Proof. Define \(v(r)=rQ(r)\), which satisfies \(v(r)\to 0\) as \(r\to\infty\) and \[v_{rr} - W(r) v = 0,\quad\text{where}\quad W(r):=1+\tfrac{a}{r^2} -\tfrac{v^2}{r^2}.\] Now, given \(\epsilon>0\), let \(R_0\geq 1\) be large enough that \(W(r)\geq 1-\epsilon\) for any \(r\geq R_0\). Let us also choose \(C_0\) large enough that \[v(R_0)\leq C_0 e^{-\lambda R_0},\quad\text{where}\quad \lambda:=\sqrt{1-\epsilon}.\]

We now introduce \[z(r) = v(r) - C_0 e^{-\lambda r},\] which satisfies \(z(R_0)\leq0\) and \(z(r)\to 0\) as \(r\to\infty\). Thus \[z''(r)-W(r)z(r) = -C_0 e^{-\lambda r}(\lambda^2-W(r))>0 \quad\text{for all}\quad r>R_0.\] We now claim that \(z(r)\leq 0\) for all \(r\geq R_0\). Indeed, suppose instead that \(z\) attains a positive maximum at \(r=r_0\geq R_0\), so that \(z''(r_0)\leq 0\). As \(W(r_0)>0\), this yields \[z''(r_0)-W(r_0)z(r_0)<0,\] a contradiction. We therefore obtain \(z(r)\leq 0\) for \(r\geq R_0\), which in turn implies \[Q(r) \lesssim \tfrac{1}{r}e^{-\lambda r},\] as desired. ◻

Proof. Let \(u_1,u_2\) be as in Lemma 9. We change variables via \(s=-\ln(r)\) and define \(v_1,v_2\) by \[v_1 = \tfrac{1}{\beta}\bigl[(-\tfrac{1}{2}+\tfrac{\beta}{2})u_1-u_2\bigr]\quad\text{and}\quad v_2 = \tfrac{1}{\beta}\bigl[(\tfrac{1}{2}+\tfrac{\beta}{2})u_1+u_2\bigr].\] Letting \('\) denote \(\tfrac{d}{ds}\), we then obtain the following system for \(v_1,v_2\): \[\begin{align} v_1' &= \tfrac{\beta}{2}v_1 + \tfrac{1}{\beta} e^{-2s}(v_1+v_2) - \tfrac{1}{\beta} e^{-s}|v_1+v_2|^2(v_1+v_2), \\ v_2' &=-\tfrac{\beta}{2}v_2 - \tfrac{1}{\beta} e^{-2s}(v_1+v_2)+\tfrac{1}{\beta} e^{-s}|v_1+v_2|^2(v_1+v_2). \end{align}\] Using Lemma 9, we have that \(v_j\to 0\) as \(s\to\infty\) for \(j=1,2\). Thus, multiplying the first equation by \(e^{-\frac{1}{2}\beta s}\) and integrating, we can obtain that \[v_1(s)=\mathcal{O}(e^{-s})\quad\text{as}\quad s\to\infty.\] Similarly, multiplying the second equation by \(e^{\frac{1}{2}\beta s}\) and integrating to infinity, we can derive that \[v_2(s) = C e^{-\frac{1}{2}\beta s}+\mathcal{O}(e^{-s})\quad\text{as}\quad s\to\infty.\]

With these bounds in hand, we can iterate to obtain \[v_1(s) = \mathcal{O}_k(e^{-ks}) \quad\text{and}\quad v_2(s) = Ce^{-\frac{1}{2}\beta s}+ \mathcal{O}_k(e^{-ks})\] for some \(k\geq \tfrac12\beta\) and some \(C\neq 0\). Returning to the variables \(u_1\) and \(u_2\), we obtain \(u_j(r(s)) \sim e^{-\frac{1}{2}\beta s}\), which translates back into the desired asymptotic behavior for \(Q\) and \(Q'\) as \(r\to 0\). ◻

Proof of Theorem 7. We continue to denote \(\beta=\sqrt{1+4a}\). Under the change of variables \(v(r) = r^{\frac{1}{2}-\frac{\beta}{2}} Q(r)\), the equation 2 takes the form \[v_{rr} + \tfrac{f_r(r)}{f(r)} v_r + g(r)v + h(r) v^3=0\] where \[f(r)=r^{1+\beta},\quad g(r) = -1,\quad\text{and}\quad h(r) = r^{-1 +\beta}.\] To obtain uniqueness, we now check conditions (1)–(7).

Condition (1) is immediate, by inspection.

For condition \((2)\), we first compute \[\begin{align} \label{Cond1} f(\tau)(|g(\tau)| + h(\tau)) = \tau^{1+\beta} + \tau^{2\beta}, \end{align}\tag{3}\] so that \[\int_0^r f(\tau)(|g(\tau)| + h(\tau)) d\tau = \tfrac{r^{2+\beta}}{2+\beta} + \tfrac{r^{1+2\beta}}{1+2\beta}.\] Dividing by \(f(r)\) and letting \(r \to 0^+\) yields \((2)\).

To verify condition \((3)\), we let \(R=1\) and first note that \((i)\) and \((iii)\) follow immediately from the fact that \(a>0\). To check \((ii)\), we first compute \[\int_\tau ^1\tfrac{d \rho}{f(\rho)} = \tfrac{1}{\beta}\left( \tau^{-\beta}-1\right).\] Combining this with 3 , we have \[\begin{align} f(\tau)(|g(\tau)| + h(\tau)) \int_\tau ^1 \tfrac{d \rho}{f(\rho)} &= \tfrac{1}{\beta}(\tau +\tau^{\beta} - \tau^{1+\beta}-\tau^{2\beta})\in L^1(0,1). \end{align}\]

For condition \((4)\), we compute \[a(r) = r^{\frac{5}{3} +\beta} \quad\text{and}\quad b(r) = (\tfrac{1}{6}-\tfrac{1}{2}\beta)r^{\frac{2}{3}+\beta},\] from which we derive that \(a,b,ag,ah\to 0\) as \(r\to 0\).

