how to obtain solution for $z$ using Frobenius method over Bessel equation

Question

Given this equation:

$$ \frac{1}{r}\frac{d}{dr}\bigg(r \frac{dT}{dr} \bigg) - \bigg(\frac{2h}{bk} \bigg)(T-T_a) = 0 $$

I can let $x=r\sqrt{(2h/bk)}$ and $z=T-T_a$ to obtain the following Bessel equation (using the appropriate substitutions):

$$ x^2 \frac{d^2z}{dx^2} + x \frac{dz}{dx} - x^2z = 0 $$

To express the solution for $z$, I read that one technique is to use the Frobenius method, but I am not sure how to apply this to be Bessel function above... Any help on how to apply the Frobenius method to the above ?

Answer 1

Starting from your last equation, using differential operator notation $$x^2\mathrm{D}^2_x[f]+x\mathrm{D}_x[f]-x^2f=0$$ Make a substitution $z=ix$ hence we can see that $\mathrm{D}_z=\frac{1}{i}\mathrm{D}_x\implies\mathrm{D}_x=i\mathrm{D}_z$ Hence, $$\frac{z^2}{i^2}i\mathrm{D}_z[i\mathrm{D}_z[f]]+\frac{z}{i}i\mathrm{D}_z[f]-\frac{z^2}{i^2}f=0$$ $$z^2\mathrm{D}_z[f]+z\mathrm{D}_z[f]+z^2f=0$$ We could apply Frobenius from here, but this is of course a usual Bessel ODE, and there are already many derivations of the Frobenius method on the Bessel ODE so I won't reproduce it here. Since this is a Bessel ODE the solutions are of course $$f(z)=c_1J_0(z)+c_2Y_0(z)$$ $$f(ix)=c_1J_0(ix)+c_2Y_0(ix)$$ The functions $J_0(ix)$ and $Y_0(ix)$ have special representations. First let $$I_\nu(x):=J_\nu(-ix)=(-ix)^\nu (ix)^{-\nu}J_{\nu}(ix)$$ $$K_\nu(x):=Y_\nu(-ix)=(-(ix)^2)^\nu ((ix)^{2\nu}Y_{\nu}(ix)+((-ix)^{2\nu}-(ix)^{2\nu})J_{\nu}(x)\cot(\pi\nu))$$ $$=x^{2\nu}(-x^{2\nu}Y_\nu(ix)+(-(-x)^{2\nu}+x^{2\nu})J_{\nu}(x)\cot(\pi\nu))$$ So in the special case $\nu=0$ as in the problem we have $$I_0(x)=J_0(ix)$$ $$K_0(x)=Y_0 (ix)$$ These (the general $\nu\in\mathbb{C}$ case) are called the modified Bessel functions of the first and second kind. Alternatively one can define $I_\nu(z),K_\nu(z)$ as the solutions $f(z)$ of the differential equation $$z^2\mathrm{D}_z^2[f]+z\mathrm{D}_z[f]-(z^2+\nu^2)f=0$$ For any $\nu\in\mathbb{C}$.

Answer 2

From the second equation in the form of

$(xz')'+xz = 0,\quad z(0)=C,$

assuming Maclaurin series of $\;z\;$ in the form of

$z=\sum\limits_{n=0}^\infty a_n z^n,$

one can get

$xz'=\sum\limits_{n=1}^\infty na_n x^n,\quad (xz')'+ xz= a_1+\sum\limits_{n=1}^\infty (n+1)^2 a_{n+1}x^{n} + \sum\limits_{n=1}^\infty a_{n-1}x^n = 0,$

with the recurrence relations in the form of

$\begin{cases} a_0 = C\\ a_1 = 0\\ (n+1)^2 a_{n+1} = a_{n-1}, \end{cases}$

which leads to the solution

$a_{2k+1} = 0,\quad a_{2k} = C\,\dfrac{(-1)^k}{(2k)!!)^2} = C\,\dfrac{(-1)^k}{4^k(k!)^2},\qquad(k=0,1,2,\dots).$

Easily to see that $\;z(x)=C J_0(x).$