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$\Phi, \Lambda$ are both scalars dependent upon, and $\mathbf u$ is a vector independent of coordinates. I'm trying to express $\Lambda$ in terms from $\mathbf U \cdot \nabla\Lambda = \Phi$ and to start with, since I'm just familiar with the basics, it looks pretty hopless. So:

  1. Get help from people more expert in this area
  2. Trawl through the list of standard vector identitites involving $\nabla$
  3. Search for an online mathematics program to solve it.
  4. Make it simpler to start with by solving in one dimension, and progress from there

what strategy would/did you use in tackling this problem? Solving it would be a bonus ;)

Edit: I've added the context of the problem since some people think this will help:

I'm trying to work out the conserved canonical momentum for a static electric field, using Noether's theorem on the relativistic electromagnetic Lagrangian $$L = \frac{m_0c^2}{\gamma} + \frac{e}{c}\mathbf{u \cdot A} - e\Phi(x,y,z)$$ $L$ needs to be independent of coordinates which can be done by transforming $\mathbf A\rightarrow \mathbf A'= \mathbf A + \nabla\Lambda$ The conserved canonical momentum $$ P = \frac\partial{\partial \mathbf u} ( \frac{-m_0c^2}{\gamma} + \frac{e}{c}\mathbf{u \cdot (A+\nabla\Lambda}) - e\Phi)$$ With no magnetic field $A=0$ $$P = -m_0\mathbf u\gamma +\frac e c \frac\partial {\partial\mathbf u}\mathbf u \cdot\nabla\Lambda$$ becomes $$P = -m_0\mathbf u\gamma +\frac e c (\nabla\Lambda + \mathbf u \cdot\frac\partial {\partial\mathbf u}\mathbf\nabla\Lambda)$$ To get any further, I need to know the form $\Lambda$ must take, which comes from making $L$ independent of the coordinates before, and so $$\nabla( \frac{e}{c}\mathbf u \nabla\Lambda - e\Phi) = 0$$

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Your question is not clear. What are $\Lambda$, $U$ and $\Phi$? –  user38268 Jul 3 '11 at 22:08
    
Looks to me like $ \Lambda, \Phi $ are scalar functions while $ \mathbf{U} $ is a vector function on $ \mathbb{R}^n $. –  anon Jul 3 '11 at 22:15
    
I'm guessing that your question is "I want to find a scalar field $\Lambda$ such that $\mathbf U \cdot \nabla \Lambda = \Phi$ for a given vector field $\mathbf U$ and a given scalar field $\Phi$. How should I begin trying to solve this problem?" Is that an accurate interpretation of what you are asking? –  Rahul Jul 3 '11 at 23:06
    
@D Lim on the left, the scalar product operator $'\cdot'$ only operates on vectors to give scalars, and $\nabla$ only operates on scalars to give vectors. So anon's conclusion is correct. –  user10389 Jul 3 '11 at 23:09
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@user10389: it is usually best if you provide enough information in your question so that people reading it do not have to go deducing and guessing what you meant. You want to make their job easier, as you want them to do something for you. –  Mariano Suárez-Alvarez Jul 4 '11 at 3:04
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2 Answers 2

up vote 5 down vote accepted

The equation $U\cdot\nabla \Lambda=\Phi$ says precisely that the derivative of scalar function $\Lambda$ in the direction of the vector field $U$ is $\Phi$. To solve it then, locally, near points where $U$ is not zero, you find the integral curves of the field $U$ and integrate $\Phi$ along them. (Solving globally, or near a zero of $U$ requires rather more elaborate analyses in this generality...)

If, for example, $U$ is the constant field $(1,0,0)$, then the equation is $$\frac{\partial}{\partial x}\Lambda(x,y,z)=\Phi(x,y,z),$$ and it obviously can be solved by integrating. The general case reduces to this one, because near a non-zero point of $U$ coordinates can be changed to that the equation becomes the above one.

If you provide concrete details about the equations you are trying to solve, one can give more concrete details about the solution...

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Am I correct to interpret from "find the integral curves of the field $ U $" you mean find the flow lines and integrate $ \Phi $ from one point to another? Such a method works because the integrand $ \Phi(x) ds = U \cdot \nabla \Lambda ds = \nabla \Lambda \cdot dx $. Line integrals over conservative vector fields are easy, so we get $ \Lambda(b) - \Lambda(a) $. Nice idea. –  anon Jul 4 '11 at 9:38
    
I've edited the question to give the context of the problem I'm trying to solve. –  user10389 Jul 4 '11 at 17:17
    
@user10389: in the generality of your question, then there is very little that can be said apart from what I wrote. –  Mariano Suárez-Alvarez Jul 4 '11 at 17:19
    
Nice clear answer, thanks. Directional derivative is the key. –  user10389 Jul 4 '11 at 18:06
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Usually when faced with something like that I look to Green's identities:

http://en.wikipedia.org/wiki/Green%27s_identities

and start looking for a Gaussian surface:

http://en.wikipedia.org/wiki/Gaussian_surface

to integrate over as is commonly done in electromagnetism. The proper choice of the surface is made to ensure the integration is easy to carry out.

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