Schrödinger equations with combined non-linearity

Abstract

This paper investigates two class of Schrödinger equations with mixed source terms, containing, respectively, a local/non-local non-linearity and an inhomogeneous perturbation. In the energy sub-critical regime, one obtains some sharp thresholds of global/non-global existence dichotomy.

Appendix

This section proves the variance identity in Proposition 2.4 and the compact Sobolev embedding in Lemma 2.5.

1.1 Proof of Proposition 2.4

Let ${u}\in C_{T^*}(\Sigma )$ be a solution to (1). Define the real function on $[0,T^*)$, by

$$\begin{aligned} V:t\longmapsto \int _{{\mathbb{R}}^N}|xu(t,x)|^2\,{\mathrm{d}}x. \end{aligned}$$

Take, for simplicity $\epsilon _i=1$ and denote also the quantities

$$\begin{aligned} a(x):=|x|^2\quad \text{and}\quad {\mathcal {N}}:={\mathcal {N}}_p+{\mathcal {N}}_q:=|x|^b|u|^{2(p-1)}u +(I_\alpha *|u|^q)|u|^{q-2}u. \end{aligned}$$

Multiplying Eq. (1) by 2u and examining the imaginary parts

$$\begin{aligned} \partial _t (|u|^2) =-2\mathfrak {I}({\bar{u}} \Delta u). \end{aligned}$$

Thus

$$\begin{aligned} V'=\, & {} -2\int _{{\mathbb{R}}^N}|x|^2\mathfrak {I}({\bar{u}}\Delta u)\,{\mathrm{d}}x\\ =\, & {} 4\mathfrak {I}\int _{{\mathbb{R}}^N}(x.\nabla u){\bar{u}} \,{\mathrm{d}}x\\ =\, & {} 2\mathfrak {I}\int _{{\mathbb{R}}^N}(\partial _ka\partial _ku){\bar{u}} \,{\mathrm{d}}x. \end{aligned}$$

Compute

$$\begin{aligned} \partial _t\mathfrak {I}(\partial _k u{\bar{u}})=\, & {} \mathfrak {I}(\partial _k{\dot{u}}{\bar{u}})+\mathfrak {I}(\partial _k u\bar{{\dot{u}}})\\ =\, & {} \mathfrak {R}(i{\dot{u}}\partial _k{\bar{u}})-\mathfrak {R}(i\partial _k {\dot{u}}{\bar{u}})\\ =\, & {} \mathfrak {R}(\partial _k{\bar{u}}(-\Delta u-{\mathcal {N}}))-\mathfrak {R}({\bar{u}}\partial _k (-\Delta u-{\mathcal {N}}))\\ =\, & {} \mathfrak {R}({\bar{u}}\partial _k\Delta u-\partial _k{\bar{u}}\Delta u)+\mathfrak {R}({\bar{u}}\partial _k{\mathcal {N}}-\partial _k{\bar{u}}{\mathcal {N}}). \end{aligned}$$

Recall the identity

$$\begin{aligned} \frac{1}{2}\partial _k\Delta (|u|^2)-2\partial _l\mathfrak {R}(\partial _{k}u \partial _l{\bar{u}})=\mathfrak {R}({\bar{u}}\partial _k\Delta u-\partial _k{\bar{u}}\Delta u). \end{aligned}$$

Then

$$\begin{aligned} \int _{{\mathbb{R}}^N}\partial _ka\mathfrak {R}({\bar{u}}\partial _k\Delta u-\partial _k{\bar{u}}\Delta u)\,{\mathrm{d}}x=\, & {} \int _{{\mathbb{R}}^N}\partial _ka\left( \frac{1}{2}\partial _k\Delta (|u|^2) -2\partial _l\mathfrak {R}(\partial _ku\partial _l{\bar{u}})\right) \,{\mathrm{d}}x\\ =\, & {} 2\int _{{\mathbb{R}}^N}\partial _l\partial _ka\mathfrak {R}(\partial _ku\partial _l{\bar{u}})\,{\mathrm{d}}x\\ =\, & {} 4\Vert \nabla u\Vert ^2. \end{aligned}$$

