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Entropy-dissipating finite-difference schemes for nonlinear fourth-order parabolic equations

Marcel Braukhoff Institute for Analysis and Scientific Computing, Vienna University of Technology, Wiedner Hauptstraße 8–10, 1040 Wien, Austria [email protected]  and  Ansgar Jüngel Institute for Analysis and Scientific Computing, Vienna University of Technology, Wiedner Hauptstraße 8–10, 1040 Wien, Austria [email protected]
Abstract.

Structure-preserving finite-difference schemes for general nonlinear fourth-order parabolic equations on the one-dimensional torus are derived. Examples include the thin-film and the Derrida–Lebowitz–Speer–Spohn equations. The schemes conserve the mass and dissipate the entropy. The scheme associated to the logarithmic entropy also preserves the positivity. The idea of the derivation is to reformulate the equations in such a way that the chain rule is avoided. A central finite-difference discretization is then applied to the reformulation. In this way, the same dissipation rates as in the continuous case are recovered. The strategy can be extended to a multi-dimensional thin-film equation. Numerical examples in one and two space dimensions illustrate the dissipation properties.

Key words and phrases:
Entropy, finite differences, thin-film equation, DLSS equation, discrete chain rule, denoising.
2000 Mathematics Subject Classification:
35K30, 35Q68, 65M06, 65M12.
The authors acknowledge partial support from the Austrian Science Fund (FWF), grants F65, P30000, P33010, and W1245.

1. Introduction

The design of numerical schemes that preserve the structure of the associated partial differential equations is an important task in numerical mathematics. In this paper, we develop new finite-difference approximations conserving the mass, preserving the positivity, and dissipating the entropy of nonlinear fourth-order parabolic equations of the type

(1) tu=Jx,J=uβuxxx+auβ1uxxux+buβ2ux3in 𝕋,\partial_{t}u=-J_{x},\quad J=u^{\beta}u_{xxx}+au^{\beta-1}u_{xx}u_{x}+bu^{\beta-2}u_{x}^{3}\quad\mbox{in }{\mathbb{T}},

where 𝕋=/{\mathbb{T}}={\mathbb{R}}/{\mathbb{Z}} is the one-dimensional torus, aa, bb\in{\mathbb{R}}, and β0\beta\geq 0, together with the initial condition u(0)=u0u(0)=u^{0} in 𝕋{\mathbb{T}}. We also discuss briefly the discretization of multi-dimensional equations; see Section 4.

A special case of (1) (with a=b=0a=b=0) is the thin-film equation

(2) tu=(uβuxxx)x,\partial_{t}u=-(u^{\beta}u_{xxx})_{x},

which models the flow of a thin liquid along a solid surface with film height u(x,t)u(x,t) or the thin neck of a Hele-Shaw flow in the lubrication approximation [1]. The multi-dimensional version is given by tu=div(uβΔu)\partial_{t}u=-\operatorname{div}(u^{\beta}\nabla\Delta u) and its discretization will be discussed in Section 4. Another example (with a=2a=-2, b=1b=1, β=0\beta=0) is the Derrida–Lebowitz–Speer–Spohn (DLSS) equation

(3) tu=2(u((u)xxu)x)x=(uxxx2uxxuxu+ux3u2)x,\partial_{t}u=-2\bigg{(}u\bigg{(}\frac{(\sqrt{u})_{xx}}{\sqrt{u}}\bigg{)}_{x}\bigg{)}_{x}=-\bigg{(}u_{xxx}-2\frac{u_{xx}u_{x}}{u}+\frac{u_{x}^{3}}{u^{2}}\bigg{)}_{x},

which arises as a scaling limit of interface fluctuations in spin systems [7] and describes the evolution of the electron density u(x,t)u(x,t) in a quantum semiconductor under simplifying assumptions.

