# 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 $a(0) \neq b(0)$.