Global and Non-global Solutions for a Class of Damped Fourth-Order Schrödinger Equations

Abstract

In this note, some well-posedness issues for a class of fourth-order Schrödinger equations with a time-dependent damping and a pure power non-linearity are investigated. Indeed, global and non-global existence of solutions is obtained under suitable conditions on the damping and the source term.

Appendix

1.1 Proof of Proposition 2.15

Let us give a brief proof of the Strichartz estimate in Proposition 2.15. Take the free high-order Schrödinger-type equation

$$ (i\partial _t+\Delta ^2)u=0,\quad u(0,\cdot )=u_0. $$

Taking the Fourrier part of u, yields

$$ u(t,x)={\mathcal {F}}^{-1}\Big (y\mapsto e^{it|y|^4}\Big )*u_0:= e^{it\Delta ^2}u_0. $$

It’s known [11] that $|{\mathcal {F}}^{-1}\Big (y\mapsto e^{it|y|^4}\Big )(x)|\lesssim t^{-\frac{N}{4}}(1+t^{-\frac{1}{4}}|x|)^{-\frac{2}{3}},\quad t> 0$. This implies that

$$ \Vert e^{it\Delta ^2}u_0\Vert _{L^\infty }\lesssim t^{-\frac{N}{4}}\Vert u_0\Vert _{1}\quad \text{ and }\quad \Vert e^{it\Delta ^2}u_0\Vert \lesssim \Vert u_0\Vert . $$

By interpolation, yields $\Vert e^{it\Delta ^2}u_0\Vert _r\lesssim t^{-\frac{N}{4}(1-\frac{2}{r})}\Vert u_0\Vert _{r'}$. This implies that

$$\begin{aligned} \Vert \int _0^te^{i(t-s)\Delta ^2}h(s)\mathrm{{d}}s\Vert _r&\lesssim \int _0^t|t-s|^{-\frac{N}{2}(\frac{1}{2}-\frac{1}{r})}\Vert h(s)\Vert _{r'}\mathrm{{d}}s\\&\lesssim \int _0^t|t-s|^{-\frac{2}{q}}\Vert h(s)\Vert _{r'}\mathrm{{d}}s. \end{aligned}$$

The proof is achieved via Riesz–Potential inequality.

1.2 Proof of Proposition 2.1

Let us use a standard fixed point argument. Denote the admissible pair

$$ q:=\frac{8(p+1)}{N(p-1)},\quad r:=1+p. $$

For $T,R>0,$ define the space

$$\begin{aligned} E_{T,R}&:=\Big \{{u}\in C_T(H^2)\cap L^q_T(W^{2,r})\,\, \text{ s. } \text{ t } \,\,\Vert {u}\Vert _{L^\infty _T(L^2) L^{q}_T(L^{r})}\\&\quad \cap +\Vert \Delta {u}\Vert _{L^\infty _T(L^2)\cap L^{q}_T(L^{r})}\le R\Big \} \end{aligned}$$

endowed with the complete distance

$$ \mathrm{{d}}({u},{v}):=\Vert u-v\Vert _{L_T^\infty (L^2)}+\Vert u-v\Vert _{L^{q}_T(L^{r})}. $$

Define the function

$$ \phi ({u}(t)) :={\tilde{u}}(t):= e^{it\Delta ^2}u_0 - i\int _0^th(s)e^{i(t-s)\Delta ^2}[|u|^{p-1}u]\,\mathrm{{d}}s. $$

Let us prove the existence of some small $0<T, R<1$ such that $\phi$ is a contraction of $E_{T,R}$. Take ${u_1}, {u_2}\in E_{T,R},$ applying the Strichartz estimate, one gets

$$\begin{aligned} \mathrm{{d}}({\tilde{u}}_1,{\tilde{u}}_2)\le & {} C_{q}\Vert h\Vert _{L^\infty ([0,1])}\Vert |u_1|^{p-1}u_1-|u_2|^{p-1}u_2\Vert _{L^{q'}((0,T),L^{r'})}\\\le & {} C_{q,h}\sum _{i=1}^2\Vert (u_1-u_2)u_i^{p-1}\Vert _{L^{q'}((0,T),L^{r'})}\\\le & {} C_{q,h}\sum _{i=1}^2\Vert u_i\Vert _{L^\infty ((0,T),L^r)}^{p-1}\Vert u_1-u_2\Vert _{L^{q'}((0,T),L^r)}\\\le & {} C_{q,h}\sum _{i=1}^2\Vert u_i\Vert _{L^{\infty }((0,T),H^2)}^{p-1}T^{1-\frac{2}{q}}\mathrm{{d}}(u_1,u_2)\\\le & {} C_{q,h}R^{p-1}T^{1-\frac{2}{q}}\mathrm{{d}}(u_1,u_2). \end{aligned}$$

