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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,

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

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 $s=0$?

EditPiAf
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h3h325
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2 Answers2

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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 $$ (x,y) \in \lbrace (t + y(y-t)/2, y), 0<t<1, y<2\rbrace $$

characteristics

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 $\lbrace(t,t), 0<t<1\rbrace$ intersect. The classical solution collapses (similar case to this one).

The reason why the domain of validity is restricted to $y<2$ relates to the following Riccati ODE problem for $q = \partial_x f$: $$ \tfrac{\text d q}{\text ds} = -q^2, \quad\text{letting}\quad q(0) = (t-2)^{-1}, \quad\text{we know}\quad q = (t+s-2)^{-1} , $$ which is derived from the initial PDE problem by partial differentiation with respect to $x$. The solution rewritten as $q(x,y) = \frac{1}{y-2}$ blows up at $y=2$ and cannot be used for larger times (even if the solution function is defined for $y>2$). See also the 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

EditPiAf
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  • Why is the solution only valid on the domain in yellow when $f(x,y)$ is defined for all $x$, $y \neq 2$ ? It seems to me that the validity of the solution still holds outside although we might lose unicity since the associated system of ODE does not have initial conditions. – h3h325 Dec 05 '18 at 15:45
  • I understand why the solution is not valid for $y =2$, I also see why the idea behind the method should fail if the characteristic curves intersect. But as I see it for $y>2$ the function $f(x,y)$ satisfies the pde and also the initial condition, so why can we not take it as a solution? Thanks a lot for the answer by the way. – h3h325 Dec 05 '18 at 16:12
  • I see! So the conclusion is that the solution still holds but we cannot guarantee uniqueness, right? – h3h325 Dec 05 '18 at 16:22
  • Sorry if im beeing too picky but I still don't get your point. The system of ODEs does have a solution by Picard-Lindelöf (even if it is not unique), since 1, 1 and $f$ are lipschitz, for any $[S, \infty)$, for some $S$ corresponding to $y >2$, so why are we losing existence? Also $f(x,y)$ satisfies the PDE everywhere where it is defined, and also the initial conditions. So how is $f$ not a solution? Could you maybe refer me somewhere were this is explained? – h3h325 Dec 05 '18 at 16:43
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    @h3h325 I'm troubled by your statement about Picard Lindelof, please note that it is a local existence and uniqueness theorem – operatorerror Dec 07 '18 at 18:24
  • @h3h325 i have edited the answer for clarity about the domain of validity. Please delete old comments. – EditPiAf Dec 08 '18 at 08:49
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$$f(x,y)\frac{\partial f}{\partial x} + \frac{\partial f}{\partial y}= 1 \tag 1$$ 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 : $$f(x,y)=\frac{y^2-4y+2x}{2y-4}\qquad y\neq 2.$$ 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.

JJacquelin
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