Results 1 to 11 of 11

Math Help - Use Charpits to solve this PDE

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

    Use Charpits to solve this PDE

    Use Charpits Equations to Solve the following given that u=1 on the line x+2y=2.
    Is the solution unique?

    <br />
4u \frac{\partial {u}}{\partial x}=(\frac{\partial {u}}{\partial y})^2

     <br />
4up-q^2=0, F(x,y,u,p,q)=0<br />

    \frac{\partial F}{\partial x}=0, \frac{\partial {F}}{\partial y}=0,\frac{\partial {F}}{\partial u}=4p

    \frac{dx}{dt}=F_p, \frac{dy}{dt}=F_q, \frac{dp}{dt}=-F_x-pF_u, \frac{dq}{dt}=-F_y-qF_u, \frac{du}{dt}=pF_P+qF_q

    Initial Condition are

    t=0,u=1, x=s,y=1-\frac{s}{2}, \frac{d}{ds} (u(s,1-\frac{s}{2}))=\frac{d}{ds}(1)

    Therefore

     \frac{\partial u}{\partial x}+ \frac{\partial u}{\partial y}( \frac{-1}{2})=0

    Therefore
    p - \frac{q}{2}=0, q=2p, @t=0, u=1 \implies 4p-4p^2=0

    2 solutions p=0, p=1

    \frac{dp}{dt}=-4p^2 => p=\frac{1}{4t-A}

    No solution for p=0, therefore continuing with p=1 at t=0 to find A

    \frac{dp}{dt}=-4p^2 => p=\frac{1}{4t-A}

     <br />
\implies A=\frac{-1}{4}<br />
.

    Therefore p=\frac{1}{4t+1}. Now q=2p therefore q=\frac{2}{4t+1}.

    Since 4up=q^2 implies u is \frac{4t+1}{16t^2+1}

    How am I doing so far?
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor
    Jester's Avatar
    Joined
    Dec 2008
    From
    Conway AR
    Posts
    2,373
    Thanks
    48
    A couple of comments. First, you say there's no solution for p = 0. Why can't p = q = 0 always? (although it does lead to a trivial solution). Second, when you integrate the Chapit equation you should obtain arbitrary functions of s and when t = 0 then

    x = s, \;y = 1-s/2, \;u = 1, \;p = 1, \;q = 2.
    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
    A couple of comments. First, you say there's no solution for p = 0. Why can't p = q = 0 always? (although it does lead to a trivial solution). Second, when you integrate the Chapit equation you should obtain arbitrary functions of s and when t = 0 then

    x = s, \;y = 1-s/2, \;u = 1, \;p = 1, \;q = 2.
    What is a trivial solution?

    For this particular problem, the charpit equations do not have s in them...so obviously I have gone wrong somewhere...pardon my lack of understanding

    Thanks
    Follow Math Help Forum on Facebook and Google+

  4. #4
    MHF Contributor
    Jester's Avatar
    Joined
    Dec 2008
    From
    Conway AR
    Posts
    2,373
    Thanks
    48
    I see another problem. You say that q =2p. This is true but only when t = 0 - not in general!
    Follow Math Help Forum on Facebook and Google+

  5. #5
    MHF Contributor
    Jester's Avatar
    Joined
    Dec 2008
    From
    Conway AR
    Posts
    2,373
    Thanks
    48
    Quote Originally Posted by bugatti79 View Post
    What is a trivial solution?

    For this particular problem, the charpit equations do not have s in them...so obviously I have gone wrong somewhere...pardon my lack of understanding

    Thanks
    If p = 0 and q = 0 always then u = u_0 (constant) and the IC gives that u = 1 (always).

    When integrating say p_t = -4p^2 this gives \dfrac{1}{p} = 4t + A(s). From your IC p = 1 when t = 0 giving A(s) = 1 so

    \dfrac{1}{p} = 4t + 1\;\;\;\Rightarrow\;\;\;p = \dfrac{1}{4t+1}.

