I want to calculate the adjoint of the following operator $$ \boldsymbol{\mathcal{A}}\textbf{x} = \begin{bmatrix} 0 & -\dfrac{a_0^2}{A}\dfrac{d}{d\xi}\\ -A\dfrac{d}{d\xi} & 0 \end{bmatrix}\begin{bmatrix} x_1\\[0.5em] x_2 \end{bmatrix}, $$ with the domain
$$ D(\boldsymbol{\mathcal{A}}) = \bigg\lbrace \textbf{x} \in \mathcal{X} \Big\vert x_1, \ x_2 \ \text{abs. cont.}, \dfrac{d x_1}{d\xi}, \dfrac{d x_2}{d\xi} \in \textbf{L}_2(0,L) \\ \text{and} \ x_2(0) - c_{1,p}x_1(0) = 0, \ x_2(L) - c_{2,p}x_1(L)= 0 \bigg\rbrace, $$ where $\mathcal{X} = \textbf{L}_2(0,L) \oplus \textbf{L}_2(0,L)$. The following must hold $$ \left\langle\boldsymbol{\mathcal{S}}\textbf{x},\textbf{y} \right\rangle= \left\langle\textbf{x}, \boldsymbol{\mathcal{S}}^*\textbf{y}\right\rangle. $$ So if I substitute, I get $$ \left\langle\boldsymbol{\mathcal{A}}\textbf{x},\textbf{y} \right\rangle =\int_{0}^{L}(-\dfrac{a_0^2}{A}\dfrac{d}{d\xi}x_2 \overline{y_1}) d\xi+ \int_{0}^{L}(-A\dfrac{d}{d\xi}x_1 \overline{y_2})d\xi\\ = [-\dfrac{a_0^2}{A}x_2 \overline{y_1}]\Big|_0^L - \int_{0}^{L}(-\dfrac{a_0^2}{A}x_2 \dfrac{d}{d\xi}\overline{y_1})d\xi \\+ [-Ax_1 \overline{y_2}]\Big|_0^L - \int_{0}^{L}(-Ax_1 \dfrac{d}{d\xi}\overline{y_2})d\xi \, . $$ So I can get the adjoint operator from the integrals and the rest should be zero $$ [-\dfrac{a_0^2}{A}x_2 \overline{y_1}]\Big|_0^L + [-Ax_1 \overline{y_2}]\Big|_0^L \stackrel{!}{=} 0\\ = -\dfrac{a_0^2}{A}(x_2(L) \overline{y_1(L)}-x_2(0) \overline{y_1(0)}) -A(x_1(L) \overline{y_2(L)} - x_1(0) \overline{y_2(0)}) $$ Is this the adjoint operator? $$ \boldsymbol{\mathcal{A}}^*\textbf{x} = \begin{bmatrix} 0 & A\dfrac{d}{d\xi}\\ \dfrac{a_0^2}{A}\dfrac{d}{d\xi} & 0 \end{bmatrix}\begin{bmatrix} x_1\\[0.5em] x_2 \end{bmatrix}, $$ And what is its domain (the adjoint boundary conditions)? If I substitute the BC into the term from above, I get $$ x_1(0)(a_0^2 c_{1,p} \overline{y_1(0)}+A^2\overline{y_2(0)}) - x_1(L)(a_0^2 c_{2,p} \overline{y_1(L)}+A^2\overline{y_2(L)}) \stackrel{!}{=} 0 $$ which leads to the following adjoint BCs $$ y_2(0) -\dfrac{a_0^2c_{1,p}}{A^2}y_1(0)=0, \ y_2(L) -\dfrac{a_0^2c_{2,p}}{A^2}y_1(L)=0. $$ So the domain of the adjoint operator would be $$ D(\boldsymbol{\mathcal{A}}^*) = \bigg\lbrace \textbf{x} \in \mathcal{X} \Big\vert x_1, \ x_2 \ \text{abs. cont.}, \dfrac{d x_1}{d\xi}, \dfrac{d x_2}{d\xi} \in \textbf{L}_2(0,L) \\ \text{and} \ x_2(0) -\dfrac{a_0^2c_{1,p}}{A^2}x_1(0) = 0, \ x_2(L) -\dfrac{a_0^2c_{2,p}}{A^2}x_1(L)= 0 \bigg\rbrace, $$ Did I make a mistake?
Thanks!