Question

Reduce the following PDEs to a set of three ODEs by method of separation of variables 1/R(d/Dr(rdv/dv))+1/r(d

Answer 1

ANSWER on Question #81519 – Math – Differential Equations

QUESTION

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

$$\frac {1}{r} \frac {\partial}{\partial r} \left(r \cdot \left(\frac {\partial V}{\partial r}\right)\right) + \left(\frac {1}{r ^ {2}}\right) \cdot \frac {\partial^ {2} V}{\partial \theta^ {2}} + \frac {\partial^ {2} V}{\partial z ^ {2}} + k ^ {2} \cdot V = 0$$

SOLUTION

Assume

$$V (r, \theta , z) = R (r) \Theta (\theta) Z (z)$$

Then,

$$\begin{array}{l} \Theta Z \frac {1}{r} \frac {\partial}{\partial r} \left(r \cdot \left(\frac {\partial R}{\partial r}\right)\right) + R Z \left(\frac {1}{r ^ {2}}\right) \cdot \frac {\partial^ {2} \Theta}{\partial \theta^ {2}} + R \Theta \frac {\partial^ {2} Z}{\partial z ^ {2}} + k ^ {2} \cdot R \Theta Z = 0 \Bigg | \div \left(\frac {1}{R \Theta Z}\right) \rightarrow \\ \frac {1}{R} \frac {1}{r} \frac {\partial}{\partial r} \left(r \cdot \left(\frac {\partial R}{\partial r}\right)\right) + \frac {1}{\Theta} \left(\frac {1}{r ^ {2}}\right) \cdot \frac {\partial^ {2} \Theta}{\partial \theta^ {2}} + \frac {1}{Z} \frac {\partial^ {2} Z}{\partial z ^ {2}} + k ^ {2} = 0 \rightarrow \\ \frac {1}{R} \frac {1}{r} \frac {\partial}{\partial r} \left(r \cdot \left(\frac {\partial R}{\partial r}\right)\right) + \frac {1}{\Theta} \left(\frac {1}{r ^ {2}}\right) \cdot \frac {\partial^ {2} \Theta}{\partial \theta^ {2}} + k ^ {2} = - \frac {1}{Z} \frac {\partial^ {2} Z}{\partial z ^ {2}} \\ \end{array}$$

In the above equation the left-hand side depends on $r$ and $\theta$ , while the right-hand side depends on $z$ . The only way these two members are going to be equal for all values of $r$ , $\theta$ and $z$ is when both of them are equal to a constant. Let us define such constant as $-l^2$ .

With this choice for the constant, we obtain:

$$\frac {d ^ {2} Z}{d z ^ {2}} - l ^ {2} \cdot Z = 0$$

The general solution of this equation is:

$$Z (z) = A _ {1} e ^ {l z} + A _ {2} e ^ {- l z}$$

Such a solution, when considering the specific boundary conditions, will allow $Z(z)$ to go to zero for $z$ going to $\pm \infty$ , which makes physical sense. If we had given the constant a value of $l^2$ , we would have had periodic trigonometric functions, which do not tend to zero for $z$ going to infinity.

Once sorted the $z$-dependency, we need take care of $r$ and $\theta$.

