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By the method of characteristics, it is possible to prove the existence of local solution for the first-order linear non homogeneous equation:

$\mathcal{L}_X f+g=0$

where $X$ is a non-singular vector field on a manifold $M$, $g$ is a given function on $M$, and $f$ is the unknown one.

Question: What could I say on the existence of solutions for the similar equation:

$\mathcal{L}_{X_t} f+g_t=0$,

with the only difference that now the non-singular vector field $X_t$ and the function $g_t$ on $M$ are time-dependent, while I need a time-independent solution?

Could it be useful to know that, in the problem I am tackling, I need a solution around a point $m_0\in M$ where $X_t(m_0)=X_0(m_0)\neq 0$ and $g_t(m_0)=0$ for all $t$?

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Rather then thinking of $t$ as time, you should think of it as a free parameter. So your question is asking, whether given a family of differential equations, there can be one simultaneous solution to all of them. Seen this way, the answer is clearly no in general. Even with your additional conditions.

As an example, consider for arbitrary fixed $\epsilon$, consider the time dependent family of vector fields $$ X(t) = \partial_x + (2\epsilon^2 x - 4x^3) t \partial_y $$ on $\mathbb{R}^2$. Observe that near $m_0 = (0,0)$ your condition on $X$ is satisfied. Let the source function $$ g(t) = g(0) \begin{cases} > 0 & y < 0 \\ = 0 & y \geq 0 \end{cases} $$

Observe that for every $t$ there exists an integral curve of $X(t)$ connecting the origin $(0,0)$ to the point $(\epsilon,0)$. For $t < 0$ this curve, between the two point, lies entirely in the lower half plane. For $t \geq 0$, the corresponding segment lies in the closed upper half plane.

So we get a contradiction. Let $f_0 = f(0,0)$ of the purported solution. For any $t\geq 0$, solving the equation requires $f(\epsilon,0) = f_0$. But for any $t < 0$, solving the equation requires the strict inequality $f(\epsilon,) < f_0$.


Remember that the method of characteristics reduces to essentially solving ordinary differential equations along integral curves. Non-uniqueness as described above will always be a possible problem if in your family there exists two different integral curves (corresponding to different parameters $t_1$ and $t_2$) connecting the same two points. In fact, the situation is slightly worse even then that:

To ensure the existence of a solution, a necessary condition is a holonomy type constraint. That is, you need that for any two points $p,q\in M$, for any piecewise integral curve $\gamma$ connecting $p$ to $q$ (that is, there exist $s_0 < s_1 < s_2 < \cdots < s_n$ such that $\gamma(s_0) = 0$, $\gamma(s_n) = q$, and parameters $t_1,\ldots, t_n$ such that $\gamma|_{(s_{i-1},s_i)}$ is an integral curve for $X_{t_i}$), the integral $\int_p^q g = \sum_{i = 1}^n \int_{s_{i-1}}^{s_i} g_{t_i}(s) \mathrm{d}s$ is independent of the choice of constituent integral curve segments.

And even in this case you cannot guarantee that the solution exists for the initial value problem. In the usual case with the method of characteristics, on any non-characteristic hypersurface $\Sigma$ we can prescribe arbitrary initial data and solve the IVP. In the case you described, the family of parametrised vector fields can lead to there being a non-local constraint being necessary for data prescribed on a hypersurface $\Sigma$ that is simultaneously non-characteristic to all of the $X_t$.

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Dear Willie Wong, thank you very much for the attention. Your counterexample shows that my doubts were not only due to a defect in my preparation but to a real obstacle. My motivation was to find a local isomorphism between two tensor fields via Lie's transform (Moser's trick). So I have to reflect more deeply on the proof. Thanks again for your instructive reply. –  Giuseppe Tortorella Apr 16 '11 at 13:49
    
Dear Willie Wong, now, certainly even for your answer, I have understood how to complete the proof. In order to show you that I have really appreciate your attention, I would remark just one small thing: should not we subtitute $\varepsilon$ with $t$ in the definition of $X(t)$? In such a way, as a contradiction, we get that the restriction of $f$ to the closed right $x$-semiaxis should be constant and, at the same time, should have a strict minimum for $x=0$. This proves the inexistence of local solutions around $(0,0)$. Thanks again, bye –  Giuseppe Tortorella Apr 18 '11 at 10:15
    
No. It is not necessary for the argument (which does not show non-existence of local solutions, but rather that you cannot get a guaranteed radius in which the local solutions exist based only on "reasonable" norms of the coefficients; so in particular you cannot have a Picard-like existence theorem). But you are welcome to think of it that way; which I admit, gives a stronger result. –  Willie Wong Apr 18 '11 at 16:55
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