Equations (2) and (3) conserve the mass 𝕋u(x,t)𝑑x\int_{\mathbb{T}}u(x,t)dx and preserve the positivity of the solution, even for the corresponding multi-dimensional versions [6, 14]. They also dissipate the entropy111Strictly speaking, equation (1) produces entropy and dissipates energy, but the notion entropy dissipation seems to be common in numerical schemes. in the sense that there exists a Lyapunov functional S(t)=𝕋s(u(x,t))𝑑xS(t)=\int_{\mathbb{T}}s(u(x,t))dx with entropy density s(u)s(u) such that the entropy production dS/dt-dS/dt provides gradient estimates. For instance, (2) and (3) dissipate the entropy density s(u)=sα(u)s(u)=s_{\alpha}(u), where

sα(u)=uαα(α1)\displaystyle s_{\alpha}(u)=\frac{u^{\alpha}}{\alpha(\alpha-1)} for α>0,α0,1,\displaystyle\quad\mbox{for }\alpha>0,\ \alpha\neq 0,1,
(4) s0(u)=logu\displaystyle s_{0}(u)=-\log u for α=0,\displaystyle\quad\mbox{for }\alpha=0,
s1(u)=u(logu1)\displaystyle s_{1}(u)=u(\log u-1) for α=1,\displaystyle\quad\mbox{for }\alpha=1,

if 32α+β3\frac{3}{2}\leq\alpha+\beta\leq 3 (thin-film equation) and 0α320\leq\alpha\leq\frac{3}{2} (DLSS equation); see [15, Section 4]. These bounds are optimal. We call s1(u)s_{1}(u) the Shannon entropy (density), s0(u)s_{0}(u) the logarithmic entropy, and sα(u)s_{\alpha}(u) for α0,1\alpha\neq 0,1 a Rényi entropy. It holds that sα(u)s0(u)s_{\alpha}(u)\to s_{0}(u) pointwise for α0\alpha\to 0 and sα(u)ulogus_{\alpha}(u)\to u\log u pointwise for α1\alpha\to 1. We prefer to define s1(u)s_{1}(u) as in (4) to avoid the additional term in d(ulogu)/du=logu+1d(u\log u)/du=\log u+1 which would complicate the subsequent computations.

The proof of the dissipation of the entropy is based on suitable integrations by parts and the chain rule [15]. On the discrete level, we face the problem that the chain rule is generally not available. We are aware of two general strategies.

The first strategy is to exploit the gradient-flow structure of the parabolic equation (if it exists). It involves only one integration by parts, and the discrete chain rule can be formulated by means of suitable mean functions. This idea was elaborated as the Discrete Variational Derivative Method for finite-difference approximations [11]. The gradient-flow formulation (with respect to the L2L^{2}-Wasserstein metric) yields a natural semi-discretization in time of the evolution using the minimizing movement scheme in finite-dimensional spaces from finite-volume or finite-difference approximations. These techniques were used in [8, 13, 19, 20] for the multi-dimensional DLSS equation. It allows for the proof of entropy dissipation of the Shannon entropy and the Fisher information 𝕋(u)x2𝑑x\int_{\mathbb{T}}(\sqrt{u})_{x}^{2}dx, but not for general Rényi entropies, since no Wasserstein gradient-flow formulation seems to be available for these functionals. The thin-film equation with 0<β<10<\beta<1 is shown in [18] to be a gradient flow with respect to a weighted Wasserstein metric. In the work [22], a finite-difference scheme that dissipates the discrete H1H^{1} norm of the solution to the one-dimensional thin-film equation was analyzed.

The minimizing movement scheme is based on the implicit Euler method. We mention that higher-order time discretizations were investigated too, in the framework of semi-discrete problems; see [4] using the two-step BDF method and [17] using one-leg multistep generalizations. A generic framework for Galerkin methods in space and discontinuous Galerkin methods in time was presented in [9].

The second strategy uses time-continuous Markov chains on finite state spaces. Birth-death processes that define the Markov chain can be interpreted as a finite-volume discretization of a one-dimensional Fokker–Planck equation, and the dissipation of the discrete Shannon entropy can be proved. The nonlinear integrations by parts are reduced to a discrete Bochner-type inequality [5, 10, 16], which is obtained by identifying the Radon–Nikodym derivative of a measure involving the jump rates of the Markov chain [3, Section 2]. It seems that this idea is restricted to linear diffusion equations.

