MHB Solving PDE using laplace transforms

[TheFallen018]

[Solved] Solving PDE using laplace transforms

Hey, I'm stuck on this problem and I don't seem to be making any headway.

I took the Laplace transform with respect to t, and ended up with the following ODE:

$\frac{\partial^2 W}{\partial x^2}-W(s^2+2s+1)=0$

and the boundry conditions for $x$

$W(0,s)=\frac{30s}{(s^2+9)^2}$
and
$W(\infty,s)=0$

I tried to solve the ODE as a second order homogeneous ODE. I got the roots
$\lambda=0$
$\lambda=s^2+2s+1$

which should correspond to the solution:
$W=C_1(s)e^{(s^2+2s+1)x}+C_2(s)$

But this doesn't seem to make sense, because the I can't see how the boundry conditions could be satisfied. If x is infinity, then that means that $C_1(s)$ would have to be zero, otherwise $C_2(s)$ would have to be $-\infty$, which doesn't work anyways, since infinity take infinity is undefined. Something has to have gone wrong here

I would really appreciate it if someone could give me a hand and tell me where I've gone wrong. Thank you all so much :)

EDIT:
Turns out I found most of my problem, and that is that I messed up the characteristic equation in my second order derivative. Once I fixed that, I ended up with the result:
$W=C_1(s)e^{(s+1)x}+C_2(s)e^{-(s+1)x}$

 

[I like Serena]

Re: [Solved] Solving PDE using laplace transforms

EDIT:
Turns out I found most of my problem, and that is that I messed up the characteristic equation in my second order derivative. Once I fixed that, I ended up with the result:
$W=C_1(s)e^{(s+1)x}+C_2(s)e^{-(s+1)x}$

Hey Fallen One,

So I guess it is solved?

 

[HallsofIvy]

I tried to solve the ODE
as a second order homogeneous ODE
.​

It appears you were trying to solve it as a second order homogenous ODE with constant coefficients. But it is not!
​

 

[Ackbach]

It appears you were trying to solve it as a second order homogenous ODE with constant coefficients. But it is not!
[/LEFT]

Not sure I agree. Isn't $s$ the frequency domain equivalent of $t?$ The resulting ODE from taking the LT is in $x$, and $s$ is independent of $x$. So I think you can treat it as a linear, homogeneous second-order ODE in $x$ with constant coefficients.

 

[I like Serena]

Not sure I agree. Isn't $s$ the frequency domain equivalent of $t?$ The resulting ODE from taking the LT is in $x$, and $s$ is independent of $x$. So I think you can treat it as a linear, homogeneous second-order ODE in $x$ with constant coefficients.

Indeed.
So let's continue the calculation...

We have:
$$W(x,s)=C_1(s) e^{(s+1)x} + C_2(s)e^{-(s+1)x},\quad W(0,s)=\frac{30s}{(s^2+9)^2}, \quad W(\infty,s)=0$$
With $\Re(s)\ge 0$ it follows that $C_1(s)=0$ and:
$$W(x,s)=\frac{30s}{(s^2+9)^2}e^{-(s+1)x}$$
The inverse Laplacian with respect to $s$ is:
$$w(x,t)=5e^{-x}(t-x)\sin(3(t-x))$$
Substitute in the original equation and W/A validates that it holds.
Substitute $x=0$ and we get the boundary condition $w(0,t)=5t\sin(3t)$ as expected.