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?

 

[stevendaryl]

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]

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?

 

[stevendaryl]

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]

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)