For condition \((5)\), we compute \[c(r) = -b_r(r) +\tfrac{f_r(r)}{f(r)}b(r) = \tfrac{1-3\beta}{18} r^{-\frac{1}{3} +\beta} \to 0 \quad\text{as}\quad r\to0.\]

Finally, for \((6)\) and \((7)\), we compute \[G(r) = -\tfrac{r^{\beta - \frac{4}{3}}}{108} \left[ 36\left(3\beta+2\right) r^2 + \left(3\beta-1\right)^2 \right].\] As \(a>0\), we obtain condition \((6)\) with \(R=0\), along with condition \((7)\).

Having verified conditions \((1)\)–\((7)\), we may now apply Theorem 11 to conclude uniqueness. ◻

Proof. The proof is an adaptation of the proof of [10]. We will equivalently prove \(L^2\)-coercivity under the appropriate orthogonality conditions for the operators \(\langle \sqrt{\mathcal{H}_a} \rangle^{-1} L_{\pm} \langle \sqrt{\mathcal{H}_a} \rangle^{-1}\). Writing \(\langle \sqrt{\mathcal{H}_a} \rangle^{-1} L_{\pm} \langle \sqrt{\mathcal{H}_a} \rangle^{-1} = I - \mathcal{K}_\pm\), we first prove that \(\mathcal{K}_\pm\) are compact on \(L^2\). Since \(\langle \sqrt{\mathcal{H}_a} \rangle^{-1}\) is continuous, it suffices to prove that \(Q_a^2 \langle \sqrt{\mathcal{H}_a} \rangle^{-1}\) is compact. To this end, we first note that this operator is bounded. Indeed, by Sobolev embedding, \[\begin{align} \| Q^2_a \langle \sqrt{\mathcal{H}_a} \rangle^{-1}v\|_{L^2} &\lesssim \|Q_a^2\|_{L^3} \|\langle \sqrt{\mathcal{H}_a} \rangle^{-1}v\|_{L^6} \lesssim \|v\|_{L^2}. \end{align}\] We next observe that \[\begin{align} \| |\nabla|^{\frac{1}{2}} &(Q_a^2 \langle \sqrt{\mathcal{H}_a} \rangle^{-1}v) \|_{L^2} \\ &\lesssim \| \nabla(Q_a^2 \langle \sqrt{\mathcal{H}_a} \rangle^{-1}v) \|_{L^{\frac{3}{2}}} \\ &\lesssim \|Q_a\|_{L^{12}}^2 \|\nabla \langle \sqrt{\mathcal{H}_a} \rangle^{-1} v\|_{L^2} + \|Q_a \nabla Q_a\|_{L^2}\| \langle \sqrt{\mathcal{H}_a} \rangle^{-1} v \|_{L^6} \lesssim \|v\|_{L^2}. \end{align}\] Finally, using the decay of the ground state away from the origin, we have \[\begin{align} \| Q_a^2 \langle \sqrt{\mathcal{H}_a} \rangle^{-1} v \|_{L^2(|x|>R)} \lesssim R^{-1} \|v\|_{L^2}. \end{align}\] The desired compactness now follows from Rellich–Kondrachov.

It follows that the eigenvalues of \(I-\mathcal{K}_\pm\) are discrete, and since the identity acts as a spectral shift, they can only accumulate at 1. Moreover, \(I-\mathcal{K}_\pm\) has a finite number of eigenvalues \(\{\lambda_i^\pm\}_{i=1}^N\) in \((-\infty, \tfrac12]\), which we assume are nondecreasing. Finally, by Weyl’s Theorem, the essential spectrum is \([1, \infty)\).

We first prove that there are orthogonal conditions that guarantee that \(L_\pm\) are nonnegative. By construction, \(Q_a\) minimizes the Weinstein functional \[W(f) = \frac{\left(\displaystyle \int | \nabla f|^2\,dx + \int \frac{a |f|^2}{|x|^2 } \,dx \right)^{3/2} \left( \displaystyle \int |f|^2 \,dx\right)^{1/2}}{ \displaystyle\int |f|^4\,dx}.\] Thus \(\frac{d^2}{d \epsilon^2} W(Q_a + \epsilon v)|_{\epsilon=0} \geq 0\). Writing \(v = v_1 + iv_2\) and assuming \((v_1,\mathcal{H}_a Q_a)=0\), we apply ?? to obtain \[\begin{align} (L_+v_1, v_1) + (L_- v_2, v_2) \geq \frac{1}{\| Q_a\|_2^2} \left( \int Q_a v_1\right)^2. \end{align}\] It follows that \(L_-\) is always nonnegative and that \(L_+\) is nonnegative if \((v_1, \mathcal{H}_a Q_a) = 0\). Now, by ?? , we have that \(\lambda_1^- = 0\), and we have now shown that although \(\lambda_1^+ <0\), we have \(\lambda_2^+ \geq 0\). To finish the proof, it will therefore suffice to show that \(\lambda_2^\pm >0\). To this end, we will prove that \[\text{ker} L_+=\emptyset\quad\text{and}\quad\text{ker} L_-=\text{span}\{Q_a\}.\]

Let us first prove that \(\text{ker} L_- = \text{span}\{Q_a\}\).³ Suppose \(f_0\) were another independent radial solution to \(L_- f = 0\). Thus, \[\begin{align} (f_0)_{rr} + \tfrac{2}{r}(f_0)_r - \frac{a}{r^2} f_0 + Q_a^2 f_0 -f_0 = 0. \end{align}\] Taking the Wronskian and using Abel’s theorem, we find \[\begin{align} \label{abel} (f_0)_r Q_a - (Q_a)_r f_0 = \tfrac{C}{r^2},\quad\text{so that}\quad\bigl(\tfrac{f_0}{Q_a}\bigr)_r = \tfrac{C}{Q_a^2 r^2}. \end{align}\tag{8}\] From the asymptotics for ground state, we know that \(Q_a \sim r^{-\frac{1}{2} + \frac{1}{2}\beta}\) as \(r\to 0\), where we denote \(\beta=\sqrt{1+4a}\) as above. Thus, integrating the above equality and using 8 , we obtain \[f_0 \sim r^{-(\frac{1}{2} + \frac{1}{2}\beta)}\quad\text{and}\quad (f_0)_r \sim r^{-(\frac{3}{2} + \frac{1}{2}\beta)} \quad\text{as}\quad r\to 0.\] In particular, \(f_0\notin H^1\). It follows that \(\text{ker}(L_{-}) = \text{span}(Q_a)\).

