Brezis' exercise 8.17: the domain of $A^*$ where $A u=u''-xu'$

Question

Let $I$ be the open interval $(0, 1)$. I'm trying to solve below exercise in Brezis' Functional Analysis, i.e.,

Exercise 8.17 Let $H=L^2(I)$ and $A: D(A) \subset H \rightarrow H$ be the unbounded operator defined by $A u=u''-xu'$, whose domain $D(A)$ will be made precise below. Determine $A^{*}, D\left(A^{*}\right), N(A)$, and $N\left(A^{*}\right)$ in the following cases:

1. $D(A)=\left\{u \in H^2 (I) : u(0)=u(1)=0\right\}$.
2. $D(A)=H^2(I)$.
3. $D(A)=\left\{u \in H^2(I) : u(0)=u(1)=u'(0)=u'(1)=0\right\}$.
4. $D(A)=\left\{u \in H^2(I) : u(0)=u(1)\right\}$.

There are possibly subtle mistakes that I could not recognize in my below attempt. Could you please have a check on my computation of $D(A^*)$?

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Let $A^* : D(A^*) \subset H \to H$ be the adjoint of $A$. We have the duality $$ \int_I v (Au) = \int_I (A^*v)u, \quad v \in D(A^*), u \in D(A), $$ which is equivalent to $$ \int_I v (u''-xu') = \int_I (A^*v)u, \quad v \in D(A^*), u \in D(A). \qquad (*) $$

We have $(*)$ implies $D(A^*) \subset H^2 (I)$. By integration by parts, $(*)$ is equivalent to $$ \begin{align*} \int_I (A^*v)u &= \int_I (v''+xv'+v)u \\ &\quad + (vu'-v'u) (1) - (vu'-v'u) (0) + (vu) (1), \\ &\quad \forall v \in D(A^*), u \in D(A). \end{align*} \qquad (**) $$

It follows from $(**)$ and $C^\infty_c (I) \subset D(A)$ that $A^*v = v''+xv'+v$ and $$ D(A^*) = \{ v \in H^2 (I) : (vu'-v'u) (1) - (vu'-v'u) (0) + (vu) (1) = 0 \quad \forall u \in D(A) \}. \quad (***) $$

1. We have $(***)$ implies $$ \begin{align*} D(A^*) &= \{ v \in H^2 (I) : (vu') (1) - (vu') (0) = 0 \quad \forall u \in D(A) \} \\ &= \{ v \in H^2 (I) : v(1) = v(0) = 0 \}. \end{align*} $$

2. We have $(***)$ implies $$ \begin{align*} D(A^*) &= \{ v \in H^2 (I) : (vu'-v'u) (1) - (vu'-v'u) (0) + (vu) (1) = 0 \quad \forall u \in H^2 (I) \} \\ &= \{ v \in H^2 (I) : v(1) = v(0) = v'(1) = v'(0) = 0 \}. \end{align*} $$

3. Clearly, $D(A^*) = H^2 (I)$.

4. We have $(***)$ implies $$ \begin{align*} D(A^*) &= \{ v \in H^2 (I) : (vu')(1) - (vu')(0) -[v'(1) - v'(0)-v(1)]u(0) = 0 \quad \forall u \in D(A) \} \\ &= \{ v \in H^2 (I) : v(1) = v(0) = v'(1) - v'(0) = 0 \}. \end{align*} $$