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I'm reading "A guide to mathematical methods for physicists" by Petrini, Pradisi and Zaffaroni, volume 2. At the beginning of chapter 6, the authors tell that

a solution $G$ of the equation:

$$ L_\mathbf{x}G(\mathbf{x}) = \delta(\mathbf{x}) $$

where $\delta(\mathbf{x}) = \prod_{i=1}^d \delta(x_i)$, is called fundamental solution for $L_\mathbf{x}$ in $\mathbb{R^n}$. [...] Fundamental solutions are important because they allow to determine a particular solution of the inhomogeneous equation

$$ L_\mathbf{x}f(\mathbf{x}) = h(\mathbf{x}) $$

as the convolution

$$ f(\mathbf{x}) = (G * h)(\mathbf{x}) = \int_{\mathbb{R}^n} G(\mathbf{x - y}) h(\mathbf{y}) d\mathbf{y} $$

However in Example 6.3, in which the fundamental solution of the harmonic oscillator is considered ($G''(x) + \omega^2 G(x) = \delta(x)$) it is told:

We find the general solution

$$ G(x) = \theta(x) \frac{\sin \omega x}{\omega} + \nu_1 e^{i \omega > x} + \nu_2 e^{-i \omega x} $$

The solution to the inhomogeneous equation

$$ f''(x) + \omega^2 f(x) = g(x) $$

is then given by

$$ f(x) = \int_{\mathbb{R}} G(x - y) g(y) dy = \int_{\mathbb{R}}\theta(x - y) \frac{\sin \omega (x - y)}{\omega} g(y) dy + C_1 e^{i\omega x} + C_2 e^{-i \omega x} $$

where $C_1 = \nu_1 \int_{\mathbb{R}}e^{-i\omega x} g(y) dy$ and $C_2 =\nu_2 \int_{\mathbb{R}}e^{i\omega x} g(y) dy$. The same result can be obtained with the method of variation of arbitrary constants.

I find it contradictory that $G(x)$ is said to be the function which convolved with the source term gives a particular solution and calling the solution general in the example. Moreover in the example is said that the $f(x)$ obtained is a particular solution, but it seems to be the sum of an integral and of a linear combination of solution of the homogeneous equation, shouldn't it be the general solution then?

Moreover, in some of my exams, for example the one whose solution I've pasted below, it is given a boundary value problem based on an ODE

  1. L'equazione omogenea associata al problema sotto esame è un'equazione di Euler, con soluzioni fondamentali date da $1/x$ ed $x$. Le due soluzioni dell'omogenea che rispettano le condizioni al contorno separatamente in $x = 1 e x = 2$ sono date da,

$f_1(x)=\frac{1}{x}-x,\quad f_2(x)=\frac{1}{x}-\frac{x}{4},$

tali che $f_1(1)=0$, $f_1(2)\neq0$, $f_2(1)\neq0$ e $f_2(2)=0$. Il Wronskiano corrispondente è dato da $W(x)=f_1(x)f_2'(x)-f_1'(x)f_2(x)=\frac{3}{2x}$. L'espressione per la funzione di Green in questi casi è,

$G(x,y)=\frac{1}{a_2(y)W(y)}\left[f_1(x)f_2(y)\theta(y-x)+f_1(y)f_2(x)\theta(x-y)\right]$

$=\frac{2}{3y}\left[\left(\frac{1}{x}-x\right)\left(\frac{1}{y}-\frac{y}{4}\right)\theta(y-x)+\left(\frac{1}{y}-y\right)\left(\frac{1}{x}-\frac{x}{4}\right)\theta(x-y)\right]$

$=\frac{2}{3}\left[\left(\frac{1}{x}-x\right)\left(\frac{1}{y^2}-\frac{1}{4}\right)\theta(y-x)+\left(\frac{1}{y^2}-1\right)\left(\frac{1}{x}-\frac{x}{4}\right)\theta(x-y)\right],$

La soluzione dell'equazione differenziale non-omogenea è, quindi,

$f(x)=\int_1^2dy\,G(x,y)y^2=\frac{2}{3}\left[\left(\frac{1}{x}-x\right)\int_x^2dy\left(1-\frac{y^2}{4}\right)+\left(\frac{1}{x}-\frac{x}{4}\right)\int_x^1dy\left(1-y^2\right)\right]$

$=\frac{2}{3}\left[\left(\frac{1}{x}-x\right)\left(\frac{4}{3}-x+\frac{x^3}{12}\right)+\left(\frac{1}{x}-\frac{x}{4}\right)\left(x-\frac{x^3}{3}-\frac{2}{3}\right)\right]=\frac{4}{9x}-\frac{7x}{9}+\frac{x^2}{3}.$

The Green's function is found using:

$$ G(x, \xi) = \frac{1}{W(\xi) a_2(\xi)} \left[ \theta(\xi - x) y_2(\xi) y_1(x) + \theta(x - \xi) y_1(\xi) y_2(x) \right] \tag{1}$$

now $f(x) = \int_{1}^2 G(x, y)y^2 dy$ is said to be the solution of the ODE without specifying if general or particular, however, since the exercise asks to solve the ODE, I assume it is the general solution of the ODE.

