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 $v(x,0) = u(x,0)-\Psi(x)$.