While studying first order Semi-linear PDE and the method of characteristics, I found this part in the note that I just cannot comprehend. From what I’ve understood, the solution surface is swept out by the characteristic curves through the initial curve gamma.

Why is there a solution surface through the curve C when gamma is characteristic curve? And moreover, what does it mean that there are infinite or no solution if the data curve and characteristic projection coincides? Does it mean there are infinite number of solution surfaces? If anyone could help me out with this, I’ll be really grateful.

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Let's start with the concept of a solution. A solution to a Cauchy problem is always a surface (at least in $\mathbb{R}^3$ -- the theory generalizes to $\mathbb{R}^n$, but hypersurfaces are difficult to visualize). So, here, any solution IS a surface -- there's no distinction between the two. In other words, to answer your last question, yes, an "infinity of solutions" means that there's an infinite number of solution surfaces, with the stipulation that each surface is a possible solution. More specifically, any solution surface must be described by a function of the input variables (your author uses $f(x,y)$, but the standard notation in the literature is $u(x,y)$).

The author's explanation is a bit terse, but by "data curve" he/she must mean the projection, $\gamma_p$, of $\gamma$ onto the $(x,y)$-plane. If $\gamma_p$ coincides with the projection, $\xi_p$, of a characteristic curve, $\xi$, then

(a) If $\gamma$ is a characteristic curve, an infinite number of solutions to the given Cauchy problem exist.

(b) If $\gamma$ is not a characteristic curve, no solution to the given Cauchy problem exists.

If $\gamma_p$ and the projection of a particular characteristic curve, say $\xi_p$, coincide, that is, if $\gamma_p = \xi_p = \{(x(s), y(s)): s \in I \subset \mathbb{R}\}$, then, since $\gamma$ and $\xi$ both must run through the solution surface, and since the surface must be described by a function $z = f(x,y)$, then for any $s_0 \in I,$ we must have $\gamma(s_0) = \xi(s_0) = (x(s_0),y(s_0),z(s_0))$. But if $\gamma$ is not a characteristic curve, there exists a $s_0 \in I$ such that $\gamma(s_0) = x(s_0),y(s_0),z_\gamma(s_0)) \neq \xi(s_0) = (x(s_0),y(s_0),z_\xi(s_0)) \Rightarrow z_\gamma(s_0) \neq z_\xi(s_0)$. Yet, this implies that $f(x,y$) is not a function, since it assumes two values over a subset, $\gamma_p = \xi_p = \{(x(s), y(s)): s \in I \subset \mathbb{R}\}$, of its domain. This is case (b), above, in which no function, $f(x,y)$, can be found to satisfy the PDE and the data (the initial curve, $\gamma$) simultaneously.

On the other hand, if $\gamma$ is a characteristic curve, say $\xi$, then the data are redundant: all we have is a curve in space, through which any number of non-tangential curves, $C$, may propagate. This is case (a), in which sufficient data isn't provided to arrive at a unique solution. Or, rather, there are one too few constraints (we need two non-tangential curves in order for a unique solution to present itself). In the simplest case, where $\gamma$ is merely a line embedded in $\mathbb{R}^3$, there are an infinite number of planes which can be fit through $\gamma$. It's only after we prescribe another line, which intersects the first line at only one point (i.e. a different line which isn't parallel, nor skew, to the first) that we can fit a unique plane. (Here, I'm using the term "plane" loosely; technically, I'm referring to a linear function of two variables -- but that's less visually suggestive.)

Hope that helps.


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