Author Topic: TT1-P2  (Read 607 times)

Victor Ivrii

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TT1-P2
« on: October 19, 2016, 10:26:33 PM »
(a) Find solution $u(x,t)$ to
\begin{align}
&u_{tt}-u_{xx}= (x^2-1)e^{-\frac{x^2}{2}},\label{eq-1}\\
&u|_{t=0}=-e^{-\frac{x^2}{2}}, \quad u_t|_{t=0}=0.\label{eq-2}
\end{align}
(b) (1 pts--bonus) Find $\lim _{t\to +\infty} u(x,t)$.

Shentao YANG

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Re: TT1-P2
« Reply #1 on: October 19, 2016, 10:54:06 PM »
My solution is:
(a)
$$u(x,t) =  - {e^{ - {{{x^2}} \over 2}}}$$
As:
$$u = {1 \over 2}( - {e^{ - {{{{(x + t)}^2}} \over 2}}} - {e^{ - {{{{(x - t)}^2}} \over 2}}}) + \underbrace {{1 \over {2c}}\int_0^t {\int_{x - t + t'}^{x + t - t'} {({y^2} - 1)} } {e^{ - {{{y^2}} \over 2}}}dydt'}_{(3)}$$
The inner integral of $(3)$ yields:
$$\int_{x - t + t'}^{x + t - t'} {{y^2}} {e^{ - {{{y^2}} \over 2}}} - {e^{ - {{{y^2}} \over 2}}}dy = \underbrace {\int_{x - t + t'}^{x + t - t'} {{y^2}} {e^{ - {{{y^2}} \over 2}}}dy}_{(4)} - \int_{x - t + t'}^{x + t - t'} {{e^{ - {{{y^2}} \over 2}}}} dy$$
$$(4) = \int_{x - t + t'}^{x + t - t'} y {e^{ - {{{y^2}} \over 2}}}ydy = \left. { - {e^{ - {{{y^2}} \over 2}}}y} \right|_{y = x - t + t'}^{y = x + t - t'} + \int_{x - t + t'}^{x + t - t'} {{e^{ - {{{y^2}} \over 2}}}} dy$$
Therefore
$$\int_{x - t + t'}^{x + t - t'} {{y^2}} {e^{ - {{{y^2}} \over 2}}} - {e^{ - {{{y^2}} \over 2}}}dy = \left. { - {e^{ - {{{y^2}} \over 2}}}y} \right|_{y = x - t + t'}^{y = x + t - t'} + \int_{x - t + t'}^{x + t - t'} {{e^{ - {{{y^2}} \over 2}}}} dy - \int_{x - t + t'}^{x + t - t'} {{e^{ - {{{y^2}} \over 2}}}} dy = (x - t + t'){e^{ - {{{{(x - t + t')}^2}} \over 2}}} - (x + t - t'){e^{ - {{{{(x + t - t')}^2}} \over 2}}}$$
so $(3)$ becomes:
$${1 \over 2}\int_0^t {(x - t + t'){e^{ - {{{{(x - t + t')}^2}} \over 2}}} - (x + t - t'){e^{ - {{{{(x + t - t')}^2}} \over 2}}}} dt'\matrix{
   {} & {}  \cr

 } (c = 1)$$
$$ = {1 \over 2}\int_{x - t}^x {z{e^{ - {{{z^2}} \over 2}}}} dz + {1 \over 2}\int_{x + t}^x {z{e^{ - {{{z^2}} \over 2}}}} dz$$
$$ =  - {e^{ - {{{x^2}} \over 2}}} + {1 \over 2}{e^{ - {{{{(x - t)}^2}} \over 2}}} + {1 \over 2}{e^{ - {{{{(x + t)}^2}} \over 2}}}$$
Finally,
$$u = {1 \over 2}( - {e^{ - {{{{(x + t)}^2}} \over 2}}} - {e^{ - {{{{(x - t)}^2}} \over 2}}})- {e^{ - {{{x^2}} \over 2}}} + {1 \over 2}{e^{ - {{{{(x - t)}^2}} \over 2}}} + {1 \over 2}{e^{ - {{{{(x + t)}^2}} \over 2}}} $$
$$ \Rightarrow u(x,t) =  - {e^{ - {{{x^2}} \over 2}}}$$
(b) Since u does not depend on t, so we basically get the same equation:
$$\mathop {\lim }\limits_{t \to \infty }  = u(x,t) =  - {e^{ - {{{x^2}} \over 2}}}$$
« Last Edit: October 19, 2016, 11:27:17 PM by Shentao YANG »

