Abstract
We study the Cauchy problem of the nonlinear fourth-order Schrödinger equation with gain or loss: \(iu_{t}+\triangle^{2}u+\lambda|u|^{\alpha}u +i\varepsilon a(t)|u|^{\beta}u=0\), \(x\in R^{n}\), \(t\in R\), where \(2\leq\alpha\leq\frac{8}{n-4}\) and \(2\leq\beta\leq\frac{8}{n-4}\), ε is a real number, \(a(t)\) is a real function, and \(n>4\). We study the asymptotic properties of its local and global solutions as \(\varepsilon\rightarrow0\).
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1 Introduction
In this paper we study the following nonlinear fourth-order Schrödinger equation with gain or loss:
where \(u(x,t)\) are complex-valued function. We have \(2\leq\alpha\leq \frac{8}{n-4}\) and \(2\leq\beta\leq\frac{8}{n-4}\), ε is a real number, \(a(t)\) is a real function, and \(n>4\).
For the case \(\varepsilon=0\), the above equation is the nonlinear fourth-order Schrödinger equation,
For (1.2), in [1] we have obtained the local well-posedness result in the space \(C([-T,T], H^{2}(R^{n}))\) if \(n>4\) and \(2\leq\alpha\leq\frac{8}{n-4}\). We also get the global well-posedness result in the space \(C(R,H^{2}(R^{n}))\) if \(n>4\) and \(\lambda>0\), \(2\leq\alpha\leq\frac{8}{n-4}\) or \(\lambda<0\), \(2\leq\alpha\leq\frac{8}{n}\). For the energy-critical case, in [2] and [3], Pausader Benoit gives the global well-posedness and scattering for \(n\geq5\) and radial initial data. In [4], Miao et al. study the defocusing case and obtain the global existence for \(n\geq9\). In [5], Zhang and Zheng obtain the global solution and scattering for \(n=8\). Pausader Benoit also discusses the mass-critical case in [6].
For the case \(\varepsilon\neq0\), \(a(t)\) is the gain (loss) if \(a(t)<0\) (\(a(t)>0\)). In [7], the authors discuss the Schrödinger equation with gain. They have obtained the result: The value of \(a(t)\) will determine whether or not the solution will blow up. Feng et al. study the Schrödinger equation with gain/loss in [8] and [9]. They, respectively, give the limit behavior of solution as \(\varepsilon\rightarrow0\) and the global solution and blow-up result. As far as we know, there are fewer results about the fourth-order Schrödinger equation with gain. In this paper, we will discuss the local well-posedness and the global well-posedness of (1.1); especially, we will discuss the asymptotic behavior of the solution as \(\varepsilon\rightarrow0\).
2 The preliminary estimates
First, we denote by \(U(t)\) (\(t\in R\)) the fundamental solution operator of the fourth-order Schrödinger equation [10], i.e.,
where φ̂ denotes the Fourier transformation of φ, and \(F^{-1}\) represents the inverse Fourier transformation.
Thus the equivalent integral equations [11] of (1.1) and (1.2) are, respectively,
and
Second, we introduce the following notations. For any given \(T>0\), we define the space \(L^{q}(0,T;W^{2,r}(R^{n}))\) with the norm
For two integers \(8\leq q\leq\infty\) and \(2\leq r<\infty\), we say that \((q,r)\) is an admissible pair if the following condition is satisfied:
For simplicity, in this paper, we will use C to denote various constants which may be different from line to line.
We have the following Strichartz estimate (see [1]): For any admissible pair \((q,r)\)
and
where \((\gamma,\rho)\) is an arbitrary admissible pair, and ′ represents the conjugate number.
From Theorem 4.5 of [1], we have the following results.
