Proof of Theorem 2. We first prove an estimate for the Caputo derivative in the weighted space. Let \(\eta\in H^1_\gamma(\Omega\times\mathbb{R}_+)\) with \(\eta(\cdot,0)=0\). Since \(0<\alpha<1\), we have \[\partial_t^\alpha \eta(x,t) = \frac{1}{\Gamma(1-\alpha)} \int_0^t (t-\tau)^{-\alpha}\partial_\tau\eta(x,\tau)\,d\tau .\] Multiplying by \(e^{-\gamma t}\), we obtain \[e^{-\gamma t}\partial_t^\alpha\eta(x,t) = \frac{1}{\Gamma(1-\alpha)} \int_0^t e^{-\gamma(t-\tau)}(t-\tau)^{-\alpha} e^{-\gamma\tau}\partial_\tau\eta(x,\tau)\,d\tau .\] Viewing the right-hand side as a convolution in time, we obtain, for almost every \(x\in\Omega\), by Young’s inequality, \[\|e^{-\gamma t}\partial_t^\alpha\eta(x,\cdot)\|_{L^2(\mathbb{R}_+)} \le \frac{1}{\Gamma(1-\alpha)} \|t^{-\alpha}e^{-\gamma t}\|_{L^1(\mathbb{R}_+)} \|e^{-\gamma t}\partial_t\eta(x,\cdot)\|_{L^2(\mathbb{R}_+)} .\] Since \[\int_0^\infty t^{-\alpha}e^{-\gamma t}\,dt = \gamma^{\alpha-1}\Gamma(1-\alpha),\] we get \[\|e^{-\gamma t}\partial_t^\alpha\eta(x,\cdot)\|_{L^2(\mathbb{R}_+)} \le \gamma^{\alpha-1} \|e^{-\gamma t}\partial_t\eta(x,\cdot)\|_{L^2(\mathbb{R}_+)} .\] Squaring and integrating over \(\Omega\), we obtain \[\label{frac32esti} \|\partial_t^\alpha\eta\|_{\gamma,0} \le \gamma^{\alpha-1}\|\partial_t\eta\|_{\gamma,0} \le \gamma^{\alpha-1}\|\eta\|_{\gamma,1}.\tag{2}\]

We now decompose the solution as \(u=w+v,\) where \(w\) solves the wave equation \[\label{wave32homo} \begin{cases} \partial_t^2 w-\Delta w=0, & (x,t)\in Q,\\ w=g, & (x,t)\in \Sigma,\\ w(\cdot,0)=0,\,\,\partial_t w(\cdot,0)=0, &x\in\Omega. \end{cases}\tag{3}\] By Lemma 1, there exist constants \(\gamma_0'>0\) and \(C'>0\), depending only on \(\Omega\), such that, for every \(\gamma\ge\gamma_0'\), problem 3 has a unique solution \(w\in H^1_\gamma(\Omega\times\mathbb{R}_+)\) satisfying \[\label{wave32esti} \gamma\|w\|_{\gamma,1}^2 + \langle\partial_\nu w\rangle_{\gamma,0}^2 \le C'\langle g\rangle_{\gamma,1}^2 .\tag{4}\] Then \(v=u-w\) must satisfy \[\label{wave32f} \begin{cases} \partial_t^2 v-\Delta v = -p(x)\partial_t^\alpha(w+v)-q(x)(w+v) =:f(x,t), & (x,t)\in Q,\\ v=0, & (x,t)\in \Sigma,\\ v(\cdot,0)=0,\quad \partial_t v(\cdot,0)=0, &x\in\Omega. \end{cases}\tag{5}\] Again by Lemma 1, there exist constants \(\gamma_0''>0\) and \(C''>0\), depending only on \(\Omega\), such that, for every \(\gamma\ge\gamma_0''\) and every \(f\in H^0_\gamma(\Omega\times\mathbb{R}_+)\), the homogeneous boundary problem 5 has a unique solution \(v\in H^1_\gamma(\Omega\times\mathbb{R}_+)\) satisfying \[\label{wave32esti1} \gamma\|v\|_{\gamma,1}^2 + \langle\partial_\nu v\rangle_{\gamma,0}^2 \le \frac{C''}{\gamma}\|f\|_{\gamma,0}^2 .\tag{6}\] Therefore, the solution operator \[E:H^0_\gamma(\Omega\times\mathbb{R}_+) \longrightarrow H^1_\gamma(\Omega\times\mathbb{R}_+), \qquad f\mapsto v,\] is well-defined and satisfies \[\label{E32bound} \|Ef\|_{\gamma,1} \le \frac{\sqrt{C''}}{\gamma}\|f\|_{\gamma,0}.\tag{7}\]

