I'm studying the following problem:
Let $X$ be a reflexive and separable Banach space, $H$ a separable Hilbert space (both real) such that $X\subseteq H$ is dense and continuous; $A: X\to X'$ not necessarily linear and $B\in L(H,H')$. Suppose that $B$ is selfadjoint and monotone and $A$ is monotone, hemicontinuous and bounded by $\|Av\|_{X'}\leq K(\|v\|_X + 1)$, for all $v\in X$. Besides, there exists $\lambda,c>0$ such that $$\langle Av;v\rangle + \lambda\langle Bv,v\rangle \geq c\|v\|_X^2 \quad,\quad v\in X$$
Finally, let $u_0\in H$ and $f\in L^2(0,T;X')$ be given. My objective is determine the well-posedness of the following problem:
Find $u\in L^2(0,T;X)$ (where $L^2(0,T;X)=\{u:[0,T]\to X] : \int_0^T\|u(t)\|_X^2dt<\infty\}$) such that $$\frac{d}{dt}\langle Bu(t);v \rangle + \langle Au(t);v \rangle = \langle f(t);v \rangle \quad,\quad v\in X$$ $$\langle Bu(0);v \rangle = \langle Bu_0;v \rangle \quad,\quad v\in Y$$
The derivative $\frac{d}{dt}$ is taken in the distributional sense. Given the hypothesis above, there's a theorem that assures that there exists a solution $u$ to the problem, and I managed to prove the continuous dependence of the initial conditions. I'm having a hard time trying to prove uniqueness and continuous dependence of initial data.
Before I keep going with the problem, here are some results that I have:
A solution of the given problem also satisfies that for all $v\in L^2(0,T;X)$ with $v'\in L^2(0,T;X)$ and $v(T)=0$ $$-\int_0^T\langle Bu(t);v'(t)\rangle dt + \int_0^T\langle Au(t);v(t)\rangle dt = \int_0^T\langle f(t);v(t)\rangle dt + \langle Bu_0;v(0)\rangle$$
There's a derivative formula that states that for functions $u,v\in L^2(0,T;X)$ with $u',v'\in L^2(0,T;X)$: $$\frac{d}{dt}\langle Bu(t),v(t)\rangle = \langle Bu'(t),v(t)\rangle + \langle Bu(t),v'(t)\rangle$$
Now then, for uniqueness I'm trying to do the usual, suppose there are two solution $u,v\in L^2(0,T;X)$ and define $z(t)=u(t)-v(t)$ in order to obtain something like $$\frac{d}{dt}\langle Bz(t);v \rangle + \langle Az(t);v \rangle = 0 \quad,\quad v\in X$$ $$\langle Bz(0);v \rangle = 0 \quad,\quad v\in Y$$
and then use the integral formula that I mention above and work with that. But I won't obtain something like that because $A$ is nonlinear. I've been thinking on this several days and I don't know what to do nor I can't find a reference that can help me with this. I'll appreciate any help you can give me.