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Math Help - PDEs solvable as ODEs

  1. #1
    Member Altair's Avatar
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    PDEs solvable as ODEs

    We have started PDEs in our MAth course in BE ME. We are following Kreyszig and I had problem understanding the basic concepts. It says "PDEs SOLVABLE AS ODEs : It happens if a PDE involves derivatives with respect to one variable only (or can be transformed to such a form), so that the other variable9s) can be treasted as parameters(s)"

    There's an example I am confused in,

    u_xy = -u_x where u=u(x,y)
    The steps goes as,
    u_x = p, p_y = -1, \frac{p_y}{p}=-1, lnp=-y +c(x) and by integration with respect to x,
    u(x,y) = f(x)e^-y + g(y) where f(x) = \int c(x) dx


    I do not understand the bold part and the natural log step how does y come in ?
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  2. #2
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    Quote Originally Posted by Altair View Post
    We have started PDEs in our MAth course in BE ME. We are following Kreyszig and I had problem understanding the basic concepts. It says "PDEs SOLVABLE AS ODEs : It happens if a PDE involves derivatives with respect to one variable only (or can be transformed to such a form), so that the other variable9s) can be treasted as parameters(s)"

    There's an example I am confused in,

    u_xy = -u_x where u=u(x,y)
    The steps goes as,
    u_x = p, p_y = -1, \frac{p_y}{p}=-1, lnp=-y +c(x) and by integration with respect to x,
    u(x,y) = f(x)e^-y + g(y) where f(x) = \int c(x) dx

    I do not understand the bold part and the natural log step how does y come in ?
    Let's start with a simplier example. Suppose that u = x^2 + f(y) where f(y) is arbitary. So taking the x derivative gives u_x = 2x. Now given u_x = 2x find u. So \partial u = 2x\, \partial x (separable, just like ODEs) and integrating gives u = x^2 + c. This would be correct if it was an ODE but it's a PDE so instead of a constant of integration, we should have a function of integration and with respect to the other variable. So in our case u = x^2 + f(y)

    Now for your question, you have \frac{p_y}{p} = - 1 or \frac{\partial p}{p} = - \partial y however we typical write \frac{d p}{p} = - d y noting it's partial integration (integration wrt a certain variable)

    so \ln p = - y + c(x) or p = e^{-y + c(x)}

     <br />
u_x = e^{-y} e^{ c(x)}<br />

    so du = e^{-y} e^{ c(x)}dx \; \Rightarrow\; u = f(x) e^{-y} + g(y)
    where f(x) = \int e^{ c(x)}dx (Since c(x) is abitrary then \int e^{ c(x)}dx is abritrary which we call f(x) ).
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  3. #3
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    Quote Originally Posted by Altair View Post
    We have started PDEs in our MAth course in BE ME. We are following Kreyszig and I had problem understanding the basic concepts. It says "PDEs SOLVABLE AS ODEs : It happens if a PDE involves derivatives with respect to one variable only (or can be transformed to such a form), so that the other variable9s) can be treasted as parameters(s)"

    There's an example I am confused in,

    u_xy = -u_x where u=u(x,y)
    The steps goes as,
    u_x = p,
    p_y = -1
    That is incorrect. You meant p_y= -p didn't you?

    , \frac{p_y}{p}=-1, lnp=-y +c(x) and by integration with respect to x,
    Well, first take the exponential of each side: exp(ln p)= exp(-y+ c(x))= exp(c(x))exp(-y). And since "c(x)" is an unknown function, g(x)= exp(c(x)) is again an unknown function: p(x)= g(x)e^{-y}.
    Now, p(x)= du/dx= g(x)e^{-y}

    Integrating with respect to x gives u(x)= \left(\int g(x)dx\right)e^{-y}. And since g(x) is an unknown function, \int g(x)dx is some unknown function: call it "f(x)". That gives
    u(x,y) = f(x)e^{-y} + g(y) where f(x) = \int c(x) dx
    exept that f(x) is not " \int c(x) dx". It is \int e^{c(x)}dx

    I do not understand the bold part and the natural log step how does y come in ?
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