We now solve 5 by a fixed-point argument. The equation for \(v\) is equivalent to \[\label{fixed32point} v = E\left( -p\partial_t^\alpha(w+v)-q(w+v) \right).\tag{8}\] Set \[M:=\max\{\|p\|_{L^\infty(\Omega)},\|q\|_{L^\infty(\Omega)}\}.\] From 2 , we obtain \[\label{f32esti} \begin{align} \| f \|_{\gamma,0} &\leq M \bigl( \| \partial_t^\alpha (w+v) \|_{\gamma,0} + \| w+v \|_{\gamma,0} \bigr) \\[2pt] &\leq M \bigl( \gamma^{\alpha-1} \| w+v \|_{\gamma,1} + \| w+v \|_{\gamma,1} \bigr) \\[2pt] &= M \bigl( \gamma^{\alpha-1} + 1 \bigr) \bigl(\| w\|_{\gamma,1}+\|v\|_{\gamma,1}\bigr). \end{align}\tag{9}\]

Define \[\Phi(v) := E\left( -p\partial_t^\alpha(w+v)-q(w+v) \right), \qquad v\in H^1_\gamma(\Omega\times\mathbb{R}_+).\] Then \(\Phi\) maps \(H^1_\gamma(\Omega\times\mathbb{R}_+)\) into itself. By 7 and 9 , \[\label{mapping} \| \Phi(v) \|_{\gamma,1} \leq \frac{\sqrt{C''}}{{\gamma}} \| f \|_{\gamma,0} \leq \frac{\sqrt{C''}\,M\bigl( \gamma^{\alpha-1}+1 \bigr)}{{\gamma}} \; \bigl( \| w \|_{\gamma,1} + \| v \|_{\gamma,1} \bigr).\tag{10}\] Moreover, for any \(v_1,v_2\in H^1_\gamma(\Omega\times\mathbb{R}_+)\), \[\begin{align} \|\Phi(v_1)-\Phi(v_2)\|_{\gamma,1} &\le \frac{\sqrt{C''}}{\gamma} \left\| p\partial_t^\alpha(v_1-v_2)+q(v_1-v_2) \right\|_{\gamma,0} \\ &\le \frac{\sqrt{C''}M(\gamma^{\alpha-1}+1)}{\gamma} \|v_1-v_2\|_{\gamma,1}. \end{align}\] Set \[A_\gamma := \frac{\sqrt{C''}M(\gamma^{\alpha-1}+1)}{\gamma}.\] Since \(0<\alpha<1\), we have \(A_\gamma\to0\) as \(\gamma\to\infty\). Choose \(\gamma_0\ge \max\{\gamma_0',\gamma_0'',1\}\) large enough such that \[A_\gamma\le \frac{1}{2}, \qquad \gamma\ge\gamma_0.\] Then \(\Phi\) is a contraction on the Banach space \(H^1_\gamma(\Omega\times\mathbb{R}_+)\). Hence there exists a unique fixed point \(v\in H^1_\gamma(\Omega\times\mathbb{R}_+)\) solving 8 . Consequently, \(u=w+v\) is a solution of the original problem.

It remains to derive the a priori estimate. Since \(v=\Phi(v)\), by 10 we have \[\|v\|_{\gamma,1} \le A_\gamma\left(\|w\|_{\gamma,1}+\|v\|_{\gamma,1}\right).\] Using \(A_\gamma\le1/2\), we obtain \[\label{v32esti} \|v\|_{\gamma,1} \le \frac{A_\gamma}{1-A_\gamma}\|w\|_{\gamma,1} \le 2A_\gamma\|w\|_{\gamma,1}.\tag{11}\] Next, by 6 and 9 , \[\begin{align} \gamma\|v\|_{\gamma,1}^2 + \langle\partial_\nu v\rangle_{\gamma,0}^2 &\le \frac{C''}{\gamma} M^2(\gamma^{\alpha-1}+1)^2 \left(\|w\|_{\gamma,1}+\|v\|_{\gamma,1}\right)^2 . \end{align}\] Using 11 and \(A_\gamma\le1/2\), we have \[\|w\|_{\gamma,1}+\|v\|_{\gamma,1} \le (1+2A_\gamma)\|w\|_{\gamma,1} \le 2\|w\|_{\gamma,1}.\] Therefore, from 4 , we get \[\label{v32energy32estimate} \gamma\|v\|_{\gamma,1}^2 + \langle\partial_\nu v\rangle_{\gamma,0}^2 \le \frac{4C'C''M^2(\gamma^{\alpha-1}+1)^2}{\gamma^2} \langle g\rangle_{\gamma,1}^2\le C\langle g\rangle_{\gamma,1}^2,\tag{12}\] where \(C>0\) is independent of \(g\). Here we use the fact that the factor \(\frac{(\gamma^{\alpha-1}+1)^2}{\gamma^2}\) is bounded uniformly for \(\gamma\ge\gamma_0\).

