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Math Help - Solve QL PDE in parametric form

  1. #1
    Senior Member bugatti79's Avatar
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    Solve QL PDE in parametric form

     <br />
\displaystyle (x+u)u_x+yu_y=u+y^2<br />


    Asked to derive the general solution in the given form which I have done and is as attached...


     <br />
x=y_0^2(s)e^{2t}+te^t (u_0(s)-y_0^2(s))+(x_0(s)-y_0^2(s)e^t<br />

     <br />
y=y_0(s)e^t<br />

     <br />
u=(u_0(s)-y_0^2(s))e^t+y_0^2(s)e^{2t}<br />

    The intial data curve is  x=x_0(s), y=y_0(s), u=u_0(s)

    For clarity the last line of attachment is as above which I know is correct...

    How do I show that \displaystyle u(x,y)=\frac{x-y^2}{1+ln(y)} +y^2 for the special case u=x when y=1?
    I believe this requires getting u in terms of x and y but I dont know how to do that with variables x_0(s), y_0(s), u_0(s) in the above equations...
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  2. #2
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    With the solution you have we find that

    u - y^2 = \left(u_0 - y_0^2\right)e^t

    x - y^2 = \left(u_0-y_0^2\right) t e^t + \left(x_0 - y_0^2)e^t

    and dividing gives

    \dfrac{u - y^2}{x - y^2} = \dfrac{u_0-y_0^2}{\left(u_0-y_0^2\right)t + x_0-y_0^2}

    or

    \dfrac{u - y^2}{x - y^2} = \dfrac{1}{t + \dfrac{x_0-y_0^2}{u_0-y_0^2}} .

    From your  y solution t = \ln y - \ln y_0.
    Thus,

    \dfrac{u - y^2}{x - y^2} = \dfrac{1}{\ln y - \ln y_0 + \dfrac{x_0-y_0^2}{u_0-y_0^2}} .

    Imposing the initial curves gives your answer.
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  3. #3
    Senior Member bugatti79's Avatar
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    Quote Originally Posted by Danny View Post

    From your  y solution t = \ln y - \ln y_0.
    Thus,

    \dfrac{u - y^2}{x - y^2} = \dfrac{1}{\ln y - \ln y_0 + \dfrac{x_0-y_0^2}{u_0-y_0^2}} .

    Imposing the initial curves gives your answer.
    rearranging your expression then

    \displaystyle u=\frac{x-y^2}{lny-lny_0(s) +\frac{x_0+y_0^2(s)}{u_0+y_0^2(s)}}+y^2

    How is the 1+ln y determined in the denominator?

    Do you mean replacing x_0(s) with x, y_0(s) with y and u_0(s) with u, because I do not get the intended answer.
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  4. #4
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    Are you not told that on the initial curve that

    u_0 = x_0 and y_0 = 1?
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  5. #5
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    As for your second question - no, I did not intend on replacing x_0 with x etc.
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  6. #6
    Senior Member bugatti79's Avatar
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    Quote Originally Posted by Danny View Post
    Are you not told that on the initial curve that

    u_0 = x_0 and y_0 = 1?
    I didnt understand that the special case of u=x and y=1 is on the intial data curve...it doesnt say that explicitly?

    Ok because  u=u_0(s)=x, y=y_0(s)=1 then

    \displaystyle u=\frac{x-y^2}{ln(y)-ln (1) +\frac{x-1}{x-1}}+y^2

    The final question is to determine the domain of definition of this solution. How is that tackled?

    Thanks alot for your help..
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