2.1 Function spaces

For $s\in (0,1)$ , the fractional Sobolev space $H^s({\mathbb R}^n)$ is defined via the norm

$$\begin{align*}\|u\|_{H^s({\mathbb R}^n)}^2 = \|u\|_{L^2({\mathbb R}^n)}^2 + [u]_{H^s}^2, \end{align*}$$

where the Gagliardo seminorm is

$$\begin{align*}[u]_{H^s}^2 = \iint_{{\mathbb R}^{2n}} \frac{|u(x)-u(y)|^2}{|x-y|^{n+2s}}\,dx\,dy. \end{align*}$$

For an open set $U\subset {\mathbb R}^n$ , we define

$$ \begin{align*} H^s(U) &= \{u|_U : u\in H^s({\mathbb R}^n)\}, \\ \widetilde{H}^s(U) &= \text{closure of } C_c^\infty(U) \text{ in } H^s({\mathbb R}^n), \\ H^s_0(U) &= \text{closure of } C_c^\infty(U) \text{ in } H^s(U). \end{align*} $$

When U is a bounded Lipschitz domain, we have $\widetilde {H}^s(U) = H^s_{\overline {U}}({\mathbb R}^n)$ . For boundary regularity of the Dirichlet problem for the standard fractional Laplacian, see [Reference Ros-Oton and Serra18].

2.2 Magnetic fractional Laplacian

Following Li [Reference Li13], we define the magnetic fractional Laplacian $\mathcal {L}_A^s$ via its bilinear form:

(2.1) $$ \begin{align} \langle \mathcal{L}_A^s u, v \rangle = \iint_{{\mathbb R}^{2n}} \big(u(x)-e^{i(x-y)\cdot A(\frac{x+y}{2})}u(y)\big)\overline{\big(v(x)-e^{i(x-y)\cdot A(\frac{x+y}{2})}v(y)\big)}K(x,y)\,dx\,dy, \end{align} $$

for $u,v\in H^s({\mathbb R}^n)$ , where $K(x,y)=K(y,x)$ satisfies

$$\begin{align*}C_1|x-y|^{-(n+2s)} \leq K(x,y) \leq C_2|x-y|^{-(n+2s)}. \end{align*}$$

2.3 Kernel specification

The kernel $K(x,y)$ in Li’s operator is defined through the heat kernel $p_t(x,y)$ of the uniformly elliptic operator $\mathcal {L}=-\nabla \cdot (M(x)\nabla )$ :

$$\begin{align*}K(x,y)=C_{n,s}\int_0^\infty p_t(x,y)\frac{dt}{t^{1+s}}, \end{align*}$$

where $M(x)$ is a smooth symmetric matrix-valued function satisfying $C^{-1}|\xi |^2\leq M(x)\xi \cdot \xi \leq C|\xi |^2$ . When $M=I$ , we recover $K(x,y)=c_{n,s}|x-y|^{-(n+2s)}$ , the kernel of the standard fractional Laplacian. Throughout, we assume that K satisfies the bounds above.

2.4 Bilinear form and Green’s formula

The magnetic fractional Laplacian $\mathcal {L}_A^s$ is defined through its bilinear form as in (2.1). For $u,v\in H^s({\mathbb R}^n)$ , we have the integration by parts formula (Green’s first identity for  $\mathcal {L}_A^s$ ):

$$ \begin{align*} \langle \mathcal{L}_A^s u, v \rangle = \iint_{{\mathbb R}^{2n}} \big(u(x)-e^{iA_{x,y}}u(y)\big)\overline{\big(v(x)-e^{iA_{x,y}}v(y)\big)}K(x,y)\,dx\,dy, \end{align*} $$

where $A_{x,y} = (x-y)\cdot A(\frac {x+y}{2})$ . When u solves $\mathcal {L}_A^s u + q(x)u + a(x,u) = 0$ in $\Omega $ and $v\in H^s({\mathbb R}^n)$ with $ {\mathrm {supp}}(v|_{\Omega _e})\subset W_2$ , we obtain the identity:

(2.2) $$ \begin{align} \langle \mathcal{L}_A^s u, v \rangle = \int_{\Omega_e} \mathcal{L}_A^s u \cdot v\,dx - \int_\Omega (q(x)u + a(x,u))v\,dx. \end{align} $$

2.5 Runge approximation and unique continuation

A key tool in our analysis is the Runge approximation property for $\mathcal {L}_A^s+q$ , proved by Li [Reference Li13, Proposition 5.4].

