1 Introduction

In this paper we study the general fourth-order parabolic equation

$$\begin{aligned} u_t =\nabla \cdot (A(u)\nabla (f(u)-\Delta u)), \end{aligned}$$
(1.1)

on a bounded (smooth) domain \(\Omega\) in \(\textbf{R}^N\). Equations of this form arise quite common from the study of pattern formation in material science, fluid dynamics and population dynamics whenever interfaces are involved.

A well-known one is the Cahn-Hilliard equation which describes the separation of phases in binary fluids(to see [1, 2]). In this model the unknown u is the concentration of one fluid, \(A(u)\equiv 1\), and \(f(u)-\Delta u\) is the chemical potential.

Another model under intensive investigation recently is the thin film type equation (see the survey Oron-Davis-Bankoff [3]) where u in (1.1) (\(N \le 2\)) stands for the height of a thin film on a flat surface. Here the mobility A and external force f are certain functions defined for \(u\ge 0\) where A is positive for positive u and vanishes at 0.

The well-posedness of (1.1) is usually studied under the Neumann and “no-flux” conditions

$$\begin{aligned} \nu \cdot \nabla u=\nu \cdot \nabla (f(u) - \Delta u) =0 \quad \hbox {on}\,\, \partial \Omega , \end{aligned}$$
(1.2)

where \(\nu\) is the unit outer normal at the boundary. In some works the periodic boundary conditions are imposed.

When the functions A and f, and the initial data are sufficiently regular, there is a unique solution for (1.1) under (1.2) in some maximal time interval \((0,T), T>0\). In many cases, T is finite and the solution demonstrates singular behavior as time approaches T.

There are at least two types of such singular behavior. First, it is blow-up; more precisely, the solution becomes unbounded at T. This happens, for instance, for some solutions to the Cahn-Hilliard equation when f is a cubic polynomial with negative coefficients (Elliott-Zheng [4]). On the other hand, rupture may develop even the solution remains bounded. Starting with a positive initial function, the thin film equation (\(A(u)=u^n, 1<n<3\) and \(f=0\)) admits a solution which touches down in finite time. Subsequently the equation losses parabolicity and the fourth derivative of the solution becomes infinite ([5]). Rupture could also happen due to the presence of driving forces. For the thin film equations with destabilizing forces, blow-up and rupture could happen simultaneously ([5]).

To proceed further, we note that two basic properties of (1.1) under (1.2). First, we have the conservation law

$$\begin{aligned} \int _{\Omega }u(x,t)dx=\int _{\Omega } u(x,0)dx. \end{aligned}$$
(1.3)

Second, there is a free energy associated to (1.1), namely,

$$\begin{aligned} \mathcal{E}(u)=\frac{1}{2}\int _{\Omega }|\nabla u|^2+\int _{\Omega }F(u), \end{aligned}$$

where F is the primitive function of f with \(F(0)=0\). We have the energy dissipation relation

$$\begin{aligned} \mathcal{E}(u(\cdot , t))+\int ^t_0\int _{\Omega }A(u)|\nabla (-f(u)+\Delta u)|^2dxdt=\mathcal{E}(u(\cdot , 0)). \end{aligned}$$
(1.4)

In particular, the energy is a Liapunov function for (1.1).

With its strong physical background, (1.1) constitutes an important class of fourth order parabolic equations that deserves a systematic study.

When A is a positive constant and f is the gradient of a double-well potential, the energy is bounded from below, and long time solvability is rather easy to establish. Subsequently most works are concerned with issues like asymptotic behavior, spinodal decomposition, and classification of the steady states. When A depends on the concentration, very few results are known. From the expression for the energy, we see that it is bounded from below when the primitive of f is sublinear. In this case one expects that the equation behaves nicely. On the other hand, the energy becomes unbounded when the primitve grows in a superlinear way, so the solution may behave badly in this case. Taking \(f=-u^p\), one may compare (1.1) with the semilinear heat equation

$$\begin{aligned} u_t =\Delta u + u^p, \quad p>1, \end{aligned}$$
(1.5)

which is the negative \(L^2\)-gradient flow of the same energy. The classical result of Fujita [6] tells us that under the Dirichlet condition some (nonnegative) solutions of (1.5) blow up in finite time. In view of this, one may guess that some solutions of (1.1) blow up in finite or infinite time in the superlinear case while all solutions exist globally in the sublinear case. In fact, such assertion is made precise by Hocherman and Rosenau in [7]. In this paper the authors propose a conjecture on the blow-up and long time existence of solutions for a large class of general fourth order parabolic equations. When applied to (1.1), it states as,

