Question

Reduce the following PDE to a set of three ODEs by the method of separation of variables. d^2u/dr^2 +(2/r)(du/dr)+(1/r^2)(d^2u/dθ ^2) +(cot θ/r^2)(du/dθ) + (1/r^2 sin^2 θ)(d^2u/d φ^2) =0

Answer 1

Answer on Question #70267 – Math – Differential Equations

Question

Reduce the following PDE to a set of three ODEs by the method of separation of variables.

$$\frac {\partial^ {2} u}{\partial r ^ {2}} + \frac {2}{r} \frac {\partial u}{\partial r} + \frac {1}{r ^ {2}} \frac {\partial^ {2} u}{\partial \theta^ {2}} + \frac {\cot \theta}{r ^ {2}} \frac {\partial u}{\partial \theta} + \frac {1}{r ^ {2} \sin^ {2} \theta} \frac {\partial^ {2} u}{\partial \varphi^ {2}} = 0$$

Solution

Represent $u = u(r, \theta, \phi)$ as a product of two functions depending on $r$ and $(\theta, \varphi)$:

$$u (r, \theta , \phi) = R (r) Y (\theta , \varphi)$$

Substituting this expression into original equation we obtain

$$Y \frac {d ^ {2} R}{d r ^ {2}} + \frac {2 Y}{r} \frac {d R}{d r} + \frac {R}{r ^ {2}} \frac {\partial^ {2} Y}{\partial \theta^ {2}} + \frac {R \cot \theta}{r ^ {2}} \frac {\partial Y}{\partial \theta} + \frac {R}{r ^ {2} \sin^ {2} \theta} \frac {\partial^ {2} Y}{\partial \varphi^ {2}} = 0$$

Multiplying this equation by $\frac{r^2}{RY}$ we get

$$\frac {r ^ {2}}{R} \frac {d ^ {2} R}{d r ^ {2}} + \frac {2 r}{R} \frac {d R}{d r} + \frac {1}{Y} \frac {\partial^ {2} Y}{\partial \theta^ {2}} + \frac {\cot \theta}{Y} \frac {\partial Y}{\partial \theta} + \frac {1}{Y \sin^ {2} \theta} \frac {\partial^ {2} Y}{\partial \varphi^ {2}} = 0$$

or

$$\frac {r ^ {2}}{R} \frac {d ^ {2} R}{d r ^ {2}} + \frac {2 r}{R} \frac {d R}{d r} = - \left(\frac {1}{Y} \frac {\partial^ {2} Y}{\partial \theta^ {2}} + \frac {\cot \theta}{Y} \frac {\partial Y}{\partial \theta} + \frac {1}{Y \sin^ {2} \theta} \frac {\partial^ {2} Y}{\partial \varphi^ {2}}\right)$$

Equating both sides of the equality to a some constant $\lambda$, we obtain two equations.

First equation for $R(r)$:

$$\frac {r ^ {2}}{R} \frac {d ^ {2} R}{d r ^ {2}} + \frac {2 r}{R} \frac {d R}{d r} = \lambda$$

or

$$r ^ {2} \frac {d ^ {2} R}{d r ^ {2}} + 2 r \frac {d R}{d r} - \lambda R = 0$$

Second equation for $Y(\theta, \varphi)$:

$$- \left(\frac {1}{Y} \frac {\partial^ {2} Y}{\partial \theta^ {2}} + \frac {\cot \theta}{Y} \frac {\partial Y}{\partial \theta} + \frac {1}{Y \sin^ {2} \theta} \frac {\partial^ {2} Y}{\partial \varphi^ {2}}\right) = \lambda$$

or

$$\frac {\partial^ {2} Y}{\partial \theta^ {2}} + \cot \theta \frac {\partial Y}{\partial \theta} + \frac {1}{\sin^ {2} \theta} \frac {\partial^ {2} Y}{\partial \varphi^ {2}} + \lambda Y = 0$$

Now we represent $Y(\theta, \varphi)$ as a product of two functions depending on $\theta$ and $\varphi$

$$Y (\theta , \varphi) = V (\theta) \Phi (\varphi)$$

and substitute it into the last equation

$$\Phi \frac {d ^ {2} V}{d \theta^ {2}} + \Phi \cot \theta \frac {d V}{d \theta} + \frac {V}{\sin^ {2} \theta} \frac {d ^ {2} \Phi}{d \varphi^ {2}} + \lambda V \Phi = 0$$

Multiplying this equation by $\frac{\sin^2\theta}{\Phi V}$ we get

$$\frac {\sin^ {2} \theta}{V} \frac {d ^ {2} V}{d \theta^ {2}} + \frac {\cos \theta \sin \theta}{V} \frac {d V}{d \theta} + \frac {1}{\Phi} \frac {d ^ {2} \Phi}{d \varphi^ {2}} + \lambda \sin^ {2} \theta = 0$$

or

$$\frac {\sin^ {2} \theta}{V} \frac {d ^ {2} V}{d \theta^ {2}} + \frac {\cos \theta \sin \theta}{V} \frac {d V}{d \theta} + \lambda \sin^ {2} \theta = - \frac {1}{\Phi} \frac {d ^ {2} \Phi}{d \varphi^ {2}}$$

Equating both sides of this equality to a some constant $m^2$, we obtain two equations.

First equation for $V(\theta)$:

$$\frac {\sin^ {2} \theta}{V} \frac {d ^ {2} V}{d \theta^ {2}} + \frac {\cos \theta}{V} \frac {\sin \theta}{d \theta} + \lambda \sin^ {2} \theta = m ^ {2}$$

or

$$\frac {d ^ {2} V}{d \theta^ {2}} + \cot \theta \frac {d V}{d \theta} + \left(\lambda - \frac {m ^ {2}}{\sin^ {2} \theta}\right) V = 0$$

The second equation for $\Phi (\varphi)$:

$$- \frac {1}{\Phi} \frac {d ^ {2} \Phi}{d \varphi^ {2}} = m ^ {2}$$

or

$$\frac {d ^ {2} \Phi}{d \varphi^ {2}} + m ^ {2} \Phi = 0$$

Thus representing the original function $u(r,\theta ,\varphi)$ as a product

$$u (r, \theta , \phi) = R (r) Y (\theta , \varphi) = R (r) V (\theta) \Phi (\varphi)$$

we obtained three equations for the functions $R(r), V(\theta)$ and $\Phi (\varphi)$:

$$r ^ {2} \frac {d ^ {2} R}{d r ^ {2}} + 2 r \frac {d R}{d r} - \lambda R = 0$$ $$\frac {d ^ {2} V}{d \theta^ {2}} + \cot \theta \frac {d V}{d \theta} + \left(\lambda - \frac {m ^ {2}}{\sin^ {2} \theta}\right) V = 0$$ $$\frac {d ^ {2} \Phi}{d \varphi^ {2}} + m ^ {2} \Phi = 0$$

where $\lambda$ and $m$ are the some constants

Answer: the original PDE is reduced to a set of three ODEs

$$r ^ {2} \frac {d ^ {2} R (r)}{d r ^ {2}} + 2 r \frac {d R (r)}{d r} - \lambda R (r) = 0$$ $$\frac {d ^ {2} V (\theta)}{d \theta^ {2}} + \cot \theta \frac {d V (\theta)}{d \theta} + \left(\lambda - \frac {m ^ {2}}{\sin^ {2} \theta}\right) V (\theta) = 0$$ $$\frac {d ^ {2} \Phi (\varphi)}{d \varphi^ {2}} + m ^ {2} \Phi (\varphi) = 0$$

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