How to Solve Laplace's Equation in Spherical Coordinates Without Azimuthal Symmetry

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Laplace's equation ${\displaystyle \nabla ^{2}f=0}$ is a second-order partial differential equation (PDE) widely encountered in the physical sciences. In particular, it shows up in calculations of the electric potential absent charge density, and temperature in equilibrium systems.

Because Laplace's equation is a linear PDE, we can use the technique of separation of variables in order to convert the PDE into several ordinary differential equations (ODEs) that are easier to solve. Linearity ensures that the solution set consists of an arbitrary linear combination of solutions. Once we have our general solution, we incorporate boundary conditions that are given to us.

Preliminaries

1. 1

2. 2

3. 3

   Solve the radial equation. After multiplying and using the product rule, we find that this is simply the Euler-Cauchy equation.

4. 4

   Solve the angular equation. This equation is the Legendre differential equation in the variable ${\displaystyle \cos \theta .}$

   • ${\displaystyle {\frac {1}{\sin \theta }}{\frac {\mathrm {d} }{\mathrm {d} \theta }}\left(\sin \theta {\frac {\mathrm {d} \Theta }{\mathrm {d} \theta }}\right)+l(l+1)\Theta =0}$
   • To see this, we begin with the Legendre equation in the variable ${\displaystyle x}$ and make the substitution ${\displaystyle x=\cos \theta ,}$ implying that ${\displaystyle \mathrm {d} x=-\sin \theta \mathrm {d} \theta .}$
     • ${\displaystyle {\begin{aligned}{\frac {\mathrm {d} }{\mathrm {d} x}}\left((1-x^{2}){\frac {\mathrm {d} \Theta }{\mathrm {d} x}}\right)+l(l+1)\Theta &=0\\{\frac {\mathrm {d} }{-\sin \theta \mathrm {d} \theta }}\left(\sin ^{2}\theta {\frac {\mathrm {d} \Theta }{-\sin \theta \mathrm {d} \theta }}\right)+l(l+1)\Theta &=0\\{\frac {1}{\sin \theta }}{\frac {\mathrm {d} }{\mathrm {d} \theta }}\left(\sin \theta {\frac {\mathrm {d} \Theta }{\mathrm {d} \theta }}\right)+l(l+1)\Theta &=0\end{aligned}}}$
   • This equation can be solved using the method of Frobenius. In particular, the solutions are Legendre polynomials in ${\displaystyle \cos \theta ,}$ which we write as ${\displaystyle P_{l}(\cos \theta ).}$ These are orthogonal polynomials with respect to an inner product, which we elaborate on shortly. This orthogonality means that we can write any polynomial as a linear combination of Legendre polynomials.
     • ${\displaystyle \Theta (\theta )=P_{l}(\cos \theta )}$
   • The first few Legendre polynomials are given as follows. Notice that the polynomials alternate between even and odd. These polynomials will be very important in the next sections.
   • It turns out that there is another solution to the Legendre differential equation. However, this solution cannot be part of the general solution because it blows up at ${\displaystyle \theta =0}$ and ${\displaystyle \theta =\pi ,}$ so it is omitted.
     • ${\displaystyle \Theta (\theta )=\ln \tan {\frac {\theta }{2}}}$

5. 5

   Construct the general solution. We now have our solutions to both the radial and angular equations. We can then write out the general solution as a series, since by linearity, any linear combination of these solutions is also a solution.

   • ${\displaystyle V(r,\theta )=\sum _{l=0}^{\infty }\left(A_{l}r^{l}+{\frac {B_{l}}{r^{l+1}}}\right)P_{l}(\cos \theta )}$

1. 1

2. 2

   Find the potential inside the sphere. Physically, the potential cannot blow up at the origin, so ${\displaystyle B_{l}=0}$ for all ${\displaystyle l.}$

3. 3

4. 4

1. 1

2. 2

   Write the potential on the surface in terms of Legendre polynomials. This step is crucial in comparing coefficients, and we can use trigonometric identities to do this. We then refer to the zeroth, second, and fourth polynomials to write ${\displaystyle V_{0}}$ in terms of them.

   • ${\displaystyle {\begin{aligned}k\cos 4\theta &=k(8\cos ^{4}\theta -8\cos ^{2}\theta +1)\\&=k\left({\frac {64}{35}}P_{4}(\cos \theta )-{\frac {16}{21}}P_{2}(\cos \theta )-{\frac {1}{15}}P_{0}(\cos \theta )\right)\end{aligned}}}$

3. 3

   Solve for the potential outside the sphere. Physically, the potential should go to 0 as ${\displaystyle r\to \infty .}$ This means that outside the sphere, ${\displaystyle A_{l}=0.}$

4. 4

   Solve for the potential inside the sphere. Since there is no charge density inside the sphere, the potential cannot blow up, so ${\displaystyle B_{l}=0.}$ Furthermore, the boundary conditions and this technique ensure that the potential is continuous - in other words, the potential infinitesimally near the surface is the same when approached from both outside and inside the sphere.

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How to Solve Laplace's Equation in Spherical Coordinates Without Azimuthal Symmetry

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