Conjecture 1.1

Hocherman-Rosenau conjecture

   Consider (1.1) (\(N=1\)) under (1.2) where \(A(z)>0, z\in \textbf{R},\) and \(f'(z)<0\) for \(z\ge M\), for some M. Then

$$\begin{aligned} \lim _{z\rightarrow \infty }\left| \frac{f(z)}{z}\right| =\left\{ \begin{array}{ll}\infty , &{} u \,\hbox {blows up in finite time,}\\ \hbox {finite,}&{} \hbox { marginally linear case,}\\ 0,&{}\hbox { the system has globally stable solution(s).} \end{array}\right. \end{aligned}$$
(1.6)

All results previously known support this conjecture. For instances, when A is a positive constant, [4] showed that when \(N\le 3\) and f behaves like \(-z^3\) for large values of |z|, all solutions of (1.1) with sufficiently negative energy blow up in finite time. Their results were further extended in \(N=1\) by Novick-Cohen [NC] to more general f. In Bernoff-Bertozzi [8], the Childress-Spiegel equation (\(A=1, f=-z-z^2\)) was shown to have finite time blow-up with a self-similar singularity. The blow-up rate of the solution was further studied in Bertozzi-Pugh [9] and Chou [10]. Very few results are known when A is non-constant. In Yin [11] global solutions were obtained when A is positive and F is bounded from below.

In particular, for one dimension the thin film type equation, for simplicity, we consider power law case, i.e., the mobility \(A(u)=u^n, n>0,\) and \(f'(u)u^n=- u^m\) in (1.1), that is

$$\begin{aligned} f(u)=-q^{-1}u^{q},\quad q=m-n+1. \end{aligned}$$

For thin liquid films, the fourth-order term of (1.1) comes from surface tension between the liquid and air and also incorporates any slippage at the liquid/solid interface [12].

Only nonnegative solution are interested. Using the conservation law (1.3) and Gagliardo-Nirenberg interpolation inequality it can be shown that the energy is bounded from below if \(q<3\). This motivated Bertozzi and Pugh [13] to propose the following modification of Hocherman-Rosenau’s conjecture for thin film type equation:

Conjecture 1.2

Modified Hocherman-Rosenau conjecture

Setting as above, assuming that \(f'\le 0\), consider (1.1) under the periodic boundary conditions. Then

$$\begin{aligned} \lim _{z\rightarrow \infty }\frac{|f'(z)|}{z^2}=\left\{ \begin{array}{ll}\infty &{} \hbox {supercritical: blow-up possible,}\\ \hbox {finite} &{} \hbox {critical case,}\\ 0&{}\hbox {subcritical: solutions are globally bounded.}\end{array}\right. \end{aligned}$$

Moreover, finite time blow-up is possible when \(\lim _{z\rightarrow \infty } \sqrt{A(z)}f'(z)=-\infty .\)

One may consult [13] for a heuristic argument leading to this conjecture. In the same paper they solved the conjecture in the subcritical case. Next, in [14] and [15] the conjecture for the special case \(A(z)=z\) and \(f(z)=-z^q,\, q \ge 3\) was studied when the periodic condition is replaced by considering initial data with compact support. They showed that every weak solution of the Cauchy problem blows up in finite time whenever its initial energy is negative.

Let us recall the definition of a weak period nonnegative solution:

Definition 1.1

A nonnegative function u, P period function, in

$$\begin{aligned} L^{\infty }(0, T; H^1([0,P]))\cap L^2(0, T; H^2([0,P])) \end{aligned}$$

a weak solution of (1.1) if for all \(\phi \in \mathcal {F},\)

$$\begin{aligned} \int _0^T\int _0^P u\phi _t =\int _0^T\int _0^P A' u_x\phi _x ( -f(u)+u_{xx}) +\int _0^T\int _0^P A\varphi _{xx}(-f(u)+u_{xx}), \end{aligned}$$
(1.7)

where

$$\begin{aligned} { \mathcal F}=\left\{ \varphi : \varphi _t \in L^2([0,P]\times (0,T)), \ \varphi \in L^2( 0, T; H^2(\Omega )), \ \varphi =0 \,\, \text{ near } \,\, t=0, \,\text{ and }\,\, T\right\} . \end{aligned}$$

There is a nonnegative period weak solution to see Bernis-Friedman [16], Bertozzi-Pugh [9, 13].