We next prove that \(\text{ker} L_+=\emptyset\). We suppose that \(L_+\) has a nonzero radial element \(v_0\) in its kernel. As \(L_+\) has a single negative eigenvalue, Sturm–Liouville theory guarantees that \(v_0\) has a single zero at some \(r=r_0\). Now observe that \(Q_a\) and \(Q_a^3\) are in the range of \(L_+\); indeed, we have \[L_+Q_a = -2Q_a^3 \quad\text{and}\quad L_+(Q_a+x\cdot\nabla Q_a)=-2Q_a.\] Thus, since \(L_+\) is self-adjoint, we have \(v_0\perp Q_a,Q_a^3\). In particular, \[\int v_0(r)[Q_a^3(r) - Q_a^2(r_0)Q_a(r)]\,dr=0.\] However, because \(Q_a\) is strictly decreasing and \(v_0\) changes sign only at \(r_0\), the integrand above has a constant sign. Thus we obtain a contradiction and conclude that \(L_+\) has no radial kernel.

Finally, let us upgrade to the coercivity statement for \(L_\pm\). Writing \(\tilde{L}_\pm = \langle \sqrt{\mathcal{H}_a} \rangle^{-1} L_\pm \langle \sqrt{\mathcal{H}_a} \rangle^{-1} = I+\mathcal{K}_\pm\), we suppose towards a contradiction that there exists an \(L^2\)-normalized sequence \(v_n\) such that \(\langle \tilde{L}_\pm v_n,v_n\rangle \to 0\), with \(v_n\) satisfying the appropriate orthogonality condition depending on the \(\pm\) sign. We may assume that there exists \(v\) such that \(v_n\rightharpoonup v\) weakly in \(L^2\), and note that \(v\) then satisfies the same orthogonality condition as the \(v_n\). Recalling that \(\mathcal{K}_\pm\) are compact on \(L^2\), we now obtain that \(\mathcal{K}_\pm v_n \to \mathcal{K}_\pm v\) strongly in \(L^2\). Using weak lower-semicontinuity of the norm, we therefore obtain \[\langle \tilde{L}_\pm v,v\rangle \leq \liminf_{n\to\infty} \langle \tilde{L}_\pm v_n,v_n\rangle =0.\] By the nonnegativity established above, this forces \(L_\pm v =0\) as well as \(\|v\|_{L^2}^2=1\). In particular, we have the strong convergence \(v_n\to v\) in \(L^2\) and \(v\neq 0\). But then in the case of \(L_-\), we obtain a contradiction due to the fact that \(v\perp Q_a\), while in the case of \(L_+\) we obtain a contradiction due to the fact that \(\text{ker}L_+=\emptyset\). ◻

Proof. Again, we follow the lead of [10]. The decomposition of \(H^1_{\text{rad}}(\mathbb{R}^3)\) and item \((1)\) follow directly from [22].

Recalling from the proof of the previous proposition that \(L\) has exactly one negative direction, [22] states that \[\dim \, E^u + k^{\leq 0}_0 = 1,\] where \(k^{\leq 0}_0\) is the number of nonpositive directions of \(\langle L \cdot, \cdot\rangle\) on the generalized kernel of \(\mathcal{L}\) modulo \(\text{ker} L\). In order to prove that \(k^{\leq 0}_0 = 0\), we suppose \(v\) satisfies \((JL)^2 v =0\) and \(Lv\neq 0\) and we must verify that \(\langle Lv,v\rangle>0\). Writing \(v=[f\;g]^T\), we therefore assume \(L_-L_+ f=0\) and \(L_+ L_-g=0\). By the analysis above, we must have \(L_+f = cQ_a\) and \(L_-g=0\), and we may assume \(g=0\). On the other hand, \(L_+f = cQ_a\) guarantees that \(c\neq 0\), and in fact we must have \(f=-\tfrac{c}{2}(Q_a+x\cdot\nabla Q_a)\) (as this is the unique solution to \(L_+f = cQ_a\)). The desired positivity then follows from directly verifying that \[(L_+(Q_a+x\cdot\nabla Q_a),Q_a+x\cdot\nabla Q_a)>0.\] Thus \(\dim E^u =1\), which yields (2).

Finally, (3) follows from [22]. It follows that there exists \(e_0 \in \mathbb{C}\) with \(\operatorname{Re}(e_0)>0\) and \(Y_+ = Y_1 + i Y_2\) in \(H^1_{\text{rad}}(\mathbb{R}^3)\) satisfying \(L_- Y_2 = e_0 Y_1\) and \(L_+Y_1 = -e_0 Y_2\). As \(L_{\pm}\) are real valued operators, we conclude that \(e_0 \in \mathbb{R}\). ◻

Proof. As this part is fairly standard, we omit most of the details. The basic idea is to write the system satisfied by \(Y_1\) and \(Y_2\) and utilize an elliptic regularity argument to demonstrate arbitrarily good decay away from zero. Indeed, we have \[\begin{align} \begin{cases} (1- \Delta)Y_1 = -\frac{a}{|x|^2} Y_1 + 3 Q_a^2Y_1 - e_0 Y_2, \\ (1- \Delta)Y_2 = -\frac{a}{|x|^2} Y_2 + Q_a^2Y_2 + e_0 Y_1. \end{cases} \end{align}\] We can therefore apply an inductive argument to obtain arbitrarily fast decay of the \(H^k\) norms of \(Y_1\) and \(Y_2\). ◻

