Inhomogeneous damped wave equation

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Homework Statement


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

Homework Equations


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

The Attempt at a Solution


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

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

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

where [tex]\Omega=\sqrt{\omega^{2}-\mu^{2}}[/tex]

Is this correct so far?
 

Answers and Replies

  • #2
stevendaryl
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Homework Statement


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

Homework Equations


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

The Attempt at a Solution


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

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

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

where [tex]\Omega=\sqrt{\omega^{2}-\mu^{2}}[/tex]

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 [itex]u_{tt}(x,t)+2u_{t}(x,t)-u_{xx}(x,t)=18\sin\left(\dfrac{3\pi x}{l}\right)[/itex]

Try writing [itex]u(x,t) = A(x,t) + B(x)[/itex] and choose [itex]A[/itex] and [itex]B[/itex] so that:

[itex]A_{tt}(x,t)+2A_{t}(x,t)-A_{xx}(x,t)=0[/itex]

[itex]-B_{xx}(x) = 18\sin\left(\dfrac{3\pi x}{l}\right)[/itex]

Then the approach you're describing will give you a solution to [itex]A(x,t)[/itex]
 
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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?
 
  • #4
stevendaryl
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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 [itex]A_n[/itex] and [itex]B_n[/itex]). Comparing with your original equation, [itex]\mu = 1[/itex] and [itex]c=1[/itex]
 
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Thanks, I wanted confirmation I had the right damping.
 
  • #6
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So with the boundary conditions the general solution of the homogeneous equation is
[tex]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}[/tex]
My course text is a bit confusing but to get the particular solution should I start with
[tex]u(x,t)=\sin\left( k_nx \right)g(t) [/tex] and then solve
[tex]g^{\prime\prime}+2g^\prime+\omega^2_ng=18\sin\left( \frac{3\pi x}{l} \right)[/tex]
 

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