Finally, since \(u=w+v\), we have \[\|u\|_{\gamma,1}^2 \le 2\|w\|_{\gamma,1}^2+2\|v\|_{\gamma,1}^2,\] and \[\langle\partial_\nu u\rangle_{\gamma,0}^2 \le 2\langle\partial_\nu w\rangle_{\gamma,0}^2 + 2\langle\partial_\nu v\rangle_{\gamma,0}^2.\] Combining 4 and 12 , we obtain \[\begin{align} \gamma\|u\|_{\gamma,1}^2 + \langle\partial_\nu u\rangle_{\gamma,0}^2 \le 2\left( \gamma\|w\|_{\gamma,1}^2 + \langle\partial_\nu w\rangle_{\gamma,0}^2 \right) + 2\left( \gamma\|v\|_{\gamma,1}^2 + \langle\partial_\nu v\rangle_{\gamma,0}^2 \right) \le C\langle g\rangle_{\gamma,1}^2, \end{align}\] where \(C>0\) depends only on \(\Omega\), \(\alpha\), \(\|p\|_{L^\infty(\Omega)}\), and \(\|q\|_{L^\infty(\Omega)}\), but is independent of \(g\) and \(\gamma\ge\gamma_0\). Moreover, uniqueness of the solution follows from the contraction principle applied to the difference of two solutions. This completes the proof. ◻

Proof of Theorem 3. We now give the proof of the conditional logarithmic stability estimate. The overall strategy is as follows. First, we apply the Fourier–Laplace transform in time, which reduces the non‑stationary problem to a family of stationary elliptic equations parameterized by the complex frequency \(\theta=\sigma-i\gamma\). In the frequency domain we construct complex geometrical optics (CGO) solutions and derive a key integral identity relating \(P=p_1-p_2\) to the difference of the DtN maps \(\Lambda_1^\theta-\Lambda_2^\theta\). By estimating the resulting terms and integrating over a suitable low‑frequency region, we obtain an upper bound for the Fourier transform \(\widetilde{P}\). Combining this with a high‑frequency decay estimate, which follows from the \(H^s\)-regularity of the zero extension of \(P\) for \(s<1/2\), and then choosing the free parameters appropriately, we balance the low‑ and high‑frequency contributions. This yields a logarithmic bound for \(\widetilde{P}\) in \(L^2(\mathbb{R}^n)\), and the desired stability estimate follows from Plancherel’s theorem.

In what follows, \(C\) denotes a generic positive constant that may change from line to line and depends only on the a priori data \(\Omega, n, \alpha, K, \gamma\) and, where applicable, on the Sobolev index \(s\).

The construction of CGO solutions relies on the following lemma, which is a generalization of a classical result by Sylvester and Uhlmann [4] (see also [36]). In the statement below, which was proved in [26], the dot denotes the “inner product without complex conjugation”; i.e., \(x\cdot y = x_1y_1+\cdots+x_ny_n\).

With this lemma at hand, we proceed to the proof.

Step 1. Fourier–Laplace reduction.

Let \(h(t)\) be a causal function defined on \(\mathbb{R}_+=(0,\infty)\), and denote by the same symbol its zero extension to \(\mathbb{R}\). For fixed \(\gamma>0\), its Fourier–Laplace transform is \[\widehat h(\theta) = \frac{1}{\sqrt{2\pi}} \int_0^\infty e^{-i\theta t} h(t)\,dt, \qquad \theta = \sigma - i\gamma,\; \sigma\in\mathbb{R} .\] Equivalently, \[\widehat h(\sigma-i\gamma) = \mathcal{F}_t\bigl(e^{-\gamma t}h(t)\bigr)(\sigma).\] By Plancherel’s theorem, for any Hilbert space \(X\), we have \[\label{eq:weighted-plancherel-main} \int_{\mathbb{R}} \|\widehat h(\sigma-i\gamma)\|_X^2\,d\sigma = \int_0^\infty e^{-2\gamma t}\|h(t)\|_X^2\,dt .\tag{13}\]

Let \(g\in \mathcal{H}_\gamma^1(\partial\Omega\times\mathbb{R}_+)\) and let \(u_j\) \((j=1,2)\) be the solution of 1 corresponding to the coefficient \(p_j\). Applying the Fourier–Laplace transform in time to \[\partial_t^2 u_j - \Delta u_j + p_j(x)\partial_t^\alpha u_j + q(x)u_j = 0,\] and using the vanishing initial data \(u_j(\cdot,0)=\partial_t u_j(\cdot,0)=0\), we obtain \[\widehat{\partial_t^2 u_j} = (i\theta)^2 \widehat u_j = -\theta^2 \widehat u_j .\] For the fractional derivative, the definition of the Caputo derivative together with the initial conditions yields \[\widehat{\partial_t^\alpha u_j}(\theta) = (i\theta)^\alpha \widehat u_j(\theta),\] where the principal branch is taken; this is well defined because \(\operatorname{Re}(i\theta)=\gamma>0\).