Proof. The first assertion can be established via the argument used to obtain Proposition 16. Let us therefore focus on proving \(f\in H^{1,\frac{30}{11}}\), and in fact let us simply show the result for the component \(f_1\), which satisfies \[\mathcal{H}_a f_1 = -f_1 + 3Q_a^2 f_1 + \lambda f_2 + F_2.\] Using Lemma 1, self-adjointness of \(\mathcal{H}_a\), and the equation above, we obtain \[\begin{align} \| |\nabla|^{\frac{7}{5}} f_1 \|_2^2 & \lesssim ( \mathcal{H}_a^{\frac{7}{10}} f_1,\mathcal{H}_a^{\frac{7}{10}}f_1) \\ & = ( \mathcal{H}_a f_1,\mathcal{H}_a^{\frac{2}{5}} f_1) \\ & \lesssim |( f_1,\mathcal{H}_a^{\frac{2}{5}}f_1)|+|(3Q_a^2 f_1,\mathcal{H}_a^{\frac{2}{5}} f_1)|+|(\lambda f_2,\mathcal{H}_a^{\frac{2}{5}}f_1)|+|(F_2,\mathcal{H}_a^{\frac{2}{5}}f_1)|\\ &\lesssim \|f_1\|_{H^1}^2 + \|Q_a^2\|_{L^3} \|f_1\|_{H^1}^2 + (\lambda \|f_2\|_{L^2} + \|F_2\|_{L^2}) \|f_1\|_{H^1}\lesssim 1. \end{align}\] To complete the argument, we use the Sobolev embedding \(f\in \dot{H}^{1+\frac{2}{5}}\hookrightarrow \dot{H}^{1,\frac{30}{11}}.\) ◻

Proof. Observe that \[\begin{align} \|u(t) - e^{it}Q \|_{H^1} &\lesssim \| u(t) - U^A(t) \|_{H^1} + \| V^A(t)\|_{H^1} \\ &\lesssim e^{-ct} + e^{-e_0 t}. \end{align}\] Therefore, by Proposition 22, there exists \(A_0\) such that \(u = U^{A_0}\). Moreover, \[\|U^A(t) - U^{A_0}(t)\|_{H^1} = \| V^A(t) - V^{A_0}(t) \|_{H^1} \sim |A - A_0| e^{-e_0 t} + \mathcal{O}(e^{-2e_0 t}).\] As \(c > e_0\), we conclude that \(A = A_0\). ◻

Proof. We estimate \[\begin{align} \| e^{iT_A} U^A(t) - U^{+1}(t+ T_A) \|_{H^1} &= \| e^{iT_A} U^A(t) - e^{i(t+ T_A)} \big( Q_a + e^{-e_0(t+T_A)} Y_+ \big) \|_{H^1} \\ & \quad\quad + \mathcal{O}(e^{-2e_0 t}) \\ &= \| U^A(t) - e^{it}(Q_a + A e^{-e_0 t} Y_+) \|_{H^1} + \mathcal{O}(e^{-2e_0 t}) \\ &= \mathcal{O}(e^{-2e_0 t}), \end{align}\] where we choose \(T_A = -\tfrac{\ln A}{e_0}.\) The previous corollary then yields the result. Arguing analogously yields the result for \(A<0\) as well. ◻

Proof. Suppose the statement were false. Then there exists \(\epsilon_0>0\) and \(\{f_n\}\subset H_{\text{rad}}^1\) satisfying \(M(f_n)=M(Q_a)\), \(E_a(f_n)=E_a(Q_a)\), \(d(f_n)\to 0\), and \[\label{nope} \inf_{\theta\in\mathbb{R}}\|f_n-e^{i\theta}Q_a\|_{H_a^1}\geq\epsilon_0.\tag{9}\] In particular, it follows that \[\label{fnQa} \|f_n\|_{L^4} \to \|Q_a\|_{L^4}.\tag{10}\]

As \(f_n\) is bounded in \(H^1\), we may pass to a subsequence to obtain a weak limit \(\phi\in H^1\). By the compactness of the embedding \(H^1_{\text{rad}}\hookrightarrow L^4\), it follows that \(f_n\to\phi\) strongly in \(L^4\). By weak lower-semicontinuity followed by the mass and energy constraints and 10 , we can derive that \[\begin{align} C_a \|\phi\|_{L^2}\|\phi\|_{\dot{H}^1_a}^3 \leq C_a \|Q_a\|_{L^2}\|Q_a\|_{\dot{H}_a^1}^3 = \|Q_a\|_{L^4}^4 = \|\phi\|_{L^4}^4. \end{align}\] It follows that \(\phi\) is an optimizer for the Gagliardo–Nirenberg inequality (whence \(\phi=e^{i\theta}Q_a\)), and moreover that the weak \(H^1\) convergence may be upgraded to strong convergence. In particular, \(f_n\to e^{i\theta}Q_a\) in \(H^1\), contradicting 9 . ◻

Proof. The arguments needed to prove Proposition 28 are by now fairly standard. This is especially true in the radial setting, in which no translation parameters appear in the linear profile decomposition (and hence the additional challenges due to the broken translation symmetry of 1 vanish completely). Therefore, we will proceed by giving a fairly brief summary of the arguments.

Let \(\{\tau_n\}\subset [0,\infty)\) be a sequence of times. By continuity of the flow, we may assume that \(\tau_n\to \infty\). Applying the linear profile decomposition, Proposition 4, we write for any finite \(J\leq J^*\) \[u(\tau_n)=\sum_{j=1}^J\phi_n^j+r_n^J,\quad\text{with}\quad\phi_n^j=e^{it_n^j\mathcal{H}_a}\phi^j(x)\] along a subsequence with all the properties detailed in Proposition 4. We will show that \(J^*=1\), \(t_n^1\equiv 0\), and that the remainder \(r_n^J\) vanishes strongly in \(H^1\) as \(n\to\infty\), which will yield the result. To this end, we will show that all other scenarios contradict ?? .

We can quickly rule out the case \(J^*=0\), for in that case an application of the stability theory with the approximate solutions \(e^{-it\mathcal{H}_a}u(t_n)\) implies space-time bounds for \(u\) on \([t_n,\infty)\) for all \(n\) large, contradicting ?? .