    Then continue in this fashion.
    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
    If p = 0 and q = 0 always then u = u_0 (constant) and the IC gives that u = 1 (always).

    When integrating say p_t = -4p^2 this gives \dfrac{1}{p} = 4t + A(s). From your IC p = 1 when t = 0 giving A(s) = 1 so

    \dfrac{1}{p} = 4t + 1\;\;\;\Rightarrow\;\;\;p = \dfrac{1}{4t+1}.

    Then continue in this fashion.
    Ok, thanks Danny..
    I proceed to find q using q_t = -q*4*[\frac{1}{4t+1}]? I will tackle it later this evening..
    Follow Math Help Forum on Facebook and Google+

  7. #7
    MHF Contributor
    Jester's Avatar
    Joined
    Dec 2008
    From
    Conway AR
    Posts
    2,373
    Thanks
    48
    Here's the solution that I got (to compare)

    Spoiler:
    u = e^{x+2y-2}
    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
    Here's the solution that I got (to compare)

    Spoiler:
    u = e^{x+2y-2}
     <br />
\frac{dq}{q}=\frac{-4}{4t+1} \implies q={e^{{ln(4t+1)^{-1}} +B'(s)} =\frac{B(s)}{(4t+1)}<br />

    At t=0 q=2p where p=\frac{1}{(4t+1)} therefore B(s)=2 \implies q=\frac{2}{(4t+1)}

     <br />
\frac{du}{dt}=\frac{4u}{(4t+1)}-\frac{8}{(4t+1)^2}<br />
.

    Using an integrating factor I get u=\frac{1}{4t+1}+(4t+1)C.

    At t=0 u=1 \implies u=\frac{1}{4t+1}

     <br />
\frac{dx}{dt}=4u and using x=s at t=0 I get x=ln(4t+1)+s

    Similarly y=-ln(4t+1)+1-\frac{s}{2}

    Eliminating s and t I get u(x,y)=\frac{1}{e^{(-x-2y+2)}-1+1}

    Finally u(x,y)=e^{x+2y-2}???
    Follow Math Help Forum on Facebook and Google+

  9. #9
    Senior Member bugatti79's Avatar
    Joined
    Jul 2010
    Posts
    461
    The question asks Is the solution unique?

    I am guessing that it is not unique because it has 2 valid solutions ie 1) p=0 and 2) p=1?

    However, Im confused because when solving for p using  <br />
\Displaystyle \frac{dp}{dt} given the condition that p=0 at t=0 I get

    A(s)=0 which is not possible since A is constant, right?
    Follow Math Help Forum on Facebook and Google+

  10. #10
    MHF Contributor
    Jester's Avatar
    Joined
    Dec 2008
    From
    Conway AR
    Posts
    2,373
    Thanks
    48
    But you have p_t = -4p^2 which has two solutions, p = 0 and p = \dfrac{1}{4t + A(s)}.
    Follow Math Help Forum on Facebook and Google+

  11. #11
    Senior Member bugatti79's Avatar
    Joined
    Jul 2010
    Posts
    461
    Quote Originally Posted by Danny View Post
    But you have p_t = -4p^2 which has two solutions, p = 0 and p = \dfrac{1}{4t + A(s)}.
    It is not unique because it has 2 solutions?

    Thanks
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Charpits: Non Linear 1st order PDE: Particular IC
    Posted in the Differential Equations Forum
    Replies: 9
    Last Post: August 22nd 2011, 12:47 PM
  2. need to solve summation equation to solve sum(x2)
    Posted in the Statistics Forum
    Replies: 2
    Last Post: July 16th 2010, 10:29 PM
  3. Solve for x
    Posted in the Algebra Forum
    Replies: 6
    Last Post: March 3rd 2010, 06:34 PM
  4. Replies: 1
    Last Post: June 9th 2009, 10:37 PM
  5. how can i solve this
    Posted in the Algebra Forum
    Replies: 2
    Last Post: August 6th 2008, 09:11 AM

/mathhelpforum @mathhelpforum