1. ## Second order PDE

So I'm working on a solution to this PDE,
$\displaystyle y * u_{xx} + 3y * u_{xy} = -3u_x;$
$\displaystyle y != 0$

I started by classifying the PDE and found it to be hyperbolic. So, I attempted to reduce it to canonical form, making the substitutions

$\displaystyle \eta = y - 3x$ and $\displaystyle \xi = y$

Then, I obtained that
$\displaystyle u_x = -3 *u_{\eta}$
$\displaystyle u_{xx} = 9*u_{\eta \eta}$
$\displaystyle u_{xy} = -3( u_{\xi \eta} + u_{\eta \eta})$

Substituting this back into the PDE, I end up with the equation

$\displaystyle u_{\eta \xi} = -u_{\eta} * \frac{1}{\xi}$

At about this point, I'm getting stuck on finding a general solution to the PDE by simplifying it further. I suspect the problem is with my substitution, specifically with $\displaystyle \xi$, but I'm not really sure what's wrong with it, other than it looking a bit suspect. Can anyone point me in the right direction?

2. Originally Posted by Math Major
So I'm working on a solution to this PDE,
$\displaystyle y * u_{xx} + 3y * u_{xy} = -3u_x;$
$\displaystyle y != 0$

I started by classifying the PDE and found it to be hyperbolic. So, I attempted to reduce it to canonical form, making the substitutions

$\displaystyle \eta = y - 3x$ and $\displaystyle \xi = y$

Then, I obtained that
$\displaystyle u_x = -3 *u_{\eta}$
$\displaystyle u_{xx} = 9*u_{\eta \eta}$
$\displaystyle u_{xy} = -3( u_{\xi \eta} + u_{\eta \eta})$

Substituting this back into the PDE, I end up with the equation

$\displaystyle u_{\eta \xi} = -u_{\eta} * \frac{1}{\xi}$

At about this point, I'm getting stuck on finding a general solution to the PDE by simplifying it further. I suspect the problem is with my substitution, specifically with $\displaystyle \xi$, but I'm not really sure what's wrong with it, other than it looking a bit suspect. Can anyone point me in the right direction?
Actually, you're almost there. Separate and integrate

$\displaystyle \frac{u_{\eta \xi}}{u_{\eta}} = - \frac{1}{\xi}$

so

$\displaystyle \ln u_{\eta} = - \ln \xi + \ln f'(\eta), f \text{\;is\; arbitrary}$

so

$\displaystyle u_{\eta} = \frac{f'(\eta)}{\xi}$

Then integrate again.

3. Thank you so much for that. It helped a lot to see that step worked out. Sorry for asking again, but could you possibly take a look at what I've done here and tell me where I'm going wrong? I'm sure it's something stupid.

I'm working on solving
$\displaystyle u_{xx} + u_{xy} - 2u_{yy} = 3( u_x + 2u_y + 6x - 3y)$

Once again, the equation is hyperbolic so I found appopriate substitutions to be

$\displaystyle \xi = y - 2x$ and $\displaystyle \eta = y +x$

So, taking the derivatives I find
$\displaystyle u_x = -2u_{\xi} + u_{\eta}$
$\displaystyle u_xx = 4(u_{\xi \xi} + u_{\xi \eta}) + u_{\eta \eta}$
$\displaystyle u_xy = u_{\eta \eta} -u_{\xi \eta} - 2u_{\xi \xi}$
$\displaystyle u_y = u_{\xi} + u_{\eta}$
$\displaystyle u_yy = u_{\xi \xi} + 2u_{\xi \eta} + u_{\eta \eta}$

Plugging these back into the pde, I end up with the equation (in canonical form),

$\displaystyle u_{\xi \eta} = \xi - u_{\eta}$
I tried integrating with respect to eta to get

$\displaystyle u_{\xi} = \xi * \eta - u + B'(\xi)$
For some arbitrary B. However, I can't solve this ODE. I tried using an integrating factor, but that got messy quickly. I might have made an algebraic mistake in the reduction, but I've gotten the same canonical form twice.

