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

• January 4th 2011, 04:15 PM
dwsmith
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.
• January 4th 2011, 06:45 PM
TheEmptySet
Quote:

Originally Posted by dwsmith
$u_{xxxx}+yu_{xxy}=0$

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

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

$\displaystyle\int\frac{\varphi^{(4)}(x)}{\varphi'' (x)}$ 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 $\lambda < 0 \text{ or } \lambda=0$
• January 4th 2011, 07:09 PM
dwsmith
After integrating with respect to x twice, I will obtain $yg(y)\Rightarrow g_2(y)$, correct?
• January 5th 2011, 06:19 AM
TheEmptySet
Quote:

Originally Posted by dwsmith
After integrating with respect to x twice, I will obtain $yg(y)\Rightarrow g_2(y)$, correct?

Not quite...

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

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

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

$\displaystyle u(x,y)= c_1y^{-\lambda}e^{-\sqrt{\lambda}x}+c_2y^{\lambda}e^{\sqrt{\lambda}x} +xf(y)+g(y)$
• January 5th 2011, 06:35 AM
Jester
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.
• January 5th 2011, 11:58 AM
dwsmith
To Empty, I don't know why I put y.

To Danny, the question just said use separation of variables.
• January 5th 2011, 12:00 PM
Chris L T521
Quote:

Originally Posted by dwsmith
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 $u(x,y)=\varphi(x)\psi(y)$.
• January 5th 2011, 04:47 PM
Chris L T521
Quote:

Originally Posted by dwsmith
$u_{xxxx}+yu_{xxy}=0$

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

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

$\displaystyle\int\frac{\varphi^{(4)}(x)}{\varphi'' (x)}$ 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}
\varphi^{(4)}-\lambda\varphi^{\prime\prime} &=0\\ y\psi^{\prime}+\lambda\psi &= 0.
\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.
• January 5th 2011, 04:50 PM
dwsmith
My problem was I thinking to integrate due to the form they were in.