# Solving $A\frac{\partial z}{\partial x} + B \frac{\partial^2 z}{ \partial x \partial y} + C \frac{\partial^3 z}{\partial x \partial^2 y} = 0$? [closed]

Non-mathematician here trying to find a hopefully analytic solution or any constructive directions for solving differential equations of this particular form: Take a function $$z(x,y)$$, is there any solution structure for

$$A\frac{\partial z}{\partial x} + B \frac{\partial^2 z}{ \partial x \partial y} + C \frac{\partial^3 z}{\partial x \partial^2 y} = 0$$?

$$A,B,C \in \Bbb{R}$$

Thanks

Here's an insight that could help you.

We could fix $$x$$ and define $$u_x(y) := \partial_xz(x,y)$$. According to Schwartz theorem, if $$z$$ is at least two times continuously differentiable, $$u_x$$ verifies:

$$Au_x + Bu_x' + Cu_x'' = 0$$

The general solutions of this equation are known and depend on the quantity $$B^2 - 4AC$$, look for "second order linear differential equation" on the web if you don't know the analytical solutions. With that, you will find a general form of $$u_x$$, but it will not be sufficient for the next step if you don't have any initial or boundary conditions on $$\partial_xz$$ as a function of $$y$$.

Provided that you found your exact analytical solution of $$u_x$$, you can integrate with respect to $$x$$ and find your function $$z$$, again with boundary or initial conditions on $$z$$ as a function of $$x$$.

• Thanks for your answer, that is what I needed. May 23 at 18:52

If you are satisfied with any solution, then you can consider $$Z(x,y)=X(x)Y(y).$$

Substituting this into the equation you will get that for Y that satisfies

$$AY+BY’+CY’’=0$$ Any X will work fine (except the constant, for a constant $$X$$ any $$Y$$ is ok).

In fact, at least for $$z$$ defined on a convex connected set that is pretty much all. To see that rewrite the equation as $$\partial_x(Az+B\partial_y z+C \partial_y^2 z)=0.$$ It follows from this that $$Az+B\partial_y z+C \partial_y^2 z = f(y),$$ for an arbitrary function $$f$$. If $$z$$ is independent on $$x$$ then we can get any dependence on $$y$$ tuning $$f(y)$$. If $$z$$ is $$x$$ dependent, then you can start solving this equation thinking about $$x$$ as about a parameter, it will enter the solution via the coefficients of homogeneous solutions, i.e. $$z$$ will be a sum of $$X(x)Y(y)$$ terms. Thus, solutions above are not just any solution but all you need to build up a general solution.

• Thank you for the answer! May 23 at 18:52