If someone could take a look and show me where I went awry, I'd be most apprecitive.

4. Originally Posted by Math Major
Thank you so much for that. It helped a lot to see that step worked out. Sorry for asking again, but could you possibly take a look at what I've done here and tell me where I'm going wrong? I'm sure it's something stupid.

I'm working on solving
$\displaystyle u_{xx} + u_{xy} - 2u_{yy} = 3( u_x + 2u_y + 6x - 3y)$

Once again, the equation is hyperbolic so I found appopriate substitutions to be

$\displaystyle \xi = y - 2x$ and $\displaystyle \eta = y +x$

So, taking the derivatives I find
$\displaystyle u_x = -2u_{\xi} + u_{\eta}$
$\displaystyle u_xx = 4(u_{\xi \xi} {\color{red}{+}\,} u_{\xi \eta}) + u_{\eta \eta}$
$\displaystyle u_xy = u_{\eta \eta} -u_{\xi \eta} - 2u_{\xi \xi}$
$\displaystyle u_y = u_{\xi} + u_{\eta}$
$\displaystyle u_yy = u_{\xi \xi} + 2u_{\xi \eta} + u_{\eta \eta}$

Plugging these back into the pde, I end up with the equation (in canonical form),

$\displaystyle u_{\xi \eta} = \xi - u_{\eta}$
I tried integrating with respect to eta to get

$\displaystyle u_{\xi} = \xi * \eta - u + B'(\xi)$
For some arbitrary B. However, I can't solve this ODE. I tried using an integrating factor, but that got messy quickly. I might have made an algebraic mistake in the reduction, but I've gotten the same canonical form twice.

If someone could take a look and show me where I went awry, I'd be most apprecitive.
I got a negative sign where there's a plus (in red above) but I agree with your final transformation. From what you have

$\displaystyle u_{\xi} + u = \xi \eta + F(\xi)\; (F\, \text{arbitrary})$

Integrating factor $\displaystyle e^{\xi}$ so

$\displaystyle \frac{\partial}{\partial \xi} \left( e^{\xi} u\right) = \xi e^{\xi} \eta + F(\xi)e^{\xi}$

Since $\displaystyle F$ is arbitrary then let $\displaystyle F(\xi) e^{\xi} = G'(\xi)$ where $\displaystyle G$ is arbitrary. Then integrate both sides.

5. Oh, duh, I was getting hung up on trying to find an anti-derivative for $\displaystyle F(\xi) e^{\xi}$ when it's just as much the derivative of some arbitrary function too.

Thank you so much for your help.

6. Are you positive on that negative sign?

$\displaystyle u_x = \xi_x u_{\xi} + \eta_x u_{\eta} = -2u_{\xi} + u_{\eta}$

From here,

$\displaystyle u_{xx} = -2 * ( \xi_x u_{\xi \xi} + \eta_x u_{\xi \eta} ) + \xi_x u_{\eta \xi} + \eta_x u_{\eta \eta}$

$\displaystyle = -2(-2u_{\xi \xi} + u_{\xi \eta}) - 2u_{\xi \eta} + u_{\eta \eta}$

$\displaystyle = 4u_{\xi \xi} - 4u_{\xi \eta} + u_{\eta \eta}$

Or did I make a mistake in that?

7. Originally Posted by Math Major
Are you positive on that negative sign?

$\displaystyle u_x = \xi_x u_{\xi} + \eta_x u_{\eta} = -2u_{\xi} + u_{\eta}$

From here,

$\displaystyle u_{xx} = -2 * ( \xi_x u_{\xi \xi} + \eta_x u_{\xi \eta} ) + \xi_x u_{\eta \xi} + \eta_x u_{\eta \eta}$

$\displaystyle = -2(-2u_{\xi \xi} + u_{\xi \eta}) - 2u_{\xi \eta} + u_{\eta \eta}$

$\displaystyle = 4u_{\xi \xi} - 4u_{\xi \eta} + u_{\eta \eta}$

Or did I make a mistake in that?
I agree in what you have in this post.