Answer to Question #177951 in Differential Equations for Ashweta Padhan

Question #177951

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


1
Expert's answer
2021-04-15T07:39:40-0400

Multiplying on x2

x2y"(x)+xy'(x)+x2y(x)=0

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

y(x)=n=0an(r+n)xr+n1y'(x)=\sum_{n=0}^{\propto}a_n(r+n)x^{r+n-1}

y"(x)=n=0an(r+n)(r+n1)xr+n2y"(x)=\sum_{n=0}^{\propto}a_n(r+n)(r+n-1)x^{r+n-2}

y(x)=n=0an(r+n)(r+n1)xr+n+n=0an(r+n)xr+n+n=0anxr+n+2y'(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}

a0r2xr+a1(r+1)2xr+1+n=2(an(r+n)2an2)xr+n=0a_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

an=an2(r+n)2a_n=\frac{a_{n-2}}{(r+n)^2} ,n=2,3...

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

a2m=a2m2(2m)2a_2m=\frac{a_{2m-2}}{(2m)^2} ,m=1,2...

a2m=(1)ma022m(m!)a_{2m}=\frac{(-1)^ma_0}{2^{2m}(m!)} ,m=1,2...

y1(x)=a0(1+n=1(1)ma022m(m!))y_1(x)=a_0(1+\sum_{n=1}^{\propto}\frac{(-1)^ma_0}{2^{2m}(m!)})

as y(1)=1, a0=1 so y1(x)=1+n=1(1)m22m(m!)y_1(x)=1+\sum_{n=1}^{\propto}\frac{(-1)^m}{2^{2m}(m!)}

J0(x)=n=1(1)m22m(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 an(r)r=0a_n'(r)|_{r=0}

a0(r)r2xr+a1(r)(r+1)2xr+1+n=2(an(r)(r+n)2an2(r))xr+n=0a_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

a1(r)=0,a1(0)=0a_1(r)=0, a_1'(0)=0

an(r)=an2(r)r+na_{n}(r)=\frac{a_{n-2}(r)}{r+n} , n=2,3..

a2m+1(0)=0a'_{2m+1}(0)=0 , m=1,2...

a2m(r)=a2m2(r)r+2m=(1)ma0(r+2)2...(r+2m)2a_{2m}(r)=\frac{a_{2m-2}(r)}{r+2m}= \frac{(-1)^ma_0}{(r+2)^2...(r+2m)^2}

a2m(r)a2m(r)=2(1r+2+1r+4...+1r+2m)\frac{a'_{2m}(r)}{a_{2m}(r)}=2(\frac{1}{r+2}+\frac{1}{r+4}...+\frac{1}{r+2m})

a2m(0)=2(12+14...+12m)a'_{2m}(0)=2(\frac{1}{2}+\frac{1}{4}...+\frac{1}{2m})

Hm=12+14...+12mH_m=\frac{1}{2}+\frac{1}{4}...+\frac{1}{2m}

y2(x)=J0(x)lnx1+m=1(1)m+1Hm22m(m!)x2m,x>0y_2(x)=J_0(x)lnx1+\sum_{m=1}^{\propto}\frac{(-1)^{m+1}H_m}{2^{2m}(m!)}x^{2m},x>0


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