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| Tags: pde, question, simple |
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#1
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For some reason I can't solve one of the simplest equations..
Is there a solution? The equation: B = curl(A) Or in coordinate form: B_i = e_ijk* d/dx_j ( A_k ) (3 dimensions OK for now, e_ijk is anti-symmetric unit tensor or Levi-Civita tensor) The problem is to solve for the vector field A given a vector field B. I know that if A is a solution, so is A' = A + grad(f) where f is any function of the coordinates. I also know that there is no solution if div(B) != 0. But how can I write some (any) solution A in terms of divergenceless B? Thanks in advance! |
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#2
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In article ,
shevek wrote: B = curl(A) The problem is to solve for the vector field A given a vector field B. I know that if A is a solution, so is A' = A + grad(f) where f is any function of the coordinates. I also know that there is no solution if div(B) != 0. But how can I write some (any) solution A in terms of divergenceless B? In other words, you want to construct a vector potential for the solenoidal vector field B. In general, you can do this on a domain D such that every closed surface in D is the boundary of a domain contained in D (i.e. D has no "holes"). I suppose your domain is all of R^3. The trick is to choose gauge conditions to reduce the amount of arbitrariness in the solution. One possible gauge condition is A_3 = 0. Then you want dA_2/dx_3 = -B_1 dA_1/dx_3 = B_2 dA_2/dx_1 - dA_1/dx_2 = B_3 The first and second conditions determine A_1 and A_2 up to functions of x_1 and x_2, i.e. A_1 = int B_2 dx_3 + a_1(x_1,x_2) A_2 = -int B_1 dx_3 + a_2(x_1,x_2) and then the third equation says da_1/dx_1 - da_2/dx_2 = g(x_1,x_2) for a certain function g (exercise: show that if div(B) = 0, this is a function of x_1 and x_2 only). Now you could e.g. take a_2 = 0 and a_1 = int g(x_1,x_2) dx_1. Robert Israel Department of Mathematics http://www.math.ubc.ca/~israel University of British Columbia Vancouver, BC, Canada V6T 1Z2 |
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#4
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shevek:
For some reason I can't solve one of the simplest equations.. Is there a solution? The equation: B = curl(A) Or in coordinate form: B_i = e_ijk* d/dx_j ( A_k ) (3 dimensions OK for now, e_ijk is anti-symmetric unit tensor or Levi-Civita tensor) The problem is to solve for the vector field A given a vector field B. I know that if A is a solution, so is A' = A + grad(f) where f is any function of the coordinates. I also know that there is no solution if div(B) != 0. But how can I write some (any) solution A in terms of divergenceless B? It depends upon what you are given _explicitly_. One way, is to take the curl of B = curl A to get J, curl V = curl curl A = grad (div A) - \grad^2 A = (4pi/c) J Use the gauge freedom to set div A = 0, then solve the three poisson equations for the components of A, -grad^2 A = (4pi/c) J Another way is to exploit some symmetry and solve for the line integral of A, B = curl A = \integral B.dS = \integral (curl A).dS Using the identity, \integral (curl A).dS = \integral A.dl, I get, \integral A.dl = \integral B.dS If you can write down A in a coordinate system that allows you to take A outside the integral sign in order to evaluate the line integral easily, then this will work. |
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#5
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shevek in article
posed the question: Given div(B)=0, how can one solve the equation curl(A)=B? In addition to the answers prior posted, here is another direct approach. Let antigrad be an R^3 into R^3 mapping defined by: (antigrad_1(G))x = Int{y3=0 to x_3: G_2(x_1,x_2,y3)} - Int{y2=0 to x_2: G_3(x_1, y2, 0)} (antigrad_2(G))x = - Int{y3=0 to x_3: G_1(x_1,x_2,y3)} (antigrad_3(G))x = 0 Hint: These formulae will look better if manually rewritten, with y3 changed to x_3 bar, etc. ------------------------------------ We now have the following result: If div(B)=0 then [A=anticurl(B) is a solution of: curl(A)=B] Alternatively; If div(B)=0 then curl(anticurl(B))=B Notes: 1) The term “anticurl” is meant to be suggestive rather than literal -- it indicates that under a special condition (but not universally), the anticurl is the right inverse of the curl. 2) The proof of the above given result follows from direct computation of curl(anticurl(B)). David Ziskind |
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#6
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shevek in article
posed the question: Given div(B)=0, how can one solve the equation curl(A)=B? Sorry, there is a mis-writing in my first reply: the four instances of “antigrad” SHOULD BE “anticurl”. Please disregard my first reply, thanks. Here is the reply text rewritten in full: In addition to the answers prior posted, here is another direct approach. Let anticurl be an R^3 into R^3 mapping defined by: (anticurl_1(G))x = Int{y3=0 to x_3: G_2(x_1,x_2,y3)} - Int{y2=0 to x_2: G_3(x_1, y2, 0)} (anticurl_2(G))x = - Int{y3=0 to x_3: G_1(x_1,x_2,y3)} (anticurl_3(G))x = 0 Hint: These formulae will look better if manually rewritten, with y3 changed to x_3 bar, etc. ------------------------------------ We now have the following result: If div(B)=0 then [A=anticurl(B) is a solution of: curl(A)=B] Alternatively; If div(B)=0 then curl(anticurl(B))=B Notes: 1) The term “anticurl” is meant to be suggestive rather than literal -- it indicates that under a special condition (but not universally), the anticurl is the right inverse of the curl. 2) The proof of the above given result follows from direct computation of curl(anticurl(B)). David Ziskind |
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#7
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"David Ziskind" wrote in message news:01c49533$953ba000$4c214b41@default...
shevek in article posed the question: Given div(B)=0, how can one solve the equation curl(A)=B? Sorry, there is a mis-writing in my first reply: the four instances of ?antigrad? SHOULD BE ?anticurl?. Please disregard my first reply, thanks. Here is the reply text rewritten in full: In addition to the answers prior posted, here is another direct approach. Let anticurl be an R^3 into R^3 mapping defined by: (anticurl_1(G))x = Int{y3=0 to x_3: G_2(x_1,x_2,y3)} - Int{y2=0 to x_2: G_3(x_1, y2, 0)} (anticurl_2(G))x = - Int{y3=0 to x_3: G_1(x_1,x_2,y3)} (anticurl_3(G))x = 0 Hint: These formulae will look better if manually rewritten, with y3 changed to x_3 bar, etc. ------------------------------------ We now have the following result: If div(B)=0 then [A=anticurl(B) is a solution of: curl(A)=B] Alternatively; If div(B)=0 then curl(anticurl(B))=B Notes: 1) The term ?anticurl? is meant to be suggestive rather than literal -- it indicates that under a special condition (but not universally), the anticurl is the right inverse of the curl. 2) The proof of the above given result follows from direct computation of curl(anticurl(B)). Thanks all for your useful responses! I was indeed after an "anticurl" operator, and a few of you gave me specific solutions, as well as good general considerations to make life easier including defining a "current" (doesn't have to be electric) and using symmetry. I should tell you my application. It is fluid mechanics.. I have been looking at the vorticity equation. If I come up with a solution for the vorticity of a fluid.. i.e. I know curl(v) (v = bulk fluid velocity).. then I was having trouble finding what the velocity is, and still more particularly what the divergence of the velocity is. Clearly (now) I need to include some boundary conditions on the flow to uniquely determine the velocity from the vorticity.. and the velocity may not have a closed form. Thanks again - |
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