By induction, one has $v(t)|_{\partial M} = (\mathrm{diag}(\Theta)f)|_{\{t\} \times \partial M}$ in $H^{s-\frac{1}{2}}(M, \mathbb{C}^N)$ for $t \in [0, T]$. In particular, this gives $v(t)|_{\partial M} \in H^{s+\frac{1}{2}}(M, \mathbb{C}^N)$. As in the case $s=1$, one has $\partial_t v = \tilde{v}$, where $\tilde{v}$ is the solution of (B.44), and this gives

$$ \|v\|_{\mathcal{C}^1([0,T], H^s) \cap \mathcal{C}^2([0,T], H^{s-1})} \lesssim \|f\|_{H^{s+1}((0,T) \times \partial M, \mathbb{C}^N)} $$

since $\tilde{v}$ fulfills (B.47). In particular, one has $\mathbf{P}_{\mathcal{D}'}^* v(t) = \partial_t^2 v(t) \in H^{s-1}(M, \mathbb{C}^N)$ for $t \in [0, T]$. Hence, for $t \in [0, T]$, Lemma B.6 gives $v(t) \in H^{s+1}(M, \mathbb{C}^N)$, and

$$ \begin{aligned} \|v(t)\|_{H^{s+1}(M, \mathbb{C}^N)} &\lesssim \|\mathbf{P}_{\mathcal{D}'}^* v(t)\|_{H^{s-1}(M, \mathbb{C}^N)} + \|v(t)|_{\partial M}\|_{H^{s+\frac{1}{2}}(\partial M, \mathbb{C}^N)} + \|v(t)\|_{L^2(M, \mathbb{C}^N)} \\ &\lesssim \|\partial_t^2 v(t)\|_{H^{s-1}(M, \mathbb{C}^N)} + \|(\mathrm{diag}(\Theta)f)|_{\{t\} \times \partial M}\|_{H^{s+\frac{1}{2}}(\partial M, \mathbb{C}^N)} + \|v\|_{L^\infty((0,T), L^2)} \\ &\lesssim \|f\|_{H^{s+1}((0,T) \times \partial M, \mathbb{C}^N)}. \end{aligned} $$

This gives $v \in L^\infty([0, T], H^{s+1}(M, \mathbb{C}^N))$. To complete the proof, it suffices to show that

$$ v \in \mathcal{C}^0([0, T], H^{s+1}(M, \mathbb{C}^N)) $$

for $f$ smooth. For such $f$, Lemma B.6 gives

$$ \begin{aligned} \|v(t + \varepsilon) - v(t)\|_{H^{s+1}(M, \mathbb{C}^N)} &\lesssim \|\partial_t^2 v(t + \varepsilon) - \partial_t^2 v(t)\|_{H^{s-1}(M, \mathbb{C}^N)} + \|f(t + \varepsilon) - f(t)\|_{H^{s+\frac{1}{2}}(\partial M, \mathbb{C}^N)} + \|v(t + \varepsilon) - v(t)\|_{L^2(M, \mathbb{C}^N)}, \end{aligned} $$

and one concludes

$$ \|v(t + \varepsilon) - v(t)\|_{H^{s+1}(M, \mathbb{C}^N)} \xrightarrow{\varepsilon \to 0} 0. $$

This gives (B.47).

**Step 3: The additional regularity result.** Here, we complete the proof of Theorem 2.14 by proving

$$ (v(T), \partial_t v(T)) \in \mathcal{K}_*^s \times \mathcal{K}_*^{s-1} \qquad (B.48) $$

for $f \in H^s((0, T) \times \partial M, \mathbb{C}^N)$ and $s \in \mathbb{N}^*$. We will also prove the duality equality of Theorem 2.14. As $\mathcal{K}_*^s \times \mathcal{K}_*^{s-1}$ is a closed subspace of $H^s(M, \mathbb{C}^N) \times H^{s-1}(M, \mathbb{C}^N)$, it suffices to prove (B.48) for $f$ smooth. We proceed by induction. The result is true for $s=0$, but one has to treat the case $s=1$ separately.