We note that the local existence of the positive solution implies that the solution is smooth, as \(A(u)\ge a_0(T)>0,\) (1.1) is parabolic equation, and by using the Gagliardo-Nirenberg inequality, the \(L^{\infty }(0,T; H^1(\Omega ))\)- and \(L^2(0,T; H^3(\Omega ))\)-norms of u can be estimated by constants depending on \(\Vert u\Vert _{L^\infty }, \mathcal{E}(u_0), T\) etc., in the case \(N=1,2\). Moreover the energy dissipation relation (1.4) holds.

Let us denote by \(T_{\max }\) the maximal existence time of the solution u,

$$\begin{aligned} T_{max}=\sup \left\{ T>0,u(x,t)\, exist\, on\, [0,T] \right\} , \end{aligned}$$

the solution on \([0,T_{max})\) is called the maximal solution. In particular, if \(T_{max}=\infty\), we set \(T_\infty =T_{max}.\)

In the present article we consider the blow-up part of the Conjecture 1.2 in the case of power law. We rewrite the following the thin film type equation, power law case,

$$\begin{aligned} h_t=-(h^nh_{xxx})_x-(h^mh_x)_x \end{aligned}$$
(1.8)

with initial date \(h|_{t=0}=h_0>0\) and periodic condition \(h(x+P,t)=h(x,t).\) We have the conservation law

$$\begin{aligned} \int h(x,t)dx=\int h_0(x)dx=c_0. \end{aligned}$$

The local existence of the positive period solution for \(h_0>0\) of (1.8) was proved by Bertozzi and Pigh, to see Theorm 3.3 in [13].

We recall the steady state solution \(h_{ss}\) of (1.8) if \(h_{ss}=h\) satisfies

$$\begin{aligned} h_{xxx}-h^{m-n}h_x=0, \end{aligned}$$
(1.9)

and

$$\begin{aligned} \int h_{ss}=\int h_0=c_0. \end{aligned}$$

Note that

$$\begin{aligned} q=m-n+1. \end{aligned}$$

The energy

$$\begin{aligned} \mathcal{E}(h)=\int _0^P\left( \frac{1}{2}h_x^2-\frac{h^{q+1}}{q(q+1)}\right) dx, \end{aligned}$$

and the energy is decreasing:

$$\frac{d\mathcal{E}(h)}{dt} =-\int _0^Ph^n(h_{xxx}+h^{m-n}h_x)^2dx <0,$$

moreover

$$\begin{aligned} \mathcal{E}(h_0)-\mathcal{E}(h)(t)=\int _0^t\int _0^Ph^n(h_{xxx}+h^{m-n}h_x)^2.\end{aligned}$$
(1.10)

For any given \(c_0>0,\) let

$$\begin{aligned} \gamma \equiv \inf \left\{ \mathcal{E}(w): w \in H^1[0,P], \int w =c_0, \,\,\hbox {and} \,\, w \,\, \hbox {is a positive steady state of (1.8)} \right\} . \end{aligned}$$

It is known (Laugesen-Pugh [17]) that the energy of every positive periodic steady state h is greater than or equal to the energy of its average. We define, \(\overline{h}=\frac{1}{P}\int _0^Phdx=\frac{1}{P}c_0,\) then

$$\begin{aligned} \gamma (P, q, c_0) \equiv \mathcal{E}(\overline{h}) \nonumber = -\frac{c_0^{q+1}}{q(q+1)|P|^q}<0. \end{aligned}$$

We shall prove the Conjecture 1.2 in the case of super-critical. Our main theorem is

Theorem 1.1

Assume \(q>3\). Then the solution of (1.8) blows up at \(T_\infty\) or at \(T_{max}<\infty\), for suitable initial date \(h_0>0\), with \(\int h_0=c_0\) and \(h_0\in H^1\).

Remark 1.1

  1. (1)

    In general, the blows up occur in \(\lim _{t\uparrow T_{max} }\Vert h_{xx}\Vert _{L^1}(t)=\infty ,\) for example, the solution h of the Modified Kuramoto-Sivashinsky equation, to see [8, 13].

  2. (2)

    For thin film equation, the blows up occur in \(\{(x,t):h=0\}\), because the equation (1.8) is degenerate parabolic equation, to see [5].

  3. (3)

    We choose initial data \(h_0\) such that

    $$\begin{aligned} \mathcal{E}(h_0)<\mathcal{E}(\overline{h_0})=\gamma , \end{aligned}$$

    then solution of (1.8) blows up.

