I been trying to find the Green's function for a particular problem. For the equation $$ q(x) \frac{\mathrm{d}^2u(x)}{\mathrm{d}x^2} + p(x) u(x) =0 \tag{1} $$ where $q(x)$ and $p(x)$ are some functions of $x$. I have solved these and have two linearly independent results. $$ u(x) = c_1 u_1(x) + c_2 u_2(x) $$ with $c_1, c_2$ being arbitrary constants and both $u_1, u_2$ solve equation (1). Now for the boundary conditions. At $x\to +\infty$ we require that the solution goes to zero, only $u_2$ satisfies, another condition is at $x\to1_+$. Only $u_1$ satisfies this. Thus our Green's function can be written as $$ G(x,y) = c_1 \mathcal{H}(y-x) u_1(x) + c_2 \mathcal{H}(x-y) u_2(x). $$ Where $\mathcal{H}$ is the Heaviside step function and $y>1$.

Now the Green's function must be continous at $x = y$, thus we have condition for (for example) $c_1$. But what is the other condition? Does the condition $$ \lim_{x\to y_-} \frac{\partial G(x,y) }{\partial x} -\lim_{x\to y_+} \frac{\partial G(x,y) }{\partial x} = 1 $$ hold even for our special ODE (1)?

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    $\begingroup$ Did you search first for the form $u_{xx}(x)+s(x)u(x) = 0$? since $p(x)$ and $q(x)$ are arbitrary using another arbitrary function $s(x) = p(x)/q(x)$ should be equivalent, right? $\endgroup$
    – Joako
    Commented Mar 17, 2023 at 23:13

1 Answer 1


If we require the Green's function to be continuous at the point $y = x$, we can just divide equation's (1) by $q(x)$ (if we assume that it's non zero around $x = y$), we then have the equation $$ \frac{\mathrm{d}^2 G(x,y)}{dx^2} + s(x) G(x,y) = \delta(x-y), $$ with $s(x) = p(x)/q(x)$ being continuous around $x = y$ we can integrate infinitesimally
$$ \int_{y-\epsilon}^{y+\epsilon}\left( \frac{\mathrm{d}^2 G(x,y)}{dx^2} + s(x) G(x,y)\right) dx = 1. $$ Limiting $\epsilon \to 0$, and utilising that the function $s(x) G(x,y)$ is continuous and also utilising the fundamental theorem of calculus we have $$ -\frac{\mathrm{d}G(x,y) }{dx}|_{x = y_-}+\frac{\mathrm{d}G(x,y) }{dx}|_{x = y_+} = 1. $$ Thus we have the ordinary just requirement. This also fixes the other constant.

Futhermore if really want to hold the equation (1) form, the condition changes to $$ -\frac{\mathrm{d}G(x,y) }{dx}|_{x = y_-}+\frac{\mathrm{d}G(x,y) }{dx}|_{x = y_+} = \frac{1}{q(y)}, $$ for $q(x)$ continous at $x = y$.


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