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Let's say we have quasi-linear first order PDE $$a(x,y,u)u_x+b(x,y,u)u_y=c(x,y,u)$$ Solving it, is equivalent to finding first integrals of following ODE system $$\frac{\partial x}{\partial t}=a(x(t),y(t),u(t))\hspace{5pt}\frac{\partial y}{\partial t}=b(x(t),y(t),u(t))\hspace{5pt}\frac{\partial u}{\partial t}=c(x(t),y(t),u(t)).$$ Given two independent first itegrals $\Phi,\Psi$ we have general solution of our equation given by $F(\Psi,\Phi),$ where $F$ is arbitrary $C^1$ function.

In practice however we write something called Lagrange-Charpit equations $$\frac{dx}{a(x,y,u)}=\frac{dy}{b(x,y,u)}=\frac{dz}{c(x,y,u)}$$ and use some algebraic wishy washy methods to find $\Phi,\Psi.$

I am aware of two methods to derive $\Phi,\Psi.$

First. Take one of element of Lagrange-Charpit equations, lets say $\frac{dx}{a(x,y,u)}=\frac{dy}{b(x,y,u)}.$ Now using algebraic methods, if possible transform it to $A(x)dx-B(y)dy=0.$ As a result $$\int A(x)dx-\int B(y)dy$$ is first integral.

Second. If maneuvering with Lagrange-Charpit equations we can find a function $G$ such that $dG=0,$ then $G$ is a first integral.

Also I heard something about finding integrating factor $\mu.$

Question 1 Are there any other methods of finding first integral (using Lagrange-Charpit equations) . If so, can you write it with an example or give me some reference?

I am asking this question because I have problem with following example. Namely in book "PDE throu examples and exercises" of Pap and Takaci I encountered the example $$xu_x+(z+u)u_y+(y+u)u_z=(y+z)$$ Authors gives following explenation

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Unfortunetely I understand nothing.

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  • $\begingroup$ Well, the key is the linearity of the $d$ operator. For example ($z >u$), $$\frac{d z - d u}{u - z} = \frac{d (z-u)}{u-z} = d \left(\log \frac{1}{z - u}\right).$$ The identities are easy to prove, then. $\endgroup$ – Pragabhava Dec 31 '15 at 19:21
  • $\begingroup$ @Pragabhava So it is special example of second method. Thx. I didn't notice that. Do you see as well how he get $$\frac{d z - d u}{u - z} = \frac{dy-dz}{z-y}?$$ $\endgroup$ – Fallen Apart Dec 31 '15 at 19:26
  • $\begingroup$ In the same manner: $$\frac{dy - dz}{z-y} = \frac{d (y-z)}{z - y} = d \left(\log \frac{1}{z - y}\right),$$ so $$d \left(\log \frac{1}{z - u} - \log \frac{1}{z - y} \right) = 0,$$ and the result follows. $\endgroup$ – Pragabhava Dec 31 '15 at 19:38
  • $\begingroup$ @Pragabhava I asked how he get $\frac{d z - d u}{u - z} = \frac{dy-dz}{z-y}$ from Lagrange-Charpit equations? $\endgroup$ – Fallen Apart Dec 31 '15 at 19:40
  • $\begingroup$ Is just a matter of playing with the equations. See my answer. $\endgroup$ – Pragabhava Dec 31 '15 at 20:57
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Let $$\frac{d y}{z + u} = \frac{d z}{y + u} = \frac{d u}{y +z}.$$ Then,

$$ \frac{dy}{z + u} - \frac{dz}{y + u} = 0, \qquad \frac{dz}{y + u} - \frac{du}{y + z} = 0, \qquad \frac{du}{y + z} - \frac{dy}{z + u} = 0 $$ so \begin{array}\, f &= (y + u)dy - (z + u) dz &= 0,\\ g &= (y + z)dz - (y + u) du &= 0,\\ h &= (z + u)du - (y + z) dy &= 0. \end{array} From here, is easy to see that $$ \frac{g + h}{y^2 - z^2} - \frac{h + f}{z^2 - u^2} = 0 $$ is identity you are looking.

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  • $\begingroup$ I see it now. However I have one more question. The main purpous of posting this question was to find some general methods for solving L-C equations. I wrote two methods I am familar with and asked about others. I can see that you are quite handy with solving this type of exercises so I have personal question. Are there any other mathods that you use to solve L-C equations? $\endgroup$ – Fallen Apart Jan 1 '16 at 18:21
  • $\begingroup$ @FallenApart In E.L. Ince's extraordinary Ordnary Differential Equations, there are some techniques to solve different types of LC equations (you might have to browse the book in different sections). In general, when it comes to odes, everything goes. What I can say is that, for quasilinear first order pdes, I prefer the characteristics method: it's the same method, but stated in a more modern approach. Two examples can be found here and here. $\endgroup$ – Pragabhava Jan 4 '16 at 16:11
  • $\begingroup$ Thx. This is what I was looking for. $\endgroup$ – Fallen Apart Jan 4 '16 at 16:36

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