In this paper, we suggest a third strategy. The idea is to write the flux JJ as a combination of derivatives of the function s(u)s^{\prime}(u). This allows for integrations by parts that can be extended to the discrete level and it avoids the application of the chain rule. More precisely, we determine two functions AA and BB depending on v:=s(u)v:=s^{\prime}(u) and its derivatives such that J=AxvBxJ=A_{x}-vB_{x}. The function s(u)s^{\prime}(u) is known in thermodynamics as the chemical potential, and the formulation of the flux in terms of the chemical potential seems to be natural from a thermodynamic viewpoint. We apply this idea to fourth-order parabolic equations for the first time. It turns out that for s=sαs=s_{\alpha} with α1\alpha\neq 1, we can write

A=uα+β(α1)2v(λ1ξ2+λ2ξ12),B=uα+β(α1)2v2(λ3ξ2+λ4ξ12),A=\frac{u^{\alpha+\beta}}{(\alpha-1)^{2}v}(\lambda_{1}\xi_{2}+\lambda_{2}\xi_{1}^{2}),\quad B=\frac{u^{\alpha+\beta}}{(\alpha-1)^{2}v^{2}}(\lambda_{3}\xi_{2}+\lambda_{4}\xi_{1}^{2}),

where ξ1=vx/v\xi_{1}=v_{x}/v, ξ2=vxx/v\xi_{2}=v_{xx}/v, and the coefficients λi\lambda_{i} depend on aa, bb, α\alpha, and β\beta; see (10) and (11). Integrating by parts twice and using equation (1) gives for Sα=𝕋sα(u)𝑑xS_{\alpha}=\int_{\mathbb{T}}s_{\alpha}(u)dx:

dSαdt=𝕋Jvx𝑑x=𝕋(vxxA(vvx)xB)𝑑x.\frac{dS_{\alpha}}{dt}=\int_{\mathbb{T}}Jv_{x}dx=-\int_{\mathbb{T}}\big{(}v_{xx}A-(vv_{x})_{x}B\big{)}dx.

The task is to show that the integrand, written as a polynomial in (ξ1,ξ2)(\xi_{1},\xi_{2}), is nonnegative for all values of (ξ1,ξ2)2(\xi_{1},\xi_{2})\in{\mathbb{R}}^{2}. It follows from the product rule (vvx)x=vvxx+vx2(vv_{x})_{x}=vv_{xx}+v_{x}^{2} that, for α1\alpha\neq 1,

dSαdt\displaystyle\frac{dS_{\alpha}}{dt} =𝕋(ξ2vA(ξ2+ξ12)v2B)𝑑x\displaystyle=-\int_{\mathbb{T}}(\xi_{2}vA-(\xi_{2}+\xi_{1}^{2})v^{2}B)dx
(5) =𝕋uα+β(α1)2((λ1λ3)ξ22+(λ2λ3λ4)ξ2ξ12λ4ξ14)𝑑x.\displaystyle=-\int_{\mathbb{T}}\frac{u^{\alpha+\beta}}{(\alpha-1)^{2}}\big{(}(\lambda_{1}-\lambda_{3})\xi_{2}^{2}+(\lambda_{2}-\lambda_{3}-\lambda_{4})\xi_{2}\xi_{1}^{2}-\lambda_{4}\xi_{1}^{4}\big{)}dx.

Under certain conditions on the parameters, we expect that the integrand is bounded from below by uα+β(ξ22+ξ14)u^{\alpha+\beta}(\xi_{2}^{2}+\xi_{1}^{4}), up to a factor, which yields some gradient estimates.

On the discrete level, we imitate this idea: The flux J=AxvBxJ=A_{x}-vB_{x} and the variables ξ1\xi_{1}, ξ2\xi_{2} of the polynomials AA and BB is discretized by central finite differences. For this, let 𝕋h=𝕋/(h){\mathbb{T}}_{h}={\mathbb{T}}/(h{\mathbb{Z}}) be a discrete torus with grid size h>0h>0 and define the scheme

(6) tui=1h(Ji+1/2Ji1/2),Ji+1/2=1h(Ai+1Ai)12h(vi+1+vi)(Bi+1Bi),\partial_{t}u_{i}=-\frac{1}{h}(J_{i+1/2}-J_{i-1/2}),\quad J_{i+1/2}=\frac{1}{h}(A_{i+1}-A_{i})-\frac{1}{2h}(v_{i+1}+v_{i})(B_{i+1}-B_{i}),

with the initial condition ui(0)=u0(i)u_{i}(0)=u^{0}(i) for i𝕋hi\in{\mathbb{T}}_{h}, where ui=u(i)u_{i}=u(i), vi=s(ui)v_{i}=s^{\prime}(u_{i}), and AiA_{i} and BiB_{i} are the polynomials AA and BB, evaluated at i𝕋hi\in{\mathbb{T}}_{h}, respectively; see (19). We show below that with the discrete entropy Sαh=hi𝕋hsα(u)S_{\alpha}^{h}=h\sum_{i\in{\mathbb{T}}_{h}}s_{\alpha}(u), the discrete analog of (5) becomes

