# Solving Partial Differential Equations with Laplace Transform

## Homework Statement

$\dfrac{\partial^2 \varphi }{ \partial x^2} - \dfrac{\partial ^2 \varphi }{\partial t^2} = 1$

Initial Conditions:

$\varphi (x, 0) = 1; \varphi_t (x, 0) = 1$

Boundary Condition:

$\varphi (0, t) = 1$

On $0 \leq x < \infty, 0 \leq t < \infty$

## Homework Equations

Let ${\Phi}$ denote the Laplace transform from t to s.

## The Attempt at a Solution

Apply Laplace Transform to PDE:

$\dfrac{\partial^2 \Phi }{ \partial x^2} - s^2\Phi + s\varphi (x, 0) + \varphi _t (x, 0) = \dfrac{1}{s}$

Apply Initial Conditions:

$\dfrac{\partial^2 \Phi }{ \partial x^2} - s^2\Phi + s + 1 - \dfrac{1}{s} = 0$

$\dfrac{\partial^2 \Phi }{ \partial x^2} - s^2\Phi + \dfrac{s^2 + s - 1}{s} = 0$

This leads to the eigenvalue being something like:

$r = \dfrac{s^3 \pm \sqrt{s^6 - 4s(s^2 + s - 1)}}{2s}$

which is always positive for both cases.

The solution in Laplace space should be something like:

$\Phi (x, s) = A(s) e ^{\dfrac{s^3 + \sqrt{s^6 - 4s(s^2 + s - 1)}}{2s}x} + B(s) e ^{\dfrac{s^3 - \sqrt{s^6 - 4s(s^2 + s - 1)}}{2s}x}$

and since the eigenvalue is always positive, as x goes to infinity the exponential terms will explode, requiring both A(s) and B(s) to be equal to 0.

What am I doing wrong? Any ideas?

I would check your final calculation for the solution Φ in Laplace space. It seems to not depend on x, which is weird. If you're still working on this, that is.

Mute
Homework Helper
$\dfrac{\partial^2 \Phi }{ \partial x^2} - s^2\Phi + \dfrac{s^2 + s - 1}{s} = 0$

This leads to the eigenvalue being something like:

$r = \dfrac{s^3 \pm \sqrt{s^6 - 4s(s^2 + s - 1)}}{2s}$

which is always positive for both cases.

I would double-check this step. It looks like you assumed a solution ##\Phi(x,s) = e^{rx}##, and then tried to solve the characteristic equation ##r^2 - s^2 r + (s^2 + s -1)/s = 0##, but this equation for r is not correct. Your purely ordinary differential equation in x is not homogeneous - only two of your terms depend on ##\Phi##. If you plug in ##\Phi(x,s) = e^{rx}##, what happens when you try to divide out the ##e^{rx}## to get the characteristic equation for r?