Scattering theory for the subcritical wave equation with inverse square potential

Abstract

We consider the scattering theory for the defocusing energy subcritical wave equations with an inverse square potential. By employing the energy flux method we establish energy flux estimates on the light cone. Then by the characteristic line method and radiation theorem, we show that the radial finite-energy solutions scatter to free waves outside of light cones. Using Morawetz estimates we then obtain the scattering theory for radial solutions with finite weighted energy initial data.

Appendix A

1.1 A.1. The asymptotic behaviour of linear solution

Let u be a finite-energy radial solution to the linear wave equation with an inverse-square potential

$$\begin{aligned} \left\{ \begin{array}{l} u_{tt} +\Big (- \Delta + \frac{a}{|x|^2} \Big ) u = 0, \qquad (x,t)\in {{\mathbb {R}}}^d \times {{\mathbb {R}}},\\ (u,u_t)|_{t=0} = (u_0,u_1) \in {\dot{H}}^1 \times L^2. \end{array}\right. \end{aligned}$$

Here $d \ge 3$ and $a > -\tfrac{(d-2)^2}{4}$. We prove that there exists a free wave v, i.e., a solution to the linear wave equation $\partial _t^2 v - \Delta v = 0$, so that

$$\begin{aligned} \lim _{t \rightarrow +\infty } \Vert (u(\cdot ,t), u_t(\cdot ,t)) - (v(\cdot ,t), v_t(\cdot ,t))\Vert _{{\dot{H}}^1 \times L^2 } = 0. \end{aligned}$$

(A.1)

The proof of (A.1) divides into three steps. We first prove a few estimates in step 1. Following the same argument as in Lemma 6.1, one can obtain

$$\begin{aligned}&\int _{|x|>|t|+r} \Big (|\nabla u(x,t)|^2 + \frac{|u(x,t)|^2}{|x|^2} + |u_t(x,t)|^2\Big ) dx \lesssim E_{0,r}, \end{aligned}$$

(A.2)

$$\begin{aligned}&|u(x,t)| \lesssim E_{0,r}^{1/2} |x|^{-\frac{d-2}{2}}, \quad |x|\ge r+|t| \end{aligned}$$

(A.3)

such that

$$\begin{aligned} E_{0,r} = \int _{|x|>r/2} \Big (|\nabla u_0(x)|^2 + |u_1(x)|^2\Big ) dx \rightarrow 0, \quad \hbox {as}\; r\rightarrow \infty . \end{aligned}$$

(A.4)

We may combine (A.2) with the universal estimate $|u(x,t)| \lesssim |x|^{-\frac{d-2}{2}}$ and obtain

$$\begin{aligned} \lim _{t\rightarrow +\infty } \int _{|x|>t} \frac{|u(x,\pm t)|^2}{|x|^2} dx = 0. \end{aligned}$$

(A.5)

We also have by a direct computation

$$\begin{aligned} |u(r_1,t) - u(r_2,t)|&= \int _{r_1}^{r_2} |u_r (r,t)| dr \\&\lesssim \Big (\int _{r_1}^{r_2} r^{d-1} |u_r(r,t)|^2 dr\Big )^{1/2} \Big (\int _{r_1}^{r_2} r^{-(d-1)} dr\Big )^{1/2} \\&\lesssim (r_2-r_1)^{1/2} r_1^{-\frac{d-1}{2}}. \end{aligned}$$

Thus if $(1-\delta )|t|\le r\le |t|+r_0\le (1+\delta )|t|$ with a small constant $\delta \ll 1$, then we have

$$\begin{aligned} |u(r,t)| \le |u(|t|+r_0,t)| + |u(r,t)- u(|t|+r_0,t)| \lesssim (E_{0,r_0}^{1/2} + \delta ^{1/2}) r^{-\frac{d-2}{2}}.\nonumber \\ \end{aligned}$$

(A.6)

It immediately follows that

$$\begin{aligned} \limsup _{t\rightarrow +\infty } \int _{|x|= (1-\delta ) t} \frac{|u(x,t)|^2}{|x|} dS(x) \lesssim \delta . \end{aligned}$$

(A.7)

In step 2 we prove that there exists a finite-energy free wave v, so that for $0<\delta <1$,

$$\begin{aligned} \limsup _{t \rightarrow +\infty } \int _{|x|>(1-\delta )t} \Big (|\nabla u(x,t) - \nabla v(x,t)|^2 + |u_t(x,t)-v_t(x,t)|^2\Big ) dx \lesssim \delta .\nonumber \\ \end{aligned}$$

