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Thread: u_{xxxx}+yu_{xxy}=0

  1. #1
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    u_{xxxx}+yu_{xxy}=0

    $\displaystyle u_{xxxx}+yu_{xxy}=0$

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

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

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

    Which way is this integrate:

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

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

    or

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

    Thanks.
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  2. #2
    Behold, the power of SARDINES!
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    Quote Originally Posted by dwsmith View Post
    $\displaystyle u_{xxxx}+yu_{xxy}=0$

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

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

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

    Which way is this integrate:

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

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

    or

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

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

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

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

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

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

    This gives the two ODE's

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

    The solution to the first equation will depend on $\displaystyle \lambda$

    So if $\displaystyle \lambda > 0$ we get

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

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

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

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

    Now just integrate twice with respect to x.

    You will get two other solutions depending of if $\displaystyle \lambda < 0 \text{ or } \lambda=0$
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  3. #3
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    After integrating with respect to x twice, I will obtain $\displaystyle yg(y)\Rightarrow g_2(y)$, correct?
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    Quote Originally Posted by dwsmith View Post
    After integrating with respect to x twice, I will obtain $\displaystyle yg(y)\Rightarrow g_2(y)$, correct?
    Not quite...

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


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

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

    $\displaystyle \displaystyle u(x,y)= c_1y^{-\lambda}e^{-\sqrt{\lambda}x}+c_2y^{\lambda}e^{\sqrt{\lambda}x} +xf(y)+g(y)$
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  5. #5
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    I believe the sign of $\displaystyle \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.
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  6. #6
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    To Empty, I don't know why I put y.

    To Danny, the question just said use separation of variables.
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  7. #7
    Rhymes with Orange Chris L T521's Avatar
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    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 $\displaystyle u(x,y)=\varphi(x)\psi(y)$.
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  8. #8
    Rhymes with Orange Chris L T521's Avatar
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    Quote Originally Posted by dwsmith View Post
    $\displaystyle u_{xxxx}+yu_{xxy}=0$

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

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

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

    Which way is this integrate:

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

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

    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 $\displaystyle u(x,y)=\varphi(x)\psi(y)$ (for short, $\displaystyle u=\varphi\psi$).

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

    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 $\displaystyle \lambda$.

    Thus, we can rewrite this as two different ODEs:

    $\displaystyle \begin{aligned}
    \varphi^{(4)}-\lambda\varphi^{\prime\prime} &=0\\ y\psi^{\prime}+\lambda\psi &= 0.
    \end{aligned}$

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

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

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

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

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

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

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

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

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

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

    Hopefully this makes some sense.
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  9. #9
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    My problem was I thinking to integrate due to the form they were in.
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