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I can't find anywhere how to solute PDE's.

  1. Feb 9, 2009 #1
    I can't find anywhere how to solute PDE's. For exaple ODE:
    [tex]\frac{du}{dt}=u\Rightarrow\ln u=t+\ln C\Rightarrow u=Ce^t[/tex]
    But this?
    [tex]\frac{du}{dt}+\frac{du}{dx}=u[/tex]

    [tex]dudx+dudt=udtdx\Rightarrow udx+udt=udtdx[/tex]

    I haven't had pde's yet, but i'm interested in solving these equations :-(
     
  2. jcsd
  3. Feb 13, 2009 #2

    Ben Niehoff

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    Science Advisor
    Gold Member

    Re: pde

    For many equations, there is a technique that works called "separation of variables". To separate the equation, we assume u(x,t) can be factored into two one-variable functions:

    [tex]u(x,t) = X(x)T(t)[/tex]

    And then

    [tex]\frac{\partial u}{\partial x} = T(t)\frac{dX(x)}{dx}[/tex]

    [tex]\frac{\partial u}{\partial t} = X(x)\frac{dT(t)}{dt}[/tex]

    Then, for your equation, we can write:

    [tex]X(x)\frac{dT(t)}{dt} + T(t)\frac{dX(x)}{dx} = X(x)T(t)[/tex]

    Next, divide by X(x)T(t):

    [tex]\frac{1}{T(t)}\frac{dT(t)}{dt} + \frac{1}{X(x)}\frac{dX(x)}{dx} = 1[/tex]

    Now, notice that we've separated the variables into the form

    [tex]P(t) + Q(x) = 1[/tex]

    Since this must be true for all x and t, the functions P and Q must individually be constant. This gives

    [tex]a + b = 1[/tex]

    [tex]\frac{1}{T(t)}\frac{dT(t)}{dt} = a[/tex]

    [tex]\frac{1}{X(x)}\frac{dX(x)}{dx} = b[/tex]

    These have the solution

    [tex]X(x) = e^{bx}[/tex]

    [tex]T(t) = e^{at}[/tex]

    and so

    [tex]u(x,t) = Ce^{at}e^{bx} = Ce^{at + (1-a)x}[/tex]

    for arbitrary constants C and a. Also note that any linear combination of u's (with different values for a) are also solutions, because the equation is linear. This can be used to construct a more general solution, given some boundary conditions.
     
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