Write now

$$\begin{aligned} \partial _ka\mathfrak {R}({\bar{u}}\partial _k{\mathcal {N}}-\partial _k {\bar{u}}{\mathcal {N}})=\partial _ka\mathfrak {R}({\bar{u}}\partial _k {\mathcal {N}}_p-\partial _k{\bar{u}}{\mathcal {N}}_p) +\partial _ka\mathfrak {R}({\bar{u}}\partial _k{\mathcal {N}}_q -\partial _k{\bar{u}}{\mathcal {N}}_q). \end{aligned}$$

On the other hand

$$\begin{aligned}&\partial _ka\mathfrak {R}({\bar{u}}\partial _k{\mathcal {N}}_p- \partial _k{\bar{u}}{\mathcal {N}}_p)\,{\mathrm{d}}x\\&\quad =\int _{{\mathbb{R}}^N}\partial _ka\mathfrak {R}(\partial _k[{\bar{u}} {\mathcal {N}}_p]-2\partial _k{\bar{u}}{\mathcal {N}}_p)\,{\mathrm{d}}x\\&\quad =-\int _{{\mathbb{R}}^N}\Big (\Delta a{\bar{u}}{\mathcal {N}}_p+2\mathfrak {R}(\partial _ka\partial _k{\bar{u}}{\mathcal {N}}_p)\Big )\,{\mathrm{d}}x\\&\quad =-2N\int _{{\mathbb{R}}^N}|u|^{2p}|x|^b\,{\mathrm{d}}x-2\int _{{\mathbb{R}}^N} \partial _ka|x|^b\partial _k\left( \frac{|u|^{2p}}{2p}\right) \,{\mathrm{d}}x. \end{aligned}$$

Moreover

$$\begin{aligned} \int _{{\mathbb{R}}^N}\partial _ka|x|^b\partial _k(|u|^{2p})\,{\mathrm{d}}x= & {} -\int _{{\mathbb{R}}^N}\text{div}\Big (\partial _ka|x|^b\Big )|u|^{2p}\,{\mathrm{d}}x\\= & {} -\int _{{\mathbb{R}}^N}\Big (\Delta a|x|^b+\partial _ka\partial _k(|x|^b) \Big )|u|^{2p}\,{\mathrm{d}}x\\= & {} -\int _{{\mathbb{R}}^N}\Big (2N|x|^b+2x.bx|x|^{b-2}\Big )|u|^{2p}\,{\mathrm{d}}x\\= & {} -2(N+b)\int _{{\mathbb{R}}^N}|u|^{2p}|x|^{b}\,{\mathrm{d}}x. \end{aligned}$$

Thus

$$\begin{aligned} \partial _ka\mathfrak {R}({\bar{u}}\partial _k{\mathcal {N}}_p- \partial _k{\bar{u}}{\mathcal {N}}_p)\,{\mathrm{d}}x= & {} -2N\int _{{\mathbb{R}}^N}|u|^{2p}|x|^b\,{\mathrm{d}}x-2 \int _{{\mathbb{R}}^N}\partial _ka|x|^b\partial _k\left( \frac{|u|^{2p}}{2p}\right) \,{\mathrm{d}}x\\= & {} -2N\int _{{\mathbb{R}}^N}|u|^{2p}|x|^b\,{\mathrm{d}}x+ \frac{2(N+b)}{p}\int _{{\mathbb{R}}^N}|u|^{2p}|x|^{b}\,{\mathrm{d}}x\\= & {} -\frac{2B_p}{p}\int _{{\mathbb{R}}^N}|u|^{2p}|x|^b\,{\mathrm{d}}x. \end{aligned}$$