It follows that \(J^*\geq 1\). Now, the decoupling statement in the linear profile decomposition and Proposition 13 imply \[\limsup_{n\to\infty}\|\phi_n^j\|_{L_x^2}\|\phi_n^j\|_{\dot{H}_a^1}\leq\mathcal{K}_a.\] Thus the sharp Gagliardo–Nirenberg inequality implies that \(\liminf_n E_a(\phi_n^j)\geq 0\) for all \(j\). Similar considerations show that \(\liminf_{n} E_a(r_n^J)\geq 0\) for any finite \(J\). Thus there are two possible cases: \[\begin{align} &\text{Case 1:}\quad\sup_j\limsup_n M(\phi_n^j)E_a(\phi_n^j)=\mathcal{E}_a,\tag{11}\\ &\text{Case 2:}\quad \sup_j\limsup_n M(\phi_n^j)E_a(\phi_n^j)\leq \mathcal{E}_a-3\delta\quad\text{for some}\quad \delta>0.\tag{12} \end{align}\]

We now show that 12 is inconsistent with ?? . Arguing as above, we see that for any \(j\) we have that \[\|\phi_n^j\|_{L_x^2}\|\phi_n^j\|_{\dot{H}_a^1}< (1-\delta')\mathcal{K}_a,\] for \(n\) sufficiently large. We associate to each \(\phi^j\) a corresponding scattering solution to 1 . If \(t_n^j\equiv0\) we define \(v^j=v_n^j\) as the global scattering solution to 1 with initial data \(\phi^j\). If instead \(t_n^j\to \pm\infty\) we define \(v^j\) as the global solution to 1 that scatters to \(\phi^j\) as \(t\to\pm\infty\), and then define \(v_n^j(t,x)=v^j(t+t_n^j,x)\).

We then define an approximate solution to 1 by \[u_n^J=\sum_{j=1}^Jv_n^j+e^{it\mathcal{H}_a}r_n^J.\] By construction we have \(u_n^J(0)-u(t_n)\to 0\) in \(H^1\) as \(n\to\infty\). We claim that \[\begin{align} &\limsup_{J\to J^*}\limsup_{n\to\infty}\{\|u_n^J(0)\|_{H_a^1}+\|u_n^J\|_{L_{t,x}^5}\}\lesssim 1,\tag{13}\\ &\limsup_{J\to J^*}\limsup_{n\to\infty}\||\nabla|^\frac{1}{2}e_n^J\|_{L_{t,x}^\frac{10}{7}}=0, \tag{14} \end{align}\] where the error \(e_n^J\) is defined by \[e_n^J=(i\partial_t-\mathcal{H}_a)u_n^J+|u_n^J|^2u_n^J.\] In this case, we can apply Proposition 5 and obtain a contradiction to ?? .

Both 13 and 14 rely on the following orthogonality result, which follows from ?? and approximation by \(C_c^\infty\) functions.

Using Lemma 3 and the space-time bounds obeyed by each individual solution \(v_n^j\), we can firstly obtain 13 . To estimate the error and obtain 14 , we write \(F(z)=-|z|^2z\) and decompose \(e_n^J\) as follows \[\begin{align} e_n^J&=\sum_{j=1}^JF(v_n^j)-F\bigl(\sum_{j=1}^Jv_n^j\bigr)\tag{15}\\ &\quad+F(u_n^J-e^{-it\mathcal{H}_a}r_n^J)-F(u_n^J)\tag{16} \end{align}\] For 15 , we can use pointwise bounds and Lemma 3 to obtain vanishing in \(L_{t,x}^{\frac{10}{7}}\). Interpolating this with boundedness in the presence of slightly more than \(\tfrac12\) derivatives (which follows from the global space-time bounds of the individual solutions \(v_n^j\)), we obtain the desired estimate. For 16 , we similarly use interpolation, this time relying on ?? to obtain vanishing with no derivatives.

At this point, we have shown that 11 holds. In particular, using \[\liminf_{n}E_a(\phi_n^j)\geq 0\quad\text{and}\quad\liminf_{n}E_a(r_n^J)\geq 0,\] we derive that \(J^*=1\), and so \[u(\tau_n)=\phi_n^1+r_n^1,\quad\text{with}\quad r_n^1\to 0\quad\text{strongly in}\quad H^1.\] To conclude the argument, it remains to see that \(t_n^1\equiv 0\). Indeed, if instead \(t_n^1\to\pm\infty\), then approximation by free solutions contradicts ?? . ◻

Proof. The proof follows along the lines of previous works such as [7], [9], [14], so we will keep our presentation brief. We rely on the following virial identity (see [3]): Let \(R \in [1,\infty]\) and let \(\phi\) be a real-valued radial function satisfying \[\begin{align} \phi(x)=\begin{cases} |x|^2 & |x|\leq 1\\ \text{const.}& |x|>3, \end{cases}\quad\text{with}\quad |\nabla \phi|\leq 2|x|,\quad |\partial_{jk}\phi|\leq 2. \end{align}\] We then define \[\label{E:VirialQuantity} P_R[u]=2\,\mathrm{Im}\int \overline{u}\,\nabla u\cdot\nabla w_R\,dx,\tag{17}\] where \(w_R(x)=R^2\phi\!\left(\frac{x}{R}\right)\) and \(w_\infty(x):=|x|^2\). We have the following identities: \[\tfrac{d}{dt} P_R[u(t)] = F_R[u(t)],\] where for finite \(R\) we have \[F_R[u]=\int [(-\Delta^2 w_R)|u|^2 +4\,\mathrm{Re}\,\overline{u}_j u_k\,\partial_{jk}(w_R) +4a|u|^2\tfrac{x}{|x|^4}\cdot\nabla w_R -|u|^4\Delta w_R]\,dx\] and (using the energy constraint and ?? ) \[F_\infty[u] = 8\bigl[\|u\|_{\dot{H}^1_a}^2-\tfrac{3}{4}\|u\|_{L^4}^4\bigr] =4\,d(u(t)).\]

We note that \(P_R[e^{i\theta}Q_a]=0\) and \(F_R[ e^{i \theta}Q_a] = 0\) for all \(R \in [1, \infty]\) and \(\theta \in \mathbb{R}\) (see e.g. [8], [14]).

