Group Abstract Group Abstract

Message Boards Message Boards

0
|
8.7K Views
|
4 Replies
|
5 Total Likes
View groups...
Share
Share this post:

How to solve vector equation?

Posted 12 years ago

Hi there,

I have an equation: curl(B)=Cj, where B and j are both 3D vectors. Now I know everything in this equation, except B. I want to find it.

It doesn't work, if I use:

In[195]:= tok = sigma*Elpolje

Out[195]= {-((
  A sigma ((a + x)^2 + 
     y^2) (-((2 (a + x) ((-a + x)^2 + y^2))/((a + x)^2 + y^2)^2) + (
     2 (-a + x))/((a + x)^2 + y^2)))/((-a + x)^2 + y^2)), -((
  A sigma ((a + x)^2 + 
     y^2) (-((2 y ((-a + x)^2 + y^2))/((a + x)^2 + y^2)^2) + (
     2 y)/((a + x)^2 + y^2)))/((-a + x)^2 + y^2))}

In[202]:= Solve[Curl[B, {x, y, z}] == tokec, B]

Out[202]= {}

Any other ideas? Can Mathemtica do that?

POSTED BY: Mitja Jan?i?
4 Replies
POSTED BY: Daniel Lichtblau

Actually, Daniel, I highly appreciate your answer but... I found another formula on how to get that B. I don't know how much you know about physics but I remembered the Biot-Savart law. And I think that professor wants us to use it.

So, I hope you didn't waste to much time writing your last post, because I still need your help. :D

I have this vector:

In[13]:= tok = Simplify[sigma*Elpolje]

Out[13]= {(
 4 a A sigma (a^2 - x^2 + y^2))/((a^2 - 2 a x + x^2 + y^2) (a^2 + 
    2 a x + x^2 + y^2)), -((
  8 a A sigma x y)/((a^2 - 2 a x + x^2 + y^2) (a^2 + 2 a x + x^2 + 
     y^2)))}

And now I want to do a horrible integral. (If the integral simply can't be done, than I will of course do it with your method). The integral should be something like:

Integrate[(1/(4*
      Pi))*((tok[[2]]*z0)/((x0 - x)^2 + (y0 - y)^2 + z0^2)^(3/
       2)), {x, -Infinity, Infinity}, {y, -Infinity, Infinity}]

But.... mathematica can't do it. And now, since my knowledge of mathematica is really poor, I am asking you if I am doing something wrogn or is this really a bit too complicated for mathematica?

POSTED BY: Mitja Jan?i?

This involves calculus. Solve does not do calculus. It does not know calculus (it is a whiz at algebra and knows a bit of trig).

To get a viable result notice that the (missing) third vector component is zero, and the divergence is 0 as it must be for the field to be a curl. Since D[tok[[1]],x]+D[tok[[2]],y] vanishes, it makes sense to look for a vector field of the form {0,0,F} where D[F,y]==tok[[1]] and -D[F,x]==tok[[2]]. So integrate tok[[1]] with respect to y and see what happens.

vf = {-((A sigma ((a + x)^2 +  y^2) (-((2 (a + x) ((-a + x)^2 + y^2))/((a + x)^2 + 
               y^2)^2) + (2 (-a + x))/((a + x)^2 + y^2)))/((-a +  x)^2 + y^2)), 
-((A sigma ((a + x)^2 +  y^2) (-((2 y ((-a + x)^2 + y^2))/((a + x)^2 + 
               y^2)^2) + (2 y)/((a + x)^2 + y^2)))/((-a + x)^2 +   y^2)), 0}

Together[Curl[{0, 0, Integrate[vf[[1]], y]}, {x, y, z}] - vf]
(* Out[30]= {0, 0, 0} *)

That worked. So the field is {0, 0, Integrate[vf[[1]], y]}.

Simplify[{0, 0, Integrate[vf[[1]], y]}]

(* Out[32]= {0, 0, 2 A sigma (ArcTan[y/(a - x)] + ArcTan[y/(a + x)])} *)
POSTED BY: Daniel Lichtblau

The problem is similar to finding the vector potential A corresponding to a magnetic field B, where B = curl A. The result is usually not unique, but may be narrowed down with the appropriate boundary conditions. I don't think Mathematica can directly solve vector differential equations like yours. You might try reformulating the problem as a set of simultaneous partial differential equations.

POSTED BY: S M Blinder
Reply to this discussion
Community posts can be styled and formatted using the Markdown syntax.
Reply Preview
Attachments
Remove
or Discard