Below and beyond the mass–energy threshold: scattering for the Hartree equation with radial data in $\pmb {d \ge 5}$

Abstract

We consider the Cauchy problem of the focusing $\dot{H}^{1/2}$-critical Hartree equation

$$\begin{aligned} i u_{t} + \Delta u = - \Bigl ( |\cdot |^{-3} *|u(t)|^{2} \Bigr )u(t,x), \quad (t,x)\in {\mathbb {R}} \times {\mathbb {R}}^{d}. \end{aligned}$$

By adapting the methods in Dodson and Murphy (Proc Am Math Soc 145(11):4859–4867, 2017), we shall prove a scattering result for solutions both below and beyond the mass–energy threshold M(Q)E(Q) and uniformly describe both cases the boundary of the scattering region by the ground state’s mass and potential energy product.

Appendix A: Construct a scattering solution with initial datum beyond the mass–energy threshold

In appendix, we shall construct a scattering solution with initial datum beyond the mass–energy threshold in Corollary A.3 following the process used in [5, Corollary 1.12].

First, we introduce the virial quantity. Let u(t, x) be a solution of (1.1) and denote the virial quantity $V(t) = \mathop {\int }\limits _{{\mathbb {R}}^{d}} |x|^{2}|u(t,x)|^{2}\,\mathrm {d}x.$ If $V(0) < \infty$, a direct computation using the equation (1.1) yields

$$\begin{aligned} V'(t)= & {} 4 {{\,\mathrm{Im}\,}}\mathop {\int }\limits _{{\mathbb {R}}^d} x \cdot \nabla u(t, x) {\bar{u}}(t, x) \,\mathrm d x. \end{aligned}$$

(A.1)

$$\begin{aligned} V''(t)= & {} 8 \mathop {\int }\limits _{{\mathbb {R}}^{d}}|\nabla u(t)|^{2}\,\mathrm {d}x - 6 \mathop {\int }\limits _{{\mathbb {R}}^{d}} (|\cdot |^{-3}*|u|^{2})|u|^{2}(t,x)\,\mathrm {d}x \nonumber \\= & {} 16 E(u) - 2P(u(t)). \end{aligned}$$

(A.2)

Yang et al. proved a general global well-posedness theorem in [20]. Here we summarize their results in the version that we need.

Theorem A.1

([20, Theorem 3.1 with $s_{c} = 1/2$]) Let u(t, x) be a solution of (1.1), assume $u_{0} \in H^{1}(R^{d})$ and $V(0) < \infty$. If

$$\begin{aligned}&M(u)E(u) \ge M(Q)E(Q), \end{aligned}$$

(A.3)

$$\begin{aligned}&\frac{M(u)E(u)}{M(Q)E(Q)}\Bigl ( 1 - \frac{(V'(0))^{2}}{32 E(u)V(0)} \Bigr ) \le 1, \end{aligned}$$

(A.4)

$$\begin{aligned}&M(u_{0}) P(u_{0})<M(Q)P(Q), \end{aligned}$$

(A.5)

and

$$\begin{aligned} V'(0)\ge 0. \end{aligned}$$

(A.6)

Then, u(t, x) exists globally, and

$$\begin{aligned} \sup _{t\in [0,\infty )}M(u(t)) P(u(t)) < M(Q)P(Q). \end{aligned}$$

(A.7)

Remark A.2

Instead of (A.7), the authors in [20] proved that

$$\begin{aligned} \limsup \limits _{t\rightarrow +\infty }M(u(t)) P(u(t)) < M(Q)P(Q), \end{aligned}$$

which is the same as Theorem 1.4 in [5] for nonlinear Schrödinger equation. Reviewing the proof in [20], one finds that (A.7) can be also obtained.

Next, we shall construct a scattering solution with initial datum beyond the mass–energy threshold by Theorem A.1.