Proposition 2.2 (Runge approximation)

Assume $\Omega \cup {\mathrm {supp}} A\subset B_r(0)$ and let ${W\subset \Omega _e}$ be open with $W\setminus \overline {B_{3r}(0)}\neq \emptyset $ . For $q\in L^\infty (\Omega )$ such that the problem $\mathcal {L}_A^s u+qu=0$ is well-posed, the set

$$\begin{align*}\mathcal{S}_A(q,W) := \{u|_\Omega : \mathcal{L}_A^s u+qu=0 \text{ in } \Omega,\ u|_{\Omega_e}\in C_c^\infty(W)\} \end{align*}$$

is dense in $L^2(\Omega )$ . Moreover, the same holds for the adjoint operator $\mathcal {L}_{-A}^s+q$ .

This is based on the strong uniqueness property of $\mathcal {L}^s$ and the fact that $\mathcal {L}_A^s u|_W = \mathcal {L}^s u|_W$ when $u\in \widetilde {H}^s(\Omega )$ and W is far from $\Omega $ .

Remark 2.3 (UCP for inhomogeneous equations)

The unique continuation property extends to inhomogeneous equations as follows: If $u_1, u_2 \in \widetilde {H}^s(\Omega )$ satisfy

$$\begin{align*}\mathcal{L}_A^s u_1 + q u_1 = f, \quad \mathcal{L}_A^s u_2 + q u_2 = f \quad \text{in } \Omega, \end{align*}$$

and $\mathcal {L}_A^s u_1|_W = \mathcal {L}_A^s u_2|_W$ for W as above, then $u_1 = u_2$ in ${\mathbb R}^n$ . The proof considers ${w = u_1 - u_2}$ , which satisfies $\mathcal {L}_A^s w + q w = 0$ in $\Omega $ and $\mathcal {L}_A^s w|_W = 0$ . Since $w \in \widetilde {H}^s(\Omega )$ , Li’s Lemma 5.2 gives $\mathcal {L}^s w|_W = 0$ , and the UCP for $\mathcal {L}^s$ implies $w \equiv 0$ .

Lemma 2.4 (Tensor product density)

Let $\mathcal {S} = \mathcal {S}_A(q,W_1)$ be as in Proposition 2.2. Then, the set

$$\begin{align*}\mathcal{T}_k := \{u_1 \ldots u_k : u_1, \dots, u_k \in \mathcal{S}\} \end{align*}$$

is dense in $L^1(\Omega )$ for each $k \geq 1$ .

Proof The case $k=1$ is Proposition 2.2. For $k \geq 2$ , we argue as follows. Since $\mathcal {S}$ is dense in $L^2(\Omega )$ and contains functions with nonvanishing product on any given compact set, the tensor products $\mathcal {T}_k$ are dense. The key observation is that if a function $\phi \in L^\infty (\Omega )$ satisfies

$$\begin{align*}\int_\Omega \phi(x) u_1(x) \ldots u_k(x) \, dx = 0 \quad \text{for all } u_1, \dots, u_k \in \mathcal{S}, \end{align*}$$

then by successively fixing $u_2, \dots , u_k$ and using density of $\mathcal {S}$ , we conclude $\phi = 0$ a.e.

4.1 Alessandrini-type identity

The following integral identity, proven by Li [Reference Li13, Proposition 4.7], is crucial for our uniqueness proof.

