Question

Separate the following partial differential equation into a set of three ODEs by the method of separation of variables :
d^2 u/dt^2 = c^2 [d^2 u/dr^2 + 1/r * du/dr + 1/r^2 * d^2/d(thita)^2 ]

Answer 1

Answer on Question #51339 – Math – Differential Calculus | Equations

Separate the following partial differential equation into a set of three ODEs by the method of separation of variables :

$$\mathrm{d}^2 \mathrm{u} / \mathrm{dt}^2 = \mathrm{c}^2 \left[ \mathrm{d}^2 \mathrm{u} / \mathrm{dr}^2 + \frac{1}{r} \cdot \frac{\partial u}{\partial r} + \frac{1}{r^2} \cdot \frac{\partial^2 u}{\partial \theta^2} \right]$$

Solution

$$\frac{\partial^2 u}{\partial t^2} = c^2 \left[ \frac{\partial^2 u}{\partial r^2} + \frac{1}{r} \frac{\partial u}{\partial r} + \frac{1}{r^2} \frac{\partial^2 u}{\partial \theta^2} \right]$$

Assume that $u(r, \theta, t) = R(r) \Theta(\theta) T(t)$

So $R(r) \Theta(\theta) T''(t) = c^2 \left[ \left( R'' + \frac{1}{r} R' \right) \Theta(\theta) T(t) + \frac{1}{r^2} \Theta''(\theta) R(r) T(t) \right]$ → divide both sides by $c^2 R(r) \Theta(\theta) T(t) \rightarrow$

$$\rightarrow \frac{1}{c^2} \frac{T''}{T} = \left( R'' + \frac{1}{r} R' \right) \frac{1}{R} + \frac{1}{r^2} \frac{\Theta''}{\Theta}$$

The right side of this equation does not depend on $T$, hence the left side of this equation must be constant.

Thus, $\frac{1}{c^2} \frac{T''}{T} = \left( R'' + \frac{1}{r} R' \right) \frac{1}{R} + \frac{1}{r^2} \frac{\Theta''}{\Theta} = \lambda$.

$$\left( R'' + \frac{1}{r} R' \right) \frac{1}{R} + \frac{1}{r^2} \frac{\Theta''}{\Theta} = \lambda \rightarrow -\frac{\Theta''}{\Theta} = \left( R'' + \frac{1}{r} R' \right) \frac{r^2}{R} - \lambda r^2$$

Because each side only depends on one independent variable, both sides of this equation must be constant. This gives the second separation constant, which we call $\mu$.

The equation with respect to $\Theta$ can then be written as

$$\Theta'' + \mu \Theta = 0$$

And equation with respect to $R$:

$$\left(R ^ {\prime \prime} + \frac {1}{r} R ^ {\prime}\right) \frac {r ^ {2}}{R} - \lambda r ^ {2} = \mu \rightarrow r ^ {2} R ^ {\prime \prime} + r R ^ {\prime} - (\lambda r ^ {2} + \mu) R = 0$$

Finally we have 3 ODEs:

$$\begin{array}{l} T ^ {\prime \prime} - \lambda c ^ {2} T = 0; \\ \boldsymbol {\Theta} ^ {\prime \prime} + \mu \boldsymbol {\Theta} = \mathbf {0}; \\ r ^ {2} R ^ {\prime \prime} + r R ^ {\prime} - (\lambda r ^ {2} + \mu) R = 0. \end{array}$$

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