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Laplace transform

Markov

Member
Feb 1, 2012
149
Solve

$\begin{eqnarray*}
{{u}_{t}}&=&{{u}_{xx}},\text{ }0<x<\infty ,\text{ }0<t<\infty \\
u(0,t)&=&\sin t,\text{ }0<t<\infty \\
u(x,0)&=&0,\text{ }0\le x<\infty .
\end{eqnarray*}$

I need to apply the Laplace transform to solve this, so by applying it I get $su(x,s)-u(x,0)=u_{xx}(x,s)$ and $u(0,s)=\frac1{1+s^2}.$ I need to treat the first equation as a second order ODE right? Then the solution would be $u(x,s)=c_1e^{x\sqrt s}+c_2e^{-x\sqrt s},$ here's where I'm getting stuck, I've been told that I must found bounded solutions, so in order to get it bounded I should consider $u(x,s)=c_2e^{-x\sqrt s},$ by applying the initial conditions I get $u(0,s)=c_2=\frac1{1+s^2}$ so we have $u(x,s)=\frac1{1+s^2}e^{-x\sqrt s},$ now I must apply the inverse Laplace transform, so first $u(x,s)=\frac s{1+s^2}\frac{e^{-x\sqrt s}}s=\mathcal L(\cos t)\mathcal L^{-1}\left(\frac{e^{-x\sqrt s}}s\right)$
I can't find the inverse of the remaining function.

Is this correct?
Thanks!
 

Ackbach

Indicium Physicus
Staff member
Jan 26, 2012
4,197
Solve

$\begin{eqnarray*}
{{u}_{t}}&=&{{u}_{xx}},\text{ }0<x<\infty ,\text{ }0<t<\infty \\
u(0,t)&=&\sin t,\text{ }0<t<\infty \\
u(x,0)&=&0,\text{ }0\le x<\infty .
\end{eqnarray*}$

I need to apply the Laplace transform to solve this, so by applying it I get $su(x,s)-u(x,0)=u_{xx}(x,s)$ and $u(0,s)=\frac1{1+s^2}.$ I need to treat the first equation as a second order ODE right? Then the solution would be $u(x,s)=c_1e^{x\sqrt s}+c_2e^{-x\sqrt s},$ here's where I'm getting stuck, I've been told that I must found bounded solutions, so in order to get it bounded I should consider $u(x,s)=c_2e^{-x\sqrt s},$ by applying the initial conditions I get $u(0,s)=c_2=\frac1{1+s^2}$ so we have $u(x,s)=\frac1{1+s^2}e^{-x\sqrt s},$ now I must apply the inverse Laplace transform, so first $u(x,s)=\frac s{1+s^2}\frac{e^{-x\sqrt s}}s=\mathcal L(\cos t)\mathcal L^{-1}\left(\frac{e^{-x\sqrt s}}s\right)$
There are two problems with this approach.

1. Why insert the additional s's? You already know the inverse LT of $1/(s^{2}+1)$.
2. Second of all, the inverse LT of a product is NOT the product of the inverse LT's. Instead, you must use the convolution theorem, which says that

$$\mathcal{L}^{-1}\{F(s)\cdot G(s)\}=\int_{0}^{t}f(\tau)g(t-\tau)\,d\tau\equiv(f* g)(t).$$

As it turns out, the inverse LT of the remaining piece is

$$\mathcal{L}^{-1}\{e^{-x\sqrt{s}}\}=\frac{xe^{-\frac{x^{2}}{4t}}}{2\sqrt{\pi}\,t^{3/2}}.$$

Can you finish from here?
 

Markov

Member
Feb 1, 2012
149
Oh yes, so it's actually $\mathcal L(\sin t)\mathcal L\left( {\frac{{x{e^{ - \frac{{{x^2}}}{{4t}}}}}}{{2\sqrt \pi {t^{\frac{3}{2}}}}}} \right),$ so the solution is $u(x,t)=(\sin (t) *{\frac{{x{e^{ - \frac{{{x^2}}}{{4t}}}}}}{{2\sqrt \pi {t^{\frac{3}{2}}}}}}),$ is it correct now?
 

Ackbach

Indicium Physicus
Staff member
Jan 26, 2012
4,197
Oh yes, so it's actually $\mathcal L(\sin t)\mathcal L\left( {\frac{{x{e^{ - \frac{{{x^2}}}{{4t}}}}}}{{2\sqrt \pi {t^{\frac{3}{2}}}}}} \right),$ so the solution is $u(x,t)=(\sin (t) *{\frac{{x{e^{ - \frac{{{x^2}}}{{4t}}}}}}{{2\sqrt \pi {t^{\frac{3}{2}}}}}}),$ is it correct now?
If by
$$\sin (t) *{\frac{{x{e^{ - \frac{{{x^2}}}{{4t}}}}}}{{2\sqrt \pi {t^{\frac{3}{2}}}}}}$$
you mean $\sin(t)$ convolved with
$${\frac{{x{e^{ - \frac{{{x^2}}}{{4t}}}}}}{{2\sqrt \pi {t^{\frac{3}{2}}}}}},$$
I would agree. You might want to write out what that is in terms of the integral. If the integral is tractable, you might even compute it.
 

Markov

Member
Feb 1, 2012
149
Yes, I meant the convolution, thanks!
 

Ackbach

Indicium Physicus
Staff member
Jan 26, 2012
4,197
You're welcome. Have a good one!
 

Markov

Member
Feb 1, 2012
149

Ackbach

Indicium Physicus
Staff member
Jan 26, 2012
4,197
Wait a sec, is this step correct?
It's correct. You took the LT correctly, applied the initial conditions correctly.
 

Ackbach

Indicium Physicus
Staff member
Jan 26, 2012
4,197