Solving a wave equation with seperation of variables.

[Particle Head]

Homework Statement

I am trying to solve the given wave equation using separation of variables,

[itex] u_{tt} - 4u_{xx} = 4 [/itex] for [itex]0 < x < 2[/itex] and [itex]t > 0[/itex]
(BC) [itex] u(0,t) = 0[/itex] , [itex]u(2,t) = -2[/itex], for [itex]t>0 [/itex]
(IC) [itex] u(x,0)=x-x^2[/itex] , [itex] u_t(x,0)=0 [/itex] for [itex] 0\leq x \leq2 [/itex]

Homework Equations

We are told we will need to use,

[tex] x = \frac{2L}{\pi} \sum_{n\geq1}^{} \frac{(-1)^{n+1}}{n} \sin{\frac{n\pi x}{L}} [/tex]
[tex] x^2 = \frac{2L^2}{\pi} \sum_{n\geq1}^{} [\frac{(-1)^{n+1}}{n} + \frac{2}{n^3 \pi^2} ((-1)^n -1)] \sin{\frac{n\pi x}{L}} [/tex]

The Attempt at a Solution

[/B]
I first assumed a solution of the form,

[tex] u(x,t) = X(x)T(t) [/tex]

Plugging this back into the PDE this suggests that,

[tex] XT''-4X''T=4 [/tex]

With the homogeneous case we got a relation where in general [itex] \frac{T''}{c^2T} = \frac{X''}{X} = -\lambda [/itex] and this is where I am unsure because I cannot seem to separate [itex] XT''-4X''T=4 [/itex] in order to get a constant ratio between T and X.

I have a feeling I am supposed to solve the homogeneous case first however when progressing through that I ended up finding that [itex] \lambda = 0 [/itex] satisfied my boundary conditions. This is because in the homogeneous case we want to solve [itex] X''+\lambda X = 0 [/itex] and in the case where [itex] \lambda = 0 [/itex] we have [itex] X = Ax+B [/itex] and imposing the boundary conditions this seemed to imply [itex] X=-x [/itex]

Just not sure how to go from here ?

 

[vancouver_water]

Can you find any other [itex]\lambda[/itex] which satisfy the B.C.s?

 

[LCKurtz]

Homework Statement

I am trying to solve the given wave equation using separation of variables,

[itex] u_{tt} - 4u_{xx} = 4 [/itex] for [itex]0 < x < 2[/itex] and [itex]t > 0[/itex]
(BC) [itex] u(0,t) = 0[/itex] , [itex]u(2,t) = -2[/itex], for [itex]t>0 [/itex]
(IC) [itex] u(x,0)=x-x^2[/itex] , [itex] u_t(x,0)=0 [/itex] for [itex] 0\leq x \leq2 [/itex]

Make the substitution ##u(x,t) = v(x,t) + \Psi(x)##. Put that into your equation and initial conditions and see if you can make a homogeneous problem in ##v(x,t)## by choosing ##\Psi(x)## in such a way to take care of the non-homogeneous terms in the DE and boundary conditions.

 

[Particle Head]

Thank you for your response,

I believe if I make the substitution [itex] u(x,t) = v(x,t) -\frac{x^2}{2} [/itex] this enables me to take care of the non-homogeneous terms in the DE and BC. Substituting into the PDE I get [itex] v_{tt} - 4v_{xx} +4 = 4 \implies v_{tt} - 4v_{xx} =0 [/itex] and the boundary conditions becomes [itex] u(0,t)=v(0,t)=0[/itex] and [itex]u(2,t) = v(2,t) -2 =- 2 \implies v(2,t) = 0 [/itex]

I presume from here we solve the homogeneous case in [itex] v(x,t) [/itex] and then use our substitution to get the solution for the original PDE.

 

[LCKurtz]

Thank you for your response,

I believe if I make the substitution [itex] u(x,t) = v(x,t) -\frac{x^2}{2} [/itex] this enables me to take care of the non-homogeneous terms in the DE and BC. Substituting into the PDE I get [itex] v_{tt} - 4v_{xx} +4 = 4 \implies v_{tt} - 4v_{xx} =0 [/itex] and the boundary conditions becomes [itex] u(0,t)=v(0,t)=0[/itex] and [itex]u(2,t) = v(2,t) -2 =- 2 \implies v(2,t) = 0 [/itex]

I presume from here we solve the homogeneous case in [itex] v(x,t) [/itex] and then use our substitution to get the solution for the original PDE.

Yes, that looks good. Don't overlook that fact that the initial condition now becomes ##v(x,0) = u(x,0)-\Psi(x)##.