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I've been reading about the method of characteristics and came across the theorem:

The general solution of a first-order, quasi-linear partial differential equation $$ a(x,y,u)u_x+b(x,y,u)u_y=c(x,y,u)\tag{2.5.1} $$ is $$ f(\phi,\psi)=0,\tag{2.5.2} $$ where $f$ is an arbitrary function of $\phi(x,y,u)$ and $\psi(x,y,u)$, and $\phi=\text{constant}=c_1$ and $\psi=\text{constant}=c_2$ are solution curves of the characteristic equations $$ \frac{dx}{a}=\frac{dy}{b}=\frac{du}{c}.\tag{2.5.3} $$ The solution curves defined by $\phi(x,y,u)=c_1$ and $\psi(x,y,u)=c_2$ are called the families of characteristic curves of equation $(2.5.1)$.

I've gone through the proof and it seems straightforward. But I'm unable to visualize what the characteristic curves $\phi=c_1$ and $\psi=c_2$ represent in the $(x,y,u)$ space(can they be any curves on the solution surface or they follow a certain property?) and why $(2.5.2)$ intuitively should give me the general solution.

Jamāl
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  • I tried to work out a few problems but i couldn't visualize the method because of the appearance of the arbitrary function. Can you suggest good sources that explain the method in a very intuitive manner? Thanks. – Jamāl Feb 23 '20 at 11:19
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    I'd suggest to learn the method based on some examples from this site or from Wikipedia – EditPiAf Feb 23 '20 at 15:09

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Consider fixed $c_1$ and $c_2$, then these two equations determine a curve. Now, let $c_1$ run free but remaining $c_2$ fixed, obviously we have a surface and it is a solution of the PDE. This gives us a tip about how the particular solutions emerge (we got one!). We can do better, we can move $c_1$ and simultaneously move $c_2$, we draw a surface that is solution too. But the simultaneous movement of $c_1$ and $c_2$ is what we call "function" and because this function is not still determined, we say that it is arbitrary. So $c_2=f(c_1)$ or $\psi=f(\phi)$

A particular solution determines $f$. Consider this example: we know the value of $u$ along the line $y=0$, for each $x$ we know $u$, i. e. $u$ is a perfectly known function $g$ of $x$ $u=g(x)$ along $y=0$ (you can imagine $x^2$ or $e^x$). Then each of the curves have to satisfy that requirement, so is, $\phi(x,0,g(x))=c_1$ and $\psi(x,0.g(x))=c_2$. Now, $x$ can be eliminated determining the needed functional relation between $c_1$ and $c_2$

Rafa Budría
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  • Can you explain why fixing $c_2$ and letting $c_1$ run freely yield the solution to the PDE? I thought about it a bit but couldn't wrap my mind around it. – Jamāl Feb 24 '20 at 16:58
  • My answer is about how a solution, a surface, emerges from a generic characteristic curve. Anyway, ψ(x,y,u)=c2 and ϕ(x,y,u)=c1 are separately solutions of the system of ODEs.(for each value of c1 and c2). Get some examples to check all these (this is good to start with) – Rafa Budría Feb 24 '20 at 18:50
  • I've been thinking about your answer and here's what I've understood so far. We have been given this pde which can be written in the form $(a,b,c)\cdot (u_x,u_y,-1)=0.$ Our objective is to generate the integral surface $u$ of the pde. So we find two surfaces $\phi =c_1$ and $\psi =c_2$ whose intersection gives a curve having tangent direction at each point $(x,y,u)$ as $(a,b,c)$. $\phi =c_1$ and $\psi =c_2$ have the property that the normal to their surfaces is orthogonal to $(a,b,c)$. – Jamāl Feb 25 '20 at 19:58
  • Now varying $c_1$ sweeps $\phi =c_1$ across $\psi =c_2$ which results in lines of intersection creating a surface having the desired properties of $u$. Likewise for $c_2$. An arbitrary function of $c_1$ and $c_2$ also has the desired properties of $u$. Is my understanding correct? – Jamāl Feb 25 '20 at 20:04
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    Effectively, the surface solution is formed, so said, "moving" a characteristic curve (the lines of intersection you mention) along, driven by $c_2=f(c_1)$. By the way, the PDE holds for each of the functions $\psi$ and $\phi$ (nevertheless has to be considered carefully the case in which one of them has not dependence on $u$) – Rafa Budría Feb 25 '20 at 22:59