(A.8)

Define $w = r^{\frac{d-1}{2}} u(r,t)$, then w satisfies

$$\begin{aligned} (\partial _t^2 -\partial _r^2)w = -(\lambda _d+a) r^{\frac{d-5}{2}} u. \end{aligned}$$

(A.9)

Note that $|u(r,t)| \lesssim r^{-\frac{d-2}{2}}$, applying method of characteristic lines, we can obtain $g_+, g_- \in L^2 ({{\mathbb {R}}})$ such that

$$\begin{aligned}&|(w_t+w_r)(r,t)-2g_-(r+t)| \lesssim r^{-1/2},{} & {} r>1;&\\&|(w_t-w_r)(r,t)-2g_+(t-r)| \lesssim r^{-1/2},{} & {} r>1.&\end{aligned}$$

Thus we have ($\delta \ll 1$, $R>0$ are constants)

$$\begin{aligned} \limsup _{t\rightarrow +\infty } \int _{(1-\delta )t}^{t+R} (|w_t(r,t)-g_+(t-r)|^2 + |w_r(r,t)+g_+(t-r)|^2) dr \lesssim \delta . \end{aligned}$$

By the identity $w_r = r^{\frac{d-1}{2}} u_r + \frac{d-1}{2} r^{\frac{d-3}{2}} u$ and the upper bound

$$\begin{aligned} \limsup _{t\rightarrow +\infty } \int _{(1-\delta )t}^{t+R} r^{d-3} |u(r,t)|^2 dr \lesssim \limsup _{t\rightarrow +\infty } \int _{(1-\delta )t}^{t+R} \frac{1}{r} dr \lesssim \delta . \end{aligned}$$

We have

$$\begin{aligned} \limsup _{t\rightarrow +\infty } \int _{(1-\delta )t}^{t+R} (|r^{\frac{d-1}{2}} u_t(r,t)-g_+(t-r)|^2 + |r^{\frac{d-1}{2}} u_r(r,t)+g_+(t-r)|^2) dr \lesssim \delta . \end{aligned}$$

By radiation fields, there exists a free wave v so that

$$\begin{aligned} \limsup _{t \rightarrow +\infty } \int _{(1-\delta )t<|x|<t+R} \Big (|\nabla u(x,t) - \nabla v(x,t)|^2 + |u_t(x,t)-v_t(x,t)|^2\Big ) dx \lesssim \delta . \end{aligned}$$

Combining this with (A.2), we obtain (A.8).

In step 3 we show

$$\begin{aligned} \limsup _{t \rightarrow +\infty } \int _{|x|<(1-\delta )t} \Big (|\nabla u(x,t) - \nabla v(x,t)|^2 + |u_t(x,t)-v_t(x,t)|^2\Big ) dx \lesssim \delta .\nonumber \\ \end{aligned}$$

(A.10)

We follow a similar argument to Proposition 4.2 and Corollary 4.3 (simply ignore the terms involving $|u|^{p+1}$), and obtain a Morawetz-type estimate

$$\begin{aligned} \int _{|x|<t} J_u^+(x,t) dx&\lesssim \frac{1}{t}\int _{-t}^t \int _{|x|> t} \Big (|\nabla u|^2 + |u_t|^2 \Big ) dxdt' + \sum _{\pm } \int _{|x|>t} \frac{|u(x,\pm t)|^2}{2|x|^2} dx. \end{aligned}$$

We claim that the right hand side converges to zero as $t\rightarrow +\infty$. In fact we may show the convergence of the first term in the same way as in Proposition 6.3 by (A.2). The convergence of the second term has been known (A.5). As a result

$$\begin{aligned} \lim _{t\rightarrow +\infty } \int _{|x|<t} J_u^+(x,t) dx = 0. \end{aligned}$$

We recall the definition of $J_u^+$ and obtain

Next we combine these estimates with (A.7), Proposition 5.2, and Lemma 4.1 to complete the proof of (A.10), as we did in the proof of Proposition 6.3. Finally we combine (A.8) and (A.10) to conclude

$$\begin{aligned} \limsup _{t \rightarrow +\infty } \int _{{{\mathbb {R}}}^d} \Big (|\nabla u(x,t) - \nabla v(x,t)|^2 + |u_t(x,t)-v_t(x,t)|^2\Big ) dx \lesssim \delta . \end{aligned}$$

We then finish the proof by letting $\delta \rightarrow 0^+$.

