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Thread: 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 $\displaystyle \frac{\partial u}{\partial x}+\frac{\partial u}{\partial y}=1$

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

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

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

    so $\displaystyle y = x + C_1$ so $\displaystyle x - y = -C_1 = C$. So $\displaystyle u(x,y) = f(x - y)$, where $\displaystyle f$ 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 $\displaystyle u = x + v$ you'll end up with $\displaystyle v_x + v_y = 0$.

    BTW - do you know the method of characteristics?
    Last edited by Jester; Sep 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 $\displaystyle u = x + v$ uou'll end up with $\displaystyle v_x + v+y = 0$.

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

    But if $\displaystyle u(x,y) = x + v$, wouldn't you get $\displaystyle u_x + u_y = (1 + v_x) + 0$? Since $\displaystyle (x + v)_y = 0$ 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 $\displaystyle u(x,y) = v(x,y) + x$. The method of characteristics will probably come up. It's a way to solve

    $\displaystyle a(x,y,u)u_x + b(x,y,u) u_y = c(x,y,u)$.
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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 $\displaystyle u(x,y) = v(x,y) + x$.
    But when you say $\displaystyle v(x,y) $ here, do you mean $\displaystyle v(x,y) = f(x - y) $ and not just any $\displaystyle v(x,y) $ which I found in my original post for the homogeneous PDE?

    And did I get $\displaystyle u_x + u_y = (1 + v_x) + 0$ 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 $\displaystyle v$ in the end. Also, if you let $\displaystyle u = v + x$ then

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