# Finding first integral in method of characteristics of PDE

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

Unfortunetely I understand nothing.

• 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. – Pragabhava Dec 31 '15 at 19:21
• @Pragabhava So it is special example of second method. Thx. I didn't notice that. Do you see as well how he get $$\frac{d z - d u}{u - z} = \frac{dy-dz}{z-y}?$$ – Fallen Apart Dec 31 '15 at 19:26
• 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. – Pragabhava Dec 31 '15 at 19:38
• @Pragabhava I asked how he get $\frac{d z - d u}{u - z} = \frac{dy-dz}{z-y}$ from Lagrange-Charpit equations? – Fallen Apart Dec 31 '15 at 19:40
• Is just a matter of playing with the equations. See my answer. – Pragabhava Dec 31 '15 at 20:57

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.