Differential equation of type $u_{tt}=u_{xx} + u$ with specific boundary conditions

Question

I need to solve the following differential equations.

$u(x, t)\;;x\in(0,2)\;;t>0$
$u_{tt} = u_{xx} + u$

Boundary conditions:
$u(x,0) = 0$
$u_t(x,0) = 0$
$u(0, t) = 2t$
$u(2, t) = 0$

I have tried to separate variables: $u(x,t) = X(x)T(t)$.

$XT'' = X''T + XT$
$\frac{T''}{T} = \frac{X''}{X} + 1 = \lambda$

But whatever lambda I choose ($<0;=0;>0$), $T$ is equal to $0$ due to boundary conditions:

$T = ax + b$
$0 = T(0)X(x) = bX(x)$
$0 = T'(0)X(x) = aX(x)$
$a = b = 0$ if $\exists x\;X(x)\neq0$

$T = ae^{kt} + be^{-kt}$
$0 = T(0)X(x) = (a+b)bX(x)$
$0 = T'(0)X(x) = k(a-b)X(x)$
$a = b = 0$ if $\exists x\;X(x)\neq0$

$T = a\sin{kt} + b\cos{kt}$
$0 = T(0)X(x) = bX(x)$
$0 = T'(0)X(x) = kaX(x)$
$a = b = 0$ if $\exists x\;X(x)\neq0$

What is wrong? Is separation of variables not a valid way for solving equations of such type? How could it be solved then?

This is a particular case of the Klein-Gordon differential equation. Have you tried using the partial Fourier Transform? — Actually Fritz, Feb 14 '21 at 16:58
You imposed too much condition at the boundary $t=0$. Either impose $u=0$ or $u_t=0$ or a linear combination but not both at the same time. — user10354138, Feb 14 '21 at 17:06
Seems something's amiss in how you've set things up. One boundary condition, the one at $x=0$, is telling you that $u(0, t) = 2t$, from which $u_t(0, t) = 2$. But that's not consistent with your initial conditions $u_t(x, 0) = 0$. — A rural reader, Feb 14 '21 at 17:27

score 1 · Answer 1 · answered Feb 14 '21 at 16:53

Hint: you are going to have to solve the following two equations:

$T'' - T(\lambda -1) = 0$

and also:

$X'' - X(\lambda -1) = 0$

And you have to considerer solutions of the form $e^{rx}$.

Obs.: it's always good to note if the solution will or not be linearly independent.

1 Answers1