Moreover, taking $u_2=0$ in the previous inequality, yields

$$ \Vert \tilde{u}\Vert _{L_T^q(L^r)}\le C_{q,h}\Big (\Vert u_0\Vert + T^{1-\frac{2}{q}}R^p\Big ). $$

With Hölder inequality

$$\begin{aligned} \Vert \Delta (f(u))\Vert _{L_T^{q'}(L^{r'})}\lesssim & {} \Vert |u|^{p-1}\Delta {u}\Vert _{L_T^{q'}(L^{r'})}+\Vert |\nabla {u}|^2|u|^{p-2}\Vert _{L_T^{q'}(L^{r'})}\\\lesssim & {} \Vert u\Vert _{L_T^\infty (L^r)}^{p-1}\Vert \Delta {u}\Vert _{L_T^{q'}(L^{r})}+\Vert |\nabla {u}|^2|u|^{p-2}\Vert _{L_T^{q'}(L^{r'})}. \end{aligned}$$

Using the interpolation inequality $\Vert \nabla \cdot \Vert _{r}^2\lesssim \Vert \cdot \Vert _{r}\Vert \Delta \cdot \Vert _{r}$, one gets

$$\begin{aligned} \Vert |\nabla {u}|^2|u|^{p-2}\Vert _{L^{r'}_x}\lesssim & {} \Vert \nabla {u}\Vert _{L^{r}_x}^2\Vert u\Vert _{L^{r}_x}^{p-2} \lesssim \Vert \Delta {u}\Vert _{L^{r}_x}\Vert {u}\Vert _{L^r_x}^{p-1}. \end{aligned}$$

This implies that

$$\begin{aligned} \Vert \Delta (f(u))\Vert _{L_T^{q'}(L^{r'})}\lesssim & {} \Vert u\Vert _{L_T^\infty (L^r)}^{p-1}\Vert \Delta {u}\Vert _{L_T^{q}(L^{r})}T^{1-\frac{2}{q}}. \end{aligned}$$

Then, thanks to Strichartz estimates, one has

$$ \Vert {\tilde{u}}\Vert _{L_T^q(W^{2,r})\cap L_T^\infty (H^2)}\le C_{q,h}\Big (\Vert e^{i.\Delta ^2}u_0\Vert _{L_T^q(W^{2,r})}+ T^{1-\frac{2}{q}}R^{p}\Big ). $$

Since $\Vert e^{i.\Delta ^2}u_0\Vert _{L_T^q(W^{2,r})}\rightarrow 0$ when $T\rightarrow 0$, Choosing $T>0$ sufficiently small and $R>2C_{q,h}\Vert e^{i.\Delta ^2}u_0\Vert _{L_T^q(W^{2,r})}$ via the fact that ${2\le }p\le p^*$, $\phi$ is a contraction of $E_{T, R}$. Local existence is proved via a classical Picard argument. Moreover,

$$ C\Vert u(t)\Vert _{H^2}+C'(T^*-t)^{1-\frac{2}{q}}R^p>R,\quad \forall R>0. $$

Thus, for $R=2C\Vert u(t)\Vert _{H^2}$,

$$ \Vert u(t)\Vert _{H^2}^{p-1}(T^*-t)^{1-\frac{2}{q}}\gtrsim 1. $$

Thus,

$$ \limsup _{T^*}\Vert u(t)\Vert _{\dot{H}^2}=\infty . $$

Now, in the critical case, arguing as previously with a Picard fixed point in a centered ball of $L_T^{a}(W^{2,\rho })$, with computation of the end of the previous section, one obtains a local solution. Moreover, arguing as in the sub-critical case, it follows that

$$ \limsup _{T^*}\Vert u(t)\Vert _{L^{a}((0,t),\dot{W}^{2,\rho })}=\infty . $$

1.3 Proof of Proposition 2.3

Define the differential operator

$$ {a}_{\varphi _R} := -i (\nabla {\varphi _R}\cdot \nabla + \nabla \cdot \nabla {\varphi _R} ) $$

which acts on functions according to

$$ {a}_{\varphi _R} f:= -i \big (\nabla {\varphi _R}\cdot \nabla f + \nabla \cdot (f\nabla {\varphi _R}) \big ). $$

Define also

$$ M_R[v(t)]:=2\mathfrak {I}(\int _{{\mathbb {R}}^N}{{\bar{v}}}\nabla \varphi _R\nabla v\,\mathrm{{d}}x)=\langle u(t), {a}_{\varphi _R}v(t) \rangle . $$

By taking the time derivative and using $i\dot{v}$ in (1.2), one gets

$$\begin{aligned} \frac{\mathrm{{d}}}{\mathrm{{d}}t}M_R[v(t)]= & {} \langle v,[\Delta ^2, i{a}_{\varphi _R}]v(t) \rangle + \langle v,[\lambda e^{-(p-1)\int _0^ta(s)\mathrm{{d}}s}|v|^{p-1}, i{a}_{\varphi _R}]v(t) \rangle \nonumber \\= & {} {\mathcal {A}} [v(t)] + {\mathcal {B}} [v(t)], \end{aligned}$$

(7.1)

where $[X, Y]= XY- YX$ denotes the commutator of X and Y.

