# Green function jump conditions for second order differential equation

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)?

• 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? Mar 17 at 23:13

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 $$x = y$$.