Study of nonlinear PDE with power nonlinearities

1. Introduction

Let \[\label{e1} u'+Au=h(u,t)+f(x,t),\, u(x,0)=u_0(x), \tag{1}\] where \(u\in H\), \(u':=u_t:=\frac {du}{dt}\), \(H\) is a Hilbert space, \((u,v)=(v,u)\) is an inner product in this space, \(\|u\|^2:=(u,u)\), \(A\) is a linear operator, \((Au,u)\ge \gamma(t)(u,u)\), where \(\gamma(t0>0\) is a continuous function, \(h(u,t)\) and \(f(x,t)\) are continuous functions of their arguments, \(t\ge 0\). The assumptions about \(\gamma(t)\) are formulated in formula (9) below. We assume that \[\label{e2} \|h(u,t)\|\le \alpha(t, \|u\|)\le a(t)\|u\|^p, \,\,p\ge 1, \tag{2}\] and \[\label{e3} \|f(x,t)\|\le \beta(t), \tag{3}\] where \(\alpha:=\alpha (t, s)\ge 0\) is a continuous function of \(t\) on \([0,\infty):={\mathbb R}_+\), non-decreasing as a function of \(s\), locally Lipschitz on \(s\in {\mathbb R}_+\), and \(\beta:=\beta(t)\) is a continuous function on \({\mathbb R}_+\). Let us denote \(\|u\|:=g(t)\).

We would like to find sufficient conditions on \(a(t)\) and \(\beta(t)\) for the solution to problem (1) to be bounded as \(t\to \infty\) or for this solution to satisfy the condition \[\label{e4} \lim_{t\to \infty}\|u(x,t)\|=0. \tag{4}\] Our basic method is based on the results in [1], [2], [3]. Let us formulate one of these results that we will use.

Consider the following inequality: \[\label{e5} g'(t)\le -\gamma(t) g(t)+\alpha(t, g(t))+\beta(t), \,\,\, t\ge 0, \,\, g\ge 0, \tag{5}\] where \(\beta=\beta(t)\) and \(\alpha=\alpha(t, g)\) satisfy the assumptions mentioned below formula (3), and \(\gamma(t)\) is a continuous function on \({\mathbb R}_+\).

A proof of Proposition 1 can be found in [3], pp. 104–109.

Assume that: \[\label{e9} 0\le \gamma(t)\le q_0(1+t)^{-q};\,\,\, 0\le \alpha(t,g)\le a_0(1+t)^{-a}g^p;\,\,\, 0\le \beta\le b_0(1+t)^{-b}, \tag{9}\] where \(q_0, q, a_0, a, b_0, b, p\) are positive constants, \(p\ge 1\).

Take the inner product of (1) with \(u\) and use the following relations: \[\label{e10} (u',u)=g'g;\,\, -(Au,u)\le -\gamma(t)g^2;\,\, |(h(t,u), u)|\le \alpha(t,g)g;\,\, |(f,u)|\le \beta(t)g. \tag{10}\]

Then one gets inequality (5) for \(g\).

Let us choose \[\label{e11} \mu(t)=m_0(1+t)^{m}, \quad m_0,m>0, \tag{11}\] where \(m_0,m\) are constants.

Let us rewrite (6) and (7) as \[\label{e12} -\gamma(t)\frac 1 {\mu(t)}+\alpha(t, \frac 1 {\mu(t)}) +\beta(t)\le \frac {d }{dt}\Big(\frac 1 {\mu(t)}\Big), \quad \mu(0)g(0)<1. \tag{12}\] We want to derive from Proposition 1 the following Theorem.

2. Proofs

Let us prove Theorem 1.

To do this, it is sufficient to check that \(m_0\) and \(m\) can be chosen as required in Theorem 1 for any fixed \(p\ge 1\).

Using estimates (9) one checks that \[\label{e13} -q_0m_0^{-1}(1+t)^{-q-m}+a_0(1+t)^{-a} [m_0(1+t)^m]^{-p}+b_0(1+t)^{-b}<-m_0^{-1}m (1+t)^{-m-1}. \tag{13}\]

Let us rewrite this inequality: \[\label{e13a} m_0^{-1}m (1+t)^{-m-1} +a_0(1+t)^{-a} [m_0(1+t)^m]^{-p}+b_0(1+t)^{-b}<q_0m_0^{-1}(1+t)^{-q-m}. \tag{14}\] The first inequality (12) is satisfied if \[\label{e14} a+mp>q+m;\,\, b>q+m;\,\, m+1>q+m;\quad \frac{a_0}{m_0^p}+b_0+\frac m {m_0}\le \frac{q_0}{m_0}. \tag{15}\]

The second inequality (12) is satisfied if \[\label{e15} m_0 g(0)<1. \tag{16}\]

Inequalities (15) and (16) are valid for any fixed \(p\ge 1\) if, for example, \(m,\,q_0,\, q,\,b,\, a_0, \,b_0\) and \(g(0)\) are small, and \(m_0,\, b,\, a,\) are not too small.

There are many choices of these parameters to satisfy the inequalities (15) and (16). For example, one may choose \(q=0,9, p=1, m=1, b=3, m_0=\frac 1 2, a_0=\frac 1 {10}, a=1, q_0=10, b_0=\frac 1 {10}\). One can check that the inequalities (15) are satisfied.

Estimate (8) shows that \(\lim_{t\to \infty}g(t)=0\) provided that \(\lim_{t\to \infty}\mu(t)=\infty\). If \(m>0\) then \(\lim_{t\to \infty}\mu(t)=\infty\).

Since \(\mu(t)\) exists for all \(t>0\), the \(g(t)\) also exists for all \(t>0\), and so is \(u(t)\), the solution to (1). Sufficient conditions for the local solvability of problem (1) are known: if \(A\) is an elliptic operator and \(h(t,u)\) is locally Lipschitz with respect to \(u\), continuous with respect to \(t\), and \(f(x,t)\) is smooth with respect to both arguments, then the solution to (1) exists locally and, by Theorem 1, globally.

Consider now the choice: \[\label{e16} \mu(t)=m_0+\frac {m_1}{(1+t)^{m_2}}, \quad m_0, m_1, m_2>0, \tag{17}\] where \(m_0, m_1, m_2\) are constants.

Then \[\label{e17} 0<c_0\le \mu(t)\le c_1, \quad \frac 1 {c_1}\le \frac 1 {\mu(t)}\le \frac 1 {c_0}, \tag{18}\] where \(c_0, c_1>0\) are constants.

One checks that \[\label{e18} \frac {d }{dt}\Big(\frac 1 {\mu(t)}\Big)= -\frac {\mu'}{\mu^2}=m_2 m_1(1+t)^{-m_2-1} \mu(t)^{-2}, \tag{19}\] for \(\mu\) defined in (17).

One chooses parameters \(m_2, m_0, m_1, q_0, q, a_0, a, b_0, b,\) to satisfy the first inequality (12). The second inequality (12) is satisfied if \(g(0)\) is sufficiently small. If these inequalities are satisfied, then estimate (8) shows that \(g(t)\) is bounded on \({\mathbb R}_+\). Therefore, \[\label{e19} \sup_{t\ge 0}\|u(x,t)\|\le c, \tag{20}\] where \(c>0\) is a constant and \(u=u(x,t)\) solves Eq. (1).

3. Conclusion

Sufficient conditions are given for the solution to (1) to exist for all \(t>0\) and to tend to zero as \(t\to \infty\), or to be bounded at infinity.