Let \(R \geq 1\) (which will be chosen large enough below) and \(\delta_1 \in (0, \delta_0)\), where \(\delta_0\) is as in Proposition 26. We define the characteristic function \[\chi = \chi_J,\quad J = \{ t \in [t_1,t_2] : d(u(t)) < \delta_1 \},\] and its complement \(\chi^c = 1- \chi\). We write \[\begin{align} \label{auxscat} \tfrac{d}{dt} P_R[u] - F_\infty[u] &= \chi^c(t) (F_R[u] - F_\infty[u]) \nonumber\\ &\quad + \chi(t) \bigl(F_R[u]-F_\infty[u] - (F_R[e^{i \theta(t)}Q_a]- F_\infty[e^{i\theta(t)}Q_a])\bigr). \end{align}\tag{18}\]

We now estimate \(P_R[u(t_j)]\) as well as the right-hand side of this equality. We begin by fixing \(j \in \{1,2\}\). If \(d(u(t_j)) \geq \delta_1\) then \[|P_R[u(t_j)] | \lesssim R \|u\|^2_{L^\infty_t H^1_x} \lesssim \tfrac{R}{\delta_1} d(u(t_j)).\] If instead \(d(u(t_j)) < \delta_1\), we use the modulation analysis and estimate as follows: \[\begin{align} |P_R[u(t_j)] | &= \biggl| 2\,\operatorname{Im}\int \bigl( \bar{u}\nabla u - e^{-i \theta(t_j)}Q_a \nabla [e^{i\theta(t_j)}Q_a]\bigr) \cdot \nabla w_R \,dx \biggr| \\ &\lesssim R\bigl(\| \bar{u}\nabla (u(t_j)- e^{i\theta(t_j)}Q_a)\|_{L^1} + \|(\bar{u}- e^{-i\theta(t_j)}Q_a) \nabla e^{i\theta(t_j)}Q_a \|_{L^1}\bigr) \\ &\lesssim R\bigl(\|u\|_{L^2} + \| \nabla Q_a\|_{L^{\frac{6}{5}}}\bigr) d(u(t_j)) \\ &\lesssim R \,d(u(t_j)). \end{align}\]

To estimate the right-hand side we begin with the \(\chi^c\) term. As \(d(u(t)) \geq \delta_1\) on the support of \(\chi^c\), we rely directly on \(H^1\) precompactness. We begin by writing \[\begin{align} F_R[u] - F_\infty[u] = &\int_{|x|>R} (-\Delta^2 w_R)|u|^2 + 4\,\mathrm{Re}\, \overline{u_j} u_k\, \partial_{jk}(w_R) + 4a |u|^2 \tfrac{x}{|x|^4} \cdot\nabla w_R \\ &-|u|^4 \Delta w_R \,dx - 8\|u(t)\|_{\dot{H}^1_a(|x|>R)}^2 + 6 \|u(t)\|^4_{L^4(|x|>R)}. \end{align}\] Now, given \(\epsilon>0\), we choose \(R>1\) large enough such that all the above integrals are bounded by \(\epsilon\) uniformly for \(t \in [t_1,t_2]\backslash J\). Thus, we find \[\bigl|\chi^c(t)[F_R[u] - F_\infty[u]] \bigr| < \tfrac{\epsilon}{\delta_1} d(u(t)).\]

Next, we write \(Q_a(t) = e^{i\theta(t)}Q_a\) and expand the term involving \(\chi\) as \[\begin{align} &-8\bigl[\|u(t)\|_{\dot{H}^1_a(|x|>R)}^2-\|Q_a(t)\|_{\dot{H}^1_a(|x|>R)}^2\bigr] + 6 \bigl[\|u(t)\|^4_{L^4(|x|>R)}-\|Q_a(t)\|^4_{L^4(|x|>R)}\bigr] \\ &\quad + \int_{|x|>R} 4\,\operatorname{Re}\bigl[\overline{u_j}u_k -\overline{Q_a(t)}_jQ_a(t)_k\bigr] \partial_{jk}(w_R) - \bigl[|u|^2-|Q_a(t)|^2\bigr]\Delta^2w_R \,dx \\ &\quad - \int_{|x|>R} \bigl[|u|^4 - |Q_a(t)|^4\bigr] \Delta w_R -4a\bigl[|u|^2-|Q_a(t)|^2\bigr] \tfrac{x}{|x|^4} \cdot \nabla w_R \,dx. \end{align}\] In each term we can exhibit the difference \(u(t)-Q_a(t)\), which we estimate in \(H^1\) by \(d(u(t))\) (cf. Proposition 26). The remaining terms contain either \(u\) or \(Q_a\) norms on \(|x|>R\). By the compactness, we can take \(R\) large enough such that these norms are all \(\mathcal{O}(\epsilon)\). For example, \[\begin{align} \left| \int_{|x|>R} \bigl[|u|^4- |Q_a(t)|^4\bigr] dx \right| &\lesssim \int_{|x|>R} (|u|^3 + |Q_a(t)|^3) |u-Q_a(t)| \,dx \\ &\lesssim\bigl(\|u\|^3_{L^4(|x|>R)} + \|Q_a(t)\|^3_{L^4(|x|>R)}\bigr) \|u-Q_a(t)\|_{L^4} \\ &\lesssim \epsilon\, d(u(t)). \end{align}\] In this way we obtain \[\bigl| \bigl(F_R[u]-F_\infty[u] - (F_R[e^{i \theta(t)}Q_a]- F_\infty[e^{i\theta(t)}Q_a])\bigr) \bigr| \lesssim \epsilon\, d(u(t))\quad\text{for}\quad t\in J.\]

We now integrate 18 over \([t_1,t_2]\), using \(F_\infty[u] = 4d(u(t))\), \(\delta_1 \ll 1\), and the estimates above to obtain \[\int_{t_1}^{t_2} d(u(t))\, dt \leq \tfrac{CR}{\delta_1} \bigl[d(u(t_1)) + d(u(t_2))\bigr] + \tfrac{\epsilon}{\delta_1} \int_{t_1}^{t_2} d(u(t))\, dt.\] Choosing \(\epsilon=\epsilon(\delta_1)\) small enough we obtain the desired estimate. ◻

Proof of Proposition 29.. Combining Lemma 4, Corollary 30, and Corollary 27, we obtain \[\int_t^\infty d(u(s)) \, ds \leq C d(u(t))\] for all \(t>0\). Thus Gronwall’s inequality implies \[\int_t^\infty d(u(s)) \, ds \lesssim e^{-ct}\] for some \(c>0\) and all \(t>0\). Finally, we apply Corollary 27 to obtain \[\|u(t) - e^{i(t+ \theta_0)} Q_a\|_{H^1} \lesssim e^{-ct}\] for some \(\theta_0\in\mathbb{R}\) and all \(t>0\). ◻