Lemma 4.1 (Alessandrini-type identity)

Let $q_j \in L^\infty (\Omega )$ be regular for $\mathcal {L}_A^s$ ( $j=1,2$ ). For $g_1, g_2 \in H^s(\mathbb {R}^n)$ , let $u_1^+ = P_{A,q_1}(g_1)$ be the unique solution to

$$\begin{align*}\mathcal{L}_A^s u_1^+ + q_1 u_1^+ = 0 \quad \text{in } \Omega, \qquad u_1^+|_{\Omega_e} = g_1, \end{align*}$$

and let $u_2^- = P_{-A,q_2}(g_2)$ be the unique solution to

$$\begin{align*}\mathcal{L}_{-A}^s u_2^- + q_2 u_2^- = 0 \quad \text{in } \Omega, \qquad u_2^-|_{\Omega_e} = g_2. \end{align*}$$

Then,

$$ \begin{align*} \langle (\Lambda_{A,q_1} - \Lambda_{A,q_2}) g_1, g_2 \rangle = \int_\Omega (q_1 - q_2) u_1^+ u_2^- \, dx. \end{align*} $$

4.2 Higher-order linearization

To recover the coefficients, we employ higher-order linearization. For $\varepsilon>0$ small and $f\in C_c^\infty (W_1)$ , let $u_\varepsilon $ solve (1.1) with $u_\varepsilon |_{\Omega _e}=\varepsilon f$ . The analyticity of $a(x,u)$ in u and the analytic implicit function theorem ensure that $\varepsilon \mapsto u_\varepsilon $ is analytic in a neighborhood of $\varepsilon =0$ with values in $C^s({\mathbb R}^n)$ [Reference Lai and Lin9, Reference Li15]. Thus, we have the convergent expansion

$$\begin{align*}u_\varepsilon = \sum_{k=1}^\infty \varepsilon^k u^{(k)}, \quad \text{where} \quad u^{(k)}(x) = \left.\frac{1}{k!}\frac{\partial^k}{\partial\varepsilon^k} u_\varepsilon(x)\right|_{\varepsilon=0}. \end{align*}$$

Differentiating (1.1) k times at $\varepsilon =0$ yields the following lemma.

Lemma 4.2 (Linearized equations)

Let $u_\varepsilon $ be the unique solution to (1.1) with exterior data $u_\varepsilon |_{\Omega _e} = \varepsilon f$ for $f \in C_c^\infty (W_1)$ . Then, the k-th order variation $u^{(k)} = \frac {\partial ^k}{\partial \varepsilon ^k} u_\varepsilon |_{\varepsilon =0}$ satisfies

$$ \begin{align*} \mathcal{L}_A^s u^{(k)} + q u^{(k)} + \mathcal{H}_k(x) = 0 \quad \text{in } \Omega, \qquad u^{(k)}|_{\Omega_e} = \delta_{k1} f, \end{align*} $$

where $\mathcal {H}_1 = 0$ and for $k \geq 2$ ,

$$\begin{align*}\mathcal{H}_k(x) = k! \sum_{m=2}^k \frac{a_m(x)}{m!} \sum_{\substack{\ell_1+\cdots+\ell_m=k \\ \ell_j \geq 1}} \frac{u^{(\ell_1)} \ldots u^{(\ell_m)}}{\ell_1! \ldots \ell_m!}. \end{align*}$$

Proof By Theorem 3.1, for sufficiently small $\varepsilon $ , the solution $u_\varepsilon $ depends analytically on $\varepsilon $ . We can thus express $u_\varepsilon $ as the power series:

(4.1) $$ \begin{align} u_\varepsilon(x) = \sum_{k=1}^\infty \frac{\varepsilon^k}{k!} u^{(k)}(x). \end{align} $$

Since $a(x, u)$ is analytic in u with $a(x, 0) = \partial _u a(x, 0) = 0$ , its Taylor expansion is

$$ \begin{align*} a(x, u_\varepsilon) = \sum_{m=2}^\infty \frac{a_m(x)}{m!} (u_\varepsilon)^m, \quad \text{where } a_m(x) = \partial_u^m a(x, 0). \end{align*} $$