**Case 1:** $s=1$. Take $f \in \mathcal{C}^\infty((0, T) \times \partial M, \mathbb{C}^N)$, and write $v$ for the associated solution. We prove that $v(T) \in H_0^1(M, \mathbb{C}^N)$. Write $v = w + \tilde{f}$ as above. One has $w \in \mathcal{C}^0([0, T], H_0^1(M, \mathbb{C}^N))$, implying

$$ v(T) = w(T) + \tilde{f}(T) = w(T) \in H_0^1(M, \mathbb{C}^N), $$

since $\tilde{f}$ is compactly supported in $(0, T) \times M$. Note that $w \in \mathcal{C}^0([0, T], \mathcal{K}_*^s)$ for $s \ge 2$ is not always true, preventing a straightforward generalization of this argument.

**Case 2:** $s$ odd. Assume that the result holds true for some odd $s \in \mathbb{N}^*$. Write $s = 2\sigma + 1$, and consider $f \in \mathcal{C}^\infty((0, T) \times \partial M, \mathbb{C}^N)$. We prove that $(v(T), \partial_t v(T)) \in \mathcal{K}_*^{2\sigma+2} \times \mathcal{K}_*^{2\sigma+1}$. By induction, one has $(v(T), \partial_t v(T)) \in \mathcal{K}_*^{2\sigma+1} \times \mathcal{K}_*^{2\sigma}$, and as $f \in H^{s+1}((0, T) \times \partial M, \mathbb{C}^N)$, we know that

$$ v \in \mathcal{C}^0([0, T], H^{s+1}(M, \mathbb{C}^N)) \cap \mathcal{C}^1([0, T], H^s(M, \mathbb{C}^N)). $$

By definition, $H^{2\sigma+2}(M, \mathbb{C}^N) \cap \mathcal{K}_*^{2\sigma+1} = \mathcal{K}_*^{2\sigma+2}$, implying $v(T) \in \mathcal{K}_*^{2\sigma+2}$. Set $\tilde{v} = \partial_t v$, solution to (B.44). By induction, one has $\partial_t v(T) = \tilde{v}(T) \in \mathcal{K}_*^{2\sigma+1}$, completing the proof in the case $s$ is odd.
**Case 3:** s even. Assume that the result holds for some even $s \in \mathbb{N}^*$. Write $s = 2\sigma$, and consider $f \in \mathcal{C}^\infty((0, T) \times \partial M, \mathbb{C}^N)$. We show that $(v(T), \partial_t v(T)) \in \mathcal{K}_*^{2\sigma+1} \times \mathcal{K}_*^{2\sigma}$. By induction, one has $(v(T), \partial_t v(T)) \in \mathcal{K}_*^{2\sigma} \times \mathcal{K}_*^{2\sigma-1}$, and we know that $(v(T), \partial_t v(T)) \in H^{s+1}(M, \mathbb{C}^N) \times H^s(M, \mathbb{C}^N)$, yielding $\partial_t v(T) \in \mathcal{K}_*^{2\sigma}$ as above. The definition of $\mathcal{K}_*^{2\sigma+1}$ reads

$$ \mathcal{K}_*^{2\sigma+1} = \{u \in H^{2\sigma+1}(M, \mathbb{C}^N)) \cap \mathcal{K}_*^{2\sigma}, P_{\mathcal{D}', \sigma}^* u \in H_0^1(M, \mathbb{C}^N))\}, $$

implying that one has $v(T) \in \mathcal{K}_*^{2\sigma+1}$, if $P_{\mathcal{D}', \sigma}^* v(T) \in H_0^1(M, \mathbb{C}^N))$. Again, write $\tilde{v} = \partial_t v$. One has $P_{\mathcal{D}', \sigma}^* v(T) = \partial_t^2 v(T) = \partial_t \tilde{v}(T)$, and by induction, one finds $\partial_t \tilde{v}(T) \in \mathcal{K}_*^{2\sigma-1}$. Thus, as $\sigma > 0$, one obtains

$$ P_{\mathcal{D}', \sigma}^* v(T) = P_{\mathcal{D}', \sigma}^{*-1} \partial_t \tilde{v}(T) \in H_0^1(M, \mathbb{C}^N), $$

completing the proof in the case s is even.