Hence, for almost every \(\sigma\in\mathbb{R}\), the Fourier–Laplace transform \(\widehat u_j(\cdot,\theta)\), with \(\theta=\sigma-i\gamma\), satisfies the stationary boundary value problem \[\label{eq:frequency-problem-main} \Delta \widehat u_j + \bigl( \theta^2-p_j(x)(i\theta)^\alpha-q(x) \bigr)\widehat u_j=0 \quad\text{in }\Omega, \qquad \widehat u_j|_{\partial\Omega}=\widehat g .\tag{14}\] By the uniqueness of the time-domain problem and the Fourier–Laplace transform, the Dirichlet problem 14 has a unique solution for almost every \(\theta=\sigma-i\gamma\), \(\sigma\in\mathbb{R}\). Therefore, for such \(\theta\), we define the corresponding stationary DtN map by \[\Lambda_j^\theta \widehat g := \partial_\nu \widehat u_j|_{\partial\Omega}.\] Since the trace and the normal trace commute with the Fourier–Laplace transform, we have \[\label{eq:DN-transform-main} \widehat{\Lambda_j g}(\cdot,\theta) = \Lambda_j^\theta \widehat g(\cdot,\theta), \qquad \text{for a.e. }\sigma\in\mathbb{R} .\tag{15}\]

We will also need a frequency-domain estimate for the operator norm. We now recall the fractional integral operator \(J^\alpha\) defined by \[(J^\alpha h)(t) = \frac{1}{\Gamma(\alpha)} \int_0^t (t-\tau)^{\alpha-1} h(\tau)\,d\tau .\] Its Fourier–Laplace transform satisfies \[\widehat{J^\alpha h}(\theta) = (i\theta)^{-\alpha} \widehat h(\theta).\] To estimate its weighted norm, write \(k(t) = \frac{1}{\Gamma(\alpha)} t^{\alpha-1}\) and \[(J^\alpha h)(t) = (k * h)(t) = \int_0^t k(t-\tau) h(\tau)\,d\tau .\] Set \(m(t) = e^{-\gamma t} h(t)\), then \(h(t) = e^{\gamma t} m(t)\) and \[e^{-\gamma t} (J^\alpha h)(t) = \int_0^t e^{-\gamma(t-\tau)} k(t-\tau) m(\tau)\,d\tau = (k_\gamma * m)(t),\] where \(k_\gamma(t) = e^{-\gamma t} k(t) = \frac{1}{\Gamma(\alpha)} e^{-\gamma t} t^{\alpha-1}\). Applying Young’s inequality for convolutions, \[\|k_\gamma * m\|_{L^2(\mathbb{R}_+)} \le \|k_\gamma\|_{L^1(\mathbb{R}_+)} \|m\|_{L^2(\mathbb{R}_+)} .\] A direct computation gives \[\|k_\gamma\|_{L^1(\mathbb{R}_+)} = \frac{1}{\Gamma(\alpha)} \int_0^\infty e^{-\gamma t} t^{\alpha-1}\,dt = \gamma^{-\alpha},\] while \(\|m\|_{L^2(\mathbb{R}_+)} = \|h\|_{\gamma,0}\). Hence \[\|J^\alpha h\|_{\gamma,0} \le \gamma^{-\alpha} \|h\|_{\gamma,0}.\]

Combining the Plancherel identity 13 , the commutation relation 15 and the above estimate for \(J^\alpha\) (applied to boundary functions), we obtain for every \(g\in\mathcal{H}_\gamma^1(\partial\Omega\times\mathbb{R}_+)\) \[\int_{\mathbb{R}} \bigl\| (i\theta)^{-\alpha} (\Lambda_1^\theta-\Lambda_2^\theta) \widehat g(\cdot,\theta) \bigr\|_{L^2(\partial\Omega)}^2\,d\sigma = \langle J^\alpha(\Lambda_1-\Lambda_2)g \rangle_{\gamma,0}^2 \le \gamma^{-2\alpha} \langle (\Lambda_1-\Lambda_2)g \rangle_{\gamma,0}^2 .\] Since \[\langle (\Lambda_1 - \Lambda_2) g \rangle_{\gamma,0} \le \langle g \rangle_{\gamma,1}\, \| \Lambda_1 - \Lambda_2 \|_{\mathcal{B} \left(\mathcal{H}_\gamma^1(\partial\Omega\times\mathbb{R}_+), H_\gamma^0(\partial\Omega\times\mathbb{R}_+)\right)} = \varepsilon \, \langle g \rangle_{\gamma,1},\] where \(\varepsilon := \|\Lambda_1-\Lambda_2\|_{\mathcal{B}(\mathcal{H}_\gamma^1(\partial\Omega\times\mathbb{R}_+), H_\gamma^0(\partial\Omega\times\mathbb{R}_+))}\), then we have \[\label{eq:frequency-DN-estimate-main} \int_{\mathbb{R}} \| (i\theta)^{-\alpha} (\Lambda_1^\theta - \Lambda_2^\theta) \widehat{g} \|_{L^2(\partial\Omega)}^2 d\sigma \le \gamma^{-2\alpha} \varepsilon^2 \langle g\rangle_{\gamma,1}^2\tag{16}\]

Step 2. The frequency-domain integral identity.

