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

In summary, the conversation involves solving a second order partial differential equation using Laplace transform. After applying the transform and considering bounded solutions, the inverse transform is applied to find the solution. There is also a discussion about the correct approach in another thread involving Fourier transform.
  • #1
Markov2
149
0
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!
 
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  • #2
Markov said:
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?
 
  • #3
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?
 
  • #4
Markov said:
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.
 
  • #5
Yes, I meant the convolution, thanks!
 
  • #6
You're welcome. Have a good one!
 
  • #7
Markov said:
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?
 
  • #8
Markov said:
Wait a sec, is this step correct?

It's correct. You took the LT correctly, applied the initial conditions correctly.
 
  • #9
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?
 
  • #10
Markov said:
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.
 

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

How do you define the Laplace transform?

The Laplace transform is a mathematical operation that transforms a function of time into a function of frequency. It is often used in solving differential equations and finding solutions to initial value problems.

Why is the Laplace transform useful in solving ODEs?

The Laplace transform allows us to convert a differential equation into an algebraic equation, making it easier to solve. It also helps us to find solutions for initial value problems, which are difficult to solve using traditional methods.

How do you solve ODEs using the Laplace transform?

To solve an ODE using the Laplace transform, we first take the Laplace transform of both sides of the equation. This transforms the differential equation into an algebraic equation, which can then be solved for the unknown function. We then use the inverse Laplace transform to find the solution in the time domain.

What is the Laplace transform of sine function?

The Laplace transform of sine function is given by 1/(s^2+1), where s is the complex frequency variable. This can be derived using the definition of the Laplace transform and the trigonometric identity for sine.

Can the Laplace transform be used for all types of ODEs?

No, the Laplace transform is mainly used for solving linear ODEs with constant coefficients. It is not applicable for nonlinear or time-varying ODEs. Additionally, care must be taken when dealing with singularities and discontinuities in the functions being transformed.

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