Inhomogeneous damped wave equation

bobred

Homework Statement

Solve
$$u_{tt}(x,t)+2u_{t}(x,t)-u_{xx}(x,t)=18\sin\left(\dfrac{3\pi x}{l}\right)$$
$$\omega=\frac{\pi n c}{l}$$
Boundary conditions
$$u(0,t)=u(l,t)=0$$
$$l<\pi$$
Initial conditions
$$u(x,0)=u_t(x,0)=0$$

Homework Equations

[/B]
The general inhomogeneous damped wave equation is
$$u_{tt}(x,t)+2\mu u_{t}(x,t)-c^{2}u_{xx}(x,t)=p(x,t)$$

The Attempt at a Solution

By separation of variables the position dependent part is
$$f(x)=a\cos(kx)+b\sin(kx)$$
and the time dependent part auxilliary equation is
$$\lambda^2+2\mu\lambda+\omega^2=0$$
which takes one of three expressions depending on the sign of $\mu-\omega$

So from what is given I would say $\mu=1$ and $c^2=1$ and with
$\omega=\frac{\pi n c}{l}$ and $l<\pi$ then $\mu<\omega$ which is weak damping

$$g_{n}(t)=\left(A_{n}\cos\left(\Omega t\right)+B_{n}\sin\left(\Omega t\right)\right)e^{-\mu t}$$

where $$\Omega=\sqrt{\omega^{2}-\mu^{2}}$$

Is this correct so far?

Staff Emeritus

Homework Statement

Solve
$$u_{tt}(x,t)+2u_{t}(x,t)-u_{xx}(x,t)=18\sin\left(\dfrac{3\pi x}{l}\right)$$
$$\omega=\frac{\pi n c}{l}$$
Boundary conditions
$$u(0,t)=u(l,t)=0$$
$$l<\pi$$
Initial conditions
$$u(x,0)=u_t(x,0)=0$$

Homework Equations

[/B]
The general inhomogeneous damped wave equation is
$$u_{tt}(x,t)+2\mu u_{t}(x,t)-c^{2}u_{xx}(x,t)=p(x,t)$$

The Attempt at a Solution

By separation of variables the position dependent part is
$$f(x)=a\cos(kx)+b\sin(kx)$$
and the time dependent part auxilliary equation is
$$\lambda^2+2\mu\lambda+\omega^2=0$$
which takes one of three expressions depending on the sign of $\mu-\omega$

So from what is given I would say $\mu=1$ and $c^2=1$ and with
$\omega=\frac{\pi n c}{l}$ and $l<\pi$ then $\mu<\omega$ which is weak damping

$$g_{n}(t)=\left(A_{n}\cos\left(\Omega t\right)+B_{n}\sin\left(\Omega t\right)\right)e^{-\mu t}$$

where $$\Omega=\sqrt{\omega^{2}-\mu^{2}}$$

Is this correct so far?

I think you're on the right track, but separation of variables is mostly useful for homogeneous partial differential equations. Here's a trick to solving the inhomogeneous case:

You want a solution to $u_{tt}(x,t)+2u_{t}(x,t)-u_{xx}(x,t)=18\sin\left(\dfrac{3\pi x}{l}\right)$

Try writing $u(x,t) = A(x,t) + B(x)$ and choose $A$ and $B$ so that:

$A_{tt}(x,t)+2A_{t}(x,t)-A_{xx}(x,t)=0$

$-B_{xx}(x) = 18\sin\left(\dfrac{3\pi x}{l}\right)$

Then the approach you're describing will give you a solution to $A(x,t)$

bobred
bobred
Hi, thanks, that's great but we are asked to solve the homogeneous part by separation of variables first. Is what I have done ok? Is it weak damping?

Staff Emeritus
Hi, thanks, that's great but we are asked to solve the homogeneous part by separation of variables first. Is what I have done ok? Is it weak damping?

Yes, you have the solution to the homogeneous equation (except for the constants $A_n$ and $B_n$). Comparing with your original equation, $\mu = 1$ and $c=1$

bobred
bobred
Thanks, I wanted confirmation I had the right damping.

bobred
So with the boundary conditions the general solution of the homogeneous equation is
$$u(x,t)=\sum_{n=1}^{\infty}b_{n}\sin(kx)\left(A_{n}\cos\left(\Omega t\right)+B_{n}\sin\left(\Omega t\right)\right)e^{- t}$$
My course text is a bit confusing but to get the particular solution should I start with
$$u(x,t)=\sin\left( k_nx \right)g(t)$$ and then solve
$$g^{\prime\prime}+2g^\prime+\omega^2_ng=18\sin\left( \frac{3\pi x}{l} \right)$$