MHB Solve Laplace Transform ODE: $\sin t, 0<t<\infty$

[Markov2]

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]

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?

 

[Markov2]

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]

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.

 

[Markov2]

Yes, I meant the convolution, thanks!

 

[Ackbach]

You're welcome. Have a good one!

 

[Markov2]

I get $u(0,s)=c_2=\frac1{1+s^2}$ so we have $u(x,s)=\frac1{1+s^2}e^{-x\sqrt s},$

Wait a sec, is this step correct? I remember I applied the Laplace transform to $u(0,t),$ and substitute in the solution which I applied the Laplace transform? Is it correct? I'm not sure, because when dealing with Fourier transform, in this post: http://www.mathhelpboards.com/showthread.php?85-Solving-PDF-by-using-Fourier-transform&p=446&viewfull=1#post446

It wasn't correct, can you verify?

 

[Ackbach]

Wait a sec, is this step correct?

It's correct. You took the LT correctly, applied the initial conditions correctly.

 

[Markov2]

I did the same on http://www.mathhelpboards.com/showthread.php?85-Solving-PDF-by-using-Fourier-transform&p=446&viewfull=1#post446

So why it wasn't correct there?

 

[Ackbach]

I did the same on http://www.mathhelpboards.com/showthread.php?85-Solving-PDF-by-using-Fourier-transform&p=446&viewfull=1#post446

So why it wasn't correct there?

Because you didn't do the same thing in the other thread. You didn't take the FT of the initial condition, you just took the initial condition straight up and applied that. ThePerfectHacker pointed out that you needed to take the FT of the initial condition.