Fix \(\theta=\sigma-i\gamma\) and set \[V_j(x,\theta) := -p_j(x)(i\theta)^\alpha - q(x).\] Then the frequency-domain equation becomes \[\Delta u_j + (\theta^2+V_j)u_j = 0 \quad\text{in }\Omega .\] Let \(u_j\) be two sufficiently regular solutions of these equations and write \(f_j := u_j|_{\partial\Omega}\). By Green’s identity, \[\int_\Omega (u_2 \Delta u_1 - u_1 \Delta u_2)\,dx = \int_{\partial\Omega} \Bigl( u_2 \frac{\partial u_1}{\partial\nu} - u_1 \frac{\partial u_2}{\partial\nu} \Bigr) dS .\] Set \(f_j := u_j|_{\partial\Omega}\) and recall that \(\partial_\nu u_j|_{\partial\Omega} = \Lambda_j^\theta f_j\). Then the right-hand side becomes \[\int_{\partial\Omega} \bigl( f_2 \Lambda_1^\theta f_1 - f_1 \Lambda_2^\theta f_2 \bigr) dS .\] On the left-hand side, using the equations \(\Delta u_j = -(\theta^2+V_j)u_j\), we obtain \[u_2\Delta u_1 - u_1\Delta u_2 = (V_2-V_1)u_1u_2 .\] Thus, \[\int_\Omega (V_2-V_1) u_1 u_2\,dx = \int_{\partial\Omega} \bigl( f_2 \Lambda_1^\theta f_1 - f_1 \Lambda_2^\theta f_2 \bigr) dS .\] Because the DtN map is symmetric with respect to the bilinear pairing without complex conjugation, \[\int_{\partial\Omega} f_2 \Lambda_1^\theta f_1\,dS = \int_{\partial\Omega} f_1 \Lambda_1^\theta f_2\,dS .\] Consequently, \[\label{eq:alessandrini-main} \int_\Omega (V_2-V_1) u_1 u_2\,dx = \int_{\partial\Omega} f_1 (\Lambda_1^\theta-\Lambda_2^\theta) f_2\,dS .\tag{17}\] Noticing that \[V_2-V_1 = -p_2(i\theta)^\alpha - q + p_1(i\theta)^\alpha + q = P(i\theta)^\alpha,\qquad P:=p_1-p_2,\] identity 17 turns into \[\label{eq:basic-identity-main} (i\theta)^\alpha \int_\Omega P(x) u_1(x) u_2(x)\,dx = \int_{\partial\Omega} f_1 (\Lambda_1^\theta-\Lambda_2^\theta) f_2\,dS .\tag{18}\]

Step 3. Construction of complex geometrical optics solutions.

We construct CGO solutions for \[\Delta u_j + (\theta^2+V_j)u_j = 0 \quad\text{in }\Omega,\qquad j=1,2.\] By assumption, \(p_j,q\in C^2(\overline{\Omega})\) and \(\|p_j\|_{C^2(\overline{\Omega})}+\|q\|_{C^2(\overline{\Omega})}\le K\). Hence \[V_j(\cdot,\theta)\in C^2(\overline{\Omega}),\qquad \|V_j(\cdot,\theta)\|_{C^2(\overline{\Omega})} \le C K (1+|\theta|^\alpha).\]

Apply Lemma 5 with \(\ell=2\) and \(a(x)=V_j(x,\theta)\). There exist constants \(C_1,C_2,C_3,C_4>0\), depending only on \(\Omega\) and \(\ell=2\), such that whenever \(\zeta_j\in\mathbb{C}^n\) satisfies \[\zeta_j\cdot\zeta_j + \theta^2 = 0,\quad |\operatorname{Im}\zeta_j|\ge\sqrt2,\quad |\zeta_j|\ge C_1\|V_j(\cdot,\theta)\|_{C^2(\overline{\Omega})},\] the equation admits a solution of the form \[u_j(x,\zeta_j,\theta) = e^{\zeta_j\cdot x}\bigl( 1 + w_j(x,\zeta_j,\theta) \bigr),\] with the estimate \[\label{eq:cgo-remainder-main} \|w_j(\cdot,\zeta_j,\theta)\|_{H^2(\Omega)} \le \frac{C_2}{|\zeta_j|} \|V_j(\cdot,\theta)\|_{C^2(\overline{\Omega})} .\tag{19}\] Inserting the bound for \(V_j\) yields \[\label{eq:cgo-remainder-main-2} \|w_j\|_{H^2(\Omega)} \le C \frac{1+|\theta|^\alpha}{|\zeta_j|}.\tag{20}\] Moreover, Lemma 5 provides the norm estimate \[\label{eq:cgo-trace-main} \|u_j(\cdot,\zeta_j,\theta)\|_{H^2(\Omega)} \le C e^{C|\zeta_j|}.\tag{21}\]

Now we choose concrete vectors \(\zeta_j\). Take \(\gamma\ge\gamma_0\) sufficiently large so that for all \(\sigma\in\mathbb{R}\), \[C_1 C K (1+|\theta|^\alpha) \le |\theta|,\] which is possible since \(|\theta|\ge\gamma\) and \(0<\alpha<1\). Fix \(\xi\in\mathbb{R}^n\), let \(\rho>0\), take \(\sigma\in[0,\rho]\) and \(r\ge\sqrt2\). Since \(n\ge3\), we can pick real vectors \(\mu,\lambda\in\mathbb{R}^n\) satisfying \[\xi\cdot\mu = 0,\quad \xi\cdot\lambda = 0,\] and \[|\mu|^2 = \sigma^2 + r^2,\qquad |\lambda|^2 = \frac{|\xi|^2}{4} + \gamma^2 + r^2,\qquad \mu\cdot\lambda = \sigma\gamma .\] The last condition is fulfilled by adjusting the angle between \(\mu\) and \(\lambda\) in the plane orthogonal to \(\xi\), because \(|\mu||\lambda|\ge\sigma\gamma\).