My question is: what do Green's functions let you find using the convolution: general or particular solutions? The Green's function found in (1), convolved, results in a general or in a particular solution?

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    $\begingroup$ Seems like a rather sloppy textbook. See also Advanced Green's function for the classical forced harmonic oscillator $\endgroup$ Commented Jul 8 at 10:59
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    $\begingroup$ It indeed is, in my opinion, at least throughout the functional analysis part $\endgroup$ Commented Jul 9 at 15:21
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    $\begingroup$ Can you translate the Italian part in English? $\endgroup$ Commented Aug 5 at 14:32
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    $\begingroup$ Just so you know a lot of Green's functions can start to be reinterpreted in the language of distributions and distributional derivatives. In other words, they can help you find more solutions to PDEs than you would initially be able to find with traditional forms of derivatives. Try not to overthink it too much. Get good at finding them. Practice taking weak or distributional derivatives outside of the context of PDEs and enjoy the new possibilities when solving PDEs. $\endgroup$ Commented Aug 5 at 15:00
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    $\begingroup$ And understand that not every measure theorist can precisely tell you what a 'distributional derivative' and a 'Green's function' is when you catch them walking down the alley. Even experts who can learn it if needed get tripped up now and then. Good luck $\endgroup$ Commented Aug 5 at 15:09

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Strictly speaking, fundamental solutions are particular solutions to the considered inhomogeneous problem only. Nevertheless, it has to be recalled that the said fundamental solutions are not unique, so that homogeneous solutions might be added freely.

Indeed, let $G$ be the (particular) Green function of an inhomogeneous problem $Lf = h$ and $g$ a solution to the associated homogeneous problem. Then, the quantity $G+g$ is still a valid Green function, because $L(g*h) = (Lg)*h = 0*h = 0$. Note that this last result implies that $g*h$ is another homogeneous solution $-$ that is why the homogeneous solution in Petrini's example reappears with new constants of integration.

Consequently, in the case where $g$ corresponds to the general homogeneous solution, then $G+g$ turns out to be the general solution to the original inhomogeneous problem, up to convolution with the source $h$. Some authors refer to $G$ alone, while others consider $G+g$ instead, as you noticed; it is thus a matter of conventions.

As a side note, let's underline that this "gauge freedom" sometimes permits to "design" the Green function in order to simplify future algebraic manipulations. For instance, in Petrini's example again, you might choose $\nu_1 = -\nu_2 = -\frac{1}{4i\omega}$, hence $G(x) = \frac{\sin\omega|x|}{2\omega}$.

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  • $\begingroup$ Thank you very much! It is much clearer now. Just a thing, could you please elaborate further on this Gauge freedom you are talking about? $\endgroup$ Commented Jul 7 at 23:32
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    $\begingroup$ @Luke__ Glad to help ;) I mentionned gauge freedom in reference to Maxwell's equations, where the electromagnetic potentials can be designed freely to a certained extent (but not at absolute will, however) by fixing a so-called gauge, in such a way that the electromagnetic fields stay the same at the end. Analogously, it is possible to choose a homogeneous solution and add it to the fundamental solution, so that the result is still a Green function. In Petrini's example, it allows us to work with a shorter/simpler and moreover even expression for the Green function. $\endgroup$ Commented Jul 8 at 6:42
  • $\begingroup$ Thank you again! $\endgroup$ Commented Jul 9 at 17:28
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The Green's function is defined with some set of "auxiliary conditions": boundary conditions, initial conditions, decay at infinity, something like that. In this situation, the problem for the Green's function only has one solution.

Given the Green's function, you can construct a particular solution to an inhomogeneous problem by convolution. This will satisfy whatever auxiliary conditions were used to define the Green's function in the first place. You can then add in a homogeneous solution in order to adjust to a different set of auxiliary conditions, or you can add in a general homogeneous solution in order to write down the general solution to the inhomogeneous problem.

I would not call the $G$ in the middle of your exposition "the Green's function". Technically it represents the family of all possible Green's functions, with different auxiliary conditions for each choice of the $\nu_i$.

("Auxiliary conditions" is not a standard term in mathematical English. Most people in my experience say "boundary conditions" when speaking generally, and pin down the relevant type of condition when speaking specifically.)

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