Derivation of Lagrange-Charpit Equations

Question

I am working through the derivation of the Lagrange-Charpit equations presented in this Wikipedia article: http://en.wikipedia.org/wiki/Method_of_characteristics#Fully_nonlinear_case

I am interested in the "fully" nonlinear case. Where we have:

$$F(x_1,...,x_n,u,p_1,...,p_n)=0 $$ And $$ p_i=\frac{\partial u}{\partial x_i} $$

I am fine with the derivation up until the point where they say that (Also, $\dot{x_i}=dx_i/ds$):

$$\sum_i(\dot{x}_idp_i-\dot{p_i}dx_i)=0 $$

Follows from taking the exterior derivative of:

$$ du-\sum_ip_idx_i=0 $$

From the little I know about exterior derivatives, it seems like doing this would give (for the two dimensional case):

$$0=\left(\frac{\partial p_2}{\partial x_1}-\frac{\partial p_1}{\partial x_2}\right)dx_1\wedge dx_2 $$

Because $ddu=0$ and the anti-symmetry of the wedge product. I don't see how this result leads to the one given. Could someone help me out? I feel like there is something very simple that I am missing.

I'm afraid I can't help you with that step, but I can show you how to derive the Lagrange-Charpit equations in a slightly different (and I think easier) way. Let me know. Cheers! — Dmoreno, Aug 21 '14 at 09:29

Dmoreno · Accepted Answer · 2014-08-21T14:54:40.707

4

$ \newcommand{\pd}[2]{ \frac{\partial #1}{\partial #2} } $ Attending to the positive answer to my request from the OP, consider the non-linear 1st order PDE:

$$F(x_i,u,p_i) = 0, \quad i = 1,\ldots, n, \quad p_i = \pd{u}{x_i}, \quad u = u(x_1,\ldots,x_n). \tag{1}$$

Assume $F \in \mathcal{C}^1_{x_i}$ and compute the partial derivative of eq. $(1)$ with respect to $x_i$:

$$\pd{F}{x_i} + \pd{F}{u}\pd{u}{x_i} + \pd{F}{p_i}\pd{p_i}{x_i} = 0,$$

or equivalently:

$$\pd{F}{x_i} + p_i \pd{F}{u} + \pd{F}{p_i}\pd{p_i}{x_i} = 0,$$

which is a quasilinear PDE for $p_i$ which can be readily solved leading the set of equations known as Lagrange-Charpit equations:

$$ \color{blue}{\large{\frac{\mathrm{d}x_i}{\pd{F}{p_i}} = -\frac{\mathrm{d}p_i}{\pd{F}{x_i}+p_i \pd{F}{u}} = \frac{\mathrm{d}u}{\sum_ip_i \pd{F}{p_i}}}} $$

where the last equality comes from the fact $\sum_i p_i \mathrm{d}x_i = \mathrm{d}u. $

Hope this helps.

Cheers!

edited Aug 21 '14 at 14:54

answered Aug 21 '14 at 13:50

Dmoreno

7,647

1

Thanks for the response mate! Should that be a $\partial F/\partial p_i$ on the last equality instead of $\partial F/\partial x_i$? – ClassicStyle Aug 21 '14 at 14:52
Of course. Thanks for noting! – Dmoreno Aug 21 '14 at 14:54
1

Yes, this way is much easier lol. Appreciate your help! I'll accept your answer if I don't get any answers to the original question soon (which is unlikely). – ClassicStyle Aug 21 '14 at 14:56

score -1 · Answer 2 · answered Dec 03 '14 at 09:03

let the general pde be f(x,y,z,p,q)......1 since z depends on (y)and(x) then, dz=dz/dx.dx+dzdy.dy dz=pdx+qdy...............................2 let another relation be Q(x,y,z,p,q)......3 on solving 1 and 3 we get the value of p and q to determine Q we differentiate w.r.t x and y. then,eliminate dp/dx

Derivation of Lagrange-Charpit Equations

2 Answers2

Linked