Answer to Question #177951 in Differential Equations for Ashweta Padhan

Question

Solve the Bessel's equation of order zero

y^''(x)+(1/x)y^'(x)+y(x)=0

With boundary conditions y^'(0)=0, y(1)=1

Answer 1

Multiplying on x²

x²y"(x)+xy'(x)+x²y(x)=0

$y(x)=\sum_{n=0}^{\propto}a_nx^{r+n}$ , for a not equal 0, x>0

$y'(x)=\sum_{n=0}^{\propto}a_n(r+n)x^{r+n-1}$

"y"(x)=\\sum_{n=0}^{\\propto}a_n(r+n)(r+n-1)x^{r+n-2}"

$y'(x)=\sum_{n=0}^{\propto}a_n(r+n)(r+n-1)x^{r+n}+\sum_{n=0}^{\propto}a_n(r+n)x^{r+n}+\sum_{n=0}^{\propto}a_nx^{r+n+2}$

$a_0r^2x^r+a_1(r+1)^2x^{r+1}+\sum_{n=2}^{\propto}(a_n(r+n)^2a_{n-2})x^{r+n}=0$

The indicial equation is r²= 0, and hence r1 = r2 = 0. 

a1 = 0; the recurrence relation is

$a_n=\frac{a_{n-2}}{(r+n)^2}$ ,n=2,3...

We conclude a₁= a₃ = a₅ = … = 0, and since r = 0,

$a_2m=\frac{a_{2m-2}}{(2m)^2}$ ,m=1,2...

$a_{2m}=\frac{(-1)^ma_0}{2^{2m}(m!)}$ ,m=1,2...

$y_1(x)=a_0(1+\sum_{n=1}^{\propto}\frac{(-1)^ma_0}{2^{2m}(m!)})$

as y(1)=1, a₀=1 so $y_1(x)=1+\sum_{n=1}^{\propto}\frac{(-1)^m}{2^{2m}(m!)}$

$J_0(x)=\sum_{n=1}^{\propto}\frac{(-1)^m}{2^{2m}(m!)}$

Since indicial equation has repeated roots, coeficients in second solution can be found using $a_n'(r)|_{r=0}$

$a_0(r)r^2x^r+a_1(r)(r+1)^2x^{r+1}+\sum_{n=2}^{\propto}(a_n(r)(r+n)^2a_{n-2}(r))x^{r+n}=0$

$a_1(r)=0, a_1'(0)=0$

$a_{n}(r)=\frac{a_{n-2}(r)}{r+n}$ , n=2,3..

$a'_{2m+1}(0)=0$ , m=1,2...

$a_{2m}(r)=\frac{a_{2m-2}(r)}{r+2m}= \frac{(-1)^ma_0}{(r+2)^2...(r+2m)^2}$

$\frac{a'_{2m}(r)}{a_{2m}(r)}=2(\frac{1}{r+2}+\frac{1}{r+4}...+\frac{1}{r+2m})$

$a'_{2m}(0)=2(\frac{1}{2}+\frac{1}{4}...+\frac{1}{2m})$

$H_m=\frac{1}{2}+\frac{1}{4}...+\frac{1}{2m}$

$y_2(x)=J_0(x)lnx1+\sum_{m=1}^{\propto}\frac{(-1)^{m+1}H_m}{2^{2m}(m!)}x^{2m},x>0$