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Math Help - Determine a relationship bewteen x,y and u for this QL PDE

  1. #1
    Senior Member bugatti79's Avatar
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    Determine a relationship bewteen x,y and u for this QL PDE

    <br />
\displaystyle (x+y+u)u_x+(x+y-u)u_y=2u .Where u(x,x)=0

    <br />
\displaystyle \frac{dx}{x+y+u}=\frac{dy}{x+y-u}=\frac{du}{2u}(=dt)<br />

    <br />
\displaystyle \frac{d(x+y)}{2x+2y}=\frac{du}{2u} \implies \frac{d(x+y)}{x+y}=\frac{du}{u}<br />
. Integrating for u we get

     \displaystyle ln(u)=ln(x+y)+A' \implies u=A(x+y) \implies A=\frac{u}{(x+y)}=\phi (x,y,u)

    Check  \displaystyle a\frac{\partial \phi}{\partial x}+b\frac{\partial \phi}{\partial y}+h\frac{\partial \phi}{\partial u}=0

    <br />
\displaystyle \frac{-u(x+y+u)-u(x+y-u)+2u(x+y)}{(x+y)^2}=0<br />

    Determining \psi(x,y,u).

    Attempt A

    <br />
\displaystyle \frac{\frac{du}{dt}}{\frac{dy}{dt}}=\frac{2u}{x+y-u}<br />

    Now

     <br />
\displaystyle y=\frac{u}{A}-x \implies \frac{2u}{\frac{u}{A} -x+x-u} =\frac{2Au}{u(1-A)}<br />

    Attempt B

    <br />
\displaystyle \frac{Adx+Ady-du}{2Ax+2Ay-(2Ax+2Ay)}=\frac{du}{2u}<br />
. Since u=Ax+Ay

    Not sure of either of these attempts are correct or how to proceed?
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  2. #2
    MHF Contributor
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    Since you added the first two to get

    \dfrac{d(x+y)}{2(x+y)} = \dfrac{du}{2u}

    why not try substracting the first two and see what that does for you.
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  3. #3
    Senior Member bugatti79's Avatar
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    Quote Originally Posted by Danny View Post
    Since you added the first two to get

    \dfrac{d(x+y)}{2(x+y)} = \dfrac{du}{2u}

    why not try substracting the first two and see what that does for you.

     <br />
\displaystyle \frac{d(x-y)}{(x+y+u)-(x+y-u)}=\frac{du}{2u} \implies \frac{d(x-y)}{2u}=\frac{du}{2u} \implies <br />
u=(x-y)

     <br />
\displaystyle \implies \frac{u}{(x-y)}=\psi(x,y,u)<br />
    Differentiating this wrt x and y and substituting satisfies the original equation, but what about \phi (x,y,u)?

    Both \phi(x,y,u) and \psi(x,y,u) must be = constant ie

     <br />
\displaystyle \frac{u}{x-y}=f(\frac{u}{x+y})<br />
. Using the IC u(x,x)=0 we get

     <br />
\displaystyle \frac{0}{2x}=f(0)<br />

    I dont understand what this means assuming its correct?
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  4. #4
    Senior Member bugatti79's Avatar
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    Quote Originally Posted by bugatti79 View Post
     <br />
\displaystyle \frac{d(x-y)}{(x+y+u)-(x+y-u)}=\frac{du}{2u} \implies \frac{d(x-y)}{2u}=\frac{du}{2u} \implies <br />
u=(x-y)

     <br />
\displaystyle \implies \frac{u}{(x-y)}=\psi(x,y,u)<br />
    Differentiating this wrt x and y and substituting satisfies the original equation, but what about \phi (x,y,u)?

    Both \phi(x,y,u) and \psi(x,y,u) must be = constant ie

     <br />
\displaystyle \frac{u}{x-y}=f(\frac{u}{x+y})<br />
. Using the IC u(x,x)=0 we get

     <br />
\displaystyle \frac{0}{2x}=f(0)<br />

    I dont understand what this means assuming its correct?
    Correction on above, I forgot the constant of integration

    Ie in the first line of post 3.

     \displaystyle u=x-y+c \implies c=u-x+y = \psi (x,y,u)

    \psi (x,y,u)= \phi(x,y,u) \implies

    \displaystyle u-x+y=f(\frac{u}{x+y})

    IC is u(x,x)=0 implies

    0=f(\frac{0}{0})

    How would you develop the particular solution based on this?
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