Define \[\zeta_j = -\frac{i\xi}{2} + (-1)^j (i\mu+\lambda),\qquad j=1,2,\] so that \(\zeta_1+\zeta_2 = -i\xi\). A direct computation gives \[\zeta_j\cdot\zeta_j = -\frac{|\xi|^2}{4} - |\mu|^2 + |\lambda|^2 + 2i\mu\cdot\lambda = \gamma^2 - \sigma^2 + 2i\sigma\gamma = -\theta^2,\] hence \(\zeta_j\cdot\zeta_j+\theta^2=0\). Moreover, \[|\operatorname{Im}\zeta_j| = \Bigl|(-1)^j\mu - \frac{\xi}{2}\Bigr| = \Bigl(|\mu|^2 + \frac{|\xi|^2}{4}\Bigr)^{1/2} \ge r \ge \sqrt2,\] and \[|\zeta_j|^2 = \frac{|\xi|^2}{2} + |\theta|^2 + 2r^2 \ge |\theta|^2 .\] Thus \(|\zeta_j|\ge |\theta| \ge C_1\|V_j\|_{C^2(\overline{\Omega})}\). All the assumptions of Lemma 5 are thus satisfied, and the CGO solutions constructed above are well defined.

Step 4. Low-frequency decomposition.

Insert the CGO ansatz into the identity 18 . Using \[u_1u_2 = e^{(\zeta_1+\zeta_2)\cdot x} (1+w_1)(1+w_2) = e^{-i\xi\cdot x} (1 + w_1 + w_2 + w_1w_2),\] we obtain \[(i\theta)^\alpha \biggl[ \int_\Omega P(x) e^{-i\xi\cdot x}\,dx + \int_\Omega P(x) e^{-i\xi\cdot x} (w_1+w_2+w_1w_2)\,dx \biggr] = \int_{\partial\Omega} f_1 (\Lambda_1^\theta-\Lambda_2^\theta) f_2\,dS .\] Define the Fourier transform of the zero extension of \(P\) by \[\widetilde{P}(\xi) := (2\pi)^{-n/2} \int_\Omega P(x) e^{-i\xi\cdot x}\,dx,\] and the remainder \[E(\xi,\theta) := \int_\Omega P(x) e^{-i\xi\cdot x} (w_1+w_2+w_1w_2)\,dx .\] The above relation rewrites as \[(i\theta)^\alpha \bigl[ (2\pi)^{n/2} \widetilde{P}(\xi) + E(\xi,\theta) \bigr] = \int_{\partial\Omega} f_1 (\Lambda_1^\theta-\Lambda_2^\theta) f_2\,dS .\] Equivalently, \[\label{eq:P-decomposition-main} \widetilde{P}(\xi) = B(\xi,\theta) + \mathcal{E}(\xi,\theta),\tag{22}\] where \[B(\xi,\theta) := \frac{(i\theta)^{-\alpha}}{(2\pi)^{n/2}} \int_{\partial\Omega} f_1 (\Lambda_1^\theta-\Lambda_2^\theta) f_2\,dS,\qquad \mathcal{E}(\xi,\theta) := -\frac{1}{(2\pi)^{n/2}} E(\xi,\theta).\] Consequently, \[\label{eq:P-square-main} |\widetilde{P}(\xi)|^2 \le C\bigl( |B(\xi,\theta)|^2 + |\mathcal{E}(\xi,\theta)|^2 \bigr).\tag{23}\]

Step 5. Estimates of the remainder and the boundary term.

We first estimate \(\mathcal{E}(\xi,\theta)\). From 20 and \(|\zeta_j|\ge r\), \[\|w_j\|_{L^2(\Omega)} \le C\frac{1+|\theta|^\alpha}{r}.\] Using \(\|P\|_{L^\infty(\Omega)}\le 2K\) and Hölder’s inequality, \[|E(\xi,\theta)| \le \|P\|_{L^\infty(\Omega)} \int_\Omega (|w_1|+|w_2|+|w_1w_2|)\,dx \le C \Bigl( \frac{1+|\theta|^\alpha}{r} + \frac{(1+|\theta|^\alpha)^2}{r^2} \Bigr).\] For \(r\ge1\) and large \(|\theta|\) the dominating term is \(\frac{(1+|\theta|^\alpha)^2}{r}\); hence \[\label{eq:E-bound-main} |\mathcal{E}(\xi,\theta)| \le C \frac{(1+|\theta|^\alpha)^2}{r}.\tag{24}\]

