First order, linear PDE with unknown inhomogeneity function

1. Nov 22, 2014

2sin54

1. The problem statement, all variables and given/known data
Solve the following IVP:

$\frac{\partial v(x,t)}{\partial x} + \frac{\partial v(x,t)}{\partial t} + v(x,t) = g(x,t)$

2. Relevant equations
The initial values: v(0,t) = a(t) and v(x,0) = b(x)

3. The attempt at a solution

I applied the Laplace transform x -> s to get:

$sV(s,t) - a(t) + \frac{\partial V(s,t)}{\partial t} + V(s,t) = G(s,t)$, here V is the laplace transform of v, and G the L.t. of g.

Applying the Laplace transform t -> p I arrive at:

$sV_{2}(s,p) - A(p) + pV_{2}(s,p) - b(s) + V_{2}(s,p)=G_{2}(s,p)$,

here V2(s,p) is the Laplace transform of V(s,t), A(p) is the transform of a(t)...

After some term grouping:

$V_{2}(s,p) (s+p+1) = G_{2}(s,p)+A(p)+b(s)$

$V_{2}(s,p)= \frac{G_{2}(s,p)+A(p)+b(s)}{s+p+1}$

I can write this as three separate fractions:

$V_{2}(s,p)= \frac{G_{2}(s,p)}{s+p+1} + \frac{A(p)}{s+p+1} + \frac{b(s)}{s+p+1}$

I can find the inverse laplace transform twice of the third term rather easily, but if I use the convolution theorem for the first two I end up with integrals inside of integrals. Any suggestions? Or maybe there's a cleaner way to solve this inhomogeneous PDE?

2. Nov 23, 2014

2sin54

Alright so I managed to get this problem down to this new one which I do not know how to deal with:

Say I have a function which was Laplace transformed twice:

$v(x,t) -> V_{1}(s,t)$
and
$V_{1}(s,t) -> V_{2}(s,p)$

AND I end up with this double transform equal to something like this:

$V_{2}(s,p) = \frac{G_{2}(s,p)}{s+p+1}$ where $G_{2}(s,p)$ is the Laplace transform applied twice to the arbitrary function $g(x,t)$

Question is, how do I get the original $v(x,t)$?
Had there been only one Laplace transform on $v$ and $g$ I could express the general solution using a simple convolution integral but what to do now?

3. Nov 23, 2014

pasmith

You do not want to use Laplace Transforms here. You want to use the method of characteristics. It may be easier to work with $u(x,t) = e^tv(x,t)$ instead of $v$, so that $$\frac{\partial u}{\partial t} + \frac{\partial u}{\partial x} = e^tg(x,t).$$ This tells you rate of change of $u$ in the direction (1,1) in the (x,t) plane. If you start at a point $(x_0,t_0)$ on the boundary, where $u(x_0,t_0)$ is known, and head along the line $(x,t) = (s + x_0, s + t_0)$ then $$\frac{du}{ds} = \frac{\partial u}{\partial t} + \frac{\partial u}{\partial x} = e^{s + t_0}g(s + x_0,s + t_0)$$
and hence you can find $u(s + x_0, s + t_0)$.

To find $u(x,t)$ you work your way back to the boundary in the direction (-1,-1) to find $(x_0,t_0)$ and then head back to $(x,t)$ as above. Given the form of the boundary, you will have to treat the cases $x > t$ and $x < t$ separately. The case $x = t$ will cause a problem if $a(0) \neq b(0)$.