Answer to Question #273508 in Differential Equations for DEBANKUR BISWAS

Question

reduce the equation

∇²ψ + [k² + f(ρ) + (1/ρ²)g(φ) + h(z)]ψ = 0

to a set of ODEs by the method of separation of variables.

Answer 1

$\nabla^{2} \psi+\left[k^{2}+f(\rho)+\left(\frac{1}{\rho^{2}}\right) g(\varphi)+h(z)\right] \psi=0$

$\frac{1}{\rho} \frac{\partial}{\partial \rho}\left[\rho \frac{\partial \psi}{\partial \rho}\right]+\frac{1}{\rho^{2} \frac{\partial^{2} \psi}{\partial \varphi^{2}}}+\frac{\partial \psi}{\partial \beta^{2}}+\left[k^{2}+f(\rho)+\left(\frac{1}{\rho^{2}}\right) g(\varphi)+h(z)\right] \psi=0$

$\frac{\partial^{2} \varphi}{\partial^{2}}+\frac{1}{\rho} \frac{\partial \psi}{\partial \rho}+\frac{1}{\rho^{2}} \frac{\partial^{2} \psi}{\partial \varphi^{2}}+\frac{\partial^{2} \varphi}{\partial \underline{ \beta}^{2}}+\left[k^{2}+f(\rho)+\left(\frac{1}{\rho^{2}}\right) g(\varphi)+h(z)\right] \psi=0$

Method of separation of variable

$\psi(\rho, Q, z)=R(\rho) H(\varphi) Z(\beta)$

$R^{\prime \prime} H z+\frac{1}{\rho} R^{\prime} H z+\frac{1}{\rho^{2}} R H^{\prime \prime} z+R H Z^{\prime \prime}+ \left[k^{2}+f(\rho)+\left(\frac{1}{\rho^{3}}\right) g(\varphi)+h(z)\right] R H z=0$

$\frac{R^{\prime \prime}}{R}+\frac{1}{\rho} \frac{R^{\prime}}{R}+\frac{1}{\rho^{2}} \frac{H^{\prime \prime}}{H}+\frac{Z^{\prime \prime}}{Z}+\left[k^{2}+f(\rho)+\left(\frac{1}{\rho^{2}}\right) g(\varphi)+h(z)\right]=0$

$\frac{R^{\prime \prime}}{R}+\frac{1}{\rho} \frac{R^{\prime}}{R}+\frac{1}{\rho^{2}} \frac{H^{\prime \prime}}{H}+\left[k^{2}+f(\rho)+\left(\frac{1}{\rho^{2}}\right) g(\varphi)\right]=\frac{-Z^{\prime \prime}}{z}-h(z) =1$

Then $\quad-\frac{z^{\prime \prime}}{z}-R(\xi)=\lambda \Rightarrow z^{\prime \prime}+(\lambda+R(3)) z=0$ and $\frac{R^{\prime \prime}}{R}+\frac{1}{\rho} \frac{R^{\prime}}{R}+\frac{1}{\rho^{2}} \frac{H^{\prime \prime}}{H}+\left[k^{2}+f(\rho)+\left(\frac{1}{\rho^{\prime}}\right) g(\varphi)\right]=\lambda$

$\begin{aligned} &\Rightarrow \quad \frac{\rho^{2} R^{\prime \prime}}{R}+\frac{\rho R^{\prime}}{R}+\frac{H^{\prime \prime}}{H}+\left(K^{2} \rho^{2}+\rho^{2} f(\rho)+g(\varphi)\right)=\lambda \rho^{2}\\ &\Rightarrow \frac{\rho^{2} R^{\prime \prime}}{R}+\frac{\rho R^{\prime}}{R}+K^{2} \rho^{2}+\rho^{2} f(\rho)-\lambda \rho^{2}=\frac{-H^{\prime \prime}}{H}-g(\varphi)=l \\ &\text { Then } \quad \frac{-H^{\prime \prime}}{H}-g(\varphi)=l \Rightarrow H^{\prime \prime}+(l+g(\varphi)) H=0\\ &\text { and } \frac{\rho^{2} R^{\prime \prime}}{R}+\frac{\rho R^{\prime}}{R}+R^{2} \rho^{2}+\rho^{2} f(\rho)-\lambda \rho^{2}=l\\ &\Rightarrow \quad \rho^{2} R^{\prime \prime}+\rho R^{\prime}+\left(K^{2} \rho^{2}+\rho^{2} f(\rho)-\lambda \rho^{2}-\ell\right) R=0\\ &\text { set of ODE's }\\ &\begin{gathered} z^{\prime \prime}+(\lambda+R(z))z=0 \\ H^{\prime \prime}+(l+g(Q)) H=0 \\ \rho^{2} R^{\prime \prime}+\rho R^{1}+\left(K^{2} \rho^{2}+\rho^{2} f(\rho)-\lambda \rho^{2}-l\right) R=0 \end{gathered}\\ \end{aligned}$