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,

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

Homework Equations

We are told we will need to use,

x = \frac{2L}{\pi} \sum_{n\geq1}^{} \frac{(-1)^{n+1}}{n} \sin{\frac{n\pi x}{L}}
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}}

The Attempt at a Solution

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

u(x,t) = X(x)T(t)

Plugging this back into the PDE this suggests that,

XT''-4X''T=4

With the homogeneous case we got a relation where in general \frac{T''}{c^2T} = \frac{X''}{X} = -\lambda and this is where I am unsure because I cannot seem to separate XT''-4X''T=4 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 \lambda = 0 satisfied my boundary conditions. This is because in the homogeneous case we want to solve X''+\lambda X = 0 and in the case where \lambda = 0 we have X = Ax+B and imposing the boundary conditions this seemed to imply X=-x

Just not sure how to go from here ?

 

[vancouver_water]

Can you find any other \lambda which satisfy the B.C.s?

 

[LCKurtz]

Homework Statement

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

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

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

I presume from here we solve the homogeneous case in v(x,t) 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 u(x,t) = v(x,t) -\frac{x^2}{2} this enables me to take care of the non-homogeneous terms in the DE and BC. Substituting into the PDE I get v_{tt} - 4v_{xx} +4 = 4 \implies v_{tt} - 4v_{xx} =0 and the boundary conditions becomes u(0,t)=v(0,t)=0 and u(2,t) = v(2,t) -2 =- 2 \implies v(2,t) = 0

I presume from here we solve the homogeneous case in v(x,t) 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 $v(x,0) = u(x,0)-\Psi(x)$.