Proof. We use the same notation for the virial quantities used in Lemma 4. We fix an interval \([t_1,t_2]\) and \(\delta_1\in (0,\delta_0)\), where \(\delta_0\) is as in Proposition 26. We define \[J=\{t\in [t_1,t_2]:d(t)<\delta_1\},\quad \chi=\chi_J,\quad\chi^c=1-\chi.\] As in Lemma 4, we have \[\begin{align} \tfrac{d}{dt}P_R[u]-F_\infty[u]&=\chi^c(t)\{F_R[u]-F_\infty[u]\}\tag{21}\\ &\quad+\chi(t)\{F_R[u]-F_\infty[u]-(F_R[e^{i\theta(t)}Q_a]-F_\infty[e^{i\theta(t)}Q_a])\}\tag{22}. \end{align}\]

For 21 , we write \[\begin{align} F_R[u]-F_\infty[u]&= \int_{|x|>R}4[\operatorname{Re}\bar{u}_j\partial_{jk}w_Ru_k]-8|\nabla u|^2\,dx\tag{23}\\ &\quad +\mathcal{O}\biggl(\int_{|x|>R}R^{-2}|u|^2+|u|^4\,dx\biggr)\tag{24}. \end{align}\] The properties of the virial weight \(\phi\) guarantee that \(\eqref{E:UncVir1461}\leq 0\). For the first term in 24 we use the support properties of \(\chi^c\) to estimate \[R^{-2}\|u\|_{L^2}^2\lesssim R^{-2}\tfrac{d(t)}{\delta_1}.\] For the second term in 24 we instead use the radial Gagliardo-Nirenberg inequality Lemma 2 to obtain \[\begin{align} \|u\|_{L^4(|x|>R)}^4&\lesssim R^{-2}\|u\|_{\dot{H}_a^1}\|u\|_{L_x^2}^3\\ &\lesssim R^{-2}\sqrt{d(u(t))+1}\lesssim R^{-2}d(u(t))(\delta_1^{-1}+\delta_1^{-2})^{\frac{1}{2}} \end{align}\] Thus \[|\eqref{E:UncVirialMain1}|\lesssim \tfrac{1}{\delta_1R^2}d(u(t)).\] We turn to 22 . We first write \[\begin{align} F_R[u]&-F_\infty[u]-(F_R[e^{i\theta(t)}Q_a]-F_\infty[e^{i\theta(t)}Q_a])\nonumber\\ &=\int_{|x|>R}4\text{Re}[\partial_{jk}w_R(\overline{u}_ju_k-\partial_jQ_a\partial_kQ_a)]-8[|\nabla u|^2-|\nabla Q_a|^2]\,dx\tag{25}\\ &\quad+\int_{|x|>R}4[|u|^2-|Q_a|^2]\tfrac{ax}{|x|^4}\cdot\nabla w_R-8[|u|^2-|Q_a|^2]\tfrac{a}{|x|^2}\,dx\tag{26}\\ &\quad+\mathcal{O}\biggl(\int_{|x|>R}R^{-2}(|u|^2-|Q_a|^2)+(|u|^4-|Q_a|^4)\,dx\biggr).\tag{27} \end{align}\]

For \(t\in J\) we may use Proposition 26 to decompose \[u(t)=e^{i\theta(t)}[(1+\alpha(t))Q_a+h(t)].\] Writing \(Q_a(t) = e^{i\theta(t)}Q_a\), we then estimate 25 as follows: \[\begin{align} |\eqref{E:UncVir2461}|&\lesssim\int_{|x|>R}(2|\alpha|+|\alpha|^2)|\nabla Q_a|^2+(1+|\alpha|)|\nabla h|\,|\nabla Q_a(t)|+|\nabla h|^2\,dx\\ &\lesssim d(u(t))\|Q_a\|_{H_x^1(|x|>R)}^2+(1+d(u(t)))d(u(t))\|Q_a\|_{H_x^1(|x|>R)}+[d(u(t))]^2\\ &\lesssim (o_R(1)+\delta_1)d(u(t)), \end{align}\] where \(o_R(1) = \|Q_a\|_{H^1(|x|>R)}\to 0\) as \(R\to\infty\). We can similarly obtain \[\begin{align} |\eqref{E:UncVir2462} | + | \eqref{E:UncVir2463}|&\lesssim R^{-2}\int_{|x|>R}(2|\alpha|+|\alpha|^2)|Q_a|^2+(1+|\alpha|)|h|\,|Q_a|+|h|^2\,dx\\ & \quad + \int_{|x|>R} |\alpha| |Q_a|^4 + |h|^4 + |Q_a|^3 |h|\,dx \\ &\lesssim o_R(1) d(u(t))+[d(u(t))]^2 + [d(u(t))]^4 \\ &\lesssim (o_R(1)+\delta_1)d(u(t)), \end{align}\] where again \(o_R(1)\) represents norms of \(Q_a\) at large radii that vanish as \(R\to\infty\).

Collecting the estimates above, we obtain \[\begin{align} |\eqref{E:UncVirialMain1}|+|\eqref{E:UncVirialMain2}|\lesssim \bigl(\tfrac{1}{\delta_1R^{2}}+o_R(1)+\delta_1\bigr)d(u(t)). \end{align}\] As \(F_\infty[u]=-4d(u(t))\), we can then choose \(\delta_1\) sufficiently small and \(R=R(\delta_1,Q_a)\) sufficiently large to guarantee that \[\label{E:UncDerivBnd} \tfrac{d}{dt}P_R[u(t)]\leq -cd(u(t))\tag{28}\] for some \(c>0\). Integrating over \([t_1,t_2]\), we obtain 19 .

It remains to prove 20 . We begin by showing that \[\label{e:UncPPos} P_R[u(t)]\geq 0\quad\text{for all}\quad t\geq 0.\tag{29}\] Suppose instead that \(P_R[u(t_0)]<-\eta\) for some \(\eta>0\) and \(t_0\geq 0\), then by 28 and Proposition 13 we can derive that \(P_R[u(t)]<-\eta\) for all \(t>t_0\). Using 1 , this implies \[\int|u(t,x)|^2w_R\,dx\leq \int|u(t_0,x)|^2w_R\,dx-\eta(t-t_0),\] yielding a contradiction if \(t\) is sufficiently large.

