How to convert the parametric solution of $z_x + z z_y = 6x$ to $z = z(x,y)$?

Question

In the book of PDE by Kumar, at page 22, it is asked that to solve

$$z_x + z z_y = 6x \quad z(0,y) = 3y.$$

I have proceeded as follows:

Let, for example, denote our coordinate as $x = x(s,t)$ s.t $s$ parametrises the initial curve, and $t$ parametrises the characteristic curve. Then $$dt = \frac{dx}{ 1} = \frac{dy}{ z} = \frac{dz}{6x },$$ with $x(s,0) = 0, \quad y(s,0) = s, \quad z(s,0) = 3s$ (the initial conditions are set by me according the the given initial conditions in the question).

These imply $$x(s,t) = C_1(s), \quad \text{for some function $C_1$ of s.}$$ $$z(s,t) = 6 C_1(s) + 3s,$$ $$y(s,t) = 3C_1 t^2 + 3st + s.$$

If I didn't do any algebra mistake, theoretically, I've solved the problem.

However, this is not the "closed form" of the question, i.e I don't know how does $z$ behaves as a function of $x,y$, so I need to solve $t,s$ in terms of $x,y,z$ but as far as I can see, there is no easy way to do that.

Question:

First of all, how can we find $z = z(x,y)$ ?

Secondly, is the given answer correct as a parametric form, and how can one check it ?

Thirdly, is there a way to solve the same PDE without appealing to the parametric form of the solution ?

Answer 1

You have correctly found the system of ODEs : $$\frac{dx}{ 1} = \frac{dy}{ z} = \frac{dz}{6x }$$ A first characteristic equation comes from $\frac{dx}{ 1} = \frac{dz}{6x } \quad;\quad dz-6xdx=0$ : $$z-3x^2=c_1$$ A second characteristic equation comes from $\frac{dx}{ 1} = \frac{dy}{ z}=\frac{dy}{ c_1+3x^2}\quad;\quad (c_1+3x^2)dx-dy=0$

$c_1x+x^3-y=c_2\quad;\quad (z-3x^2)x+x^3-y=c_2$ $$zx-2x^3-y=c_2$$ The general solution of the PDE, on the form of implicit equation, is $c_2=F(c_1)$ : $$xz-2x^3-y=F(z-3x^2)$$ where $F$ is an arbitrary function, to be determined according to the boundary condition.

Condition : $z(0,y)=3y$

$0z-2(0^3)-y=F(3y-3(0^2))\quad;\quad -y=F(3y)$

Let $X=3y$ $$F(X)=-\frac{X}{3}$$ Now the function $F$ is determined. We put it into the above general solution where $X=z-3x^2$ : $$xz-2x^3-y=-\frac{(z-3x^2)}{3}$$ Solving it for $z$ : $$z(x,y)=\frac{2x^3+x^2+y}{x+\frac13}$$ If you want to check it, compute $z_x$ and $z_y$. Put them into the PDE and simplify.

Answer 2

You made an error in the first step, $dt=dx$ implies $x=t+C_1$.

Then, $6xdx=dz$ gives $z=3x^2+C_2$ and as third identity one can take $$dy=zdx=d(xz)-xdz=d(xz)-6x^2dx$$ so that $y=xz-2x^3+C_3$.

Now use that there is only one independent parameter $s$ differentiating between the characteristic curves in the solution surface to conclude that $C_2=f(C_3)$ or $$ z-3x^2=f(y-xz+2x^3). $$ The function $f$ is fixed by the initial conditions $(x,y,z)=(0,y,3y)$, so that $$ 3y=f(y)\implies z-3x^2=3(y-xz+2x^3) $$ to get to a formula for the solution.