How to prove a certain maximum principle Suppose $b\in C^4(\bar\Omega)$ where $\bar\Omega=[0,1]$ and that $b(x)\geq \beta>0$. How do I prove the following maximum principle: Assume $\phi(0)\geq0$ and $\phi(1)\geq0$. Then $$-\epsilon\phi''(x)+b(x)\phi'(x)\geq0,\quad 0<\epsilon<<1$$ for all $x\in \Omega$ implies $\phi(x)\geq0$ for all $x\in \bar\Omega$.
 A: We can take $\epsilon=1$ without loss of generality.
Suppose first 
$$
-\phi''(x)+b(x)\phi'(x)>0
$$
Then if $\phi<0$ somewhere in $\Omega$, there exists a minimun $x_0\in \Omega$. Therefore 
$$
\phi'(x_0)=0 \mbox{ and } \phi''(x_0)\ge0
$$
which is a contradiction.
For the general case we will use barriers and the previous case:
Suppose $\phi<0$ somewhere in $\Omega$ and define $\lambda= -\min (\phi) >0$.
We need a function $\alpha$ such that:
$$
\alpha(0)=\alpha(1)=0 \; \mbox{ (actually }\alpha\ge 0 \mbox{ is suficient) }\\
\alpha(x) \le \lambda/2, \;\; \\ 
\alpha''(x)-b(x)\alpha'(x)<0 \;\; \forall x\in \Omega
$$
Suppose we have such a function. Then $v=\phi + \alpha$ satisfies
$$
v\le \phi +\lambda/2 <0 \mbox{ somewhere }
$$
For the other side
$$
v(0)=\phi(0)+\alpha(0)\ge 0 \\
v(1)=\phi(1)+\alpha(1)\ge 0 \\
$$
and 
$$
-v''+b(x) v' =-\phi''+b(x)\phi -\alpha'' +b(x)\alpha'>0 
$$
But then, by the first case, we have that $v\ge 0$. A contradiction. 
Construction of $\alpha$:
$$
\alpha(x)= \delta \left( e^{M}- e^{M x} \right)
$$
with $M$ large enough and $\delta=\delta(M)$ small enough.