We next show that \[\label{E:UncPRBnd} P_R[u(t)]\lesssim Rd(u(t))\quad\text{for all}\quad t\geq 0.\tag{30}\] If \(d(u(t))>\delta_1\) we simply estimate \[P_R[u(t)]\lesssim R\|u\|_{H^1}^2\lesssim R[d(u(t))+\|Q_a\|_{H_a^1}^2]\lesssim R\bigl(1+\tfrac{1}{\delta_1}\|Q_a\|_{H_a^1}^2\bigr)d(u(t)).\] If instead \(d(u(t))\leq d_1\), then we write \(u(t)=e^{i\theta(t)}[(1+\alpha(t))Q_a+h(t)]\) and use the fact that \(P_R[e^{i\theta(t)}Q_a]\equiv 0\) to obtain \[\begin{align} P_R[u(t)]&=|P_R[u(t)]-P_R[e^{i\theta(t)}Q_a]|\\ &\lesssim R\int|(1+\alpha(t))(Q_a\nabla h(t)+\nabla Q_ah(t))|+|\alpha(t)Q_a\nabla Q_a|\,dx\\ &\lesssim R(\|Q_a\|_{H_a^1}^2 [d(u(t))]^2+d(u(t))\|Q_a\|_{H_a^1}^2)\\ &\lesssim R(\delta_1+1)d(u(t)). \end{align}\]

Now, as \(t\mapsto P_R[u(t)]\) is positive and decreasing, we may define \(\ell=\displaystyle\lim_{t\to\infty}P_R[u(t)]\). As 19 implies \[\int_0^\infty d(u(t))\,dt<\infty,\] we must have \(\ell=0\). Combining this fact with 19 and 30 we obtain \[\int_t^\infty d(u(s))\,ds\lesssim d(u(t)),\] yielding 20 and completing the proof. ◻

Proof of Proposition 31. By Lemma 5 and Gronwall’s inequality, \[\int_t^\infty d(u(s))\,ds\lesssim e^{-ct},\] for some \(c>0\) and all \(t\) large. Combining this Corollary 27, we obtain \[\|u(t)-e^{i(t+\theta_0)}Q_a\|_{H_a^1}\lesssim e^{-ct},\] for some \(\theta_0\) and \(t\) sufficiently large, which implies the result. ◻

Proof. Define \[V_R[u(t)]=\int|u(t,x)|^2w_R\,dx.\] We proved above that for \(R\) sufficiently large that \[\tfrac{d}{dt}V_R[u(t)]=P_R[u(t)]\geq 0,\] so that \(t\mapsto V_R[u(t)]\) is nondecreasing. As \[\|u(t)-e^{i(t+\theta)}Q_a\|_{H_a^1}\to 0\quad\text{as}\quad t\to\infty,\] we have \[V_R[u(t)]\leq V_R[Q_a]\leq \int|x Q_a|^2\,dx\quad\text{for all}\quad t\geq 0.\] In particular, \(V_R[u(0)]\) is uniformly bounded in \(R\). Thus we can take the limit as \(R\to\infty\) to deduce that \(xu(0)\in L^2\), which implies the result. ◻

Existence of heteroclinic orbits. Recalling Proposition 20, we define \(Q_a^\pm=U^{\pm1}\). By construction, \[\|Q^\pm\|_{\dot{H}_a^1}^2=\|Q_a\|_{\dot{H}_a^1}^2\pm2e^{-e_0t}\|Y_+\|_{\dot{H}_a^1}^2+\mathcal{O}(e^{-2e_0t}).\] In particular \(\|Q_a^+(t)\|_{\dot{H}_a^1}>\|Q_a\|_{\dot{H}_a^1}\) and \(\|Q_a^-(t)\|_{\dot{H}_a^1}<\|Q_a\|_{\dot{H}_a^1}\) for \(t\) large.

Now, if \(Q_a^-\) does not scatter backward in time, then Proposition 28 implies that the orbit \(\{Q_a^-(t):t\in\mathbb{R}\}\) is precompact in \(H^1\). But then by time reversal symmetry, Lemma 4, Corollary 30, and Corollary 27, we have \[\int_{\mathbb{R}}d(Q_a^-(t))\,dt\lesssim \lim_{t\to\infty}d(Q_a^-(t))+\lim_{t\to-\infty}d(Q_a^-(t))=0.\] This implies that \(d(Q_a^-(t))\equiv0\), contradicting that \(\|Q_a^-(t)\|_{\dot{H}_a^1}<\|Q_a\|_{\dot{H}_a^1}\).

Next, we consider \(Q_a^+\). As \(Q_a^+\) is radial, Corollary 32 implies \(xQ_a^+\in L^2\). Moreover, by 29 we have that \[P_\infty[Q_a^+(0)]=2\operatorname{Im}\int Q_a^+(0)[x\cdot\overline{\nabla {Q_a^+}(0)}]\,dx\geq 0.\] In fact, since \(\frac{d}{dt}P_\infty[Q_a^+(t)]<0\) we have that the quantity above is strictly positive.

Now, if \(Q_a^+\) fails to blow up in negative time, then we may apply the same argument to the time-reversed solution \(\overline{Q_a^+}(t)=Q_a^+(-t)\) to obtain \[2\operatorname{Im}\int\overline{Q_a^+(0)}[x\cdot\nabla Q_a^+(0)]\,dx>0,\] a contradiction. ◻

Classification of threshold solutions. Let \(u_0\in H_{\text{rad}}^1\) satisfy the mass-energy constraint \[M(u_0)E_a(u_0)=M(Q_a)E_a(Q_a)\] Using the scaling symmetry, we may in fact assume that \[M(u_0)=M(Q_a) \quad\text{and}\quad E_a(u_0)=E_a(Q_a).\]

First suppose \(\|u_0\|_{\dot{H}_a^1}<\|Q_a\|_{\dot{H}_a^1}\). If the corresponding solution \(u\) does not scatter in both time directions, then by Proposition 28, Proposition 29, Corollary 23, and Corollary 24, we derive that \(u=Q_a^-\) up to the symmetries of the equation.

Next suppose \(\|u_0\|_{\dot{H}_a^1}>\|Q_a\|_{\dot{H}_a^1}\). If the corresponding solution \(u\) does not blow up forward (say) in time, then by Proposition 31, Corollary 23, Corollary 24, we derive that \(u=Q_a^+\) up to the symmetries of the equation. This completes the proof. ◻