I am trying to show that the following pde is well-posed by computing an energy estimate for it by analyzing the time derivative of the $L^{2}$-norm.
$$ \begin{align} u_{t} &= \alpha u_{xx} + bu,& u(x,0) &= f(x),& u(0,t) &= 0,& u(1,t) + hu_{x}(1,t) &= 0, \end{align} $$
where $\alpha$, $b$ and $h$ are real constants and $\alpha, h > 0$.
My attempt:
$$ \begin{align} &\frac{1}{2}\frac{d}{dt}\vert\vert u(\cdot,t) \vert\vert^{2} = \int_{0}^{1}u(x,t)u_{t}(x,t)dx = \{\text{Use PDE}\} = \int_{0}^{1}u(\alpha u_{xx} + bu)dx =\\[3mm] &\alpha\int_{0}^{1}uu_{xx}dx + b\int_{0}^{1}u^{2}dx = \alpha\int_{0}^{1}uu_{xx}dx + b\vert\vert u(\cdot,t)\vert\vert^{2} = \dots \end{align} $$
Where do I go from here? Can I rewrite the product $uu_{xx}$ in some way?
Edit 1: I am trying to find an upper bound for $d/dt\;\vert\vert u(\cdot,t)\vert\vert^{2}$.
Edit 2: maybe partial integration of $\int uu_{xx}$? Can that help me somehow?
Edit 3: by integration by parts of the second term I obtain
$$ \begin{align} &\alpha\int_{0}^{1}uu_{xx}dx = \alpha[uu_{x}]^{1}_{0} -\alpha\int_{0}^{1}u_{x}u_{x}dx = \alpha[u(1,t)u_{x}(1,t) - u(0,t)u_{x}(0,t)] \\[3mm] &-\alpha\int_{0}^{1}u_{x}^{2}dx = \{\text{Use BC:s}\}= - \alpha hu_{x}^{2}(1,t) - \alpha\int_{0}^{1}u_{x}^{2}dx. \end{align} $$
I realize that if all terms are negative, then we have a decreasing time derivative which is what we wanted to show. But if $b>0$ then what can I do in order to establish an energy estimate?