Green function of first order differential equation

Question

I want to find the Green's function (kernel) of the following problem \begin{equation} \begin{cases} (a(x)u')'=f, \text{ in } (-1, 1)\\[4pt] u(1)=u(-1)=0 \end{cases} \end{equation} where, $a(x)\in L^{\infty}(-1, 1)$ and $f\in L^{\infty}(-1, 1)$. I take the homogenuous part with $f=0$ and develop the problem as, setting $g(x, s)$ be the Green function, \begin{equation} g''(x, s)+\dfrac{a'(x)}{a(x)}g'(x, s)=0, \end{equation} Solving this problem, I obtain $$ g'(x, s)=\dfrac{c_1}{a(x)} $$ and consequently $$ g(x, s)=\int \dfrac{c_1}{a(x)}\,dx +c_2. $$ Is this correct? and if this is correct I get $c_1=c_2=0$.

Answer 1

It's not surprising that you get $0$ when solving a homogeneous ODE with homogeneous boundary conditions.

Green's function $g(x,s)$ for a differential operator $L$ satisfies the equation $Lg(x,s) = \delta_s$, where $\delta_s$ is the Dirac delta at $x=s$. In order for $(au')'$ to be $\delta_s$, you need $au'$ to have a jump discontnuity of size $1$ at $x=s$. This means $u'$ should have jump discontinuity of size $1/a(s)$. So, instead of $$g'(x, s)=\dfrac{c_1}{a(x)} $$ you need $$g'(x, s)=\dfrac{c_1 + H(x-s)}{a(x)}$$ where $H$ is the Heaviside function. This gives $$g(x, s)=\int_{-1}^x \dfrac{c_1 + H(t-s)}{a(t)}\,dt$$ where the constant $c_1$ is determined from the condition $g(1,s)=0$: that is, $$c_1\int_{-1}^1 \frac{1}{a(t)}\,dt + \int_s^1 \frac{1}{a(t)}\,dt = 0$$