I've been studying Green's functions and basically I've found out the book and the teacher doing some strange manipulations in integrals. Basically it has been shown the following: if we have the following differential equation for $x\in [a,b]$


With some boundary conditions then we can find it's solution by substituting $f(x)=\delta(x-\xi)$ which will yield a solution $G(x,\xi)$ of the following form:

$$G(x,\xi)=\begin{cases}c_1y_1(x)+c_2y_2(x), & x<\xi \\ d_1y_1(x) + d_2y_2(x), & x>\xi\end{cases}$$

Where both $y_1$ and $y_2$ are linearly independent solutions of the homegenous equation. Then with suitable conditions on $G$ we can find $c_i$ and $d_i$. That's fine, I understand perfectly well this derivation, why $G$ works and how to find it.

My doubt is that then it is said that for a general $f$ the solution $y$ is:

$$y(x)=\int_a^b f(\xi) G(x,\xi) d\xi$$

But this integral turns to be complicated. The reason is that $G$ is different for $x<\xi$ and for $x>\xi$, but $\xi$ varies over all $[a,b]$ so I don't know how to do this. I mean, if I suppose $x < \xi$, since $\xi$ runs all over $[a,b]$ it will pick places where $x > \xi$.

The teacher then turned this integral into

$$y(x) = \int_a^\xi f(\xi)G(x,\xi)d\xi + \int_\xi^b f(\xi) G(x,\xi)d\xi$$

But this seems nonsense. The variable $\xi$ is the free variable that runs from $a$ to $b$ it doesn't make sense to turn it into a bound for the integral. Also, there's another point: I tried to work formally with this, but then the solution $y$ depends on $\xi$.

So, how this kind of integral is manipulated correctly?

  • $\begingroup$ I think your concern is valid. Are you sure your teacher didn't mean to split the integral as $\int_a^x + \int_x^b$? You can then divide it into the two cases you wanted to work with... $\endgroup$ – Tom May 30 '14 at 18:37
  • 2
    $\begingroup$ It's probably $$y(x) = \int_a^x f(\xi)G(x,\xi)\,d\xi + \int_x^b f(\xi)G(x,\xi)\,d\xi.$$ Then you can in each of the two integrals insert the formula for $G$, and compute. $\endgroup$ – Daniel Fischer May 30 '14 at 18:37
  • $\begingroup$ Yes, you're right @DanielFischer, I didn't think about that. In this way everything works fine. Thanks for the answer. $\endgroup$ – user1620696 May 30 '14 at 18:58

As pointed out by commenters, the intended formula was $$y(x) = \int_a^x f(\xi)G(x,\xi)d\xi + \int_x^b f(\xi) G(x,\xi)d\xi$$

As an aside, I find it more convenient to write $G$ using Heaviside function:

$$G(x,\xi)= c_1y_1(x)+c_2y_2(x) +( b_1y_1(x) + b_2y_2(x))H ( x-\xi)$$ The advantage is that this decouples the problem of finding the coefficients. Namely, $b_1,b_2$ are determined by the desired singularity at $x=\xi$. Once they are found, one can solve for $c_1,c_2$ from the boundary conditions.


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