If we have an inhomogeneous boundary value problem
$x^2 y'' + xy' + (x^2 -1)y = x,$ $y(0) = y(b) = 0,$ where $b>0$
How to use Green's Funtion to Solve this problem. I am facing issues with equations and the number of variables. Please help me solving this
So I'll admit, I didn't know nearly enough to answer this question. However curiosity got the best of me, so here it is.
Indeed, the given ODE is a Bessel's equation of the first kind, we can write this in sturmian form as
$$ \begin{cases} (xy)'+(x-\frac{1}{x})y = 1\\ y(0)=y(b)=0 \end{cases} $$
In order to construct a Green's Function for the problem, consider the homogeneous version above. Further note that there is a singularity at $x = 0$. This means we're going to have to handle constructing our green's function differently.
Note that a general solution to the homogeneous version of our problem is
$$y(x) = A J_{1}(x) + BY_{1}(x)$$
where $J_{1}(x)$ and $Y_{1}(x)$ are Bessel Functions of the first and second kind of order one respectively, and $A,B$ are constants. We wish to construct functionf $y_{1}(x)$ and $y_{2}(x)$ so that $y_{1}(x)$ satisfies the left boundary condition and $y_{2}(x)$ satisfies the right boundary condition.
To deal with the singularity, it will be impossible to find a solution to the homogeneous version as it is. In order to fix this, we will require that $y_{1}, y'_{1}$ to be bounded as $x \rightarrow 0$. Thinking about this for a second, we see that $$y_{1}(x) = J_{1}(x)$$ works. This is because $Y_{1}(x)$ becomes unbounded as $x \rightarrow 0$.
Now, to find $y_{2}(x)$, this is pretty standard and can be found as follows. $y_{2}(x)$ must satisfy $y_{2}(b) = \lambda$ and $b y'_{2}(b)=\mu$ for $(\lambda,\mu) \neq 0$ but $(1)\lambda + (0) \mu =0$, where the coefficients in the last expression come from the coefficients of the right boundary condition, $\beta_1 y(1) + \beta_2 y'(1) = 0 $. Here, choose $\lambda = 0$ and $\mu = 1$, and solve the resulting system:
$$\begin{cases} A J_{1}(b) = B Y_{1}(b) = 0 \\ bA J'_{1}(b) + b B Y'_{1}(b) = 1 \end{cases} $$
This results in $$y_{2}(x) = \frac{2 J_1(b) Y_1(x)-2 Y_1(b) J_1(x)}{b ((J_2(b)-J_0(b)) Y_1(b)+J_1(b) (Y_0(b)-Y_2(b)))}$$
Finally, it must be that $c = x W(y_{1}(x),y_{2}(x))$ is constant, where $W$ is the Wronskian.
We see that $c = J_{1}(b)$.
Thus, our green's function to the problem is
$$G(x,\xi)= \frac{1}{c} \begin{cases} y_{1}(\xi) y_{2}(x) & 0 \leq \xi \leq x \leq b \\ y_{1}(x) y_{2}(\xi) & 0 \leq x \leq \xi \leq b \end{cases} $$
Where $y_{1}(x)$ and $y_{2}(x)$ are defined above. Hence the solution to the problem is given by
$$y(x) = \int_{0}^{b} G(x,\xi) d \xi $$