# Laplace transform

#### Markov

##### Member
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
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
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
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

#### Markov

##### Member
Yes, I meant the convolution, thanks!

#### Ackbach

##### Indicium Physicus
Staff member
You're welcome. Have a good one!

#### Ackbach

##### Indicium Physicus
Staff member
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
• Markov