Consider the solution $b(u) \in L^2(0,T;H^1)\cap H^1(0,T;H^{-1})$ with $u \in L^2(0,T;H^1)$ to $$\frac{\partial}{\partial t}b(u) - \Delta u = f$$ where $b$ is continuous, increasing and locally Lipschitz and $f \in L^2(0,T;L^2)$.
How to obtain a comparison principle for this equations of this form? So I want to show that if $f_1 \geq f_2$ and $b(u_{01}) \geq b(u_{02})$ that $b(u_1) \geq b(u_2)$ a.e.
When the solution is smoother, the standard technique is: actually the solution $b(u)$ is constructed from a series of approximate problems where we consider the equation with $b_m$ replacing $b$ where $b_m$ is a smooth approximation of $b$. These approximations have solutions $b(u^m) \in H^1(0,T;L^2) \cap L^2(0,T;H^1)$ and we multiply the equation (which now holds pointwise) by $s_\epsilon(u^m_1 - u^m_2)$ where $s_\epsilon$ is an approximation of the function $\text{sign}^+(x) = \chi_{(0,\infty)}(x)$. Then we can pass to the limit in $\epsilon$ (only require convergence in $L^2(0,T;L^2)$) and in $m$.
But in my particular case I only have $\partial_t b(u_m) \in L^2(0,T;H^{-1})$ so the equation does not hold pointwise and so we need to take duality pairings instead of multiplying with the $s_\epsilon(\cdot)$. And we cannot pass to the limit in $\epsilon$ because we would need a convergence of the form $s_\epsilon(u^m_1-u^m_2) \to \text{sign}^+(u^m_1-u^m_2)$ in $L^2(0,T;H^1)$ which doesn't hold. So how do you get around this?