Proposition 2.1
(subcritical case)
Assume that \(n>4\), \(a\in L^{\infty}(0,\infty)\), \(2\leq\alpha<\frac{8}{n-4}\), and \(2\leq\beta<\frac{8}{n-4}\), \((\gamma_{1},\rho_{1})=(\alpha+2,\frac{2n(\alpha+2)}{n(\alpha+2)-8})\), \((\gamma_{2},\rho_{2})=(\frac{8(\beta+2)}{n\beta},\beta+2)\). For any \(u_{0}\in H^{2}(R^{n})\), there exists δ such that the Cauchy problem (1.1) has a unique solution \(u_{\varepsilon}\) in the space \(L^{\infty}(0,\delta;H^{2}(R^{n}))\cap L^{\gamma_{1}}(0,\delta;W^{2,\rho_{1}}(R^{n})) \cap L^{\gamma_{2}}(0,\delta;W^{2,\rho_{2}}(R^{n}))\). Moreover,
Proposition 2.2
(critical case)
Assume that \(n>4\), \(a\in L^{\infty}(0,\infty)\), \(\alpha=\frac{8}{n-4}\), \(2\leq\beta<\frac{8}{n-4}\), \((\gamma^{*},\rho^{*})=(\frac{2n}{n-4},\frac{2n^{2}}{n^{2}-4n+16})\), \((\gamma_{2},\rho_{2})=(\frac{8(\beta+2)}{n\beta},\beta+2)\). For any \(u_{0}\in H^{2}(R^{n})\), there exists δ such that the Cauchy problem (1.1) has a unique solution \(u_{\varepsilon}\) in the space \(L^{\infty}(0,\delta;H^{2}(R^{n}))\cap L^{\gamma^{*}}(0,\delta;W^{2,\rho^{*}}(R^{n})) \cap L^{\gamma_{2}}(0,\delta;W^{2,{\rho_{2}}}(R^{n}))\). Moreover,
3 Main results
Lemma 3.1
Let n, α, β, \((\gamma_{1},\rho_{1})\), \((\gamma _{2},\rho_{2})\) be as in Proposition 2.1. Assume that u is the solution of (1.2), defined on a maximal time interval \([0,T^{*})\), \(0< l< T^{*}\), and \(u_{\varepsilon}\) exists on \([0,l]\). If \(\lim\sup_{\varepsilon\rightarrow 0}\|u_{\varepsilon}\|_{L^{\infty}(0,l;H^{2})\cap L^{\gamma_{1}}(0,l;W^{2,\rho _{1}})\cap L^{\gamma_{2}}(0,l;W^{2,\rho_{2}})}<+\infty\), then we have \(u_{\varepsilon}\rightarrow u\) in \(L^{q}(0,l;W^{2,r}(R^{n}))\) as \(\varepsilon\rightarrow0\), where \((q,r)\) is arbitrary admissible pair.
Proof
First, we prove
From (2.1) and (2.2), using Strichartz estimates, we have
where \(J(t)=i\lambda\int_{0}^{t}U(t-s)(|u_{\varepsilon}|^{\alpha}u_{\varepsilon}-|u|^{\alpha}u)(s)\,ds\), \(K(t)=-\varepsilon\int_{0}^{t}U(t-s)a(s)(|u_{\varepsilon}|^{\beta}u_{\varepsilon})(s)\,ds\), \((\gamma,\rho)=(\frac{8(\alpha+2)}{n\alpha},\alpha+2)\).
Since \(\lim\sup_{\varepsilon\rightarrow0}\|u_{\varepsilon}\|_{L^{\infty }(0,l;H^{2})\cap L^{\gamma_{1}}(0,l;W^{2,\rho_{1}})\cap L^{\gamma _{2}}(0,l;W^{2,\rho_{2}})}<+\infty\), there exist \(N_{1}, \varepsilon_{0}\) such that
Let \(N_{2}=\|u\|_{L^{\infty}(0,l;H^{2})}\), it is obvious that \(N_{2}<+\infty\). Using the Hölder inequality and the Sobolev embedding [12], we have
where \(a=\frac{4\alpha(\alpha+2)}{8-(n-4)\alpha}\).
Similarly, we have
where \(b=\frac{4\beta(\beta+2)}{8-(n-4)\beta}\).
Let \(N_{3}=C\|a\|_{L^{\infty}}N_{1}^{\beta+1}\). Substituting (3.2) and (3.3) into (3.1), we have
In the following we will prove that \(\|u_{\varepsilon}-u\|_{L^{\gamma}(0,l;L^{\rho})}\rightarrow0\) as \(\varepsilon\rightarrow 0\).
Noting that \(N_{1}, N_{2}<\infty\), we can divide the time interval \([0,l]\) into subintervals \([t_{i},t_{i+1}]\), \(i=0, 1, \ldots, J-1\), where \(t_{0}=0\), \(t_{J-1}=l\) such that in each part \(C(\|u_{\varepsilon}\|^{\alpha}_{L^{a}(t_{i},t_{i+1};L^{\alpha+2})} +\|u\|^{\alpha}_{L^{a}(t_{i},t_{i+1};L^{\alpha+2})})=\frac{1}{2}\).
On \([t_{0},t_{1}]\), since \(u_{\varepsilon}(t_{0})=u(t_{0})=u_{0}\), we have
which means
By (3.4), we have
On \([t_{1},t_{2}]\), we have
from which we can obtain
Especially, we have \(\|u_{\varepsilon}-u\|_{L^{\infty}(t_{1},t_{2};L^{2})}\leq 6\varepsilon N_{3}\).
By induction, we have
So we have
Furthermore, we have
Second, we prove
Let \(g_{1}(u)=|u|^{\alpha}u\), \(g_{2}(u)=|u|^{\beta}u\). Then, using Strichartz estimates, we have
Using the Hölder inequality, the Sobolev embedding, and the Young inequality, we obtain
and
where \(d_{1}=\frac{2n(\alpha+2)(\alpha-1)}{24-(n-4)(\alpha+2)}\), \(e_{1}=\frac {2n(\alpha+2)}{(n-2)(\alpha+2)-8}\).
Substituting (3.6)-(3.8) into (3.5), we have
Similar to the proof in the first step, we have
At last, we prove
By simple computing, we have
where \(K_{1}=\lambda\int_{0}^{t}U(t-s)A_{1}(u_{\varepsilon},u)(s)\,ds\), \(K_{2}=\lambda\int_{0}^{t}U(t-s)A_{2}(u_{\varepsilon},u)(s)\,ds\), \(K_{3}=-\varepsilon\int_{0}^{t}U(t-s)a(s)A_{3}(u_{\varepsilon})(s)\,ds\). The arrays \(A_{1}(u_{\varepsilon},u)=g_{1}^{\prime}(u_{\varepsilon})D^{2}(u_{\varepsilon}-u)+g_{1}^{\prime\prime}(u_{\varepsilon})D(u_{\varepsilon}-u)\times Du\), \(A_{2}(u_{\varepsilon},u)=Du\times[g_{1}^{\prime\prime}(u_{\varepsilon})Du_{\varepsilon}-g_{1}^{\prime\prime}(u)Du]+ [g_{1}^{\prime}(u_{\varepsilon})-g_{1}^{\prime}(u)]D^{2}u\), \(A_{3}(u_{\varepsilon})= g_{2}^{\prime\prime}(u_{\varepsilon})Du_{\varepsilon}\times Du_{\varepsilon}+ g_{2}^{\prime}(u_{\varepsilon})D^{2}u_{\varepsilon}\).
By the Hölder inequality and the Sobolev embedding, we have
and
Thus we have from (3.10) and (3.11)
Similar to the proof of (3.11), we obtain
Noting that \(\alpha\geq2\), we have
where \(e_{2}=\frac{2n(\alpha+2)}{(n-2)(\alpha+2)-8}\), \(\frac{1}{{\rho_{1}}^{\prime}}=\frac{(\rho_{1}-2)(\alpha-2)}{\rho_{1}\alpha}+\frac {1}{d_{2}}+\frac{2}{e_{2}}\).
Similarly, using the Hölder inequality and the Sobolev embedding, we obtain
Thus we have from (3.13) and (3.15)
Similar to the proof of (3.3), we obtain
and
From (3.17) and (3.18), we immediately obtain
Taking, respectively, \((q,r)=(\gamma,\rho)\) and \((q,r)=(\gamma_{1},\rho_{1})\) in (3.9), (3.12), (3.16), and (3.19), similar to the method of the first step, we can obtain
□
Noting that if \(\alpha=\frac{8}{n-4}\), a in (3.2) will be meaningless. So we will need the following lemma for the critical case.
Lemma 3.2
Let n, α, β, \((\gamma^{*},\rho^{*})\), \((\gamma _{2},\rho_{2})\) be as in Proposition 2.2. Assume that u is the solution of (1.2), defined on a maximal time interval \([0,T^{*})\), \(0< l< T^{*}\), and \(u_{\varepsilon}\) exists on \([0,l]\). If \(\lim\sup_{\varepsilon\rightarrow 0}\|u_{\varepsilon}\|_{L^{\infty}(0,l;H^{2})\cap L^{\gamma ^{*}}(0,l;W^{2,\rho^{*}})\cap L^{\gamma_{2}}(0,l;W^{2,\rho_{2}})}<+\infty\), then we have \(u_{\varepsilon}\rightarrow u\) in \(L^{q}(0,l;W^{2,r}(R^{n}))\) as \(\varepsilon\rightarrow0\), where \((q,r)\) is arbitrary admissible pair.
Proof
Using the Hölder inequality and a Sobolev embedding, we have
From (2.1) and (2.2), using Strichartz estimates, we have
similarly as in Lemma 3.1, we can obtain
Noting that for \((\gamma_{1},\rho_{1})\) in Lemma 3.1 in the case \(\alpha =\frac{8}{n-4}\), \(2\leq\beta<\frac{8}{n-4}\), we have
thus obviously
and
for all admissible pairs \((q,r)\). □
Remark 3.1
For the critical case \(2\leq\alpha<\frac{8}{n-4}\), \(\beta=\frac{8}{n-4}\), we only take the working space as \(L^{\infty}(0,\delta;H^{2}(R^{n})) \cap L^{\gamma_{1}}(0,\delta;W^{2,\rho_{1}}(R^{n})) \cap L^{\gamma^{*}}(0,\delta;W^{2,\rho^{*}}(R^{n}))\).
For the case \(\alpha=\beta=\frac{8}{n-4}\), we take the working space as \(L^{\infty}(0,\delta;H^{2}(R^{n})) \cap L^{\gamma^{*}}(0,\delta; W^{2,\rho^{*}}(R^{n}))\).
Theorem 3.1
Assume that \(n>4\), \(a\in L^{\infty}(0,\infty)\), \(2\leq\alpha\leq\frac{8}{n-4}\), and \(2\leq\beta\leq\frac{8}{n-4}\). Assume that u is the solution of (1.2) with initial value \(u_{0}\in H^{2}(R^{n})\), defined on a maximal time interval \([0,T^{*})\). Then we have:
-
(1)
For any given \(0< T< T^{*}\), there is a solution \(u_{\varepsilon}\) on \([0,T]\).
-
(2)
\(u_{\varepsilon}\rightarrow u\) in \(L^{q}(0,T;W^{2,r}(R^{n}))\) as \(\varepsilon\rightarrow0\), where \((q,r)\) is an arbitrary admissible pair.
Proof
(1) The case \(2\leq\alpha<\frac{8}{n-4}\) and \(2\leq\beta<\frac{8}{n-4}\).
From Proposition 2.1, we find that there exists \(u_{\varepsilon}\) on \([0,\delta]\) such that
So for small ε, we have
Using Lemma 3.1, we have \(u_{\varepsilon}\rightarrow u\) in \(L^{q}(0,\delta ;W^{2,r}(R^{n}))\) as \(\varepsilon\rightarrow0\), for any arbitrary admissible pair \((q,r)\).
Especially, we have \(\|u_{\varepsilon}(\delta)\|_{H^{2}}\leq2\|u_{0}\|_{H^{2}}\). Again using Proposition 2.1, there exists \(u_{\varepsilon}\) on \([\delta ,2\delta]\) such that
By a continuation extension method, we obtain the solution \(u_{\varepsilon}\) on \([0,T] \) (\(0< T< T^{*}\)) such that
So
using Lemma 3.1, we immediately have \(u_{\varepsilon}\rightarrow u\) in \(L^{q}(0,T;W^{2,r}(R^{n}))\) as \(\varepsilon\rightarrow0\), for any arbitrary admissible pair \((q,r)\).
(2) Case 1: \(\alpha=\frac{8}{n-4}\), \(2\leq\beta<\frac{8}{n-4}\).
From Proposition 2.2, we find that there exists \(u_{\varepsilon}\) on \([0,\delta]\) such that
So for small ε, we have
Using Lemma 3.2, we have \(u_{\varepsilon}\rightarrow u\) in \(L^{q}(0,\delta ;W^{2,r}(R^{n}))\) as \(\varepsilon\rightarrow0\), for any arbitrary admissible pair \((q,r)\).
Noting that
so, again using Proposition 2.2, there exists \(u_{\varepsilon}\) on \([\delta,2\delta]\) such that
By continuation extension method, we obtain the solution \(u_{\varepsilon }\) on \([0,T]\) (\(0< T< T^{*}\)) such that
So
using Lemma 3.2, we immediately have \(u_{\varepsilon}\rightarrow u\) in \(L^{q}(0,T;W^{2,r}(R^{n}))\) as \(\varepsilon\rightarrow0\), for any arbitrary admissible pair \((q,r)\).
Case 2: \(\beta=\frac{8}{n-4}\), \(2\leq\alpha<\frac{8}{n-4}\) or \(\alpha =\beta=\frac{8}{n-4}\).
See Remark 3.1, the proof is similar; here we omit it. □
Lemma 3.3
Assume that u is the global solution of (1.2) with the initial valve \(u_{0}\in H^{2}(R^{n})\) and \(u\in L^{q}_{loc}(0,\infty ;W^{2,r}(R^{n}))\). Then we have:
-
(1)
The solution \(u_{\varepsilon}\) of (1.1) with the initial valve \(u_{0}\) is global for sufficiently small ε.
-
(2)
\(u_{\varepsilon}\rightarrow u\) in \(L^{q}(0,\infty;W^{2,r}(R^{n}))\) as \(\varepsilon\rightarrow0\), where \((q,r)\) is an arbitrary admissible pair.
Proof
(1) We will prove that \(u_{\varepsilon}\) is also global for small ε if u is global.
From Theorem 3.1, we can see
for all \(T<\infty\).
Since u is global, for any \(\eta>0\), there exists sufficient large T such that
\((\gamma_{1},\rho_{1})\) is the same as in Theorem 3.1.
Case 1: \(2\leq\alpha<\frac{8}{n-4}\), \(2\leq\beta\leq\frac{8}{n-4}\).
From (2.2), (2.3)-(2.4), using a continuity argument we can obtain
Thus we have
Obviously \(\|U(t)u_{\varepsilon}(T)\|_{L^{q}(0,\infty;W^{2,r})}\leq\eta\) for suitable T and any admissible pair \((q,r)\).
Furthermore we define the working space as follows:
where \((\gamma_{2},\rho_{2})\) is the same as in Theorem 3.1.
Using the Hölder inequality, the interpolation inequality [13], and the Sobolev embedding, we have
Similarly, we can obtain
For the case \(4< n<8\), we have
For the case \(8\leq n<12\), we have
thus we have
Noting that \((\frac{2(n+4)}{n+8},\frac{2(n+4)}{n+8})\) is an admissible pair, using Strichartz estimates, we can obtain
Using (2.1), we have
Using a continuity argument, we immediately have
which means that \(\|u_{\varepsilon}\|_{X(T,\infty)}\leq M\), where M is a constant.
Furthermore, we have \(\|u_{\varepsilon}\|_{L^{q}(T,\infty;W^{2,r})}\leq M\), for any admissible pair \((q,r)\). Thus \(u_{\varepsilon}\) is global.
Case 2: \(\alpha=\frac{8}{n-4}\), \(2\leq\beta\leq\frac{8}{n-4}\).
We need the following working space:
The process of proof is similar to the case 1, so here we omit the detailed proof.
(2) In the sequel, we prove \(u_{\varepsilon}\rightarrow u\) in \(L^{q}(0,\infty;W^{2,r}(R^{n}))\) as \(\varepsilon\rightarrow0\), for any admissible pair \((q,r)\).
Using (2.1) and (2.2), we have
Thus we have
□
Theorem 3.2
Assume that \(n>4\), \(a\in L^{\infty}(0,\infty)\), \(2\leq\alpha\leq\frac{8}{n-4}\), and \(2\leq\beta\leq\frac{8}{n-4}\). One of the following conditions holds:
-
(i)
\(\lambda<0\),
-
(ii)
\(\lambda>0\), \(\|u_{0}\|_{H^{2}}\) is small.
Then we have
-
(1)
The solution \(u_{\varepsilon}\) of (1.1) is global for small ε.
-
(2)
\(u_{\varepsilon}\rightarrow u\) in \(L^{q}(0,\infty;W^{2,r}(R^{n}))\) as \(\varepsilon\rightarrow0\), where \((q,r)\) is arbitrary admissible pair.
Proof
Note that the solution u of (1.2) is global provided the conditions (i) \(\lambda<0\) or (ii) \(\lambda>0\), \(\|u_{0}\| _{H^{2}}\) is small hold. Combing Lemma 3.3, the proof of Theorem 3.2 immediately is complete. □
4 Conclusions
The appearance of gain/loss does not affect the local well-posedness of the solution. Moreover, the solution \(u_{\varepsilon}\) will converge to u in the space \(L^{q}(0,T;W^{2,r}(R^{n}))\) as ε converges to 0. Furthermore, if (i) \(\lambda<0\), or (ii) \(\lambda>0\), \(\|u_{0}\|_{H^{2}}\) is small, then we have found that the global solution \(u_{\varepsilon}\) will converge to u in the space \(L^{q}(0,\infty;W^{2,r}(R^{n}))\) as ε converges to 0.
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Acknowledgements
This work is supported by Natural Science of the Shanxi province (No. 2013011003-2) and the Natural Science Foundation of China (Nos. 61473180, 11571209, 61503230).
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Guo, C. The asymptotic property for nonlinear fourth-order Schrödinger equation with gain or loss. Bound Value Probl 2015, 177 (2015). https://doi.org/10.1186/s13661-015-0442-1
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DOI: https://doi.org/10.1186/s13661-015-0442-1