Given a heat equation of $u$ with spatial dependent diffusion coefficient $\alpha(x)\ge 0$ $$\frac{\partial u_\alpha}{\partial t}=\alpha(x) \frac{\partial^2 u_\alpha}{\partial x^2}$$ where $(t,x)\in [0,T]\times[0,1]$, $u_\alpha(0,x)$ is a positive concave function, and $u_\alpha(t,0)=u_\alpha(t,1)=0$. If $\alpha(x)<\beta(x), \forall x\in [0,1]$, is it true that $u_\alpha(t,x)>u_\beta(t,x)$, $\forall t>0, x\in (0,1)$?
I now realize the answer to the last question is in the negative. Consider $\alpha$ and $\beta$ are interspersed with intervals with zero value.
Yes, it is true that $u_\alpha > u_\beta$ for $t > 0, 0 < x < 1$.
Consider first the problem $$ \partial_t u = \gamma(x) \partial^2_x u, \, 0 < x < 1, \, u(0) = u(1) = 1 \, . $$ with smooth $\gamma$ that is positive on $[0,1]$.
Lemma. If $u(0,\cdot)$ is concave, then $\partial^2_x u(t, \cdot) \le 0$ for all $t > 0$.
Proof. Without loss of generality, $u(0,\cdot)$ is smooth. Set $w = \partial^2_x u$. Then $w(0,\cdot) \le 0$ and $w$ satisfies $$ \partial_t w = \gamma(x)\partial^2_x w + 2\gamma'(x) \partial_x w + \gamma''(x)w $$ Also, for $x \in \{0,1\}$, we know that $\gamma(x) w(t,x) = \gamma(x) \partial_t u(t,x) = \gamma(x) \cdot 0 = 0$ and thus $w(t,x) = 0$ for $x \in \{0,1\}$. By the usual comparison principle for solutions of parabolic equations, $u_{xx}(t,x) = w(\cdot, x) \le 0$ for $t > 0, \, 0 < x < 1$.
Now consider the two problem $$ \partial_t u_\alpha = \alpha(x) \partial^2_x u_\alpha, \quad \partial_t u_\beta = \beta(x) \partial^2_x u_\beta $$ with $\alpha < \beta$ on $[0,1]$. Also assume that $\alpha, \beta$ are both continuous. Find a smooth $\gamma$ such that $\alpha < \gamma < \beta$. Let $u_\gamma$ the solution of the heat equation with this coefficient function and set $v = u_\alpha - u_\gamma$, with the same initial and boundary data. Then $v$ satisfies $$ \partial_t v = \alpha(x) \partial^2_x v + (\alpha(x) - \gamma(x)) \partial^2_x u_\gamma $$ with data $v(0,x) = 0, \, v(t,0) = v(t,1) = 0$. Since $\partial^2_x u_\gamma \le 0$ and $\alpha - \gamma > 0$, it follows that $v > 0$ for $t > 0, \, 0 < x < 1$. Thus $u_\alpha > u_\gamma$ in this set. By the same argument also $u_\gamma > u_\beta$ in this set.
Therefore $u_\alpha > u_\beta$ for $t > 0, 0 < x < 1$.