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Math Help - Help on solving first-order NONhomogeneous PDE

  1. #1
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    Help on solving first-order NONhomogeneous PDE

    Hello everybody. I'm trying to solve this first-order NONhomogenous PDE. But because my awfully bad textbook only gave examples on how to solve homogeneous PDEs, I'm not sure what to do. Here's my scratch work.

    I also know the theorem that if two solutions solve linear NONhomogeneous PDE, then their difference solve linear homogeneous PDE. But I don't think that's going to help here.


    Problem: Solve \frac{\partial u}{\partial x}+\frac{\partial u}{\partial y}=1

    Scratch work:
    I know how to solve the HOMOgeneous PDE: \frac{\partial u}{\partial x}+\frac{\partial u}{\partial y}=0

    The LHS is just the directional derivative of u in the direction of [1, 1].

    So if a curve in 3D has [1,1] as a tangent vector, the curve's equation is just \frac{dy}{dx} = 1

    so y = x + C_1 so x - y = -C_1 = C. So u(x,y) = f(x - y), where f is differentiable.

    But how do I solve the NONhomogeneous question from this?

    Thanks everybody.
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  2. #2
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    Re: Help on solving first-order NONhomogeneous PDE

    One solution is u = x. If you let u = x + v you'll end up with v_x + v_y = 0.

    BTW - do you know the method of characteristics?
    Last edited by Jester; September 26th 2011 at 03:12 PM.
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    Re: Help on solving first-order NONhomogeneous PDE

    Quote Originally Posted by Danny View Post
    One solution is u = x. If you let u = x + v uou'll end up with v_x + v+y = 0.

    BTW - do you know the method of characteristics?
    Hello and thanks Danny. So u(x,y) = x could've been just guessed?

    But if u(x,y) = x + v, wouldn't you get  u_x + u_y = (1 + v_x) + 0? Since (x + v)_y = 0 because there is no y in here.

    And I don't think so since I'm taking a first course in PDEs.
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  4. #4
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    Re: Help on solving first-order NONhomogeneous PDE

    Well, you assme that u(x,y) = v(x,y) + x. The method of characteristics will probably come up. It's a way to solve

    a(x,y,u)u_x + b(x,y,u) u_y = c(x,y,u).
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    Re: Help on solving first-order NONhomogeneous PDE

    Hello and thanks Danny.

    Quote Originally Posted by Danny View Post
    Well, you assme that u(x,y) = v(x,y) + x.
    But when you say  v(x,y) here, do you mean  v(x,y) = f(x - y) and not just any  v(x,y) which I found in my original post for the homogeneous PDE?

    And did I get  u_x + u_y = (1 + v_x) + 0 right?
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  6. #6
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    Re: Help on solving first-order NONhomogeneous PDE

    Well, you would end up with that v in the end. Also, if you let u = v + x then

    u_x + u_y = 1 gives \left( v_x + 1\right) + \left( v_y + 0\right) = 1 gives v_x + v_y = 0.
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