$$\frac {1}{R} \frac {1}{r} \frac {\partial}{\partial r} \left(r \cdot \left(\frac {\partial R}{\partial r}\right)\right) + \frac {1}{\Theta} \left(\frac {1}{r ^ {2}}\right) \cdot \frac {\partial^ {2} \Theta}{\partial \theta^ {2}} + k ^ {2} = - l ^ {2} \rightarrow$$ $$\frac {1}{R} \frac {1}{r} \frac {\partial}{\partial r} \left(r \cdot \left(\frac {\partial R}{\partial r}\right)\right) + \frac {1}{\Theta} \left(\frac {1}{r ^ {2}}\right) \cdot \frac {\partial^ {2} \Theta}{\partial \theta^ {2}} = - (k ^ {2} + l ^ {2}) \rightarrow$$ $$\frac {1}{R} \frac {1}{r} \left(\frac {\partial R}{\partial r} + r \cdot \frac {\partial^ {2} R}{\partial r ^ {2}}\right) + \frac {1}{\Theta} \left(\frac {1}{r ^ {2}}\right) \cdot \frac {\partial^ {2} \Theta}{\partial \theta^ {2}} = - (k ^ {2} + l ^ {2}) \rightarrow$$ $$\frac {1}{R} \frac {\partial^ {2} R}{\partial r ^ {2}} + \frac {1}{R} \frac {1}{r} \frac {\partial R}{\partial r} + \frac {1}{\Theta} \left(\frac {1}{r ^ {2}}\right) \cdot \frac {\partial^ {2} \Theta}{\partial \theta^ {2}} = - (k ^ {2} + l ^ {2}) \Bigg | \times (r ^ {2}) \rightarrow$$ $$\frac {r ^ {2}}{R} \frac {\partial^ {2} R}{\partial r ^ {2}} + \frac {r}{R} \frac {\partial R}{\partial r} + (k ^ {2} + l ^ {2}) r ^ {2} = \frac {1}{\Theta} \cdot \frac {\partial^ {2} \Theta}{\partial \theta^ {2}}$$

Again we are in a situation where the only way a solution can be found for the above equation is when both members are equal to a constant. This time we select a positive constant, which we call $m^2$. The equation for $\Theta$ becomes, for:

$$\frac {d ^ {2} \Theta}{d \theta^ {2}} + m ^ {2} \Theta = 0$$

This solution is well suited, to describe the variation for an angular coordinate like $\theta$. Had we chosen the set both members equal to a negative number, we would have ended up with exponential functions with a different value assigned to $\Theta(\theta)$ for each $360{}^{\circ}$ turn, a clear non-physical solution.

Last to be examined is the $r$-dependency. We have:

$$\frac {r ^ {2}}{R} \frac {d ^ {2} R}{d r ^ {2}} + \frac {r}{R} \frac {d R}{d r} + (k ^ {2} + l ^ {2}) r ^ {2} = m ^ {2} \rightarrow$$ $$r ^ {2} \cdot \frac {d ^ {2} R}{d r ^ {2}} + r \cdot \frac {d R}{d r} + [ (k ^ {2} + l ^ {2}) r ^ {2} - m ^ {2} ] \cdot R = 0$$

This equation is a well-known equation of mathematical physics called parametric Bessel's equation. With sample linear transformation of variable, $x = r \cdot \sqrt{(k^2 + l^2)}$ , equation is readily changed into a Bessel's equation:

$$\frac {d R}{d r} = \frac {d R}{d x} \cdot \frac {d x}{d r} = \sqrt {k ^ {2} + l ^ {2}} \cdot R ^ {\prime}$$ $$\frac {d ^ {2} R}{d r ^ {2}} = \frac {d}{d x} \Big (\sqrt {k ^ {2} + l ^ {2}} \cdot R ^ {\prime} \Big) \cdot \frac {d x}{d r} = (k ^ {2} + l ^ {2}) \cdot R ^ {\prime \prime}$$

Then,

$$r ^ {2} \cdot \frac {d ^ {2} R}{d r ^ {2}} + r \cdot \frac {d R}{d r} + [ (k ^ {2} + l ^ {2}) r ^ {2} - m ^ {2} ] \cdot R = 0 \rightarrow$$ $$\frac {x ^ {2}}{(k ^ {2} + l ^ {2})} \cdot (k ^ {2} + l ^ {2}) \cdot R ^ {\prime \prime} + \frac {x}{\sqrt {k ^ {2} + l ^ {2}}} \cdot \sqrt {k ^ {2} + l ^ {2}} \cdot R ^ {\prime} + [ x ^ {2} - m ^ {2} ] \cdot R = 0 \rightarrow$$ $$x ^ {2} \cdot R ^ {\prime \prime} + x \cdot R ^ {\prime} + [ x ^ {2} - m ^ {2} ] \cdot R = 0$$

where $R''$ and $R'$ indicate the first and second derivatives with respect to $x$ .

In what follows we will assume that $m$ is a real, non-negative number.

Linearly independent solutions are typically denote $J_{m}(x)$ (Bessel Functions) and $N_{m}(x)$ (Neumann Functions).

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