Consider $\left(E_0\right): L y:=-y^{\prime \prime}+q(x) y=0$ with non-negative function $q \in C(\mathbb{R})$. Let $y_1$ be the solution of $\left(E_0\right)$ satisfying $y_1(0)=0, y_1^{\prime}(0)=1$; and $y_2$ be the solution of $\left(E_0\right)$ satisfying $y_2(1)=0, y_2^{\prime}(1)=-1$. By ODE theory, we know that $y_1$ and $y_2$ are uniquely determined $C^2$ functions over $\mathbb{R}$.
- Let $H$ denote the standard Heaviside function, and $$ G(x, t)=-\frac{1}{W_0}\left[H(x-t) y_1(t) y_2(x)+H(t-x) y_1(x) y_2(t)\right], \quad \forall t, x \in \mathbb{R} . $$
- Prove that for any $t \in(0,1), G(0, t)=G(1, t)=0$.
- Prove that for any $x, t \in(0,1)$, there holds $G(x, t)>0$. I have searched on the internet that we are using variation of parameters to find the particular solutions, and the Wronskian is constant, but in this case in which q(x) is non-negative, I don't know how to prove that this Wronskian is negative, and also I don't know how to prove the above properties, I know if q(x) is nonnegative, then the eigenvalues are non-negative, but then I don't know how to proceed
Claim: $y_1$ can't have a zero on $(0,1)$ since $y_1(0)=0$. Suppose there is $t_0\in(0,1)$ and then $y_1$ reaches the max/min in $(0,t_0)$. If $y_1$ reaches the max at $t_1\in(0,t_0)$, then $y_1(t_1)>0$ and $y_1''(t_0)$ which are against each other. Since $y_1(0)=0,y_1'(0)=1>0$, one must have $y_1(1)>0$. Now $$ (W_0(y_1,y_2))'=(y_1y_2'-y_1'y_2)'=y_1y_2''-y_1''y_2=0 $$ which implies that $W_0(y_1,y_2)$ is constant and hence $$ W_0(y_1,y_2)=(y_1y_2'-y_1'y_2)(1)=-y_1(1)<0. $$