Finally, we prove the duality equality of Theorem 2.14 for $s \ge 1$. Consider $s \ge 1$, $f \in H^s((0, T) \times \partial M, \mathbb{C}^N)$ and $(u^0, u^1) \in \mathcal{K}^{-s+1} \times \mathcal{K}^{-s}$. Write $u$ for the solution of (B.38). We show that

$$ \langle u^1, v(T) \rangle_{\mathcal{K}^{-s+1}, \mathcal{K}_*^{s-1}} - \langle u^0, \partial_t v(T) \rangle_{\mathcal{K}^{-s}, \mathcal{K}_*^s} = \langle \partial_\nu u, \operatorname{diag}(\Theta) f \rangle_{H^{-s}, H_0^s}. \quad (B.49) $$

With the approximation result of Theorem 2.12, consider a sequence $((u_k^0, u_k^1))_{k \in \mathbb{N}}$ of elements of $\mathcal{K}^2 \times \mathcal{K}^1$ such that

$$ (i_{\mathcal{K}^2 \to \mathcal{K}^{-s+1}} u_k^0, i_{\mathcal{K}^1 \to \mathcal{K}^{-s}} u_k^1) \underset{k \to \infty}{\longrightarrow} (u^0, u^1) \text{ in } \mathcal{K}^{-s+1} \times \mathcal{K}^{-s}. $$

If $u_k$ is the solution of (B.38) associated with $(u_k^0, u_k^1)$, then one has

$$ i_{\mathcal{K}^2 \to \mathcal{K}^{-s+1}} u_k \underset{k \to \infty}{\longrightarrow} u \text{ in } \mathcal{C}^0([0, T], \mathcal{K}^{-s+1}) \cap \mathcal{C}^1([0, T], \mathcal{K}^{-s}), $$

and

$$ \partial_\nu (i_{\mathcal{K}^2 \to \mathcal{K}^{-s+1}} u_k) \underset{k \to \infty}{\longrightarrow} \partial_\nu u \text{ in } H^{-s}((0, T) \times \partial M, \mathbb{C}^N). $$

The duality equality of Theorem 2.14 for $s = 0$ gives

$$ \langle u_k^1, v(T) \rangle_{L^2(M, \mathbb{C}^N)} - \langle u_k^0, \partial_t v(T) \rangle_{L^2(M, \mathbb{C}^N)} = \langle \operatorname{diag}(\Theta) \partial_\nu u_k, f \rangle_{L^2((0,T) \times \partial M, \mathbb{C}^N)}. $$

Since $(v(T), \partial_t v(T)) \in \mathcal{K}_*^s \times \mathcal{K}_*^{s-1}$, one finds

$$ \langle i_{\mathcal{K}^2 \to \mathcal{K}^{-s+1}} u_k^1, v(T) \rangle_{\mathcal{K}^{-s+1}, \mathcal{K}_*^{s-1}} - \langle i_{\mathcal{K}^1 \to \mathcal{K}^{-s}} u_k^0, \partial_t v(T) \rangle_{\mathcal{K}^{-s}, \mathcal{K}_*^s} = \langle i_{H^1 \to H^{-s}} \partial_\nu u_k, \operatorname{diag}(\Theta) f \rangle_{H^{-s}, H_0^s}. $$

From Theorem 2.12, one has $i_{H^1 \to H^{-s}} \partial_\nu u_k = \partial_\nu i_{\mathcal{K}^2 \to \mathcal{K}^{-s+1}} u_k$, yielding (B.49).