dSαhdt=hi𝕋h(ξ2,iviAi(viξ2,i+ξ1,i2)vi2Bi),\frac{dS_{\alpha}^{h}}{dt}=-h\sum_{i\in{\mathbb{T}}_{h}}\big{(}\xi_{2,i}v_{i}A_{i}-(v_{i}\xi_{2,i}+\xi_{1,i}^{2})v_{i}^{2}B_{i}\big{)},

where ξ1,i\xi_{1,i}, ξ2,i\xi_{2,i} approximate vx(i)/v(i)v_{x}(i)/v(i), vxx(i)/v(i)v_{xx}(i)/v(i), respectively. This yields exactly the polynomial of the continuous case. Thus, we obtain the same conditions on the parameters as for the continuous equation.

We still need a discrete analog of the product rule (vvx)x=vvxx+vx2(vv_{x})_{x}=vv_{xx}+v_{x}^{2} to conclude. This is done by carefully choosing ξ1,i\xi_{1,i} and ξ2,i\xi_{2,i}. Definition (20) ensures that vi2(ξ2,i+ξ1,i2)=(vi+122vi+vi12)/(2h2)v_{i}^{2}(\xi_{2,i}+\xi_{1,i}^{2})=(v_{i+1}^{2}-2v_{i}+v_{i-1}^{2})/(2h^{2}) which approximates 12(v2)xx=vvxx+vx2\frac{1}{2}(v^{2})_{xx}=vv_{xx}+v_{x}^{2}. This choice is used in the central scheme (6) for Ji+1/2J_{i+1/2}. Noncentral schemes require different definitions of ξ1,i\xi_{1,i} and ξ2,i\xi_{2,i}; see Remark 7.

A drawback of our technique is that scheme (6) depends on the entropy to be dissipated. The scheme does not dissipate all admissible entropies. In applications, however, one usually wants to dissipate only that entropy which is physically relevant.

Our main results can be sketched as follows:

  • Lyapunov functional: Let α0\alpha\geq 0, α1\alpha\neq 1 and assume that

    (7) K(α,β):=2α2+(3a4β+9)α2β2+(3a+9)β9(a+b+1)0.K(\alpha,\beta):=-2\alpha^{2}+(3a-4\beta+9)\alpha-2\beta^{2}+(3a+9)\beta-9(a+b+1)\geq 0.

    Then the continuous entropy SαS_{\alpha} and the discrete entropy SαhS_{\alpha}^{h} are dissipated in the sense that dSα/dt0dS_{\alpha}/dt\leq 0 and dSαh/dt0dS_{\alpha}^{h}/dt\leq 0, i.e., SαS_{\alpha} and SαhS_{\alpha}^{h} are Lyapunov functionals for u(t)u(t) and ui(t)u_{i}(t), respectively; see Theorems 2 and 5. Condition (7) is optimal for the thin-film and DLSS equations.

  • Entropy dissipation: Let α0\alpha\geq 0, α1\alpha\neq 1 and assume that K(α,β)>0K(\alpha,\beta)>0. Then there exists a constant c(α,β)>0c(\alpha,\beta)>0 such that for all t>0t>0,

    (8) dSαdt+c(α,β)𝕋uα+β(ξ22+ξ14)𝑑x\displaystyle\frac{dS_{\alpha}}{dt}+c(\alpha,\beta)\int_{\mathbb{T}}u^{\alpha+\beta}(\xi_{2}^{2}+\xi_{1}^{4})dx 0,\displaystyle\leq 0,
    (9) dSαhdt+c(α,β)hi𝕋hu¯iα+β(ξ2,i2+ξ1,i4)\displaystyle\frac{dS_{\alpha}^{h}}{dt}+c(\alpha,\beta)h\sum_{i\in{\mathbb{T}}_{h}}\bar{u}_{i}^{\alpha+\beta}(\xi_{2,i}^{2}+\xi_{1,i}^{4}) 0,\displaystyle\leq 0,

    where ξ1=vx/v\xi_{1}=v_{x}/v, ξ2=vxx/v\xi_{2}=v_{xx}/v, v=sα(u)v=s_{\alpha}(u), u¯i\bar{u}_{i} is an arbitrary average of (ui)(u_{i}), and ξ1,i\xi_{1,i}, ξ2,i\xi_{2,i} are defined in (20). This result is proved in Theorems 2 and 5.