2 The proof of theorem 1.1

First we note that the energy is decreasing:

$$\begin{aligned} \frac{d\mathcal{E}(h)}{dt} =-\int _0^Ph^n(h_{xxx}+h^{m-n}h_x)^2dx <0, \end{aligned}$$

moreover

$$\begin{aligned} \mathcal{E}(h_0)-\mathcal{E}(h)(t)=\int _0^t\int _0^Ph^n(h_{xxx}+h^{m-n}h_x)^2.\end{aligned}$$
(2.1)

We prove the theorem by contrary. If the maximal solution of (1.8) does not blow up in finite time. Further suppose that it is globally uniformly bounded, or equivalently, there is some \(M>0\) such that

$$\begin{aligned} ||h||_{L^{\infty }}(t)\le M\quad \hbox {for}\, t\in (0,\infty ), \end{aligned}$$

then the fact that \(\mathcal{E}(h)(t)\le \mathcal{E}(h_0)\) and \(||h||_{L^{\infty }}(t)\le M\) implies

$$\begin{aligned} ||h||_{H^1}(t)\le C(\mathcal{E}(h_0),M, P). \end{aligned}$$

So by (2.1) we have

$$\begin{aligned} \int _0^Ph^n(h_{xxx}+h^{m-n}h_x)^2(t_j)\rightarrow 0 \end{aligned}$$

for some series \(t_j\rightarrow \infty .\)

Suppose \(h(\cdot , t_j)\rightarrow h_{ss}(\cdot )\), a steady state. Then \(h_{ss}\) satisfies

$$\begin{aligned} h_{xxx}+h^{m-n}h_x=0, \end{aligned}$$

and \(h_{ss}\) is not equivelent zero as \(\int _0^Ph_{ss}=\int _0^Ph_0>0,\) and \(h_{ss}\) is a nonnegative periodic steady state. Integrating, \(h_{ss}\) satisfies

$$\begin{aligned} h_{xx}+\frac{h^q-D}{q}=0\end{aligned}$$
(2.2)

with \(D=\frac{1}{P}\int _0^Ph^q>0\).

On the other hand, since \(h_{ss}\in L^{\infty }\) and \(||h_{ss}||_{H^1}\le C,\) we know that \(h_{ss}\in C^{1/2},\) moreover by (2.2), \(h_{ss}\in C^{2,1/2},\) by elliptic equation regularity. So that there are two cases: (i) \(h_{ss}\) is positive periodic function; (ii) \(h_{ss}\) is a droplet with zero contact angle.

Case (i): If \(h_{ss}\) is a positive periodic nonconstant steady state, then by Theorem 6 of [17] we have

$$\begin{aligned} \mathcal{E}(h_{ss})>\mathcal{E}(\overline{h_{ss}}). \end{aligned}$$

where \(\overline{h_{ss}}=\frac{1}{P}\int _0^Ph_{ss},\) and

$$\begin{aligned} \mathcal{E}(\overline{h_{ss}})=-P\frac{\overline{h_{ss}}^{q+1}}{q(q+1)} =-\frac{P\overline{h_0}^{q+1}}{q(q+1)}=\gamma (P,q,c_0). \end{aligned}$$

Case (ii): If \(h_{ss}\) is a droplet with zero contact angle, then \(h_{ss}\) satisfies (2.2). Since \(D>0\) for \(q\ge 3\), we can rescale the solution \(h_{ss}\) as follows

$$\begin{aligned} K(x)=D^{-1/q}h_{ss}(D^{\frac{1}{2q}-\frac{1}{2}}x) \end{aligned}$$

and K satisfies

$$\begin{aligned} K_{xx}+\frac{K^q-1}{q}=0. \end{aligned}$$

We fix the solution by requiring that K attains its minimum height 0 at \(x=0\), \(K'(0)=0\). Then we have

$$\begin{aligned} \frac{1}{2}K_x^2+\frac{1}{q}[\frac{K^{q+1}}{q+1}-K]=0. \end{aligned}$$
(2.3)

Define

$$\begin{aligned} H(K)=\frac{1}{q}[\frac{K^{q+1}}{q+1}-K]. \end{aligned}$$

We have

$$\begin{aligned} \mathcal{E}(K)=\frac{1}{q}\int _0^PK-\frac{2}{q(q+1)}\int _0^PK^{q+1}. \end{aligned}$$

