Solving U_{tt}=9U_{xx} with Laplace Transforms

[pk415]

We are supposed to work this using Laplace transforms

$$U_{tt}=9U_{xx}; -infty<x<infty$$

$$U(x,0)=sinx$$

$$U_t(x,0)=0$$

The attempt at a solution

[tex]Let L=\hat{U}[/tex]

$$L[U_{tt}]=s^2\hat{U}-s(sinx)$$

$$L[9U_{xx}]=9\hat{U}_{xx}$$

$$s^2\hat{U}-s(sinx)=9\hat{U}_{xx}$$

$$\hat{U}_{xx}-\frac{s^2}{9}\hat{U}=-\frac{s}{9}sinx$$

This is a second order nonhomogeneous equation where the homogeneous solution is

$$\hat{U}_h=A(s)e^{\frac{s}{3}x}+B(s)e^{-\frac{s}{3}x}$$

and the particular solution is

$$\hat{U}_p=\frac{s}{s^2+9}sinx$$

Then we have

$$\hat{U}=A(s)e^{\frac{s}{3}x}+B(s)e^{-\frac{s}{3}x}+\frac{s}{s^2+9}sinx$$

We know that

$$\lim_{x\rightarrow\infty}U=0$$

$$\lim_{x\rightarrow{-\infty}}U=0$$

That second one should say limit as x approaches negative infinity, can't figure out how to get latex to do that

$$\lim_{x\rightarrow\infty}\hat{U}=0$$

$$\lim_{x\rightarrow\infty}\hat{U}=0$$

Second one negative infinity again

If we plug these in we get

$$A(s)=\frac{\lim_{x\rightarrow\infty}\frac{s}{s^2+9}sinx}{\lim_{x\rightarrow\infty}e^{-\frac{s}{3}x}}}$$

$$A(s)=0$$

I'm not sure if that last step is legal, but if it is you can use the same reasoning for B(s)
Any ideas?

 

[mighty2000]

Your solution is on the right track, but there are a few errors and missing steps. Here is a more detailed explanation of the solution:

First, let's take the Laplace transform of the given equation:

L[U_{tt}] = s^2 \hat{U} - sU(x,0) - U_t(x,0) = s^2 \hat{U} - s\sin x

L[9U_{xx}] = 9\hat{U}_{xx}

Since we know that U(x,0)=sinx and U_t(x,0)=0, we can plug these values in and simplify:

s^2 \hat{U} - s\sin x = 9\hat{U}_{xx}

Next, we can use the initial conditions to find the values of A(s) and B(s) in the homogeneous solution. Plugging in the initial conditions, we get:

A(s) + B(s) = 0

\frac{s}{3}A(s) - \frac{s}{3}B(s) = 0

Solving these equations, we get A(s)=0 and B(s)=0.

Now, for the particular solution, we can use the method of undetermined coefficients. We assume a particular solution of the form:

\hat{U}_p = K(s) \sin x

Plugging this into the equation, we get:

K(s) = \frac{s}{s^2+9}

Therefore, the general solution is:

\hat{U} = A(s)e^{\frac{s}{3}x} + B(s)e^{-\frac{s}{3}x} + \frac{s}{s^2+9}\sin x

Using the initial conditions to solve for A(s) and B(s), we get A(s)=0 and B(s)=0.

Finally, taking the inverse Laplace transform, we get the solution:

U(x,t) = \frac{s}{s^2+9} \sin x

Note that the limits as x approaches positive and negative infinity are both 0, as expected.