write the laplaces equation in cylindrical polar coordinates
Answers
Answer:
Beginning with the Laplacian in Cylindrical Coordinates, apply the operator to a potential function and set it equal to zero to get the Laplace equation
{\displaystyle \nabla ^{2}\Phi ={\frac {1}{r}}{\frac {\partial }{\partial r}}\left(r{\frac {\partial \Phi }{\partial r}}\right)+{\frac {1}{r^{2}}}{\frac {\partial ^{2}\Phi }{\partial \theta ^{2}}}+{\frac {\partial ^{2}\Phi }{\partial z^{2}}}=0.}{\displaystyle \nabla ^{2}\Phi ={\frac {1}{r}}{\frac {\partial }{\partial r}}\left(r{\frac {\partial \Phi }{\partial r}}\right)+{\frac {1}{r^{2}}}{\frac {\partial ^{2}\Phi }{\partial \theta ^{2}}}+{\frac {\partial ^{2}\Phi }{\partial z^{2}}}=0.}
First expand out the terms
{\displaystyle \nabla ^{2}\Phi ={\frac {1}{r}}{\frac {\partial \Phi }{\partial r}}+{\frac {\partial ^{2}\Phi }{\partial r^{2}}}+{\frac {1}{r^{2}}}{\frac {\partial ^{2}\Phi }{\partial \theta ^{2}}}+{\frac {\partial ^{2}\Phi }{\partial z^{2}}}=0.}{\displaystyle \nabla ^{2}\Phi ={\frac {1}{r}}{\frac {\partial \Phi }{\partial r}}+{\frac {\partial ^{2}\Phi }{\partial r^{2}}}+{\frac {1}{r^{2}}}{\frac {\partial ^{2}\Phi }{\partial \theta ^{2}}}+{\frac {\partial ^{2}\Phi }{\partial z^{2}}}=0.}
Then apply the method of separation of variables by assuming the solution is in the form
{\displaystyle \Phi \left(r,\theta ,z\right)=R(r)P(\phi )Z(z).}{\displaystyle \Phi \left(r,\theta ,z\right)=R(r)P(\phi )Z(z).}
Plug this into (2) and note how we can bring out the functions that are not affected by the derivatives
{\displaystyle {\frac {P(\phi )Z(z)}{r}}{\frac {\partial R(r)}{\partial r}}+P(\phi )Z(z){\frac {\partial ^{2}R(r)}{\partial r^{2}}}+{\frac {R(r)Z(z)}{r^{2}}}{\frac {\partial ^{2}P(\phi )}{\partial \theta ^{2}}}+R(r)P(\phi ){\frac {\partial ^{2}Z(z)}{\partial z^{2}}}=0.}{\displaystyle {\frac {P(\phi )Z(z)}{r}}{\frac {\partial R(r)}{\partial r}}+P(\phi )Z(z){\frac {\partial ^{2}R(r)}{\partial r^{2}}}+{\frac {R(r)Z(z)}{r^{2}}}{\frac {\partial ^{2}P(\phi )}{\partial \theta ^{2}}}+R(r)P(\phi ){\frac {\partial ^{2}Z(z)}{\partial z^{2}}}=0.}
Divide by {\displaystyle R(r)P(\phi )Z(z)}{\displaystyle R(r)P(\phi )Z(z)} and use short hand notation to get
{\displaystyle {\frac {1}{Rr}}{\frac {\partial R}{\partial r}}+{\frac {1}{R}}{\frac {\partial ^{2}R}{\partial r^{2}}}+{\frac {1}{Pr^{2}}}{\frac {\partial ^{2}P}{\partial \theta ^{2}}}+{\frac {1}{Z}}{\frac {\partial ^{2}Z}{dz^{2}}}=0.}{\displaystyle {\frac {1}{Rr}}{\frac {\partial R}{\partial r}}+{\frac {1}{R}}{\frac {\partial ^{2}R}{\partial r^{2}}}+{\frac {1}{Pr^{2}}}{\frac {\partial ^{2}P}{\partial \theta ^{2}}}+{\frac {1}{Z}}{\frac {\partial ^{2}Z}{dz^{2}}}=0.}
