Verifying the general solution of the form $F(u,v)=0,~F \in C^1(\Bbb R)$ for a quasi-linear partial differential equation

Question

Have a look at this M.S.Q question:

Q. The general solution of the P.D.E $$(y+zx)z_x-(x+zy)z_y=x^2-y^2,$$

1. $F(x^2+y^2+z^2,xy+z)=0,~F \in C^1(\Bbb R^2)$,

2. $F(x^2+y^2-z^2,xy+z)=0,~F \in C^1(\Bbb R^2)$,

3. $F(x^2-y^2-z^2,xy+z)=0,~F \in C^1(\Bbb R^2)$,

4. $F(x^2+y^2+z^2,xy-z)=0,~F \in C^1(\Bbb R^2)$,

I solved the problem using the characteristics system $$\frac{dx}{y+zx}=\frac{dy}{-(x+zy)}=\frac{dz}{x^2-y^2},$$ to get the general solution $$F(u,v)=0, u=x^2+y^2-z^2,~v=xy+z~~~~~~~~~~~~~~~~~~~~~~~~(a)$$ or $$F(u,v)=0, u=x^2+y^2-z^2,~v=(x+y)^2-(z-1)^2~~~~~~~~~~~(b)$$But I am unable to verify the option 2 as the system (for $(a)$) $$F_x=(2x-2zz_x)F_u+(y+z_x)F_v=0,$$ $$F_y=(2y-2zz_y)F_u+(x+z_y)F_v=0,$$ $$F_z=-2zF_u+F_v=0,$$ provides, $$x+zy=0,$$ $$y+zx=0.$$ Also, I could not eliminate other options by the same procedure. Any help to find the erroneous steps in the concern would be appreciated.

I dont remember the source, where I have seen that the PDE satisfied by $$F(u(x,y,z),v(x,y,z))=0,$$ is given by $$\det \begin{pmatrix}u_x&v_x\\u_y&v_y\end{pmatrix}=0~~~~~~~~~~(d)$$ If so, I am interested in the justification along with the uniqueness of such PDE? Can we see the uniqueness up to some multiple?

Interestingly, Is this determinant equation, a consequence of the non-vanishing nature of gradient indicated in this post?

In addition, I feel the influence of Implicit function theorem here, while I couldn't connect.

Answer 1

You have correctly found two independent characteristic equations : $$x^2+y^2-z^2=c_1$$ $$xy+z=c_2$$ The solution of the PDE is : $$F\big((x^2+y^2-z^2)\:,\:(xy+z)\big)=0$$ Note that $z=c_3$ is not a correct characteristic equation.

Thus forget $F\big((x^2+y^2-z^2)\:,\:(xy+z)\:,\:(z)\big)=0$ .

Of course $z$ is involved into the function $F(u,v)$ but only through $u=x^2+y^2-z^2$ and $v=xy+z$.

NOTE :

Any linear combination of two characteristic equations is also a characteristic equation ( but not independent of the two first ones).

For example if we linearly combine $\quad x^2+y^2-z^2=c_1\quad$ with $\quad xy+z=c_2\quad$ such as : $$c_1+2c_2-1=c_3$$ $$(x^2+y^2-z^2)+2(xy+z)-1=c_3$$ $$(x+y)^2-(z-1)^2=c_3$$ We get what you call "option (b)" and we can define a third function : $$w=(x+y)^2-(z-1)^2$$

In fact the solution of the PDE can be expressed on many equivalent manners : $$F(u,v)=0$$ $$G(u,w)=0$$ $$H(v,w)=0$$ where $F$ , $G$ , $H$ are arbitrary functions.

Of course many other forms of the solution can be presented. But they are equivalent.

If $F(u,v)=0$ was checked and found exact there is no need to check $G(u,w)=0$ and to check $H(v,w)=0$.

Nethertheless we will check $G(u,w)=0$

$$u=x^2+y^2-z^2\quad\implies\quad u_x=2x-2zz_x \quad ; \quad u_y=2y-2zz_y $$

$$w=(x+y)^2-(z-1)^2\quad\implies\quad w_x=2(x+y)-2(z-1)z_x \quad ; \quad w_y=2(x+y)-2(z-1)z_y $$

$$\frac{\partial G(u,w))}{\partial x}=G_u u_x+G_w w_x=G_u(2x-2zz_x)+G_w(2(x+y)-2(z-1)z_x)=0$$ $$\frac{\partial G (u,w)}{\partial y}=G_u u_y+G_w w_y=G_u(2y-2zz_y)+G_w(2(x+y)-2(z-1)z_y)=0$$ $$\frac{G_u}{G_w}=-\frac{2(x+y)-2(z-1)z_x}{2x-2zz_x}=-\frac{2(x+y)-2(z-1)z_y}{2y-2zz_y}$$ $$(2(x+y)-2(z-1)z_x)(2y-2zz_y)-(2(x+y)-2(z-1)z_y)(2x-2zz_x)=0$$ Expend and simplify. The remaining terms are : $$(y+zx)z_x-(x+zy)z_y-x^2+y^2=0$$ This is exactly the PDE. Thus the checking is OK.