Corollary A.3

Let $\gamma > 0$ and $Q^{\gamma }(t,x)$ be the solution of (1.1) with initial datum

$$\begin{aligned} Q_{0}^{\gamma } = e^{i \gamma |x|^{2}} Q(x), \end{aligned}$$

where Q is the ground state solution of (1.2), which is a real positive function. Then, there exists a small positive $t_{0}$ such that

$$\begin{aligned} Q_{t_{0}}^{\gamma }(t,x) = Q^{\gamma }(t_{0} + t,x) \end{aligned}$$

is a forward global solution with initial datum beyond the mass–energy threshold

$$\begin{aligned} M(Q^{\gamma }(t_{0})) E(Q^{\gamma }(t_{0})) > M(Q)E(Q), \end{aligned}$$

and

$$\begin{aligned} \sup _{t\in [0,\infty )}M(Q_{t_{0}}^{\gamma }(t)) P(Q_{t_{0}}^{\gamma }(t)) < M(Q)P(Q). \end{aligned}$$

(A.8)

Proof

We shall check the conditions in Theorem A.1 step by step.

By the local theory and the identity

$$\begin{aligned} {{\,\mathrm{Im}\,}}\mathop {\int }\limits _{{\mathbb {R}}^{d}} x \cdot \nabla Q_{0}^{\gamma } \overline{Q_{0}^{\gamma }} \,\mathrm {d}x = 2 \gamma \mathop {\int }\limits _{{\mathbb {R}}^{d}} |xQ|^{2}\,\mathrm {d}x, \end{aligned}$$

we see that for $\gamma > 0$,

$$\begin{aligned} {{\,\mathrm{Im}\,}}\mathop {\int }\limits _{{\mathbb {R}}^{d}} x \cdot \nabla Q_{t_{0}}^{\gamma } \overline{Q_{t_{0}}^{\gamma }} \,\mathrm {d}x > 0 \quad \text { for small } t_{0}. \end{aligned}$$

Thus, by the formula (A.1), $V'(0) > 0$ for the solution $Q^{\gamma }_{t_{0}}(t,x)$. And (A.6) is satisfied.

Let $P(t) = P(u(t))$. If $u$ is a solution of (1.1), then by formal calculation,

$$\begin{aligned} \frac{\mathrm {d}P}{\mathrm {d}t}&= \mathop {\int }\limits _{{\mathbb {R}}^{d}} [|\cdot |^{-3}*(u_{t} \bar{u} + \overline{u}_{t} u) ] |u(t,x)|^{2} \,\mathrm {d}x + \mathop {\int }\limits _{{\mathbb {R}}^{d}} [|\cdot |^{-3}*|u|^{2} ] (u_{t} \bar{u} + \overline{u}_{t} u)(t,x) \,\mathrm {d}x \\&= 2 \mathop {\int }\limits _{{\mathbb {R}}^{d}} [|\cdot |^{-3}|u|^{2} ] (u_{t} \bar{u} + \overline{u}_{t} u)(t,x) \,\mathrm {d}x \\&= -4\mathop {\int }\limits _{{\mathbb {R}}^{d}} (|\cdot |^{-3} *|u(t)|^{2}) {{\,\mathrm{Im}\,}}(\Delta u {\bar{u}})(t,x)\,\mathrm {d}x. \end{aligned}$$

Let $u(t) = Q^{\gamma }(t)$. Since Q(x) is real, we have ${{\,\mathrm{Im}\,}}(\Delta u(0) \overline{u(0)}) = 2 \gamma (d Q^{2} + 2x Q \nabla Q) = 2 \gamma \nabla \cdot (x Q^{2})$.

