User:Egm6322.s09.bit.gk/r5

$$f\left (r,\theta \right )=\frac{1}{rR}\frac{\mathrm{d} }{\mathrm{d} r}\left (r\frac{\mathrm{d} R}{\mathrm{d} r} \right )+\frac{1}{r^2\Theta }\frac{\mathrm{d^2\Theta } }{\mathrm{d} \theta ^2}=-\lambda (const)$$

$$\Rightarrow \frac{r}{R}\frac{\mathrm{d} }{\mathrm{d} r}\left (r\frac{\mathrm{d} R}{\mathrm{d} r} \right )+\frac{1}{\Theta }\frac{\mathrm{d^2\Theta } }{\mathrm{d} \theta ^2}+\lambda r^2 =0 $$

let,

$$m(\theta)=\frac{1}{\Theta }\frac{\mathrm{d^2\Theta } }{\mathrm{d} \theta ^2}$$

and

$$l(r)=\frac{r}{R}\frac{\mathrm{d} }{\mathrm{d} r}\left (r\frac{\mathrm{d} R}{\mathrm{d} r} \right )+\lambda r^2$$

$$\Rightarrow l(r)+m(\theta)=0$$

$$\Rightarrow l(r)=-m(\theta)=\rho$$

$$\Rightarrow \frac{1}{\Theta }\frac{\mathrm{d^2\Theta } }{\mathrm{d} \theta ^2}=-\rho$$

$$\Rightarrow {\Theta}''+\rho\Theta=0$$

Assume,

$$ \Theta(\theta)=e^{\alpha\theta}$$, Where $$\alpha$$ is unknown

$$\Rightarrow {\Theta }'\left (\theta \right )=\frac{\mathrm{d} \Theta}{\mathrm{d} \theta}=\alpha e^{\alpha\theta}=\alpha\Theta$$

and,

$${\Theta}''=\alpha^2\Theta$$

$$\therefore$$ making the above substitutions in the ODE containing $$\Theta$$

We have,

$$\alpha^2\Theta +\rho\Theta=0$$

$$\Rightarrow \alpha^2+\rho=0$$

$$\Rightarrow \alpha=\pm i\sqrt{\rho}$$

$$\Rightarrow \Theta \left (\theta \right )=k_3e^{i\sqrt{\rho}}+k_4e^{-i\sqrt{\rho}} $$

$$\underline V=v_i \underline e_i=\overline v_i \underline \bar e_i$$

$$\underline e_i$$ linearly independent

$$\left \{\underline \bar e_i \right \}$$ linearly independent

Similarly for functions:linearly independent function. $$ \left \{ e^{i\sqrt\rho},e^{-i\sqrt\rho} \right \}$$ linearly independent.

We have from De Moivre's theorem

$$e^{i\theta}=cos(\theta)+i sin(\theta)$$

$$\Rightarrow e^{-i\theta}=cos(\theta)-i sin(\theta)$$

$$\Rightarrow cos(\theta)=\frac{1}{2}\left(e^{i\theta}+e^{-i\theta}\right)$$

$$\Rightarrow sin(\theta)=\frac{1}{2i}\left(e^{i\theta}-e^{-i\theta}\right)$$

$$\therefore$$ We have,

$$\Theta \left (\theta \right )=k_3\left [cos(\sqrt\rho\theta)+i sin(\sqrt\rho\theta ) \right ]+k_4\left [cos(\sqrt\rho\theta)-i sin(\sqrt\rho\theta ) \right ]$$