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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

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2 Answers 2

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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 $x$.

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  • $\begingroup$ Thanks for your answer, that is what I needed. $\endgroup$
    – Mwazr A
    May 23 at 18:52
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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.

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  • $\begingroup$ Thank you for the answer! $\endgroup$
    – Mwazr A
    May 23 at 18:52

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