How can

$\qquad a_2\nabla^2 f(r,\theta) - \vec{g} \cdot \nabla f(r,\theta)= 0$

be solved for $f$? There is a symmetry in $\phi$ in spherical coordinates, so the equation is a 2D equation (depends on $r$ and $\theta$ which is angle with $z$ axis).

$\vec{g}$ is definded as below:

$\qquad g_\theta(r,\theta) = \sin \theta (1-\frac{\sin \theta }{2 r^2})-\frac{\alpha \sin ^2 \theta \cos \theta}{r^4}$

$\qquad g_r(r,\theta) = \cos \theta (-1+\frac{1}{ r^3})+\frac{3\alpha}{2r^2}(\frac{1}{r^2}-1)(\cos^2 \theta-\frac{1}{3})$

Boundary conditions are:

$\qquad \partial_r f(r=1,\theta)= -a_3$

$\qquad f(r=\infty,\theta)= a_1$

$a_1, a_2, a_3,\alpha$ are defined constants for example 10000 ,0.5,1,1 , respectively.

Could anyone please help me?

$\theta$ changes from $0$ to $2 \pi$ and boundary condition doesn't depend on it.

  • $\begingroup$ And what is $\alpha$? $\endgroup$ – m_goldberg Feb 19 '18 at 4:59
  • $\begingroup$ It is also a coefficient. I add its value. Thank you for pointing out that. @m_goldberg $\endgroup$ – Wat Watson Feb 19 '18 at 5:13
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    $\begingroup$ What have you tried? Since the equation is linear, "FiniteElement" method of NDSolve should be able to handle the problem without (much) difficulty. $\endgroup$ – xzczd Feb 19 '18 at 10:14
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    $\begingroup$ I mean, you've only given 2 b.c. in $r$ direction, but you still need b.c. in $\theta$ direction. (usually it's periodic b.c.) $\endgroup$ – xzczd Feb 19 '18 at 11:31
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    $\begingroup$ That's the range of $\theta$, not the boundary condition (b.c.) in $\theta$ direction, which is necessary for determing a particular solution, because your equation involves derivative of $\theta$. $\endgroup$ – xzczd Feb 19 '18 at 11:55

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