  • Case α=1\alpha=1: For the case α=1\alpha=1, we need the formulation J=AxwBxJ=A_{x}-wB_{x} instead of J=AxvBxJ=A_{x}-vB_{x}, where w=s0(u)=u1w=s_{0}^{\prime}(u)=-u^{-1} and v=loguv=\log u, since JJ generally does not depend on the logarithm. For details, see Proposition 8.

  • Case α=0\alpha=0: We show that scheme (6) with α=0\alpha=0 possesses global positive solutions. This result is a consequence of the discrete entropy inequality and mass conservation, which imply that hi𝕋h(ui(t)logui(t))h\sum_{i\in{\mathbb{T}}_{h}}(u_{i}(t)-\log u_{i}(t)) is bounded for all t>0t>0. Consequently, ui(t)logui(t)u_{i}(t)-\log u_{i}(t) is bounded for all i𝕋hi\in{\mathbb{T}}_{h} and t>0t>0, and since the function sslogss\mapsto s-\log s tends to infinity if either s0s\to 0 or ss\to\infty, this proves that ui(t)u_{i}(t) is bounded from below and above. We refer to Proposition 6 for details.

  • Multi-dimensional case: In principle, the multi-dimensional case can be treated using functions AA and BB with many variables. Practically, however, the computations are becoming too involved and it may be unclear how to discretize mixed derivatives. One idea to overcome this issue is to use scalar variables only, like ξ1=|v|/v\xi_{1}=|\nabla v|/v and ξ2=Δv/v\xi_{2}=\Delta v/v. This allows us to treat the multi-dimensional thin-film equation; see Proposition 10.

The paper is organized as follows. We prove the continuous entropy inequality (8) in Section 2 and the discrete entropy inequality (9) in Section 3. A scheme for the multi-dimensional thin-film equation is proposed and analyzed in Section 4. Numerical simulations for the thin-film and DLSS equations in one space dimension and for a thin-film equation in two space dimensions are presented in Section 5.

2. General continuous equation

To prepare the discretization, we need to analyze the entropy dissipation properties of the continuous equation (1). We show first that JJ can be written as AxvBxA_{x}-vB_{x} with v=sα(u)v=s_{\alpha}^{\prime}(u) and functions AA and BB which depend on vv, vxv_{x}, and vxxv_{xx}.

Lemma 1.

Let JJ be given as in (1) and sαs_{\alpha} as in (4) and let α0\alpha\geq 0 satisfy α1\alpha\neq 1. Then J=AxvBxJ=A_{x}-vB_{x}, where

A\displaystyle A =uα+β(λ1u22α(sα(u))xx+λ2(α1)u33α(sα(u))x2)\displaystyle=u^{\alpha+\beta}\big{(}\lambda_{1}u^{2-2\alpha}(s_{\alpha}^{\prime}(u))_{xx}+\lambda_{2}(\alpha-1)u^{3-3\alpha}(s_{\alpha}^{\prime}(u))_{x}^{2}\big{)}
=uα+β(α1)2v(λ1vxxv+λ2(vxv)2),\displaystyle=\frac{u^{\alpha+\beta}}{(\alpha-1)^{2}v}\bigg{(}\lambda_{1}\frac{v_{xx}}{v}+\lambda_{2}\bigg{(}\frac{v_{x}}{v}\bigg{)}^{2}\bigg{)},
B\displaystyle B =uα+β(λ3(α1)u33α(sα(u))xx+λ4(α1)2u44α(sα(u))x2)\displaystyle=u^{\alpha+\beta}\big{(}\lambda_{3}(\alpha-1)u^{3-3\alpha}(s_{\alpha}^{\prime}(u))_{xx}+\lambda_{4}(\alpha-1)^{2}u^{4-4\alpha}(s_{\alpha}^{\prime}(u))_{x}^{2}\big{)}
=uα+β(α1)2v2(λ3vxxv+λ4(vxv)2)\displaystyle=\frac{u^{\alpha+\beta}}{(\alpha-1)^{2}v^{2}}\bigg{(}\lambda_{3}\frac{v_{xx}}{v}+\lambda_{4}\bigg{(}\frac{v_{x}}{v}\bigg{)}^{2}\bigg{)}

