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First order, linear PDE with unknown inhomogeneity function

  1. Nov 22, 2014 #1
    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. jcsd
  3. Nov 23, 2014 #2
    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?
     
  4. Nov 23, 2014 #3

    pasmith

    User Avatar
    Homework Helper

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

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