Cfrgtkky

Suppose we have a PDE that can be solved with the method of characteristics

begin{align}
F(nabla u, u , x) = 0 text{ in $U$}\
u|_Gamma = g text{ on $Gamma$ }
end{align}

Where $Gamma subset partial U$. Suppose the characteristic curves starting on $Gamma$ don't span the whole set $U$, what can we say about the solution in this points outside the span? Do we lack information to solve the problem or can this be somehow avoided?

To have a precise example I solved the following problem,

$$f(x,y)frac{partial f}{partial x} + frac{partial f}{partial y}= 1 qquadtext{with}qquad f(t,t) = frac{t}{2} quadtext{for}quad 0 < t < 1.$$

In the above notation we have $Gamma = {(t,t): 0 < t < 1}$ and $g(t) = t/2$. I got the solution $f(x,y) = z(s) + g(x(0)) =z(s) + frac{x_0}{2} = y - frac{x - 1/2y^2}{2-y}$ where $y neq 2$, what can I say about the validity of my solution? Is it valid for $mathbb{R^2}$ or just for points $x, y$ such that the characteristic curve trough them, $(x(s),y(s))$ lies on $Gamma$ for $s=0$?

This PDE is a non-homogeneous inviscid Burgers' equation. Let us apply the method of characteristics:

• 
  $frac{text d y}{text ds} = 1$, letting $y(0) = t$, we know $y = t + s$
  


• 
  $frac{text d f}{text ds} = 1$, letting $f(0) = t/2$, we know $f = t/2 + s$
  


• 
  $frac{text d x}{text ds} = f$, letting $x(0) = t$, we know $x = t + s(t + s)/2$
  

Therefore, the characteristics are the curves $x = t + y(y-t)/2$ for $0<t<1$, along which $f$ is given by
$$
f(x,y) = y - frac{x-y^2/2}{2-y} , .
$$
One can note that this expression is not defined at $y=2$, where the denominator vanishes. The solution is only valid over the domain in yellow
$$
(x,y) in lbrace (t + y(y-t)/2, y), 0<t<1, y<2rbrace
$$

Indeed, we start following characteristic curves at some point $(t,t)$ with $0<t<1$. On the one hand, we can decrease $y$ without restriction. But on the other hand, we cannot increase $y$ further than $y=2^-$. At the point $(x,y) = (2,2)$, all characteristic curves coming from $lbrace(t,t), 0<t<1rbrace$ intersect. The classical solution collapses (similar case to this one).

The reason why the domain of validity is restricted to $y<2$ can be compared to the case of the Riccati ODE initial-value problem
$$
x'(t) = -x(t)^2 qquadtext{with}qquad x(0) = -1, .
$$
Indeed, the solution $x(t) = frac{1}{t-1}$ blows up at $t=1$ and cannot be used for larger times (even if the solution function is defined for $t>1$). Note that this simple ODE problem can be linked to the PDE problem by following a derivation by Lax, leading to Eq. (6.10) p. 36 of (1).

(1) P.D. Lax, Hyperbolic systems of conservation laws and the mathematical theory of shock waves, SIAM, 1973 doi:10.1137/1.9781611970562.ch1

$$f(x,y)frac{partial f}{partial x} + frac{partial f}{partial y}= 1 tag 1$$
Charpit-Lagrange equations :
$$frac{dx}{f}=frac{dy}{1}=frac{df}{1}$$
First family of characteristic curves, from $frac{dy}{1}=frac{df}{1}$ :
$$f-y=c_1$$
Second family of characteristic curves, from $frac{dx}{f}=frac{df}{1}$
$$frac12 f^2-x=c_2$$
General solution of the PDE expressed on the form of implicit equation :
$$frac12 f^2-x=Phi(f-y) tag 2$$
Where $Phi$ is an arbitrary function, to be determined according to boundary condition.

Condition : $f(x,x)=frac{x}{2}$

$frac12 (frac{x}{2})^2-x=Phi(frac{x}{2}-x)quad;quad frac{x^2}{8}-x=Phi(-frac{x}{2})$

Let $X=-frac{x}{2}quad;quad x=-2X$

$ frac{(-2X)^2}{8}-(-2X)=Phi(X)$
$$Phi(X)=frac12 X^2+2X$$
So, the function $Phi$ is determined. We put it into the general solution Eq.$(2)$ where $X=f-y$ :
$$frac12 f^2-x=frac12 (f-y)^2+2(f-y)$$
After simplification :
$$f(x,y)=frac{y^2-4y+2x}{2y-4}qquad yneq 2.$$
This is the solution of the PDE which satisfies the boundary condition.

One can check that this solution is valid in putting it into the PDE.