On the other hand

$$\begin{aligned}&\int _{{\mathbb{R}}^N}\partial _ka\mathfrak {R}({\bar{u}}\partial _k{\mathcal {N}}_q -\partial _k{\bar{u}}{\mathcal {N}}_q)\,{\mathrm{d}}x\\&\quad =\int _{{\mathbb{R}}^N}\partial _ka\mathfrak {R}(\partial _k[{\bar{u}} {\mathcal {N}}_q]-2\partial _k{\bar{u}}{\mathcal {N}}_q)\,{\mathrm{d}}x\\&\quad =-\int _{{\mathbb{R}}^N}\Big (\Delta a{\bar{u}}{\mathcal {N}}_q +2\mathfrak {R}(\partial _ka\partial _k{\bar{u}}{\mathcal {N}}_q)\Big )\,{\mathrm{d}}x\\&\quad =-2N\int _{{\mathbb{R}}^N}|u|^q(I_\alpha *|u|^{q})\,{\mathrm{d}}x -2\int _{{\mathbb{R}}^N}\partial _ka\mathfrak {R}(\partial _k{\bar{u}}{\mathcal {N}}_q)\,{\mathrm{d}}x\\&\quad =-2N\int _{{\mathbb{R}}^N}|u|^q(I_\alpha *|u|^{q})\,{\mathrm{d}}x -\frac{2}{q}\int _{{\mathbb{R}}^N}\partial _ka\partial _k(|u|^q)(I_\alpha *|u|^q)\,{\mathrm{d}}x. \end{aligned}$$

Moreover

$$\begin{aligned}&\int _{{\mathbb{R}}^N}\partial _ka\partial _k(|u|^{q})(I_\alpha *|u|^q)\,{\mathrm{d}}x\\&\quad =-2\int _{{\mathbb{R}}^N}\text{div}(x(I_\alpha *|u|^q))|u|^{q}\,{\mathrm{d}}x\\&\quad =-2N\int _{{\mathbb{R}}^N}(I_\alpha *|u|^q)|u|^q\,{\mathrm{d}}x-2 \int _{{\mathbb{R}}^N}x(\nabla I_\alpha *|u|^q)|u|^q\,{\mathrm{d}}x\\&\quad =-2N\int _{{\mathbb{R}}^N}(I_\alpha *|u|^q)|u|^q\,{\mathrm{d}}x -2(\alpha -N)\int _{{\mathbb{R}}^N}x\left( \frac{.}{|\cdot |^2}I_\alpha *|u|^q\right) |u|^q\,{\mathrm{d}}x. \end{aligned}$$

Moreover

$$\begin{aligned} (B):= & {} \int _{{\mathbb{R}}^N}x\left( \frac{.}{|\cdot |^2}I_\alpha *|u|^q\right) |u|^q\,{\mathrm{d}}x\\= & {} \int _{{\mathbb{R}}^N}(x-y+y)(x-y)\frac{I_\alpha (x-y)}{|x-y|^2}|u(y)|^q|u(x)|^q\,{\mathrm{d}}y\,{\mathrm{d}}x\\= & {} \int _{{\mathbb{R}}^N}{I_\alpha (x-y)}|u(y)|^q|u(x)|^q\,{\mathrm{d}}y\,{\mathrm{d}}x+\int _{{\mathbb{R}}^N}y(x-y)\frac{I_\alpha (x-y)}{|x-y|^2}|u(y)|^q|u(x)|^q\,{\mathrm{d}}y\,{\mathrm{d}}x\\= & {} \int _{{\mathbb{R}}^N}{I_\alpha (x-y)}|u(y)|^q|u(x)|^q\,{\mathrm{d}}y\,{\mathrm{d}}x-\int _{{\mathbb{R}}^N}x(x-y)\frac{I_\alpha (x-y)}{|x-y|^2}|u(y)|^q|u(x)|^q\,{\mathrm{d}}y\,{\mathrm{d}}x\\= & {} \int _{{\mathbb{R}}^N}{I_\alpha (x-y)}|u(y)|^q|u(x)|^q\,{\mathrm{d}}y\,{\mathrm{d}}x-(B)\\= & {} \frac{1}{2}\int _{{\mathbb{R}}^N}(I_\alpha *|u|^q)|u|^q\,{\mathrm{d}}x. \end{aligned}$$

