Solving PDE with Laplace Transform

tommyhakinen

1. The problem statement, all variables and given/known data
[tex]\frac{\partial^{2}u}{\partial t^{2}} = a^{2} \frac{\partial^{2}u}{\partial x^{2}}[/tex] (x>0, t>0)

with u(0,t) = t, u(x,0) = 0, u_t(x,0) = A.

Solve the PDE using laplace transform.

3. The attempt at a solution
I have managed to get the transform:
[tex]\frac{\partial^{2}U(x,s)}{\partial x^{2}} - \frac{s^{2}}{a^{2}}U(x,s) = - \frac{A}{a^{2}}[/tex]

How should I continue from here? How do I find the solution for U(x,s)? If using Integrating factor, how should I do it? thanks..

HallsofIvy

U depends on the single variable x- s is a parameter. So that is an ordinary equation, U'+ (s²/a²)U= -A/s². That' s a linear equation with constant coefficients. If you are working with partial differential equations and the Laplace transform you should no quicker ways of solving such an equation than finding an integrating factor. Actually you should also by now have learned a general formula for an integrating factor of a linear equation.

By the way, please don't use [itex]\delta[/itex] for partial derivatives. Use [itex]\partial[/itex] using the Latex symbol "\partial'.

tommyhakinen

I know for first order ODE, we can use U' + p(x)U = q(x) by first finding the integrating factor. IF = e^∫p(x)dx. then u = 1/IF ∫ q(x) IF dx. However, this problem is a second order ODE and non-homogenous. How should I get the solution for the U(x,s)?

jambaugh

You could carry out a second Laplace transform on the x variable. You then solve the algebraic equation and take (look up) the inverse transform. Otherwise you can solve the ODE directly (remembering s is a constant w.r.t. x) and then inverse transform it back to get the solution.

The general idea is Laplace transforms convert differential equations to algebraic equations. Transform-->Solve-->Inverse Transform.

tommyhakinen

thanks jambaugh. I have tried it. Basically my problem is on solving non-homogenous second order ODE. I have reached the transformation stage. I just need to solve the ODE. I still do not understand how to solve this:

[tex]
\frac{\partial^{2}U(x,s)}{\partial x^{2}} - \frac{s^{2}}{a^{2}}U(x,s) = - \frac{A}{a^{2}}
[/tex]

thanks.

jambaugh

First find the homogeneous solution to
[tex] U'' - \frac{s^2}{a^2}U = 0[/tex]
then find a particular solution to the inhomogeneous equation and add.

The homogeneous equation is linear with constant coefficient so try the form [tex]U=Ce^{rx}[/tex] and you get the characteristic equation (a quadratic equation in r). You get, I believe, two distinct roots so take an arbitrary linear combination of these.

[tex] U_h = C_1 e^{r_1 x} + C_2 e^{r_2 x}[/tex]

The particular solution is most easily obtained by the method of Undetermined Coefficients.
Start with your function [tex] f(x) = -\frac{A}{a^2}[/tex]. Factor out any constants (the whole thing here) to get your starting function,( 1 in your case). Differentiate repeatedly and see if you get a finite set of linearly independent functions. (In this case you get trivially the set { 1 }. If you don't get a finite set (up to constant multipliers) you must use variation of parameters but that's not the case here.

If for example if your function had been [tex] kx^3[/tex] you'd get [tex] \{ x^3, x^2, x,1\} [/tex]
Or for example if your function had been [tex] 7\sin(5x)[/tex] you'd get [tex] \{ \sin(5x), \cos(5x) \}[/tex]

Now try an arbitrary linear combination from your basis set in the inhomogeneous equation and solve for the coefficients. (In your case take [tex] U_p = K\cdot 1[/tex] and so its second derivative is zero and you can solve for K trivially. (I think [tex] U_p = As^2 [/tex] but double check.)

Now that you have the general = particular + homogeneous solution
[tex] U = C_1 e^{r_1 x} + C_2 e^{r_2 x} + K[/tex]
you must apply the initial conditions to find some or all of the constants which occur in the homogeneous part.

I believe you are short one initial condition to get a unique solution. Since the x-eqn is 2nd order you'll need two x-value initial conditions. If so then that's OK you'll have an extra parameter in your solution.

Now find the reverse Laplace transform of your solution (you'll note that the constants in the solution depend on s.)

But again you could also do this by executing a second Laplace transform in the x-variable:
[tex] \mathbf{L}: U(x,s) \mapsto \tilde{U}(r,s)[/tex]
Solve the algebraic equation and then double inverse Laplace transform. I suspect that was the intent in order to avoid the above ODE method.