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If curl of a vector is $xy \hat{i} + xy \hat{j} + -(x+y)z \hat{k}$ what is the vector ? How to approach the problem. Is there specific method to get the vector function ? Only thing I get is the usual curl matrix or curl formula. How would I get the the vector by comparing $(\partial A_z/\partial y - \partial A_x/\partial z) $ = xy $(\partial A_z/\partial x - \partial A_x /\partial z)$ = xy and $(\partial A_y/\partial x - \partial A_x/\partial y )$ = -(x+y)z .

user187604
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  • Have you tried using the usual curl matrix or curl formula? I suspect you should be able to obtain something from the 3 equations – Calvin Khor Jul 31 '18 at 09:13
  • @CalvinKhor $(\partial A_z/\partial y - \partial A_x/\partial z) $ = xy $(\partial A_z/\partial x - \partial A_x /\partial z)$ = xy and $(\partial A_y/\partial x - \partial A_x/\partial y )$ = -(x+y)z . Would it help me too much? – user187604 Jul 31 '18 at 09:24
  • I guess not. $$\nabla \times (\nabla \times u) = \nabla(\nabla\cdot u) - \Delta u$$ may be useful? – Calvin Khor Jul 31 '18 at 09:32
  • @CalvinKhor please do the problem with the formula you mentioned. It'll help me a lot. – user187604 Jul 31 '18 at 09:34
  • I don't know if it can be done with it. I'm hoping it leads to an answer. – Calvin Khor Jul 31 '18 at 09:36
  • @CalvinKhor oh, Sorry I thought you have solved it. – user187604 Jul 31 '18 at 09:36
  • my initial thought was to assume that $u$ was divergence free, and then compute $ u = (-\Delta)^{-1} \nabla \times (\nabla \times u) $ (i.e. Biot-Savart law) but I'm not sure how to invert the Laplacian on polynomials. – Calvin Khor Jul 31 '18 at 09:40
  • @CalvinKhor the divergence would be zero for a curl of a vector. – user187604 Jul 31 '18 at 09:41
  • yes, but I meant $u$ in my formula, what you are saying would correspond to $\nabla \cdot \nabla\times u = 0$ – Calvin Khor Jul 31 '18 at 09:42
  • @CalvinKhor what is - delta inverse? In your formula? – user187604 Jul 31 '18 at 09:44
  • Oh, but the curl of a gradient is 0. So if $u = u_1 + \nabla u_2$ then $\nabla \times u = \nabla \times u_1$. The Biot-Savart law, if it applied, would then tell you how to recover $u_1$, and you would get a family of solutions by adding any gradient – Calvin Khor Jul 31 '18 at 09:45
  • $u = (-\Delta)^{-1}f $ if $u$ solves the equation $-\Delta u = f$ – Calvin Khor Jul 31 '18 at 09:45
  • @CalvinKhor man please write an answer now. I think now you can solve it. – user187604 Jul 31 '18 at 09:46
  • Well, I don't know how to invert the Laplacian on polynomials. I can't check right now if the green's function solution is well defined, you can check this and if it works you can post an answer :) – Calvin Khor Jul 31 '18 at 09:47
  • @CalvinKhor anyway thanks for helping. – user187604 Jul 31 '18 at 09:48
  • Oh, I suspect you can easily compute the solution of the Poisson equation by hand. Do try it and tell me if it works – Calvin Khor Jul 31 '18 at 09:57
  • Just by trial and error (using the differential form language) I found that $xyz \hat i - xyz \hat j$ works. – Daniel Schepler Jul 31 '18 at 22:32

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One way to approach this is to convert to an equivalent differential form and then apply the algorithm described in this answer to find an antiderivative of it. In detail, apply the Hodge star operator to the 1-form $A\,dx+B\,dy+C\,dz$ to get $\eta = A\,dy\wedge dz + B\,dz\wedge dx + C\,dx\wedge dy$. Steps 1 and 2 of the algorithm produce the form $$\left[tzB(tx,ty,tz)-tyC(tx,ty,tz)\right]\,dt\wedge dx + \left[txC(tx,ty,tz)-tzA(tx,ty,tz)\right]\,dt\wedge dy + \left[tyA(tx,ty,tz)-txB(tx,ty,tz)\right]\,dt\wedge dz$$ and so a vector field with curl $(A,B,C)$ is $$\int_0^1 \left[zB(tx,ty,tz)-yC(tx,ty,tz),xC(tx,ty,tz)-zA(tx,ty,tz),yA(tx,ty,tz)-xB(tx,ty,tz)\right]\,t\,dt.$$

In this problem, we have $A:(x,y,z)\mapsto xy$, $B:(x,y,z)\mapsto xy$ and $C:(x,y,z)\mapsto-(x+y)z$, so, after some simplification, $$\mathbf F(x,y,z) = \left[(2x+y)yz\,\mathbf i - (x+2y)xz\,\mathbf j - (x-y)xy\,\mathbf k\right]\int_0^1 t^3\,dt \\ = \frac14\left[(2x+y)yz\,\mathbf i - (x+2y)xz\,\mathbf j - (x-y)xy\,\mathbf k\right].$$

There are, of course, other antiderivatives. Just as you can add an arbitrary constant to the antiderivative of a single-variable scalar function to get another antiderivative, you can add any irrotational vector field to $\mathbf F$ to get another vector field with the same curl.

amd
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  • As an example of the “constant of integration,” if you compute the difference between this answer and the one obtained by trial and error ($xyz,\mathbf i - xyz,\mathbf j$) you’ll find that its curl does indeed vanish. – amd Aug 03 '18 at 01:48