No, this is not true. There is no backward uniqueness in general.

What you need is the theory of operator semigroups, and here is a simple example.

Consider the operator $Af=f'$ in the space $X=L^2[0,1]$ with domain $$D(A)=\{f\in H^1[0,1]\,,\, f(1)=0\}.$$

Then the solution to the initial value problem is given by $x(t)=T(t)x_0$ with $$T(t)f(s):=\begin{cases} f(t+s), t+s\leq 1,\\ 0, \quad t+s>1.\end{cases}$$

Clearly, $T(t)f=0$ for all $t>1$, but the initial value problem is well-posed in the classical sense ($A$ closed, classical solutions from a dense set of initial values, continuous dependence on initial values).

No, this is not true. There is no backward uniqueness in general.

What you need is the theory of operator semigroups, and here is a simple example.

Consider the operator $Af=f'$ in the space $X=L^2[0,1]$ with domain $$D(A)=\{f\in H^1[0,1]\,,\, f(1)=0\}.$$

Then the solution to the initial value problem is given by $x(t)=T(t)x_0$ with $$T(t)f(s):=\begin{cases} f(t+s), t+s\leq 1,\\ 0, \quad t+s>1.\end{cases}$$

Clearly, $T(t)f=0$ for all $t>1$, but the initial value problem is well-posed in the classical sense ($A$ closed, classical solutions from a dense set of initial values, continuous dependence on initial values).

ADDED: After clarifications on the question, let me add the following: If $A$ is sectorial and hence generates an analytic semigroup, then finite time extinction of soultions cannot happen. Other conditions are difficult to formulate. If you say more on your operator, it can help...

No, this is not true. There is no backward uniqueness in general.

What you need is the theory of operator semigroups, and here is a simple example.

Consider the operator $Af=f'$ in the space $X=L^2[0,1]$ with domain $$D(A)=\{f\in H^1[0,1]\,,\, f(1)=0\}.$$

Then teh solution to the initial value problem is given by $x(t)=T(t)x_0$ with $$T(t)f(s):=\begin{cases} f(t+s), t+s\leq 1,\\ 0, \quad t+s>1.\end{cases}$$

Clearly, $T(t)f=0$ for all $t>1$, but the initial value problem is well-posed in the classical sense ($A$ closed, classical solutions from a dense set of initial values, continuos dependence on initial values).

No, this is not true. There is no backward uniqueness in general.

What you need is the theory of operator semigroups, and here is a simple example.

Consider the operator $Af=f'$ in the space $X=L^2[0,1]$ with domain $$D(A)=\{f\in H^1[0,1]\,,\, f(1)=0\}.$$

Then the solution to the initial value problem is given by $x(t)=T(t)x_0$ with $$T(t)f(s):=\begin{cases} f(t+s), t+s\leq 1,\\ 0, \quad t+s>1.\end{cases}$$

Clearly, $T(t)f=0$ for all $t>1$, but the initial value problem is well-posed in the classical sense ($A$ closed, classical solutions from a dense set of initial values, continuous dependence on initial values).