Now we turn to the boundary term \(B(\xi,\theta)\). By the trace theorem and 21 , \[\|f_1\|_{L^2(\partial\Omega)} \le C\|u_1\|_{H^1(\Omega)} \le C e^{C|\zeta_1|},\] so that \[\label{eq:B-pre-main} |B(\xi,\theta)| \le C e^{C|\zeta_1|} \bigl\| (i\theta)^{-\alpha} (\Lambda_1^\theta-\Lambda_2^\theta) f_2 \bigr\|_{L^2(\partial\Omega)} .\tag{25}\] To exploit 16 , we construct a normalized boundary function. For fixed \(\xi\in\mathbb{R}^n\) and \(r\ge\sqrt2\), define \[\widehat g_{\xi,r}(\cdot,\sigma) := \frac{f_2(\cdot,\sigma)}{e^{C|\zeta_2|}(1+\sigma^2)}, \quad \sigma\in \mathbb{R},\] where \(C>0\) is chosen large enough so that, by the trace theorem and the CGO estimates, \[\|\widehat g_{\xi,r}(\cdot,\sigma)\|_{H^1(\partial\Omega)} \le \frac{C}{1+\sigma^2}.\] Then \[\int_{\mathbb{R}} (1+\sigma^2) \|\widehat g_{\xi,r}(\cdot,\sigma)\|_{H^1(\partial\Omega)}^2 d\sigma \le C.\] Therefore, the inverse Fourier–Laplace transform \(g_{\xi,r}\) belongs to \(\mathcal{H}_\gamma^1(\partial\Omega\times\mathbb{R}_+)\), and \(\langle g_{\xi,r}\rangle_{\gamma,1}\le C\). Applying 16 to \(g_{\xi,r}\) gives \[\int_0^\rho \bigl\| (i\theta)^{-\alpha} (\Lambda_1^\theta-\Lambda_2^\theta) \widehat g_{\xi,r}(\cdot,\sigma) \bigr\|_{L^2(\partial\Omega)}^2 d\sigma \le C\varepsilon^2 .\] Since \(f_2 = e^{C|\zeta_2|}(1+\sigma^2)\widehat g_{\xi,r}\) and \(|\zeta_2|\le C\sqrt{|\xi|^2+\rho^2+r^2}\) for \(\sigma\in[0,\rho]\), we obtain \[\int_0^\rho \bigl\| (i\theta)^{-\alpha} (\Lambda_1^\theta-\Lambda_2^\theta) f_2 \bigr\|_{L^2(\partial\Omega)}^2 d\sigma \le C e^{C\sqrt{|\xi|^2+\rho^2+r^2}} (1+\rho^2)^2 \varepsilon^2 .\] Inserting this into 25 and integrating in \(\sigma\) yields \[\label{eq:B-sigma-main} \int_0^\rho |B(\xi,\sigma-i\gamma)|^2 d\sigma \le C e^{C\sqrt{|\xi|^2+\rho^2+r^2}} (1+\rho^2)^2 \varepsilon^2 .\tag{26}\]

Step 6. Low-frequency accumulation and high-frequency cut-off.

Consider the region \[D_\rho := \{ (\xi,\sigma)\in\mathbb{R}^n\times\mathbb{R} : |\xi|\le\rho,\; 0\le\sigma\le\rho \}.\] Since \(\widetilde{P}(\xi)\) is independent of \(\sigma\), integrating 23 over \(D_\rho\) gives \[\rho \int_{|\xi|\le\rho} |\widetilde{P}(\xi)|^2 d\xi \le C \iint_{D_\rho} \bigl( |B|^2 + |\mathcal{E}|^2 \bigr) d\sigma d\xi .\] Using 26 , \[\int_{|\xi|\le\rho}\int_0^\rho |B|^2 d\sigma d\xi \le C e^{C\sqrt{2\rho^2+r^2}} (1+\rho^2)^2 \rho^n \varepsilon^2 .\] From 24 , for \(\rho\ge\gamma\) and \(0\le\sigma\le\rho\) we have \(|\theta|\le C\rho\), hence \[|\mathcal{E}|^2 \le C \frac{(1+\rho^\alpha)^4}{r^2}, \qquad \int_{|\xi|\le\rho}\int_0^\rho |\mathcal{E}|^2 d\sigma d\xi \le C \frac{(1+\rho^\alpha)^4}{r^2} \rho^{n+1}.\] Therefore, \[\rho \int_{|\xi|\le\rho} |\widetilde{P}(\xi)|^2 d\xi \le C \Bigl( e^{C\sqrt{2\rho^2+r^2}} (1+\rho^2)^2 \rho^n \varepsilon^2 + \frac{(1+\rho^\alpha)^4}{r^2} \rho^{n+1} \Bigr).\] Dividing by \(\rho\) we obtain the low-frequency estimate \[\label{eq:low-frequency-main} \int_{|\xi|\le\rho} |\widetilde{P}(\xi)|^2 d\xi \le C \Bigl( e^{C\sqrt{2\rho^2+r^2}} (1+\rho^2)^2 \rho^{n-1} \varepsilon^2 + \frac{(1+\rho^\alpha)^4}{r^2} \rho^n \Bigr).\tag{27}\]