An integral by parts and a change of variables yield

$$\begin{aligned} \frac{\mathrm {d}P(0)}{\mathrm {d}t}&= -8\gamma \mathop {\int }\limits _{{\mathbb {R}}^{d}} (|\cdot |^{-3} *Q^{2}) \nabla _{x} \cdot (xQ^{2}) \,\mathrm {d}x \nonumber \\&= -8 \gamma \iint _{{\mathbb {R}}^{d}\times {\mathbb {R}}^{d}} \frac{Q^{2}(y)}{|x-y|^{3}} \nabla _{x} \cdot (xQ^{2}(x))\,\mathrm {d}x \mathrm {d}y\nonumber \\&= 8 \gamma \iint _{{\mathbb {R}}^{d}\times {\mathbb {R}}^{d}} \Bigl ( \nabla _{x} \frac{1}{|x-y|^{3}} \Bigr ) Q^{2}(y) (xQ^{2}(x))\,\mathrm {d}x \mathrm {d}y\nonumber \\&= -24 \gamma \iint _{{\mathbb {R}}^{d}\times {\mathbb {R}}^{d}} \frac{(x-y)\cdot x}{|x-y|^{5}} Q^{2}(y) Q^{2}(x)\,\mathrm {d}x \mathrm {d}y\nonumber \\&= -12 \gamma \Bigl ( \iint _{{\mathbb {R}}^{d}\times {\mathbb {R}}^{d}} \frac{(x-y)\cdot x}{|x-y|^{5}} Q^{2}(y) Q^{2}(x)\,\mathrm {d}x \mathrm {d}y \nonumber \\&\qquad + \iint _{{\mathbb {R}}^{d}\times {\mathbb {R}}^{d}} \frac{(y-x)\cdot y}{|x-y|^{5}} Q^{2}(x) Q^{2}(y)\,\mathrm {d}x \mathrm {d}y \Bigr ) \nonumber \\&\quad = -12\gamma \iint _{{\mathbb {R}}^{d}\times {\mathbb {R}}^{d}} \frac{Q^{2}(x) Q^{2}(y) }{|x-y|^{3}} \,\mathrm {d}x \mathrm {d}y \nonumber \\&\quad < 0. \end{aligned}$$

(A.9)

Thus, for small positive $t_{0}$,

$$\begin{aligned} P(t_{0}) < P(0) = P (Q). \end{aligned}$$

So for the solution $Q^{\gamma }_{t_{0}}$, the condition (A.5) is fulfilled.

By the Pohozaev identities (2.1), $E(Q) = \frac{1}{2}\mathop {\int }\limits _{{\mathbb {R}}^{d}}|Q|^{2}\,\mathrm {d}x > 0$. By conservation of mass and energy,

$$\begin{aligned} M(Q^{\gamma }_{t_{0}}) = M(Q^{\gamma }_{0}) = M(Q), \end{aligned}$$

and

$$\begin{aligned} E(Q^{\gamma }_{t_{0}}) = E(Q^{\gamma }_{0})&= E(Q) + 2 \gamma ^{2} \mathop {\int }\limits _{{\mathbb {R}}^{d}} |xQ(x)|^{2}\,\mathrm {d}x \nonumber \\&> E(Q)>0. \end{aligned}$$

(A.10)

Thus, $M(Q^{\gamma }_{t_{0}}) E(Q^{\gamma }_{t_{0}}) > M(Q)E(Q)$, which is (A.3).

To show that $Q_{t_{0}}^{\gamma }$ obeys (A.4), we define

$$\begin{aligned} F(t) = M(Q^{\gamma }) \left( E(Q^{\gamma }) - \frac{\bigl ({{\,\mathrm{Im}\,}}\mathop {\int }\limits _{{\mathbb {R}}^{d}} x \cdot \nabla Q^{\gamma } (t) \overline{Q^{\gamma }} (t)\,\mathrm {d}x\bigr )^{2}}{2 \mathop {\int }\limits _{{\mathbb {R}}^{d}} |x|^{2}|Q^{\gamma }(t)|^{2} \,\mathrm {d}x} \right) -M(Q) E(Q). \end{aligned}$$

We claim that

$$\begin{aligned} F(0) = F'(0) = 0 \quad \text {and}\quad F''(0)<0. \end{aligned}$$

(A.11)