Thus

$$\begin{aligned} (A):= & {} \int _{{\mathbb{R}}^N}\partial _ka\mathfrak {R}({\bar{u}}\partial _k{\mathcal {N}}_q-\partial _k{\bar{u}}{\mathcal {N}}_q)\,{\mathrm{d}}x\\= & {} -2N\int _{{\mathbb{R}}^N}|u|^q(I_\alpha *|u|^{q})\,{\mathrm{d}}x-\frac{2}{q}\int _{{\mathbb{R}}^N}\partial _ka\partial _k(|u|^q)(I_\alpha *|u|^q)\,{\mathrm{d}}x\\= & {} -2N\int _{{\mathbb{R}}^N}|u|^q(I_\alpha *|u|^{q})\,{\mathrm{d}}x-\frac{2}{q}\left( -2N\int _{{\mathbb{R}}^N}(I_\alpha *|u|^q)|u|^q\,{\mathrm{d}}x \right. \\&-2\left. (\alpha -N)\int _{{\mathbb{R}}^N}x\left( \frac{.}{|\cdot |^2}I_\alpha *|u|^q\right) |u|^q\,{\mathrm{d}}x \right) \\= & {} -2N\left( 1-\frac{2}{q}\right) \int _{{\mathbb{R}}^N}|u|^q(I_\alpha *|u|^{q})\,{\mathrm{d}}x+\frac{4(\alpha -N)}{q}\int _{{\mathbb{R}}^N}x \left( \frac{.}{|\cdot |^2}I_\alpha *|u|^q\right) |u|^q\,{\mathrm{d}}x \\= & {} \left( -2N\left( 1-\frac{2}{q}\right) +\frac{2(\alpha -N)}{q}\right) \int _{{\mathbb{R}}^N}|u|^q(I_\alpha *|u|^{q})\,{\mathrm{d}}x\\= & {} -\frac{2B_q}{q}\int _{{\mathbb{R}}^N}|u|^q(I_\alpha *|u|^{q})\,{\mathrm{d}}x. \end{aligned}$$

Finally

$$\begin{aligned} V''=8\Vert \nabla u\Vert ^2-\frac{4B_p}{p}\int _{{\mathbb{R}}^N}|u|^{2p}|x|^b\,{\mathrm{d}}x-\frac{4B_q}{q}\int _{{\mathbb{R}}^N}(I_\alpha *|u|^q)|u|^q\,{\mathrm{d}}x. \end{aligned}$$

The variance identity is proved.

1.2 Proof of Lemma 2.5

Take $\epsilon >0$ and $1+\frac{b\chi _{b>0}}{N-1}< p<p^*$.

• A. First case $-2<b<0$.

  Let $(u_n)$ a bounded sequence of $H^1.$ Without loss of generality, one assumes that $(u_n)$ converges weakly to zero in $H^1.$ The purpose is to prove that $\Vert u_n\Vert _{L^{2p}(|x|^{b}\,{\mathrm{d}}x)}\rightarrow 0.$ Take $\epsilon >0$ and $R>\epsilon ^\frac{1}{b}$. Using Hölder estimate, for a couple $(q,q')$ satisfying $q'|b|<N$, which is equivalent to $q>\frac{N}{N+b}$, one gets

  $$\begin{aligned} \int _{|x|\le R}|x|^{b}|u_n|^{2p}\,{\mathrm{d}}x\le & {} \Vert u_n\Vert _{L^{2pq}(|x|\le R)}^{2p}\Vert |x|^{b}\Vert _{L^{q'}(|x|\le R)}\\\le & {} C\Vert u_n\Vert _{L^{2pq}(|x|<R)}^{2p}\int _0^R\frac{{\mathrm{d}}t}{t^{-q'b-N+1}}\\\le & {} C\Vert u_n\Vert _{L^{2pq}(|x|<R)}^{2p}R^{N+q'b}. \end{aligned}$$

  • 1. First sub-case $N\ge 3$.

