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Math Help - Another Charpit PDE

  1. #1
    Senior Member bugatti79's Avatar
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    Another Charpit PDE

    Write down the Lagrange-Charpit equations for the equation

     <br />
\displaystyle \frac{\partial u}{\partial x}\frac{\partial u}{\partial y}-y\frac{\partial u}{\partial x}-x<br />
\frac{\partial u}{\partial y}=0<br />
and use them to show that

     <br />
\displaystyle \frac{d^2 p}{dt^2}=p, \frac{d^2 q}{dt^2}=q,\frac{d^2 x}{dt^2}=x,\frac{d^2 y}{dt^2}=y<br />

    For the first term, I attempted the folllowing
     <br />
\displaystyle \frac{d}{dt}\frac{dp}{dt}=\frac{d}{dt}(-F_x-pF_u)=\frac{d}{dt}(-(-q))=\frac{dq}{dt}<br />

    Cannot see any link from this to continue on....

    Thanks
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  2. #2
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    Since p_t = q and q_t = p, then eliminating q, i.e. p_{tt} = q_t = p.
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  3. #3
    Senior Member bugatti79's Avatar
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    Quote Originally Posted by Danny View Post
    Since p_t = q and q_t = p, then eliminating q, i.e. p_{tt} = q_t = p.
    your answer made me relook at my derivation...I made a an error in signs, following on from my first message just for clarity....

     <br />
\displaystyle \frac{dp}{dt}=-F_x-pF_u=q \implies \frac{d}{dt}\frac{dq}{dt}=q<br />
similarly  \displaystyle \frac{d}{dt}\frac{dp}{dt}=p

     <br />
\displaystyle \frac{d}{dt}\frac{dx}{dt}=\frac{d}{dt} (q-y)=\frac{dq}{dt}-\frac{dy}{dt}=p-(p-x)=x<br />
and similarly for \displaystyle \frac{d}{dt}\frac{dy}{dt}

    Thanks!
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  4. #4
    Senior Member bugatti79's Avatar
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    The final part of this involves solving this non linear PDE

    \displaystyle \frac{\partial u}{\partial x}\frac{\partial u}{\partial y}-y\frac{\partial u}{\partial x}-x\frac{\partial u}{\partial y}=0 with IC's u(x,0)=x^2

    I have attached my solution as far as determining x and y in terms of s and t. Just hoping some one could check my work before I proceed?

    Thanks
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  5. #5
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    You have made a mistake which has snowballed. You have

    p_t = q which you said p = qt + A(s).  However, what you really have is

    p_t = q and q_t = p or

    p_{tt} - p = 0 which has the solution p = A(s)e^t + B(s)e^{-t} which in turn gives

    q = p_t = A(s)e^t - B(s)e^{-t}

    Using your IC you can determine A(s) and  B(s).  The rest of the problem goes similarly.
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  6. #6
    Senior Member bugatti79's Avatar
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    Quote Originally Posted by Danny View Post
    You have made a mistake which has snowballed. You have

    p_t = q which you said p = qt + A(s).  However, what you really have is

    p_t = q and q_t = p or

    p_{tt} - p = 0 which has the solution p = A(s)e^t + B(s)e^{-t} which in turn gives

    q = p_t = A(s)e^t - B(s)e^{-t}

    Using your IC you can determine A(s) and  B(s).  The rest of the problem goes similarly.
    Ok, I should have realised the first part of question wouldbe needed in the second part. Here attached is my latest x and y in terms of s and t. Solving for du/dt looks hairy....
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  7. #7
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    You made the same mistake in solving for x and you did earlier. Find x and y the same way as p and q.
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  8. #8
    Senior Member bugatti79's Avatar
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    Quote Originally Posted by Danny View Post
    You made the same mistake in solving for x and you did earlier. Find x and y the same way as p and q.
    Mistakes and me go hand in hand!

    However,


    \displaystyle \frac{d}{dt}\frac{dx}{dt}=\frac{d}{dt} (q-y)=\frac{dq}{dt}-\frac{dy}{dt}=p-(p-x)=x (1)

    \displaystyle \frac{d}{dt}\frac{dy}{dt}=\frac{d}{dt} (p-x)=\frac{dp}{dt}-\frac{dx}{dt}=q-(q-y)=y (2)

    Solving these 2nd order linear ODE's I will end up with

    \displaystyle x= C(s)e^t +D(s)e^{-t} and \displaystyle y= E(s)e^t+ F(s)e^{-t} IC's would give me s=C+D for x equation and 0=E+F for y equation??

    Unless I try (from 1 and 2 above)

    \displaystyle \frac{d}{dt} (q-y)=x

    \displaystyle \frac{d}{dt} (p-x)=y because we know p and q....?
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  9. #9
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    What you have is good! Use your expressions for x and y (you know p and q already) and subs into

    x_t = q - y,\;\;\;y_t = p - x and equate terms involving e^t and e^{-t}. This will more conditions on your functions C, D, E and  F.
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  10. #10
    Senior Member bugatti79's Avatar
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    Quote Originally Posted by Danny View Post
    What you have is good! Use your expressions for x and y (you know p and q already) and subs into

    x_t = q - y,\;\;\;y_t = p - x and equate terms involving e^t and e^{-t}. This will more conditions on your functions C, D, E and  F.
    \displaystyle x_t=s(e^t-e^{-t})-E(s) e^t- F(s)e^{-t}

    \displaystyle y_t=s(e^t+e^{-t})-C(s) e^t- D(s)e^{-t}

    Not sure how to equate terms involving powers ofe considering we have a derivative of x and y on the LHS..?
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  11. #11
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    Now subs your expressions for x and y into these.
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  12. #12
    Senior Member bugatti79's Avatar
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    Quote Originally Posted by Danny View Post
    Now subs your expressions for x and y into these.
    These expressions?

    \displaystyle \frac{d}{dt} (q-y)=x


    \displaystyle \frac{d}{dt} (p-x)=y

    Im lost at this stage
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  13. #13
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    This is what I'd do. You have x_{tt} = x which you solved giving

    x = C(s)e^{t} + D(s)e^{-t}.

    From x_t = q - y then y = q - x_t and since you know x and q then

    y = se^t - se^{-t} - C(s)e^t +D(s)e^{-t}.

    If you use your IC for x and y this will give C and D.
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  14. #14
    Senior Member bugatti79's Avatar
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    Quote Originally Posted by Danny View Post
    This is what I'd do. You have x_{tt} = x which you solved giving

    x = C(s)e^{t} + D(s)e^{-t}.

    From x_t = q - y then y = q - x_t and since you know x and q then

    y = se^t - se^{-t} - C(s)e^t +D(s)e^{-t}.

    If you use your IC for x and y this will give C and D.
    Ok, so since we have x we need to differentiate it with respect to t and then sub into eqn for y.

    At t=0 the IC's give D(s)=C(s) implies

    \displaystyle y = se^t - se^{-t} + C(s)(e^t+ e^{-t})

    For the x eqn I get th IC's to give s=E(s)-F(s)
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  15. #15
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    From post #13, when t = 0 then x = s and y = 0 so

    y = -C + D = 0, \; \text{and}\; x = C + D = s.

    Solve for C and  D and sub into x and y.
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