Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

1st order PDE, seems easy but still confusing

  1. Mar 4, 2014 #1
    I am doing some physics and I end up with this PDE:

    [tex] \frac{\partial q(x,y,t)}{\partial t} = -(x^2 + y^2)q(x,y,t) + ax\frac{\partial q(x,y,t)}{\partial y}[/tex]
    where q(x,y,t) is the scalar field to determine and a is a parameter. I need to consider two types of initial conditions: q(x,y,t=0) = 1; and q(x,y,t=0) = delta(x,y).

    I have found two tentative solutions:

    [tex] q(x,y,t) = C\exp \left(-\lambda s + \frac{y}{ax}(x^2 + y^2/3 - \lambda) \right) [/tex]
    where lambda is any (?) number. Another solution is

    [tex] q(x,y,t) = C\exp \left(-sx^2 + -y^3/3ax\right) [/tex]

    They both seem to satisfy the PDE, but I can't make them satisfy the required initial condition (either 1 or delta function). Any ideas or experience with this kind of equations?
  2. jcsd
  3. Mar 5, 2014 #2
    I don't see a 't' in either solution. Did you substitute it with s?
  4. Mar 5, 2014 #3
    Oops, sorry, I did.
    Actually I think I figured out how to find the solution. It's by the Method of Characteristics. No more help needed...
  5. Mar 5, 2014 #4


    User Avatar
    Gold Member

    Substitute [itex] q(x,y,t)=X(x)Y(y)T(t) [/itex] and treat x and X(x) as constants and then try to separate the equation to two ordinary ones.Then solve those two ODEs.The general answer is the linear combination of the answers to the ODEs but the coefficients will depend on x.
  6. Mar 28, 2014 #5
    The general solution to your PDE is

    [itex]q(x,y,t) =C(x,axt+y)\exp(\frac{xy}{a}+\frac{y^3}{3ax})[/itex],

    where [itex]C(x,y)[/itex] is an arbitrary function.


    [itex]q(x,y,t) =δ(x,axt+y)\exp(-\frac{t(a^2t^2x^2+3atxy+3x^2+3y^2)}{3})[/itex]

    is the solution with

    [itex]q(x,y,0) =δ(x,y)[/itex].
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook