Results 1 to 9 of 9

Math Help - u_{xxxx}+yu_{xxy}=0

  1. #1
    MHF Contributor
    Joined
    Mar 2010
    From
    Florida
    Posts
    3,093
    Thanks
    5

    u_{xxxx}+yu_{xxy}=0

    u_{xxxx}+yu_{xxy}=0

    \varphi^{(4)}(x)\psi(y)+y\varphi''(x)\psi'(y)=0

    \displaystyle\frac{\varphi^{(4)}(x)}{\varphi''(x)}  =-\frac{y\psi'(y)}{\psi(y)}

    \displaystyle\int\frac{\varphi^{(4)}(x)}{\varphi''  (x)} How is this integrated?

    Which way is this integrate:

    \displaystyle -\int\frac{y\psi'(y)}{\psi(y)}

    \displaystyle -y\int\frac{\psi'(y)}{\psi(y)}=-y\ln{|\psi(y)|}

    or

    If the y isn't treated as a constant, how is integrated?

    Thanks.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Behold, the power of SARDINES!
    TheEmptySet's Avatar
    Joined
    Feb 2008
    From
    Yuma, AZ, USA
    Posts
    3,764
    Thanks
    78
    Quote Originally Posted by dwsmith View Post
    u_{xxxx}+yu_{xxy}=0

    \varphi^{(4)}(x)\psi(y)+y\varphi''(x)\psi'(y)=0

    \displaystyle\frac{\varphi^{(4)}(x)}{\varphi''(x)}  =-\frac{y\psi'(y)}{\psi(y)}

    \displaystyle\int\frac{\varphi^{(4)}(x)}{\varphi''  (x)} How is this integrated?

    Which way is this integrate:

    \displaystyle -\int\frac{y\psi'(y)}{\psi(y)}

    \displaystyle -y\int\frac{\psi'(y)}{\psi(y)}=-y\ln{|\psi(y)|}

    or

    If the y isn't treated as a constant, how is integrated?

    Thanks.
    Try the sub v(x,y)=u_{xx}(x,y) this gives

    v_{xx}+yv_{y}=0 and now assume that

    v(x,y)=\varphi(x)\psi(y) this gives

    \varphi_{xx}\psi+y\varphi \psi_{y}=0

    \displaystyle \frac{\varphi_{xx}}{\varphi}=-\frac{y\psi_{y}}{\psi}=\lambda

    This gives the two ODE's

    \displaystyle \varphi ''-\lambda \varphi=0 and \frac{d \psi}{\psi}=-\frac{\lambda}{y}

    The solution to the first equation will depend on \lambda

    So if \lambda > 0 we get

    \varphi(x)=c_1e^{-\sqrt{\lambda}x}+c_2e^{\sqrt{\lambda}x}

    and \psi(y)=y^{-\lambda}

    So v(x,y)=c_1y^{-\lambda}e^{-\sqrt{\lambda}x}+c_2y^{\lambda}e^{\sqrt{\lambda}x}

    So u_{xx}(x,y)=c_1y^{-\lambda}e^{-\sqrt{\lambda}x}+c_2y^{\lambda}e^{\sqrt{\lambda}x}

    Now just integrate twice with respect to x.

    You will get two other solutions depending of if \lambda < 0 \text{ or } \lambda=0
    Follow Math Help Forum on Facebook and Google+

  3. #3
    MHF Contributor
    Joined
    Mar 2010
    From
    Florida
    Posts
    3,093
    Thanks
    5
    After integrating with respect to x twice, I will obtain yg(y)\Rightarrow g_2(y), correct?
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Behold, the power of SARDINES!
    TheEmptySet's Avatar
    Joined
    Feb 2008
    From
    Yuma, AZ, USA
    Posts
    3,764
    Thanks
    78
    Quote Originally Posted by dwsmith View Post
    After integrating with respect to x twice, I will obtain yg(y)\Rightarrow g_2(y), correct?
    Not quite...

    \displaystyle \int u_{xx}(x,y)dx= \int \left(c_1y^{-\lambda}e^{-\sqrt{\lambda}x}+c_2y^{\lambda}e^{\sqrt{\lambda}x} \right) dx


    \displaystyle  u_{x}(x,y)=  c_1y^{-\lambda}e^{-\sqrt{\lambda}x}+c_2y^{\lambda}e^{\sqrt{\lambda}x} +f(y)

    \displaystyle \int u_{x}(x,y)dx= \int \left(c_1y^{-\lambda}e^{-\sqrt{\lambda}x}+c_2y^{\lambda}e^{\sqrt{\lambda}x} +f(y)\right) dx

    \displaystyle  u(x,y)=  c_1y^{-\lambda}e^{-\sqrt{\lambda}x}+c_2y^{\lambda}e^{\sqrt{\lambda}x} +xf(y)+g(y)
    Follow Math Help Forum on Facebook and Google+

  5. #5
    MHF Contributor
    Jester's Avatar
    Joined
    Dec 2008
    From
    Conway AR
    Posts
    2,368
    Thanks
    43
    I believe the sign of \lambda on both powers of y should be the same. @dwsmith - did the question specifically ask for solutions of the form you gave? The reason I ask is because the solution that TheEmptySet gives are not separable.
    Follow Math Help Forum on Facebook and Google+

  6. #6
    MHF Contributor
    Joined
    Mar 2010
    From
    Florida
    Posts
    3,093
    Thanks
    5
    To Empty, I don't know why I put y.

