with $E(z)=\int_0^z e^{-y^2}\,dy$ which could be expressed via $\operatorname{erf}$ easily.