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I am reading the wikipedia page about applying the method of characteristics in the fully nonlinear case. We have the fully nonlinear equation $$ \tag{1} F(x_1, \cdots, x_n , u, p_1, \cdots, p_n) = 0,$$ here $$\tag{2} p_i = \frac{\partial u}{\partial x_i}$$ is the partial derivative of $u$ with respect to $x_i$.

In the method of characteristics, we wish to reduce the PDE to a family of ODE. Let assume that $u$ is a solution to (1). $$s\mapsto (x_1(s), \cdots, x_n(s), u(s), p_1(s), \cdots, p_n(s))$$ be a curve so that (1) is satisfied for all $s$. Then it is claimed that the following holds:

\begin{equation} \begin{split} \sum_i (F_{x_i} +F_up_i)\dot x_i + \sum_i F_{p_i}\dot p_i &=0\\ \dot u - \sum_i p_i\dot x_i &=0\\ \sum_i ( \dot x_i dp_i - \dot p_i dx_i) &= 0. \end{split} \end{equation}

I can see that the first two equations follow from taking total derivative with respect to $s$ of (1) and the expression $u(s) = u(x_1(s),\cdots, x_n(s))$. In the wiki page, it is claimed that

... the third follows by taking an exterior derivative of the relation $du - \sum p_i dx_i = 0$.

Unfortunately, I fail to see how the third equation are derived using exterior derivative. Could you give me the steps so I could check my work, please?

P.S.: I’ve been struggling a lot lately on Frobenius theorem and systems of total differential equations. If you could explain that, I would deeply appreciate it.

James S. Cook
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Dwight Dinkins
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  • Please let us know about your background: what did you know about exterior derivative, what texts are you using, where did you encounter these notations ... . –  Mar 03 '18 at 07:11
  • @JohnMa The second and the third equation (under the “Fully nonlinear case” of the article) that which I’ve been seeing constantly in my analysis of Frobenius’s Theorem. My original thought was that it would go to zero due to one of the properties of exterior differentiation, but apparently not; calling into question:”What is it that I’m missing? Do I need to go back and start over?” I’ve been studying tensor calculus, differential geometry for about 2 yrs because that’s just what I enjoy doing. It was only recently (6 months ago) that I came across Frobenius, and it has became a huge hassle. – Dwight Dinkins Mar 03 '18 at 09:08
  • This question is clear. It should not be closed. It shows significant effort on the part of the OP and deserves attention. – James S. Cook Mar 03 '18 at 14:47
  • @JamesS.Cook Thank you so much! I go through this a lot, where I feel like people don’t want to answer my questions or whatever, but that’s first world problems – Dwight Dinkins Mar 03 '18 at 15:24
  • People don't know the answer to your question is the real problem... I'm still confused by the Wikipedia article at the present. I also have trouble making sense of that line being the result of exterior differentiation. But, I'm not certain it's wrong. – James S. Cook Mar 03 '18 at 22:47

1 Answers1

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Indeed, the last equality is a consequence of taking exterior derivative, but the wikipedia could have been more precise.

First, think of $\alpha = du - \sum_i p_i dx_i$ as a one form on $\mathbb R^{2n+1}$, the exterior derivative is $$ d\alpha= \sum_i dp_i \wedge dx_i.$$

Now, for any function $u : \mathbb R^n \to \mathbb R$, we define the mapping $$ I_u : \mathbb R^n\to \mathbb R^{2n+1}, \ \ \ I_u(x) = \left(x_1, \cdots, x_n, u(x), \frac{\partial u}{\partial x_1}, \cdots, \frac{\partial u}{\partial x_n} \right).$$

Then we have $I_u^* \alpha = 0$ since $$ I_u^* \alpha = I_u^* (du-\sum_i p_i dx_i) = du(x) - \sum_i\frac{\partial u}{\partial x_i} dx_i = du-du = 0.$$

(So it is really the (usual) abuse of notations: $\alpha$ is not a zero form, by "$du - \sum_i p_i dx_i = 0$" it really means $I_u^* \alpha = 0$). Now since pullback commutes with exterior derivative,

$$0=d(I_u^*\alpha) = I_u^* (d\alpha) = I_u^* \left( \sum_i dp_i \wedge dx_i\right).$$

So the two form $d\alpha$ also restrict to zero under the map $I_u$. In particular, the tangent vector $X=(\dot x(s), \dot u(s), \dot p(s))$ also satisfies

$$\sum_i dp_i \wedge dx_i (X, \cdot)= 0\Rightarrow \sum_i \left( \dot p_i dx_i - \dot x_i dp_i\right) = 0,$$

which is the last equation. Note that this last equation (thus, this answer) has completely nothing to do with the equation $F$.

  • Thank you for this. The main point of confusion for me was distinguishing between abstract identities in $(x,u,p)$ space verses the curve in that space which corresponds to a solution in $(x,u)$ space via the map $I_u$. These two things are not clearly separate in the Wiki article and this was the reason for my own personal confusion. Much appreciated +1 – James S. Cook Mar 04 '18 at 13:33
  • Well done! I really appreciate it – Dwight Dinkins Mar 04 '18 at 15:40
  • @JamesS.Cook : Yes that is confusing (and that took me a while to understand what the wiki page was referring to). But on the other hand, I guess it would be equally confusing if another set of coordinates are used to distinguish those two spaces. –  Mar 05 '18 at 05:21
  • @DwightD It seems that some users also found that line confusing (here). There is an answer without using the exterior derivative, which you might find it useful. –  Mar 05 '18 at 05:23
  • @JohnMa What method is it then? – Dwight Dinkins Mar 05 '18 at 05:49
  • @DwightD I mean the answer posted in the question that I linked (Now I read that again, it doesn't seem to be too helpful...). –  Mar 05 '18 at 05:58
  • I’m a bit curious why you implemented the tangent vector. – Dwight Dinkins Mar 05 '18 at 07:43