    To Danny, the question just said use separation of variables.
    Follow Math Help Forum on Facebook and Google+

  7. #7
    Rhymes with Orange Chris L T521's Avatar
    Joined
    May 2008
    From
    Chicago, IL
    Posts
    2,844
    Thanks
    3
    Quote Originally Posted by dwsmith View Post
    To Empty, I don't know why I put y.

    To Danny, the question just said use separation of variables.
    Danny is saying that the solution TES gave you isn't of the form u(x,y)=\varphi(x)\psi(y).
    Follow Math Help Forum on Facebook and Google+

  8. #8
    Rhymes with Orange Chris L T521's Avatar
    Joined
    May 2008
    From
    Chicago, IL
    Posts
    2,844
    Thanks
    3
    Quote Originally Posted by dwsmith View Post
    u_{xxxx}+yu_{xxy}=0

    \varphi^{(4)}(x)\psi(y)+y\varphi''(x)\psi'(y)=0

    \displaystyle\frac{\varphi^{(4)}(x)}{\varphi''(x)}  =-\frac{y\psi'(y)}{\psi(y)}

    \displaystyle\int\frac{\varphi^{(4)}(x)}{\varphi''  (x)} How is this integrated?

    Which way is this integrate:

    \displaystyle -\int\frac{y\psi'(y)}{\psi(y)}

    \displaystyle -y\int\frac{\psi'(y)}{\psi(y)}=-y\ln{|\psi(y)|}

    or

    If the y isn't treated as a constant, how is integrated?

    Thanks.
    Here's my two cents on how to do this problem. I will be skipping some steps; you should be able to fill them in without a problem.

    I would separate variables right away. Let u(x,y)=\varphi(x)\psi(y) (for short, u=\varphi\psi).

    So we end up with \varphi^{(4)}\psi+y\varphi^{\prime\prime}\psi^{\pr  ime}=0\implies \dfrac{\varphi^{(4)}}{\varphi^{\prime\prime}} = -\dfrac{y\psi^{\prime}}{\psi}

    Since each side of the equation is defined in a different variable when compared to one another, we can conclude that LHS = RHS only when they're equal to the same constant, call it \lambda.

    Thus, we can rewrite this as two different ODEs:

    \begin{aligned}<br />
   \varphi^{(4)}-\lambda\varphi^{\prime\prime} &=0\\ y\psi^{\prime}+\lambda\psi &= 0.<br />
\end{aligned}

    Consider the first equation. To make life easier, let \alpha = \varphi^{\prime\prime}. Then we see that the equation becomes

    \alpha^{\prime\prime}-\lambda\alpha = 0 (*)

    As TES mentioned in his post, you're going to get different solutions depending on the value of \lambda. I will do the \lambda>0 case and leave the other two cases for you to work out.

    So the characteristic equation for (*) is r^2-\lambda = 0. Assuming \lambda>0, we see that r=\pm\sqrt{\lambda}\in\mathbb{R}. So, we see that \alpha(x) = c_1e^{-\sqrt{\lambda}x}+c_2e^{\sqrt{\lambda}x}.

    But \alpha(x) = \varphi^{\prime\prime}(x) \implies \varphi(x) = \dfrac{c_1}{\lambda}e^{-\sqrt{\lambda}x}+\dfrac{c_2}{\lambda}e^{\sqrt{\lam  bda}x}+c_3x+c_4.

    Now, we do something similar for the second equation. The nice thing is that this equation is separable:

    y\psi^{\prime}+\lambda\psi = 0 \implies \dfrac{\psi^{\prime}}{\psi} = -\dfrac{\lambda}{y}\implies \ln\left|\psi(y)\right|=-\lambda\ln\left|y\right|+C

    So we see that \psi(y) = c_5 \left|y\right|^{-\lambda} (aside question: is it ok to remove absolute values at this point?)

    Thus, your solution should be u(x,y)=\varphi(x)\psi(y) = c_5\left|y\right|^{-\lambda}\left(\dfrac{c_1}{\lambda}e^{-\sqrt{\lambda}x}+\dfrac{c_2}{\lambda}e^{\sqrt{\lam  bda}x}+c_3x+c_4\right).

    Reiterating what TES said, you will get other solutions for testing the other cases.

    Hopefully this makes some sense.
    Follow Math Help Forum on Facebook and Google+

  9. #9
    MHF Contributor
    Joined
    Mar 2010
    From
    Florida
    Posts
    3,093
    Thanks
    5
    My problem was I thinking to integrate due to the form they were in.
    Follow Math Help Forum on Facebook and Google+


/mathhelpforum @mathhelpforum