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One can show that \begin{align} G(x;\xi)= \begin{cases} & \bigg(\frac{\xi^4-16}{60\xi^4}\bigg)\bigg(x^3-\frac{1}{x}\bigg),\qquad 1\le x<\xi\\ & \bigg(\frac{\xi^4-1}{60\xi^4}\bigg)\bigg(x^3-\frac{16}{x}\bigg),\qquad \xi< x\le 2 \end{cases} \end{align} is the Green's function for the operator $$L[y]=x^2y''-xy'-3y$$ with boundary conditions $y(1)=y(2)=0$. Hence, one wants to find the solution $y(x)$ to the differential equation $L[y]=x-3$ subject to the above conditions.

The Green's function can be written as $$G(x;\xi)=H(\xi-x)\bigg(\frac{\xi^4-16}{60\xi^4}\bigg)\bigg(x^3-\frac{1}{x}\bigg)+H(x-\xi)\bigg(\frac{\xi^4-1}{60\xi^4}\bigg)\bigg(x^3-\frac{16}{x}\bigg)$$ and the solution of the boundary value problem can be written as $$y(x)=\int_{1}^{2}G(x;\xi)f(\xi)d\xi=\int_{1}^{2}G(x;\xi)(\xi-3)d\xi.$$

But I struggle to evaluate correctly the integral since I do not understand what the functions $H(x-\xi)$ and $H(\xi-x)$ exactly represent and how to write them down precisely.

I would appreciate any help or suggestion. Thank you.

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  • $\begingroup$ From the notes this example comes from, $H(x)$ is defined so that for any "good" function $F(x):$ one has $\int_{-\infty}^{\infty}H(x)F(x)dx=\int_{0}^{\infty}F(x)dx$ $\endgroup$
    – johnny09
    Dec 21 '15 at 21:08
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If this is indeed your Green's function, then the solution is

$$y(x) = \left (x^3-\frac{16}{x} \right ) \int_1^x d\xi \, \left (\frac{\xi^4-1}{60 \xi^4} \right ) (\xi-3) + \left (x^3-\frac{1}{x} \right ) \int_x^2 d\xi \, \left (\frac{\xi^4-16}{60 \xi^4} \right ) (\xi-3)$$

As is usually the case, the evaluation of these integrals is more messy than difficult. The solution I get is

$$y(x) = 1-\frac{11}{15 x} - \frac{x}{4} - \frac{x^3}{60} $$

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  • $\begingroup$ May I ask why you put $\frac{\xi^4-1}{60 \xi^4}$ in the first integral? Is this because of $H(\xi-x)$? $\endgroup$
    – johnny09
    Dec 21 '15 at 21:17
  • $\begingroup$ @johnny09: exactly. I was hoping you would see the pattern once it was written out for you explicitly. $\endgroup$
    – Ron Gordon
    Dec 21 '15 at 21:18
  • $\begingroup$ Oh right. Is there a particular reason on why we do this? $\endgroup$
    – johnny09
    Dec 21 '15 at 21:20
  • $\begingroup$ @johnny09: It is nice after all that work to get a nice, single formula for $G(x;x_i)$. The Heaviside step function allows us to do that. And it is really easy to see how to use $G$ to compute the solution. $\endgroup$
    – Ron Gordon
    Dec 21 '15 at 21:22
  • $\begingroup$ Sorry for asking too many questions, but what is the operation or formula of the Heaviside step function in this particular example? $\endgroup$
    – johnny09
    Dec 21 '15 at 22:17

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