1.2 A.2 Square function estimates

In this subsection, employing the Stein-Tomas restriction theorem in Lorentz space, we show

$$\begin{aligned} \Big \Vert \int _{{{\mathbb {R}}}} e^{is|\nabla |} F(\cdot ,s) ds \Big \Vert _{\dot{H}^{-\frac{1}{2}({{\mathbb {R}}}^d)}} \le C \Vert F\Vert _{L^{\tfrac{2d}{d+2},2 }_x L^2_t({{\mathbb {R}}}\times {{\mathbb {R}}}^d) }, \end{aligned}$$

(A.11)

for radial function $F\in {L^{\tfrac{2d}{d+2},2 }_x L^2_t({{\mathbb {R}}}\times {{\mathbb {R}}}^d) }$.

Proof

The proof follows the arguments of [31]. By duality, it suffices to show

$$\begin{aligned} \Vert |\nabla |^{-\gamma }e^{it|\nabla |} f \Vert _{L^{q,2}_x L^2_t({{\mathbb {R}}}\times {{\mathbb {R}}}^d)} \le C \Vert f\Vert _{L^2({{\mathbb {R}}}^d)}, \end{aligned}$$

(A.12)

where $\gamma =\frac{d}{2}-\frac{1}{2}-\frac{d}{q}$ and $q\ge \frac{2(d+1)}{d-1}$. Utilizing the polar coordinates, we have

$$\begin{aligned} |\nabla |^{-\gamma }e^{it|\nabla |} f(x) =\int _0^\infty e^{it r} r^{-\gamma } \int _{|\xi |=r} e^{ix \xi } {{\hat{f}}}(\xi ) d\sigma ^{n-1}_r(\xi ) dr. \end{aligned}$$

By Plancherel theorem with respective to variable t, we have

$$\begin{aligned} \Vert |\nabla |^{-\gamma }e^{it|\nabla |} f(x) \Vert _{L^{q,2}_xL^2_t} =&\Big \Vert \Big ( \int _0^\infty \Big | \int _{|\xi |=r} e^{ix \xi } {{\hat{f}}}(\xi ) d\sigma ^{n-1}_r(\xi ) \Big |^2 r^{-2\gamma } dr \Big )^\frac{1}{2}\Big \Vert _{L^{q,2}_x}\\ \le&\Big ( \int _0^\infty \Big \Vert \int _{|\xi |=r} e^{ix \xi } {{\hat{f}}}(\xi ) d\sigma ^{n-1}_r(\xi ) \Big \Vert _{L^{q,2}_x}^2 r^{-2\gamma } dr \Big )^\frac{1}{2} \\ \le&\Big ( \int _0^\infty \int _{|\xi |=r} | {{\hat{f}}}(\xi ) |^2 d\sigma ^{n-1}_r(\xi ) dr \Big )^\frac{1}{2} \\ =&\Vert f\Vert _{L^2_x({{\mathbb {R}}}^d)}. \end{aligned}$$

where we used the Stein-Tomas restriction theorem of the Lorentz spaces in [1] and the following Minkowski inequality.

$$\begin{aligned} \Vert f(x,y)\Vert _{L^{p,r}_x L^r_y} \le \Vert f(x,y)\Vert _{ L^r_y L^{p,r}_x}, \;\; 1\le r< p <\infty . \end{aligned}$$

(A.13)

This completes the proof the square function estimate in Lorentz space.

For the sake of completeness, we give a proof of the Minkowski inequalities in Lorentz space. In fact, for $1\le r< p<\infty$, we have

$$\begin{aligned} \Vert f(x,y)\Vert _{L^{p,r}_x L^r_y} =&\Big \Vert |f(x,y)|^r \Big \Vert ^{\frac{1}{r}}_{L^{\frac{p}{r},1}_x L^1_y} \\ \approx&\Big ( \sup _{\Vert g\Vert _{L^{(\frac{p}{r})',\infty }} \le 1 } \int |g(x)|\int |f(x,y)|^r dy dx \Big )^\frac{1}{r} \\ =&\Big ( \sup _{\Vert g\Vert _{L^{(\frac{p}{r})',\infty }} \le 1 } \int \int |f(x,y)|^r |g(x)| dx dy \Big )^\frac{1}{r} \\ \le&\Vert |f|^r\Vert _{L^1_y L^{\frac{p}{r},1}_x}^\frac{1}{r} = \Vert f(x,y)\Vert _{ L^r_y L^{p,r}_x}. \end{aligned}$$

$\square$