Page 1 of 2 12 LastLast
Results 1 to 15 of 23

Thread: Another Charpit PDE

  1. #1
    Senior Member bugatti79's Avatar
    Joined
    Jul 2010
    Posts
    461

    Another Charpit PDE

    Write down the Lagrange-Charpit equations for the equation

    $\displaystyle
    \displaystyle \frac{\partial u}{\partial x}\frac{\partial u}{\partial y}-y\frac{\partial u}{\partial x}-x
    \frac{\partial u}{\partial y}=0
    $ and use them to show that

    $\displaystyle
    \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
    $

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

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

    Thanks
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor
    Jester's Avatar
    Joined
    Dec 2008
    From
    Conway AR
    Posts
    2,470
    Thanks
    83
    Since $\displaystyle p_t = q$ and $\displaystyle q_t = p$, then eliminating q, i.e. $\displaystyle p_{tt} = q_t = p.$
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Senior Member bugatti79's Avatar
    Joined
    Jul 2010
    Posts
    461
    Quote Originally Posted by Danny View Post
    Since $\displaystyle p_t = q$ and $\displaystyle q_t = p$, then eliminating q, i.e. $\displaystyle p_{tt} = q_t = p.$
    your answer made me relook at my derivation...I made a an error in signs, following on from my first message just for clarity....

    $\displaystyle
    \displaystyle \frac{dp}{dt}=-F_x-pF_u=q \implies \frac{d}{dt}\frac{dq}{dt}=q
    $ similarly $\displaystyle \displaystyle \frac{d}{dt}\frac{dp}{dt}=p$

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

    Thanks!
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Senior Member bugatti79's Avatar
    Joined
    Jul 2010
    Posts
    461
    The final part of this involves solving this non linear PDE

    $\displaystyle \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 $\displaystyle u(x,0)=x^2$

    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
    Follow Math Help Forum on Facebook and Google+

  5. #5
    MHF Contributor
    Jester's Avatar
    Joined
    Dec 2008
    From
    Conway AR
    Posts
    2,470
    Thanks
    83
    You have made a mistake which has snowballed. You have

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

    $\displaystyle p_t = q$ and $\displaystyle q_t = p$ or

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

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

    Using your IC you can determine $\displaystyle A(s)$ and $\displaystyle B(s). $ The rest of the problem goes similarly.
    Follow Math Help Forum on Facebook and Google+

  6. #6
    Senior Member bugatti79's Avatar
    Joined
    Jul 2010
    Posts
    461
    Quote Originally Posted by Danny View Post
    You have made a mistake which has snowballed. You have

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

    $\displaystyle p_t = q$ and $\displaystyle q_t = p$ or

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

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

    Using your IC you can determine $\displaystyle A(s)$ and $\displaystyle 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....
    Follow Math Help Forum on Facebook and Google+

  7. #7
    MHF Contributor
    Jester's Avatar
    Joined
    Dec 2008
    From
    Conway AR
    Posts
    2,470
    Thanks
    83
    You made the same mistake in solving for $\displaystyle x$ and you did earlier. Find $\displaystyle x$ and $\displaystyle y$ the same way as $\displaystyle p$ and $\displaystyle q$.
    Follow Math Help Forum on Facebook and Google+

  8. #8
    Senior Member bugatti79's Avatar
    Joined
    Jul 2010
    Posts
    461
    Quote Originally Posted by Danny View Post
    You made the same mistake in solving for $\displaystyle x$ and you did earlier. Find $\displaystyle x$ and $\displaystyle y$ the same way as $\displaystyle p$ and $\displaystyle q$.
    Mistakes and me go hand in hand!

    However,


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

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

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

    $\displaystyle \displaystyle x= C(s)e^t +D(s)e^{-t}$ and $\displaystyle \displaystyle y= E(s)e^t+ F(s)e^{-t}$ 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)

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

    $\displaystyle \displaystyle \frac{d}{dt} (p-x)=y$ because we know p and q....?
    Follow Math Help Forum on Facebook and Google+

  9. #9
    MHF Contributor
    Jester's Avatar
    Joined
    Dec 2008
    From
    Conway AR
    Posts
    2,470
    Thanks
    83
    What you have is good! Use your expressions for $\displaystyle x$ and $\displaystyle y$ (you know $\displaystyle p$ and $\displaystyle q$ already) and subs into

    $\displaystyle x_t = q - y,\;\;\;y_t = p - x $ and equate terms involving $\displaystyle e^t$ and $\displaystyle e^{-t}$. This will more conditions on your functions $\displaystyle C, D, E$ and $\displaystyle F.$
    Follow Math Help Forum on Facebook and Google+

  10. #10
    Senior Member bugatti79's Avatar
    Joined
    Jul 2010
    Posts
    461
    Quote Originally Posted by Danny View Post
    What you have is good! Use your expressions for $\displaystyle x$ and $\displaystyle y$ (you know $\displaystyle p$ and $\displaystyle q$ already) and subs into

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

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

    Not sure how to equate terms involving powers ofe considering we have a derivative of x and y on the LHS..?
    Follow Math Help Forum on Facebook and Google+

  11. #11
    MHF Contributor
    Jester's Avatar
    Joined
    Dec 2008
    From
    Conway AR
    Posts
    2,470
    Thanks
    83
    Now subs your expressions for x and y into these.
    Follow Math Help Forum on Facebook and Google+

  12. #12
    Senior Member bugatti79's Avatar
    Joined
    Jul 2010
    Posts
    461
    Quote Originally Posted by Danny View Post
    Now subs your expressions for x and y into these.
    These expressions?

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


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

    Im lost at this stage
    Follow Math Help Forum on Facebook and Google+

  13. #13
    MHF Contributor
    Jester's Avatar
    Joined
    Dec 2008
    From
    Conway AR
    Posts
    2,470
    Thanks
    83
    This is what I'd do. You have $\displaystyle x_{tt} = x$ which you solved giving

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

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

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

    If you use your IC for $\displaystyle x$ and $\displaystyle y$ this will give $\displaystyle C$ and $\displaystyle D$.
    Follow Math Help Forum on Facebook and Google+

  14. #14
    Senior Member bugatti79's Avatar
    Joined
    Jul 2010
    Posts
    461
    Quote Originally Posted by Danny View Post
    This is what I'd do. You have $\displaystyle x_{tt} = x$ which you solved giving

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

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

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

    If you use your IC for $\displaystyle x$ and $\displaystyle y$ this will give $\displaystyle C$ and $\displaystyle D$.
    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

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

    For the x eqn I get th IC's to give s=E(s)-F(s)
    Follow Math Help Forum on Facebook and Google+

  15. #15
    MHF Contributor
    Jester's Avatar
    Joined
    Dec 2008
    From
    Conway AR
    Posts
    2,470
    Thanks
    83
    From post #13, when $\displaystyle t = 0$ then $\displaystyle x = s$ and $\displaystyle y = 0$ so

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

    Solve for $\displaystyle C$ and $\displaystyle D$ and sub into $\displaystyle x$ and $\displaystyle y$.
    Follow Math Help Forum on Facebook and Google+

Page 1 of 2 12 LastLast

/mathhelpforum @mathhelpforum