I have the following PDE: Let $G = (a, b)$ and $J = (0, 1)$. Consider the Dirichlet problem with general coefficient functions $\alpha(x)$, $\beta(x)$, $\gamma(x)$ \begin{equation} \begin{cases} \partial_t u - \partial_x (\alpha(x) \partial_x u) + \beta(x) \partial_x u + \gamma(x) u = f(t, x), \text{ in $J \times G$}\\ u = 0, \text{ on $J \times \partial G$}\\ u|_{t=0} = u_0, \text{ in $G$} \end{cases} \end{equation} where $u_0 \in L^2(G)$, $f(t,x) \in L^2(J,H^{-1}(G))$, $\alpha,\gamma \in C(G)$, and $\beta \in C^1(G)$ such that with some $\bar \alpha > 0$ the bound $\alpha(x) > \bar \alpha$ holds for all $x \in G$.
Additionally assume that $\alpha(x) \in C^1(G)$, $\gamma(x) > 0$ for all $x \in G$, and that $f(t,x) \equiv 0$. Assume that we have a strong solution $u \in C^2(\overline{J \times G})$. I am supposed to show that $$ u(t,x) \leq \max\{0, \max_{x \in \overline G} u_0(x)\} $$ for any $(t,x) \in J \times G$.
My attempt: Staring from $\partial_t u = \partial_x (\alpha(x) \partial_x u) - \beta(x) \partial_x u - \gamma(x) u$, I can integrate the LHS w.r.t. $t$ which yields $$ u(x,t) - u_0(x) = \int_0^t \partial_x (\alpha(x) \partial_x u) - \beta(x) \partial_x u - \gamma(x) u \;ds $$ Rearranging and using that $\gamma > 0$, I have $$ u(x,t) \leq u_0(x) + \int_0^t \partial_x (\alpha(x) \partial_x u) - \beta(x) \partial_x u\;ds. $$ However, I do not know how to continue from here. My other attempt would be to isolate $u$ from $\partial_t u = \partial_x (\alpha(x) \partial_x u) - \beta(x) \partial_x u - \gamma(x) u$ where I can use that $\gamma > 0$. This would yield $$ u = \frac1\gamma \left( -\partial_t u + \partial_x (\alpha(x) \partial_x u) - \beta(x) \partial_x u\right). $$ However I also do not know how this helps me, since I do not know how I could leverage that the solution satisfies the PDE.
Question: What would be the right way to tackle this problem. I think I am just missing the starting point to make some progress. I would be glad if someone could steer me into the right direction. (Or maybe this should be done using a more general theorem?)
Thanks in advance!
Edit: Thanks to the comment I try to multiply the PDE with $u$ and integrate over $x$. This yields (using partial integration) $$ \int u\partial_t u + \beta(x) u \partial_x u + \gamma(x) u^2dx = \int u\partial_x (\alpha(x) \partial_x u) dx \\ =\alpha(y)u \partial_x u \Big|_0^x - \int \alpha(x) (\partial_x u)^2dx = \alpha(x)u(t,x) \partial_x u(t,x) - \int \alpha(x) (\partial_x u)^2dx $$ Estimating $\gamma$ and $\alpha$ by zero yields (I could only estimate $\alpha(x)$ by the constant $\bar \alpha$ but I do not know how this would help) $$ \int u\partial_t u + \beta(x) u \partial_x u dx \leq \alpha(x)u(t,x) \partial_x u(t,x) $$ However, I do not know if this helps me?
What you need here is the maximum principle for parabolic equations. I suggest you look for the proof for the maximum principle for the heat equation and then see if you are able to apply the proof to your problem.
The general idea is to prove the maximum of the solution on the compact set $J\times G$ must be attained on the parabolic boundary $\{t=0\}\cup\{x=a,b\}$. To show this is true think about what you know about the derivatives of a function at an inner point (or a boundary point at $t=1$) of maximal value, and use the equation to get a contradiction.
Notice that the proof can work for $\gamma\geq 0$ (rather than strictly positive) as well, but it will be a bit more subtle.