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Math Help - Understanding method of characteristics for first order equations

  1. #1
    Senior Member bugatti79's Avatar
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    Understanding method of characteristics for first order equations

    Hi Folks,
    I am new to the method of characteristics for solving 1st order diff eqns. I am looking at an example in the book, which i do not quite follow..here goes

    x^2u_x+yu_y+xyu=1
    which is of the form a(x,y)u_x+b(x,y)u_y+c(xy)u=f(x,y)

    The characteristic equation is \frac{dy}{dx}=\frac{b}{a}=\frac{y}{x^2}
    which becomes ln(y) + 1/x = k = \eta = \varphi(x,y)

    The bit I dont understand is making the transformation with

    \xi = x\text{and}\eta = ln(y) + 1/x to the partial differential eqn as above with \omega(\xi, \eta)=u(x,y)

    u_x = \omega_\xi\xi+\omega_\eta\eta_x = \omega_\xi+\omega_\eta(\frac{-1}{x^2})

    How is this u_x calculated?
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  2. #2
    MHF Contributor

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    Using the chain rule.

    \frac{\partial u}{\partial x}= \frac{\partial u}{\partial\xi}\frac{\partial \xi}{\partial x}+ \frac{\partial u}{\partial\nu}\frac{\partial\nu}{\partial x}
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  3. #3
    Senior Member bugatti79's Avatar
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    Understanding method of characteristics for first order equations

    Quote Originally Posted by HallsofIvy View Post
    Using the chain rule.

    \frac{\partial u}{\partial x}= \frac{\partial u}{\partial\xi}\frac{\partial \xi}{\partial x}+ \frac{\partial u}{\partial\nu}\frac{\partial\nu}{\partial x}
    But what is u equal to? All i know its a function of x and y etc, i dont know how omega comes into it?

    Thanks
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  4. #4
    Senior Member bugatti79's Avatar
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    Understanding method of characteristics for first order equations

    To me it appears that
    \frac{\partial u}{\partial\xi} = \omega_\xi (not sure why)and
    \frac{\partial \xi}{\partial x} =1

    \frac{\partial u}{\partial\eta} = \omega_\eta (not sure why)and
    \frac{\partial \eta}{\partial x} =\frac{-1}{x^2}

    Thanks
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  5. #5
    MHF Contributor
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    Let me see if I can help here.

    You have the PDE

    x^2u_x + y u_y = 1 -  x y u

    If you introduce a change of coordinates r = r(x,y), s = s(x,y) (I'm using these instead of \xi and \eta)
    then using the chain rule stated by HallsofIvy

    u_x = u_r r_x + u_s s_x
    u_y = u_r r_y + u_s s_y

    then your PDE, after regrouping terms becomes

    \left(x^2r_x + y r_y\right) u_r + \left(x^2s_x + y s_y\right) u_s = 1 - x y u.

    The idea is to choose r and s such that one of the two terms vanish. The method of characteristics gives you this. If you choose

    r = x \;(\text{really anything})\; \text{and}\; s = \ln y + \frac{1}{x}

    Then substituting these into your original PDE gives an ODE (since there's no u_s, the s is treated as constant).
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  6. #6
    Senior Member bugatti79's Avatar
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    Thumbs up

    Thank you Danny and Hallsofivy,

    That makes it clearer.

    Regards
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