Note that $t=0$, ${{\,\mathrm{Im}\,}}\mathop {\int }\limits _{{\mathbb {R}}^{d}} x \cdot \nabla Q^{\gamma } (0) \overline{Q^{\gamma }} (0)\,\mathrm {d}x = 2 \gamma \mathop {\int }\limits _{{\mathbb {R}}^{d}} |xQ|^{2}\,\mathrm {d}x$. Therefore, $F(0) = 0$ by (A.10). Let

$$\begin{aligned} V(t) = \mathop {\int }\limits _{{\mathbb {R}}^{d}}|xQ^{\gamma }(t)|^{2}\,\mathrm {d}x, \quad z(t) = \sqrt{V(t)}. \end{aligned}$$

F(t) can be rewritten as

$$\begin{aligned} F(t) = M(Q^{\gamma }) \Bigl ( E(Q^{\gamma }) - \frac{1}{8} (z_{t})^{2}\Bigr ) -M(Q) E(Q). \end{aligned}$$

By conservation of mass and energy, we have

$$\begin{aligned} \frac{\mathrm {d}F(t)}{\mathrm {d}t} = -\frac{M(Q)}{4}z_{t}z_{tt}, \end{aligned}$$

and

$$\begin{aligned} \frac{\mathrm {d^{2}}F(t)}{\mathrm {d}t^{2}} = -\frac{M(Q)}{4}(z^{2}_{tt} + z_{t}z_{ttt}). \end{aligned}$$

By invoking (A.1), (A.2) and the Pohozaev identities $\mathop {\int }\limits _{{\mathbb {R}}^{d}} |\nabla Q|^{2} \,\mathrm {d}x = 3\mathop {\int }\limits _{{\mathbb {R}}^{d}}|Q|^{2}\,\mathrm {d}x$ and $P(Q) = 4 \mathop {\int }\limits _{{\mathbb {R}}^{d}}|Q|^{2}\,\mathrm {d}x$, we get

$$\begin{aligned} V_{t}(0) = 8 \gamma \mathop {\int }\limits _{{\mathbb {R}}^{d}} |xQ|^{2}\,\mathrm {d}x, \quad \text {and} \quad V_{tt}(0) = 32 \gamma ^{2}\mathop {\int }\limits _{{\mathbb {R}}^{d}} |xQ|^{2}\,\mathrm {d}x. \end{aligned}$$

Then, we get for $\gamma > 0$

$$\begin{aligned} z_{t}(0) > 0, \quad z_{tt}(0) = \frac{1}{4}V^{-3/2} \Bigl ( 2V(0)V_{tt}(0) - (V_{t}(0))^{2} \Bigr ) = 0. \end{aligned}$$

Thus,

$$\begin{aligned} F'(0)= 0, \end{aligned}$$

and

$$\begin{aligned} {{\,\mathrm{sgn}\,}}(F''(0)) = -{{\,\mathrm{sgn}\,}}(z_{ttt}(0)), \end{aligned}$$

(A.12)

where ${{\,\mathrm{sgn}\,}}(s) = \frac{s}{|s|}$ for $s \ne 0$ and ${{\,\mathrm{sgn}\,}}(0) = 0$.

Since $V(t) = (z(t))^{2}$, we have $V_{ttt} = 6 z_{t}z_{tt} + 2 z z_{ttt}$, which together with $z_{tt}(0) = 0$ and (A.12) implies that ${{\,\mathrm{sgn}\,}}(-V_{ttt}(0)) = {{\,\mathrm{sgn}\,}}(F''(0))$. Meanwhile, we get from (A.2) that

$$\begin{aligned} {{\,\mathrm{sgn}\,}}(-V_{ttt}(0)) = {{\,\mathrm{sgn}\,}}\left( \frac{\mathrm {d}P(0)}{\mathrm {d}t}\right) . \end{aligned}$$

Thus, $F''(0) < 0$ due to (A.9). The claim (A.11) is proved. As a consequence, there exists $t_{0} > 0$ such that $F(t_{0}) < 0$. Taking the initial datum $u_{0}(x) = Q^{\gamma }(t_{0},x)$, which fulfills the assumptions of Theorem A.1, we get (A.8). $\square$

By (A.8) and Theorem 1.1, $Q_{t_{0}}^{\gamma }(t,x)$ scatters forward in time.