# How can a single Green's function be used in both the inhomogeneous and homogeneous versions of a differential equation?

Consider the differential equation $$\frac{d^2}{dx^2}=f(x)$$ with boundary conditions $y(0)=0$ and $y'(1)=0$. If we then let $f(x)=0$ we have a homogeneous differential equation. Then we know that the Green's function will be $$G(x,x')=Ax+B, x<x'$$ $$G(x,x')=Cx+D, x>x'.$$ Using the given boundary conditions, the boundary condition that $G(x,x')$ must be continuous at $x=x'$ and the boundary condition that the derivative with respect to $x$ of $G(x,x')$ must be discontinuous at $x=x'$ then it can be shown that $b=0$, $c=0$, $a=-1$ and $d=-x'$. Then Green's function is $$G(x,x')=-x, x<x'$$ $$G(x,x')=-x',x>x'.$$ Choose now $f(x)=x$ with the same boundary conditions so that our differential equation is inhomogeneous. Using $$y(x)=\int_0^1G(x,x')f(x')dx'$$ it can be shown that the solution to the inhomogeneous differential equation is $$y(x)=\frac{x^3}{6}-\frac{x}{2},$$ which is clearly the correct solution. The Green's function found from the homogeneous differential equation can be used to find the solution to any inhomogeneous equation of this same form, i.e. $f(x)\ne0$.

How can this work? I don't understand how this single Green's function found from a homogeneous differential equation can be used to solve any inhomogeneous differential equation of the same form as the original homogeneous one.

• Green's function is not found by solving homogeneous equation. Instead, it's the general solution. In physics, you can think of it as solving for continuum basis(deltas). – Qiyu Wen Feb 24 '16 at 19:47
• Yes, that is generally how we think of it in physics. I am a physics student and our interpretation is that the Green's function provides a response to an impulse (i.e. the delta function) at x=x', and thus integrating gives the response across impulses at all x=x'. – ODP Feb 24 '16 at 21:17

When finding the Green's Function for the operator $\mathscr{L}=\frac{d^2}{dx^2}$, with the prescribe boundary conditions $y(0)=y'(1)=0$, we are not finding the solution to

$$\mathscr{L}\{y(x)\}=0 \tag 1$$

with those boundary conditions. The solution to $(1)$ with those boundary conditions is simply $y(x)=x$. And $G(x,x')\ne x$.

Rather, if we extend the analysis to include Generalized Functions, or Distributions, then the Green's Function $G(x,x')$ is a solution to

$$\mathscr{L}\{G(x,x')\}=\delta(x-x') \tag 2$$

with $G(0,x')=G_1(1,x')=0$, where $\delta(x)$ denotes the Dirac Delta, which I discussed in THIS ANSWER and THIS ONE.

Then, $(2)$ can be written as

$$\mathscr{L}\{G(x,x')\}=0,\,\,\text{for}\,\,x\ne x'$$

with the continuity condition and discontinuity of the first derivative condition provided by

\begin{align} \lim_{\epsilon \to 0^+}\left.\left(G(x,x')\right)\right|_{x=x'-\epsilon}^{x=x'+\epsilon}&=0\\\\ \lim_{\epsilon \to 0^+}\left.\left(\frac{\partial G(x,x')}{\partial x}\right)\right|_{x=x'-\epsilon}^{x=x'+\epsilon}&=1 \end{align}

As already established in the OP, we find $G(x,x')=-\min(x,x')$. Note, that while classically $\frac{d^2}{dx^2}G=0$ almost everywhere, in the context of distributions, $\frac{d^2}{dx^2}G=\delta(x-x')$.

To see how this reconciles classically, note that the solution to

$$\frac{d^2y(x)}{dx^2}=f(x)$$

with $y(0)=y'(1)=0$ is given by

$$y(x)=\int_0^1 G(x,x')f(x')\,dx'\tag 3$$

Taking the first derivative of $(3)$ yields

\begin{align} y'(x)&=\frac{d}{dx}\left(\int_0^x (-x')f(x')\,dx'+\int_x^1 (-x)f(x')\,dx'\right)\\\\ &=-xf(x)+xf(x)-\int_x^1f(x')\,dx'\\\\ &=-\int_x^1 f(x')\,dx' \end{align}

Proceeding to take a second derivative reveals

$$y''(x)=f(x)$$

as expected. The reason why we cannot simply interchange the order of the integral with the derivatives is that the Green's Function does not have a continuous partial derivative. And it is precisely that discontinuity that gives rise to Dirac Delta contribution.