3
$\begingroup$

I'm attempting to solve the non-homogenous quasi-linear PDE below:

$$z\frac{\partial z}{\partial x} - \alpha y\frac{\partial z}{\partial y} = \frac{\beta}{x^3y}$$

From what I've read in texts, the general form of a quasi-linear PDE is defined as

$$a(x,y,z)\frac{\partial z}{\partial x} + b(x,y,z)\frac{\partial z}{\partial y} - c(x,y,z) = 0$$

with solutions (called characteristic curves) $\phi(x,y,z) = C_1$ and $\psi(x,y,z) = C_2$ given by the characteristic equations

$$\frac{dx}{a} = \frac{dy}{b} = \frac{dz}{c}$$

When I set up these equations for my problem, I find

$$a(x,y,z) = z$$ $$b(x,y,z) = -\alpha y$$ $$c(x,y,z) = \frac{\beta}{x^3 y}$$

which leads to

$$\frac{dx}{z} = -\frac{dy}{\alpha y} = \frac{x^3ydz}{\beta}$$

Due to the coupling in the last term I cannot find a way to separate these to get two expressions containing a total derivative.

Could anyone help?

$\endgroup$

2 Answers 2

0
$\begingroup$

Where does this equation come from? I'm actually leaning towards there being no solution, for two reasons - one, Maple doesn't return one, and more seriously, I've tried solving it a couple of different ways and run into seemingly insurmountable problems in all of them.

An example - If you take the second two terms $\frac{\mathrm{d}y}{b}=\frac{\mathrm{d}z}{c}$ and divide through you get \begin{equation} \frac{\mathrm{d}z}{\mathrm{d}y} = -\frac{\beta}{\alpha}\frac{1}{x^3y^2}, \end{equation} Which you can solve easily enough by integrating with respect to $y$. Since we're integrating with respect to only one variable, this gives us the answer up to an undetermined function of the other - in this case performing the integration gives us \begin{equation} z(x,y) = \frac{\beta}{\alpha}\frac{1}{x^3y} + f(x). \end{equation}

If you try the same thing with $\frac{\mathrm{d}x}{a}=\frac{\mathrm{d}z}{c}$, you get \begin{equation} \frac{\mathrm{d}z}{\mathrm{d}x} = -\frac{\beta}{\alpha}\frac{1}{x^3yz}, \end{equation} which you can solve by separation of variables - taking the $z$ over to the other side and integrating with respect to x, you end up with \begin{equation} z^2(x,y) = -\frac{\beta}{x^2y} + g(y) \end{equation} for some unknown function $g(y)$. However, this is clearly not going to agree with what we got out of our first calculation. It doesn't seem like this impediment can be removed. I suspect the problem arises due to the combination of the nonlinearity of the equation in combination with the non-analyticity of the $\frac{\beta}{x^3y}$ term, however I'm not completely comfortable with this still and would be happy to be wrong.

$\endgroup$
3
  • $\begingroup$ This equation is part of a much bigger Hamilton-Jacobi problem, and it turns out I made a tiny mistake which snowballed into a huge unsolvable problem (what I posted here). Thanks for the advice though! $\endgroup$ May 7, 2015 at 20:29
  • $\begingroup$ @Baron Mingus: The important thing is once you go into the ODE issue, you must not go back into the PDE issue. $\endgroup$ May 9, 2015 at 22:14
  • $\begingroup$ @doraemonpaul: I'm not quite sure I follow - the ODE issue seems to be where I'm having the problem? This one's been eating away at me a bit, do you know what's going on with it? $\endgroup$ May 9, 2015 at 23:50
0
$\begingroup$

In fact the main reason of why I get struck in How to Separate Charpit Equations is that when I directly apply the method of characteristics of the last equation, it involves a very difficult type of second-order nonlinear ODE:

Follow the method in http://en.wikipedia.org/wiki/Method_of_characteristics#Example:

$\dfrac{dy}{dt}=-\alpha y$ , letting $y(0)=1$ , we have $y=e^{-\alpha t}$

$\begin{cases}\dfrac{dx}{dt}=z\\\dfrac{dz}{dt}=\dfrac{\beta}{x^3y}=\dfrac{\beta}{x^3e^{-\alpha t}}\end{cases}$

$\therefore\dfrac{d^2x}{dt^2}=\dfrac{\beta}{x^3e^{-\alpha t}}$

Which belongs to the ODE of the form http://eqworld.ipmnet.ru/en/solutions/ode/ode0313.pdf

In fact the PDE is similar to the Burger's equation with source term.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.