For the high frequencies, fix \(s\in(0,1/2)\). Since \(P=p_1-p_2\in C^2(\overline{\Omega})\), its zero extension to \(\mathbb{R}^n\), still denoted by \(P\), belongs to \(H^s(\mathbb{R}^n)\) and \[\|P\|_{H^s(\mathbb{R}^n)}^2 = \int_{\mathbb{R}^n} (1+|\xi|^2)^s |\widetilde{P}(\xi)|^2 d\xi \le C K^2 .\] Hence, for \(\rho\ge1\), \[\label{eq:high-frequency-main} \int_{|\xi|>\rho} |\widetilde{P}(\xi)|^2 d\xi \le (1+\rho^2)^{-s} \int_{\mathbb{R}^n} (1+|\xi|^2)^s |\widetilde{P}(\xi)|^2 d\xi \le C \rho^{-2s}.\tag{28}\]

Combining 27 and 28 , we get \[\label{eq:global-main} \int_{\mathbb{R}^n} |\widetilde{P}(\xi)|^2 d\xi \le C \Bigl( e^{C\sqrt{2\rho^2+r^2}} (1+\rho^2)^2 \rho^{n-1} \varepsilon^2 + \frac{(1+\rho^\alpha)^4}{r^2} \rho^n + \rho^{-2s} \Bigr).\tag{29}\]

Step 7. Logarithmic stability estimate.

We now choose the free parameters to balance the right-hand side. Let \[r = \rho^\delta,\qquad \delta := \frac{n+4\alpha+2s}{2}.\] Since \(n\ge3\), \(0<\alpha<1\) and \(s>0\), we have \(\delta>1\). For large \(\rho\), \[\sqrt{2\rho^2+r^2} \le C\rho^\delta,\qquad (1+\rho^2)^2\rho^{n-1} \le C\rho^{n+3},\] and \[\frac{(1+\rho^\alpha)^4}{r^2} \rho^n \le C \rho^{n+4\alpha-2\delta} = C \rho^{-2s}.\] Thus 29 simplifies to \[\label{eq:pre-log-main} \int_{\mathbb{R}^n} |\widetilde{P}(\xi)|^2 d\xi \le C \Bigl( e^{C\rho^\delta} \rho^{n+3} \varepsilon^2 + \rho^{-2s} \Bigr).\tag{30}\]

Now relate \(\rho\) to \(\varepsilon\) by requiring \[C_0 \rho^\delta + (n+3+2s)\log\rho = \log\frac{1}{\varepsilon^2},\] where \(C_0>0\) is a sufficiently large constant (larger than the constant appearing in the exponent of 30 ). Then \(\rho\to\infty\) as \(\varepsilon\to0\) and \(\rho^\delta \sim \frac{1}{C_0}\log\frac{1}{\varepsilon^2}\). This choice guarantees \[e^{C\rho^\delta} \rho^{n+3} \varepsilon^2 \le C \rho^{-2s},\] so that 30 reduces to \[\int_{\mathbb{R}^n} |\widetilde{P}(\xi)|^2 d\xi \le C \rho^{-2s}.\] Because \(\rho^\delta \ge c\log\frac{1}{\varepsilon}\) for small \(\varepsilon\), we have \(\rho^{-2s} \le C \bigl(\log\frac{1}{\varepsilon}\bigr)^{-2s/\delta}\). With \(\delta\) as above, \[\frac{2s}{\delta} = \frac{4s}{n+4\alpha+2s}.\] Hence \[\label{eq:fourier-final-main} \int_{\mathbb{R}^n} |\widetilde{P}(\xi)|^2 d\xi \le C \Bigl(\log\frac{1}{\varepsilon}\Bigr)^{-\frac{4s}{n+4\alpha+2s}} .\tag{31}\]

Finally, let \(P=p_1-p_2\) be extended by zero to \(\mathbb{R}^n\). By Plancherel’s theorem and by the definition of \(\widetilde{P}\), we have \[\|p_1-p_2\|_{L^2(\Omega)}^2 = \|P\|_{L^2(\mathbb{R}^n)}^2 = \int_{\mathbb{R}^n}|\widetilde{P}(\xi)|^2\,d\xi .\] Combining this with 31 and taking square roots, we obtain \[\|p_1-p_2\|_{L^2(\Omega)} \le C \Bigl(\log\frac{1}{\varepsilon}\Bigr)^{-\tau}, \qquad \tau = \frac{2s}{n+4\alpha+2s}.\] Therefore, \[\|p_1-p_2\|_{L^2(\Omega)} \le \omega(\varepsilon),\] where \(\omega(\varepsilon) \sim C \bigl(\log (1/\varepsilon)\bigr)^{-\tau}\) as \(\varepsilon\to 0\). This is precisely the conditional logarithmic stability estimate stated in Theorem 3. ◻