Substituting (4.1) into the equation $\mathcal {L}_A^s u_\varepsilon + q u_\varepsilon + a(x, u_\varepsilon ) = 0$ , we obtain

$$ \begin{align*} \sum_{k=1}^\infty \frac{\varepsilon^k}{k!} (\mathcal{L}_A^s u^{(k)} + q u^{(k)}) + \sum_{m=2}^\infty \frac{a_m(x)}{m!} \left( \sum_{j=1}^\infty \frac{\varepsilon^j}{j!} u^{(j)} \right)^m = 0. \end{align*} $$

To find the equation for a specific $u^{(k)}$ , we apply the Faà di Bruno formula to the nonlinear term. The coefficient of $\varepsilon ^k$ in the expansion of $a(x, u_\varepsilon )$ is exactly

$$ \begin{align*} \left. \frac{1}{k!} \frac{\partial^k}{\partial \varepsilon^k} a(x, u_\varepsilon) \right|_{\varepsilon=0} = \sum_{m=2}^k \frac{a_m(x)}{m!} \sum_{\substack{\ell_1+\cdots+\ell_m=k \\ \ell_j \geq 1}} \frac{u^{(\ell_1)} \ldots u^{(\ell_m)}}{\ell_1! \ldots \ell_m!}. \end{align*} $$

Equating the terms of order $\varepsilon ^k$ to zero yields the desired recursive system of linear equations. For the boundary conditions, $u_\varepsilon |_{\Omega _e} = \varepsilon f$ implies $u^{(1)}|_{\Omega _e} = f$ and ${u^{(k)}|_{\Omega _e} = 0}$ for $k> 1$ .

5.1 Step 1: Recovery of q

From the equality of DN maps, their first derivatives coincide: $\Lambda ^{(1)}_{A,q_1,a_1} = \Lambda ^{(1)}_{A,q_2,a_2}$ . But $\Lambda ^{(1)}_{A,q_j,a_j} = \Lambda _{A,q_j}$ , the DN map for the linear magnetic Schrödinger equation.

Take $f_1, f_2 \in C_c^\infty (W_1)$ . By Lemma 4.1 (Li’s Proposition 4.7), we have

$$\begin{align*}0 = \langle (\Lambda_{A,q_1} - \Lambda_{A,q_2}) f_1, f_2 \rangle = \int_\Omega (q_1 - q_2) u_1^+ u_2^- \, dx, \end{align*}$$

where $u_1^+ \kern1.3pt{=}\kern1.3pt P_{A,q_1}(f_1)$ solves $\mathcal {L}_A^s u_1^+ + q_1 u_1^+ \kern1.3pt{=}\kern1.3pt 0$ in $\Omega $ with $u_1^+|_{\Omega _e}\kern1.3pt{=}\kern1.3pt f_1$ , and ${u_2^- = P_{-A,q_2}(f_2)}$ solves $\mathcal {L}_{-A}^s u_2^- + q_2 u_2^- = 0$ in $\Omega $ with $u_2^-|_{\Omega _e}=f_2$ . Let $\mathcal {S}_A(q_j,W_1)$ and $\mathcal {S}_{-A}(q_j,W_1)$ be as in Proposition 2.2. Since these sets are dense in $L^2(\Omega )$ , the set of products $\{u_1^+ u_2^- : f_1, f_2 \in C_c^\infty (W_1)\}$ is dense in $L^1(\Omega )$ (by Lemma 2.4 with $k=2$ ). Therefore, we conclude $q_1 = q_2$ a.e. in $\Omega $ . Denote $q:=q_1=q_2$ .