Since the maximum value \(\beta\) of K must satisfy

$$\begin{aligned} 2H(\beta )=K'(0)^2=0 \end{aligned}$$

we have \(\beta =(1+q)^{\frac{1}{q}}.\) From \(\frac{dx}{dK}=\frac{1}{\sqrt{-2H(K)}}\) we obtain

$$\begin{aligned} q\mathcal{E}(K) & = 2\left[ \frac{-2}{q+1}\int _0^{\beta }K^{q+1}\frac{dx}{dK}dK +\int _0^{\beta }K\frac{dx}{dK}dK\right] \\ & = \sqrt{2q}\left[ \frac{-2}{q+1}\int _0^{\beta } \frac{K^{q+1}dK}{\sqrt{K-\frac{K^{q+1}}{q+1}}}+\int _0^{\beta } \frac{K dK}{\sqrt{K-\frac{K^{q+1}}{q+1}}}\right] \\ & = \sqrt{\frac{2}{q}}(1+q)^{3/(2q)}\int _0^1 \frac{x^{\frac{3}{2q}}(\frac{1}{x}-2)}{\sqrt{1-x}}. \end{aligned}$$

Define

$$\begin{aligned} I_q=\int _0^1\frac{x^{\frac{3}{2q}}(\frac{1}{x}-2)}{\sqrt{1-x}}. \end{aligned}$$

Then

$$\begin{aligned} I_3=\int _0^1\frac{x^{1/2}(\frac{1}{x}-2)}{\sqrt{1-x}}=0. \end{aligned}$$

We shall prove that \(I_q\ge I_3=0\) for \(q\ge 3\) in following.

In deed,

$$\begin{aligned} \frac{dI_q}{dq}= & {} \frac{3}{2q^2}\int _0^1\frac{x^{\frac{3}{2q}}|\ln x|(\frac{1}{x}-2)}{\sqrt{1-x}}dx\\ {}= & {} \frac{3}{2q^2}\left[ \int _0^{1/2}\frac{x^{\frac{3}{2q}}|\ln x|(\frac{1}{x}-2)}{\sqrt{1-x}}dx-\int _{1/2}^1\frac{x^{\frac{3}{2q}}|\ln x|(2-\frac{1}{x})}{\sqrt{1-x}}dx\right] \\ {}= & {} \frac{3}{2q^2}[A-B]. \end{aligned}$$

For \(x\in (0,\frac{1}{2}]\) we first have

$$\begin{aligned} -\ln x\ge 2(1-x)^2. \end{aligned}$$

So

$$\begin{aligned} A & = \int _0^{1/2}\frac{x^{\frac{3}{2q}}|\ln x|(\frac{1}{x}-2)}{\sqrt{1-x}}dx\\ & \ge \int _0^{1/2}x^{\frac{3}{2q}}(\frac{1}{x}-2)2(1-x)^{3/2}dx\\ & \ge \frac{1}{\sqrt{2}}\int _0^{1/2}x^{3/(2q)}\left(\frac{1}{x}-2\right)dx\\ &= \frac{1}{\sqrt{2}}\frac{4q^2}{3(2q+3)}\left(\frac{1}{2}\right)^{3/(2q)}. \end{aligned}$$

For B we use the inequality

$$\begin{gathered} \ln x \le 2(1 - x)\quad {\text{for }}{\mkern 1mu} {\mkern 1mu} x \in [\frac{1}{2},1]. \hfill \\ B = \int_{{1/2}}^{1} {\frac{{x^{{\frac{3}{{2q}}}} |\ln x|(2 - \frac{1}{x})}}{{\sqrt {1 - x} }}} dx \hfill \\ \quad \le 2\int_{{1/2}}^{1} {x^{{3/(2q)}} } \sqrt {1 - x} (2 - \frac{1}{x})dx\quad \hfill \\ \quad \le \sqrt 2 \int_{{1/2}}^{1} {x^{{3/(2q)}} } (2 - \frac{1}{x})dx\quad \hfill \\ \quad = \sqrt 2 \left[ {\frac{{2q(3 - 2q)}}{{3(3 + 2q)}} + \frac{{4q^{2} }}{{3(3 + 2q)}}(\frac{1}{2})^{{3/(2q)}} } \right]. \hfill \\ \end{gathered}$$

