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We have to find the complete solution of

$z=px+qy+p^2+q^2$

We take

$f(x,y,z,p,q)=px+qy+p^2+q^2-z$

My doubt is that when we differentiate $f$ with respect to x to find auxiliary equations, we treat p,q and z as independent of x, but they are not.

Why will $\frac{\partial f}{\partial x}$ be $p$ and not $p+x\frac{\partial p}{\partial x}+\frac{\partial q}{\partial x}+2q\frac{\partial q}{\partial x}+2p\frac{\partial p}{\partial x}-p $ ?

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  • $\begingroup$ $f$ has 5 arguments (inputs) viz. $x,y,z,p,q$ and when you do $\frac {\partial f}{\partial x}$, it is understood that only $x$ is to be varied while keeping other arguments constant/as they are/unchaged. $\endgroup$
    – Koro
    Apr 1 at 9:33
  • $\begingroup$ @Koro but p is dependent upon x. Here p is the first partial derivative of z with respect to x, where z is a function of x and y. $\endgroup$
    – Raghav
    Apr 1 at 9:55
  • $\begingroup$ You need to research the difference between partial and total derivatives. What is required here are the partial derivatives of $f$ as a function of 5 variables, what you have in mind is the total derivative of $f$ along some (solution) curve. $\endgroup$ Apr 1 at 10:21
  • $\begingroup$ @Raghav: Indeed it is! But the point I am trying to make here is that when you apply Charpit's method, $f_x$ is not our usual $f_x$ (the one which you have mentioned) and in your case it is $p$ and this is due to the way Charpit's method is derived. If you see the proof, the usual $f_x$ has already been taken into account. And since that has already been taken into consideration, it is safe to apply Charpit's method on $f$ as if $x,y,z,p,q$ were all independent (although they aren't!) $\endgroup$
    – Koro
    Apr 1 at 10:28
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You need to disentangle the notation. You are ultimately looking for a solution $z=u(x,y)$. This solution has then derivatives $p=u_x(x,y)$ and $q=u_y(x,y)$.

The solution method that is followed here is to find the characteristic curves via the Lagrange-Charpit equations. These give curves $x(t),y(t)$ with then $z(t)=u(x(t),y(t))$ and $p(t)=u_x(x(t),y(t))$, $q(t)=u_y(x(t),y(t))$.

The PDE $0=f(x,y,u(x,y),u_x, u_y)$ has to be true on all of the surface. Here the total derivatives give $$ 0=f_x+f_zu_x+f_pu_{xx}+f_qu_{xy},\\ 0=f_y+f_zu_y+f_pu_{xy}+f_qu_{yy}.\\ $$ This can then be compared with the dynamic of the characteristic curves (to-be) $$ \dot z = u_x\dot y+u_y\dot y\\ \dot p = u_{xx}\dot x+u_{xy}\dot y\\ \dot q = u_{xy}\dot x+u_{yy}\dot y\\ $$ So if one sets $\dot x=f_p$ and $\dot y=f_q$, then these equations simplify to $$ \dot z=pf_p+qf_q\\ -\dot p=f_x+f_zp\\ -\dot q=f_y+f_zq $$ That this simplification is possible at all justifies the attention given to this construction and naming the resulting curves "characteristic".

The time parameter can be exchanged for another, this results in a common factor in all time derivatives. Expressing the equality of this factor in all the equation results in the fraction form of the Lagrange-Charpit equations $$ \frac{dx}{f_p}=\frac{dy}{f_q}=\frac{dz}{pf_p+qf_q}=-\frac{dp}{f_x+pf_z}=-\frac{dq}{f_y+qf_z}, $$ with the understanding that the differentials are along the tangent direction of the characteristic curves only.

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