5.2 Step 2: Recovery of $a_2(x)=\partial _u^2 a(x,0)$

From the equality of DN maps, their second derivatives also coincide. Let ${f \in C_c^\infty (W_1)}$ and let $u^{(1)}$ solve $\mathcal {L}_A^s u^{(1)} + q u^{(1)} = 0$ with $u^{(1)}|_{\Omega _e} = f$ . By Lemma 4.2, the second variations $u^{(2)}_j$ satisfy for $j=1,2$ :

$$\begin{align*}\mathcal{L}_A^s u^{(2)}_j + q u^{(2)}_j + a_{2,j}(x) (u^{(1)})^2 = 0 \quad \text{in } \Omega, \qquad u^{(2)}_j|_{\Omega_e}=0. \end{align*}$$

Since $\Lambda ^{(2)}_{A,q,a_1} = \Lambda ^{(2)}_{A,q,a_2}$ , we have $\mathcal {L}_A^s u^{(2)}_1|_W = \mathcal {L}_A^s u^{(2)}_2|_W$ on $W_2$ . By the unique continuation property for inhomogeneous equations (Remark 2.3), we deduce $u^{(2)}_1 = u^{(2)}_2$ in ${\mathbb R}^n$ . Subtracting the equations gives

$$\begin{align*}(a_{2,1}(x) - a_{2,2}(x)) (u^{(1)})^2 = 0 \quad \text{in } \Omega. \end{align*}$$

Now, consider the integrated form: For any $f_1, f_2 \in C_c^\infty (W_1)$ with corresponding solutions $u_1, u_2$ of $\mathcal {L}_A^s u + q u = 0$ , we have

$$\begin{align*}\int_\Omega (a_{2,1}(x) - a_{2,2}(x)) u_1(x) u_2(x) \, dx = 0. \end{align*}$$

Let $\mathcal {S} = \mathcal {S}_A(q,W_1)$ . By Lemma 2.4 with $k=2$ , the set $\{u_1 u_2 : u_1, u_2 \in \mathcal {S}\}$ is dense in $L^1(\Omega )$ . Therefore, $a_{2,1}(x) = a_{2,2}(x)$ a.e. in $\Omega $ .

5.3 Step 3: Induction for higher-order coefficients

We proceed by induction. Assume for some $m \geq 2$ , we have shown

$$\begin{align*}a_{k,1}(x) = a_{k,2}(x) \quad \text{in } \Omega, \quad \text{for } k=2,\dots,m, \end{align*}$$

and consequently $u^{(k)}_1 = u^{(k)}_2$ for $k=1,\dots ,m$ . Consider the $(m+1)$ -th linearization. By Lemma 4.2 and the induction hypothesis, the difference $w = u^{(m+1)}_1 - u^{(m+1)}_2$ satisfies

$$\begin{align*}\mathcal{L}_A^s w + q w + (a_{m+1,1} - a_{m+1,2}) (u^{(1)})^{m+1} = 0 \quad \text{in } \Omega, \qquad w|_{\Omega_e}=0. \end{align*}$$

Since $\Lambda ^{(m+1)}_{A,q,a_1} = \Lambda ^{(m+1)}_{A,q,a_2}$ , we have $\mathcal {L}_A^s w|_W = 0$ on $W_2$ . By Remark 2.3, $w \equiv 0$ , so

$$\begin{align*}(a_{m+1,1}(x) - a_{m+1,2}(x)) (u^{(1)})^{m+1} = 0 \quad \text{in } \Omega. \end{align*}$$

Now, for any $f_1, \dots , f_{m+1} \in C_c^\infty (W_1)$ with corresponding solutions $u_1, \dots , u_{m+1} \in \mathcal {S}$ , we have

$$\begin{align*}\int_\Omega (a_{m+1,1}(x) - a_{m+1,2}(x)) u_1(x) \ldots u_{m+1}(x) \, dx = 0. \end{align*}$$

By Lemma 2.4 with $k=m+1$ , the set $\{u_1 \ldots u_{m+1} : u_1, \dots , u_{m+1} \in \mathcal {S}\}$ is dense in $L^1(\Omega )$ . Therefore, $a_{m+1,1}(x) = a_{m+1,2}(x)$ a.e. in $\Omega $ . This completes the induction step.

5.4 Step 4: Recovery of the full nonlinearity

Since $a_j(x,u)$ are analytic in u and all Taylor coefficients at $u=0$ coincide, we conclude

$$\begin{align*}a_1(x,u) = a_2(x,u) \quad \text{in } \Omega\times{\mathbb R}. \end{align*}$$

This completes the proof of Theorem 1.1.