Thus

$$\begin{aligned} A-B & \ge \frac{1}{\sqrt{2}}\frac{4q^2}{3(2q+3)}\left(\frac{1}{2}\right)^{3/(2q)}- \sqrt{2}\left[ \frac{2q(3-2q)}{3(3+2q)}+\frac{4q^2}{3(3+2q)}\left(\frac{1}{2}\right)^{3/(2q)}\right] \\ & = \frac{2q}{3(2q+3)}\left[ \left( \sqrt{2}-(\sqrt{2}-\frac{1}{\sqrt{2}})\left(\frac{1}{2}\right)^{3/(2q)}\right) 2q-3\sqrt{2}\right] \\ & > \frac{2q}{3(2q+3)}\left[ \left( \sqrt{2}-(\sqrt{2}-\frac{1}{\sqrt{2}})\right) 2q-3\sqrt{2}\right] \\ & = \frac{2q}{3(2q+3)}\sqrt{2}(q-3)\\ & \ge \,0. \end{aligned}$$

Turing back to the energy \(\mathcal{E}(h_{ss}),\) we get

$$\begin{aligned} \mathcal{E}(K)=D^{-(\frac{1}{2}+\frac{3}{2q})}\mathcal{E}(h_{ss}). \end{aligned}$$

So

$$\begin{aligned} \mathcal{E}(h_{ss})\ge 0, \quad \hbox {for}\,\, q\ge 3. \end{aligned}$$

If we can choose the initial date \(h_0>0\) such that

$$\begin{aligned} \mathcal{E}(h_0)< \mathcal{E}(\overline{h_0})<0, \end{aligned}$$

(see following Lemma 2.1) then we get a contrary.

We have

Lemma 2.1

There exist two positive constants \(l_0\) and \(\alpha _0\) large enough such that \(h_0(x)=l_0\exp (-(x-P/2)\alpha _0)\) satisfies \(\mathcal{E}(h_0)<\mathcal{E}(\overline{h_0}).\)

Proof

Suppose \(h_0(x)=l\exp (-\alpha (x-P/2)).\) Then

$$\begin{aligned}{} & {} \int _0^P\frac{1}{2}h_{0x}^2=\frac{l^2\alpha }{4}(e^{\alpha P}-e^{-\alpha P}),\\{} & {} \int _0^Ph_0^{q+1}=\frac{l^{q+1}}{\alpha (q+1)}(e^{(q+1)P\alpha /2}-e^{-(q+1)\alpha /2}).\\{} & {} \overline{h_0}=\left( \frac{l(e^{\alpha P/2}-e^{-\alpha P/2})}{\alpha P}\right) ^{q+1}. \end{aligned}$$

Then

$$\begin{aligned}{} & {} \mathcal{E}(h_0)=\frac{\alpha l^2}{4}(e^{P\alpha }-e^{-P\alpha }) -\frac{l^{q+1}}{q(q+1)^2\alpha }(e^{(q+1)P\alpha /2}-e^{-(q+1)P\alpha /2}),\\{} & {} \mathcal{E}(\overline{h_0})=-\frac{P}{q(q+1)}\left( \frac{l}{P}\right) ^{q+1}\left( \frac{e^{P\alpha /2}-e^{P\alpha /2}}{\alpha }\right) ^{q+1}. \end{aligned}$$

Define

$$\begin{aligned} A(\alpha )= & {} \frac{1}{q(q+1)P^q}\left( \frac{e^{P\alpha /2}-e^{-P\alpha /2}}{\alpha }\right) ^{q+1},\\ B(\alpha )= & {} \frac{1}{q(q+1)}\frac{e^{(q+1)P\alpha /2}-e^{-(q+1)P\alpha /2}}{\alpha }. \end{aligned}$$

Then

$$\begin{aligned} \lim _{\alpha \rightarrow \infty }\frac{A(\alpha )}{B(\alpha )}=0. \end{aligned}$$

So there exists a \(\alpha _0>0\) such that \(A(\alpha _0)<\frac{1}{2}B(\alpha _0).\) Now for fixed \(\alpha _0\),

$$\begin{aligned} \lim _{l\rightarrow \infty }\frac{\mathcal{E}(\overline{h_0})}{\mathcal{E}(h_0)}=\frac{A(\alpha _0)}{B(\alpha _0)}>2. \end{aligned}$$

Thus we can take \(l_0>\alpha _0\) such that for \(h_0(x)=l_0\exp (-\alpha _0 (x-P/2))\)

$$\begin{aligned} \mathcal{E}(h_0)<\mathcal{E}(\overline{h_0}). \end{aligned}$$

The proof is finished. \(\hfill\square\)