    Now, since $p<\frac{b+N}{N-2}$, there exists $q>\frac{N}{b+N}$, such that $1<qp<\frac{N}{N-2}$. Thus, with compact Sobolev injections

    $$\begin{aligned} \int _{|x|\le R}|x|^{b}|u_n|^{2p}\,{\mathrm{d}}x\rightarrow 0. \end{aligned}$$

  • 2. Second sub-case $1\le N\le 2$.

    In such a case, the condition $\frac{N}{b+N}<\frac{N}{p(N-2)}$ is always satisfied. Therefore, one has the same conclusion as above.

  Since $1<p<\frac{N}{N-2}$, one has for a sub-sequence

  $$\begin{aligned} \int _{|x|\ge R}|x|^{b}|u_n|^{2p}\,{\mathrm{d}}x \le \epsilon \Vert u_n\Vert _{L^{2p}(|x|>R)}^{2p}\le C\epsilon . \end{aligned}$$

  Since $\epsilon$ is arbitrary, one obtains $\int _{{\mathbb{R}}^N}|x|^{b}|u_n|^{2p}\,{\mathrm{d}}x\rightarrow 0$ as $n\rightarrow \infty .$ The proof is complete.

• B. Second case $b\ge 0$.

  Take $(u_n)$ a bounded sequence of $H^1_{rd}.$ Without loss of generality, one assumes that $(u_n)$ converges weakly to zero in $H^1.$ The purpose is to prove that $\Vert u_n\Vert _{L^{2p}(|x|^{b}\,{\mathrm{d}}x)}\rightarrow 0.$

  Let $\epsilon >0$. Thanks to Strauss inequality and Sobolev injections via the assumption $1+\frac{b}{N-1}<p<p^*$, write

  $$\begin{aligned} \int _{|x|\le \epsilon }|x|^{b}|u_n|^{2p}\,{\mathrm{d}}x= & {} \int _{|x|\le \epsilon }\left( |x|^{\frac{N-1}{2}}|u_n|\right) ^{\frac{2b}{N-1}}|u_n|^{2p-\frac{2b}{N-1}}\,{\mathrm{d}}x\\\le & {} C\Vert u_n\Vert _{H^1}^{\frac{2b}{N-1}}\Vert u_n\Vert _{2p-\frac{2b}{N-1}}^{2p-\frac{2b}{N-1}}\rightarrow 0. \end{aligned}$$

  Moreover, with Strauss inequality and Rellich Theorem

  $$\begin{aligned} \int _{\epsilon \le |x|\le \frac{1}{\epsilon }}|x|^{b}|u_n|^{2p}\,{\mathrm{d}}x\le C\Vert u_n\Vert _{L^\infty \left( \epsilon \le |x|\le \frac{1}{\epsilon }\right) }^{2(p-1)}\int _{\epsilon \le |x|\le \frac{1}{\epsilon }}|u_n|^2\,{\mathrm{d}}x\rightarrow 0. \end{aligned}$$

  Now, with Strauss inequality

  $$\begin{aligned} \int _{|x|\ge \frac{1}{\epsilon }}|x|^{b}|u_n|^{2p}\,{\mathrm{d}}x=\, & {} \int _{|x|\ge \frac{1}{\epsilon }}|x|^{b-(p-1)(N-1)}\left( |x|^{\frac{N-1}{2}}|u_n|\right) ^{2(p-1)}|u_n|^2\,{\mathrm{d}}x\\ \le\, & {} C\Vert u_n\Vert _{H^1}^{2(p-1)}\int _{|x|\ge \frac{1}{\epsilon }}|x|^{b-(p-1)(N-1)}|u_n|^2\,{\mathrm{d}}x\\ \le\, & {} C\epsilon ^{(p-1)(N-1)-b}\Vert u_n\Vert _{H^1}^{2p}\\ \le\, & {} C\epsilon ^{(p-1)(N-1)-b}. \end{aligned}$$

  The proof is achieved, because $b-(p-1)(N-1)<0.$

Summary In conclusion, this paper gives a discussion of the existence of global/non-global solutions to the inhomogeneous Schrödinger problems (1) and (2), depending on the competition between the free associated propagator and the two parts of the combined source term.