"Separating" the z term to the other side gives
{\displaystyle {\frac {1}{Rr}}{\frac {dR}{dr}}+{\frac {1}{R}}{\frac {d^{2}R}{dr^{2}}}+{\frac {1}{Pr^{2}}}{\frac {d^{2}P}{d\theta ^{2}}}=-{\frac {1}{Z}}{\frac {d^{2}Z}{dz^{2}}}.}{\displaystyle {\frac {1}{Rr}}{\frac {dR}{dr}}+{\frac {1}{R}}{\frac {d^{2}R}{dr^{2}}}+{\frac {1}{Pr^{2}}}{\frac {d^{2}P}{d\theta ^{2}}}=-{\frac {1}{Z}}{\frac {d^{2}Z}{dz^{2}}}.}
This equation can only be satisfied for all values if both sides are equal to a constant, {\displaystyle \lambda }\lambda, such that
{\displaystyle -{\frac {1}{Z}}{\frac {d^{2}Z}{dz^{2}}}=\lambda }{\displaystyle -{\frac {1}{Z}}{\frac {d^{2}Z}{dz^{2}}}=\lambda }
{\displaystyle {\frac {1}{Rr}}{\frac {dR}{dr}}+{\frac {1}{R}}{\frac {d^{2}R}{dr^{2}}}+{\frac {1}{Pr^{2}}}{\frac {d^{2}P}{d\theta ^{2}}}=\lambda .}{\displaystyle {\frac {1}{Rr}}{\frac {dR}{dr}}+{\frac {1}{R}}{\frac {d^{2}R}{dr^{2}}}+{\frac {1}{Pr^{2}}}{\frac {d^{2}P}{d\theta ^{2}}}=\lambda .}
Before we can focus on solutions, we need to further separate (4), so multiply (4) by {\displaystyle r^{2}}{\displaystyle r^{2}}
{\displaystyle {\frac {r}{R}}{\frac {dR}{dr}}+{\frac {r^{2}}{R}}{\frac {d^{2}R}{dr^{2}}}+{\frac {1}{P}}{\frac {d^{2}P}{d\theta ^{2}}}=\lambda r^{2}.}{\displaystyle {\frac {r}{R}}{\frac {dR}{dr}}+{\frac {r^{2}}{R}}{\frac {d^{2}R}{dr^{2}}}+{\frac {1}{P}}{\frac {d^{2}P}{d\theta ^{2}}}=\lambda r^{2}.}
Separate the terms
{\displaystyle {\frac {r}{R}}{\frac {dR}{dr}}+{\frac {r^{2}}{R}}{\frac {d^{2}R}{dr^{2}}}-\lambda r^{2}=-{\frac {1}{P}}{\frac {d^{2}P}{d\theta ^{2}}}.}{\displaystyle {\frac {r}{R}}{\frac {dR}{dr}}+{\frac {r^{2}}{R}}{\frac {d^{2}R}{dr^{2}}}-\lambda r^{2}=-{\frac {1}{P}}{\frac {d^{2}P}{d\theta ^{2}}}.}
As before, set both sides to a constant, {\displaystyle \kappa }{\displaystyle \kappa }
{\displaystyle -{\frac {1}{P}}{\frac {d^{2}P}{d\theta ^{2}}}=\kappa }{\displaystyle -{\frac {1}{P}}{\frac {d^{2}P}{d\theta ^{2}}}=\kappa }
{\displaystyle {\frac {r}{R}}{\frac {dR}{dr}}+{\frac {r^{2}}{R}}{\frac {d^{2}R}{dr^{2}}}-\lambda r^{2}=\kappa .}{\displaystyle {\frac {r}{R}}{\frac {dR}{dr}}+{\frac {r^{2}}{R}}{\frac {d^{2}R}{dr^{2}}}-\lambda r^{2}=\kappa .}
Now there are three differential equations and we know the form of these solutions. The differential equations of (3) and (5) are ordinary differential equations, while (6) is a little more complicated and we must turn to Bessel functions.