# Using Laplace transform to solve this (simple) PDE

1. Oct 30, 2011

### math2011

1. The problem statement, all variables and given/known data

Solve $\frac{\partial^2 v}{\partial x^2} = \frac{\partial^2 v}{\partial t^2}, x > 0, t > 0$ subject to
$v(x,0) = 0$
$v_t(x,0) = 0$
$v(0,t) = f(t)$
and where $v$ is bounded for all $x > 0$, by taking Laplace transforms with respect to $t$.

2. Relevant equations

3. The attempt at a solution

Take Laplace transforms with respect to $t$:
$\frac{\partial^2}{\partial x^2} V = s^2 V(x,s) - s v(x,0) - \frac{dv}{dt}(x,0)$
$\frac{\partial^2}{\partial x^2} V = s^2 V(x,s)$
where $V = \mathcal{L} \{ v \}$.

Solve this ODE.
$\lambda^2 - s^2 = 0$
$\lambda = \pm s$

Here I start to get confused. By Laplace transform definition $s$ is a complex number. So is this ODE solution correct?

$V(x,s) = A(s) e^{sx} + B(s) e^{-sx}$

Because $v$ is bounded the transform $V$ is also bounded for all $x > 0$ hence $A(s) = 0$.

$V(x,s) = B(s) e^{-sx}$
$V(0,s) = F(s) = B(s)$
$V(x,s) = F(s) e^{-sx}$

Invert using given formula $\mathcal{L} \{ H(t - c) f(t - x) \} = e^{-cs} F(s), c > 0$ where $H$ is the Heaviside function we get
$v(x,t) = H(t - x)f(t - x)$.

Is this correct?