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?


2 Answers 2


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.

  • $\begingroup$ 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

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.


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