Solving non linear pde $z_x z_y = \frac{x^2 z^3}{y}$

Question

solve non linear pde $$z_x z_y = \frac{x^2 z^3}{y}$$

By trial I assumed $$z = kx^p y^q$$, which on putting in equation and comparing powers will give : $k = pq$, $p = -3$ and $q = 0$, this is trivial solution $z = 0$, but this is not what was sought

Is there any way to proceed in such case of non linear pde?

Answer 1

Hint

As you noticed, you choice was not the best.

By analogy with the one dimension case, let $z=\frac 1 {u^2}$ which, after simplification, will give $$ u_x u_y=\frac {x^2}{4y} $$ which is much more pleasant.

Answer 2

Solution

The partial differential equation to be solved for $z(x,y)$ is

$$\frac{\partial z}{\partial x} \frac{\partial z}{\partial y} = \frac{x^2}{y} z^3\tag{1}$$

As there are no boundary conditions given we understand the problem as to present a particular non trivial solution.

Here is one such solution

$$z(x,y)=\frac{1}{\left(q+\frac{p x^3}{3}+\frac{\log (y)}{4 p}\right)^2}\tag{2}$$

where $p$ and $q$ are arbitrary parameters in a "reasonable" range.

$(2)$ can be verified by direct calculation. But we prefer to show a heuristic path to find it in the next section.

Derivation

To begin with, we try to simplify the r.h.s. $z^3$ by the ansatz $z=u^a$ which after inserting this into $(1)$ gives $a=-2$ as was pointed out by Claude Leibovici in his solution.

The resulting PDE becomes

$$ \frac{\partial u}{\partial x} \frac{\partial u}{\partial y} = \frac{x^2}{4 y}\tag{2}$$

Notice that the r.h.s. is independent of $u$.

Next we turn to the factor of the independent variables $\frac{x^2}{y}$. This leads to consider a specific dependence of $u$ in the form

$$u(x,y) = v(x^3, \log_y)= v(r,s)\tag{3}$$

with $r=x^3$, $s=\log(y)$.

Inserting $(3)$ into $(2)$ gives

$$3 x^2 \left(\frac{\partial}{\partial r} v(r,s)\right) \left( \frac{1}{y}\frac{\partial }{\partial s} v(r,s)\right)=\frac{x^2}{4 y}$$

or

$$\left(\frac{\partial }{\partial r} v(r,s)\right) \left( \frac{\partial }{\partial s} v(r,s)\right)=\frac{1}{12}\tag{4}$$

Note that now the r.h.s. is a simple constant.

Attempting a solution in the form of a sum

$$v(r,s) = A(r) + B(s)$$

gives after reinserting

$$z = \frac{1}{\left(q+\frac{a x^3}{3}+\frac{\log (y)}{4 a}\right)^2}$$

This is the particular solution $(2)$.

Answer 3

Another class of solutions can be obtained using the method of separation of variables: if $z(x,y)=X(x)Y(y)$, then \begin{align} z_xz_y=\frac{x^2z^3}{y}&\implies X'YXY'=\frac{x^2X^3Y^3}{y} \\ &\implies\frac{X'}{x^2X^2}=\frac{Y^2}{yY'}=\lambda \\ &\implies\begin{cases} X(x)=-\frac{3}{\lambda x^3+a}, \\ Y(y)=-\frac{\lambda}{\ln|y|+b}, \end{cases} \end{align} hence $$ z(x,y)=\frac{3\lambda}{(\lambda x^3+a)(\ln|y| + b)}, $$ where $\lambda$, $a$ and $b$ are arbitrary constants.