XinYu Zheng

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Re: TT1-P2
« Reply #2 on: October 19, 2016, 11:11:10 PM »
Before we begin, we need one result: $\int x^2e^{-x^2/2}\,dx$. Changing variables, this can be written as
$$-\int x\,d(e^{-x^2/2})$$
Now integrate by parts, set $u=x$, $du=dx$, $dv=d(e^{-x^2/2})$, $v=e^{-x^2/2}$, we obtain
$$-xe^{-x^2/2}+\int e^{-x^2/2}\,dx$$
Now we begin the problem.
By D'Alembert's formula and Duhamel integral, the solution is
$$u(x,t)=\frac{1}{2}\left(-e^{(x-t)^2/2}-e^{(x+t)^2/2}\right)+\frac{1}{2}\int_0^t\int_{x-(t-t')}^{x+(t-t')} x'^2e^{-x'^2/2}-e^{-x'^2/2}\,dx'dt'$$
Now consider the inside integral
$$\int_{x-(t-t')}^{x+(t-t')} x'^2e^{-x'^2/2}-e^{-x'^2/2}\,dx'\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,(*)$$
Let $a=x-(t-t')$ and $b=x+(t-t')$ for notational simplicity.
Using the result above, the first term can be integrated:
$$\int_{a}^{b} x'^2e^{-x'^2/2}\,dx'=ae^{-a^2/2}-be^{-b^2/2}+\int_a^b e^{-x'^2/2}\,dx'$$
But note that the second integral here cancels with the integral of the second term in $(*)$. So we just have
$$\int_a^b x'^2e^{-x'^2/2}-e^{-x'^2/2}\,dx'=ae^{-a^2/2}-be^{-b^2/2}$$
Now we need to integrate this with respect to $t'$, from $0$ to $t$. Consider the first term first. Note that $da=dt'$. When $t'=0$, $a=x-t$, When $t'=t$, $a=x$. Thus the first term integrates
$$\int_{x-t}^x ae^{-a^2/2}\,da=e^{-(x-t)^2/2}-e^{-x^2/2}$$
Now for the second term, note that $db=-dt'$. When $t'=0$, $b=x+t$. When $t'=t$, $b=x$. Thus we have
$$-\int_{x+t}^x be^{-b^2/2}\,db=e^{-x^2/2}-e^{-(x+t)^2/2}$$
Now putting together all the results:
$$u(x,t)=\frac{1}{2}\left(-e^{(x-t)^2/2}-e^{(x+t)^2/2}\right)+\frac{1}{2}\left(e^{-(x-t)^2/2}-e^{-x^2/2}-e^{-x^2/2}+e^{-(x+t)^2/2}\right) =-e^{-x^2/2}$$
Since this does not depend on $t$, we have $\lim_{t\to\infty} u(x,t)=-e^{-x^2/2}$.
« Last Edit: October 19, 2016, 11:18:14 PM by XinYu Zheng »

Victor Ivrii

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Re: TT1-P2
« Reply #3 on: October 20, 2016, 04:49:49 AM »
:D

Actually my intention was to have initial condition $u|_{t=0}=e^{-\frac{1}{2}x^2}$, $u_t|_{t=0}=0$.

Solve (a), (b) as a Bonus

Tianyi Zhang

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Re: TT1-P2
« Reply #4 on: October 20, 2016, 04:22:17 PM »
For the new question,
(a)We can just apply the formula here to get:
$$u = {1 \over 2}(  {e^{ - {{{{(x + t)}^2}} \over 2}}} + {e^{ - {{{{(x - t)}^2}} \over 2}}}) + {{1 \over {2c}}\int_0^t {\int_{x - t + t'}^{x + t - t'} {({y^2} - 1)} } {e^{ - {{{y^2}} \over 2}}}dydt'}$$

I skip all the integral parts because Shentao had it.

Lastly we can get,
$$u = {e^{ - {{{{(x + t)}^2}} \over 2}}} + {e^{ - {{{{(x - t)}^2}} \over 2}}}- {e^{ - {{{x^2}} \over 2}}} $$

(b)
Since the first and second items will vanish as t goes to infinity
$$ u = - {e^{ - {{{x^2}} \over 2}}} $$
« Last Edit: October 20, 2016, 04:24:12 PM by Tianyi Zhang »