Using computations done in [3], one concludes that

$$\begin{aligned} {\mathcal {A}}[v(t)]\le 8\int _{R^N}|\Delta v|^2 + {\mathcal {O}}\Big (R^{-4} + R^{-2}\Vert \nabla v\Vert ^2 \Big ). \end{aligned}$$

(7.2)

Now, for the last terms in (2.3), an integration by parts yields

$$\begin{aligned} {\mathcal {B}}[v(t)]= & {} - \langle v, [\lambda e^{-(p-1)\int _0^ta(s)\,\mathrm{{d}}s}|v|^{p-1}, \nabla \varphi _R\cdot \nabla + \nabla \cdot \nabla \varphi _R]v\rangle \\= & {} -\lambda e^{-(p-1)\int _0^ta(s)\,\mathrm{{d}}s} \langle v, [|v|^{p-1}, \nabla \varphi _R\cdot \nabla + \nabla \cdot \nabla \varphi _R]v\rangle \\= & {} 2\lambda e^{-(p-1)\int _0^ta(s)\,\mathrm{{d}}s}\int _{{\mathbb {R}}^N}|v|^2\nabla (|v|^{p-1})\nabla \varphi _R\,\mathrm{{d}}x\\= & {} -\frac{2(p-1)}{p+1}\lambda e^{-(p-1)\int _0^ta(s)\,\mathrm{{d}}s}\int _{{\mathbb {R}}^N}|v|^{p+1}\Delta (\varphi _R)\,\mathrm{{d}}x. \end{aligned}$$

Since $\varphi _R(r) = \frac{r^2}{2}$ for $r\le R$ and hence $\Delta \varphi _R(r) -N=0$ for $r\le R,$ one obtains

$$\begin{aligned} {\mathcal {B}}[v(t)]= & {} -\lambda e^{-(p-1)\int _0^ta(s)\,\mathrm{{d}}s}\\&\left( \frac{2N(p-1)}{1+p}\int _{{\mathbb {R}}^N}|v|^{1+p}\,\mathrm{{d}}x - \frac{2(p-1)}{1+p}\int _{|x|\ge R}(\Delta \varphi _R -N)|v|^{1+p}\,\mathrm{{d}}x\right) . \end{aligned}$$

Applying Strauss inequality,

$$\begin{aligned} \int _{|x|\ge R}|v|^{1+p}\,\mathrm{{d}}x\lesssim & {} \Vert v\Vert ^2\Vert v\Vert _{L^\infty (|x|\ge R)}^{p-1}\\\lesssim & {} R^{-(N-1)\frac{p-1}{2}}\Vert v\Vert ^{2+\frac{p-1}{2}}\Vert \nabla v\Vert ^{\frac{p-1}{2}}. \end{aligned}$$

Taking account of $\Vert \nabla {\varphi _R}- N\Vert _{L^\infty }\lesssim 1$, one obtains

$$\begin{aligned} {\mathcal {B}}[v(t)]&=-\lambda e^{-(p-1)\int _0^ta(s)\,\mathrm{{d}}s}\Bigg (\frac{2N(p-1)}{1+p}\int _{{\mathbb {R}}^N}|v|^{p-1}\,\mathrm{{d}}x\\&\quad + {\mathcal {O}} ( R^{-(N-1)\frac{p-1}{2}}\Vert \nabla v\Vert ^{\frac{p-1}{2}}) \Bigg ). \end{aligned}$$

Finally, thanks to (7.2) and the previous equality, one gets

$$\begin{aligned} \frac{\mathrm{{d}}}{\mathrm{{d}}t}M_R[v(t)]\le & {} 8{}\Vert \Delta v(t)\Vert ^2-\frac{2N\lambda (p-1)}{1+p} e^{-(p-1)\int _0^ta(s)\,\mathrm{{d}}s}\int _{{\mathbb {R}}^N}|v|^{p-1}\,\mathrm{{d}}x\\&+ {\mathcal {O}} \Big (R^{-4} + R^{-2}{}\Vert \nabla v\Vert ^2 + R^{-(N-1)\frac{p-1}{2}}\Vert \nabla v(t)\Vert ^{\frac{p-1}{2}} \Big ) \\\le & {} 2N(p-1)E(v(t)) - (N(p-1)-8){}\Vert \Delta v(t)\Vert ^2 \\&+ {\mathcal {O}} \Big (R^{-4} + R^{-2}{}\Vert \nabla v\Vert ^2 + R^{-(N-1)\frac{p-1}{2}}\Vert \nabla v(t)\Vert ^{\frac{p-1}{2}} \Big ). \end{aligned}$$

This completes the proof of Proposition 2.3.