and

λ1\displaystyle\lambda_{1} =2α2+(β+5)α+aββ2a2b3(α1)(β2α+3)+2(α1)β2α+3λ4,\displaystyle=\frac{-2\alpha^{2}+(\beta+5)\alpha+a\beta-\beta^{2}-a-2b-3}{(\alpha-1)(\beta-2\alpha+3)}+\frac{2(\alpha-1)}{\beta-2\alpha+3}\lambda_{4},
(10) λ2\displaystyle\lambda_{2} =2α2(a+7)α+2a+b+6(α1)(β2α+3)+β3α+4β2α+3λ4,\displaystyle=\frac{2\alpha^{2}-(a+7)\alpha+2a+b+6}{(\alpha-1)(\beta-2\alpha+3)}+\frac{\beta-3\alpha+4}{\beta-2\alpha+3}\lambda_{4},
λ3\displaystyle\lambda_{3} =(a+1)ββ2a2b(1α)(β2α+3)+2(α1)β2α+3λ4,\displaystyle=\frac{(a+1)\beta-\beta^{2}-a-2b}{(1-\alpha)(\beta-2\alpha+3)}+\frac{2(\alpha-1)}{\beta-2\alpha+3}\lambda_{4},

with λ4\lambda_{4}\in{\mathbb{R}} being a free parameter.

The lemma shows in particular that the formulation J=AxvBxJ=A_{x}-vB_{x} is not unique. This fact is used to optimize later the range of admissible parameters α\alpha, β\beta, aa, and bb.

Proof.

Let α>0\alpha>0 with α1\alpha\neq 1. A direct computation yields

A\displaystyle A =λ1uβuxx+((λ1+λ2)α2λ1λ2)uβ1ux2,\displaystyle=\lambda_{1}u^{\beta}u_{xx}+\big{(}(\lambda_{1}+\lambda_{2})\alpha-2\lambda_{1}-\lambda_{2}\big{)}u^{\beta-1}u_{x}^{2},
B\displaystyle B =λ3(α1)uβα+1uxx+((λ3+λ4)α2λ3λ4)uβαux2.\displaystyle=\lambda_{3}(\alpha-1)u^{\beta-\alpha+1}u_{xx}+\big{(}(\lambda_{3}+\lambda_{4})\alpha-2\lambda_{3}-\lambda_{4}\big{)}u^{\beta-\alpha}u_{x}^{2}.

Inserting these expressions into AxvBxA_{x}-vB_{x} and simplifying leads to

Ax\displaystyle A_{x} uα1α1Bx=(λ1λ3)uβuxxx\displaystyle-\frac{u^{\alpha-1}}{\alpha-1}B_{x}=(\lambda_{1}-\lambda_{3})u^{\beta}u_{xxx}
+((2λ1+2λ2λ32λ4)α+(λ1λ3)β4λ12λ2+3λ3+2λ4)uβ1uxxux\displaystyle\phantom{xx}{}+\big{(}(2\lambda_{1}+2\lambda_{2}-\lambda_{3}-2\lambda_{4})\alpha+(\lambda_{1}-\lambda_{3})\beta-4\lambda_{1}-2\lambda_{2}+3\lambda_{3}+2\lambda_{4}\big{)}u^{\beta-1}u_{xx}u_{x}
+((λ3+λ4)α2+(λ1+λ2λ3λ4)αβ(λ1+λ2+2λ3+λ4)α\displaystyle\phantom{xx}{}+\big{(}(\lambda_{3}+\lambda_{4})\alpha^{2}+(\lambda_{1}+\lambda_{2}-\lambda_{3}-\lambda_{4})\alpha\beta-(\lambda_{1}+\lambda_{2}+2\lambda_{3}+\lambda_{4})\alpha
+(2λ1λ2+2λ3+λ4)β+2λ1+λ2)uβ2ux3.\displaystyle\phantom{xx}{}+(-2\lambda_{1}-\lambda_{2}+2\lambda_{3}+\lambda_{4})\beta+2\lambda_{1}+\lambda_{2}\big{)}u^{\beta-2}u_{x}^{3}.

We identify the coefficients with those in the expression (1) for JJ:

1\displaystyle 1 =λ1λ3,\displaystyle=\lambda_{1}-\lambda_{3},
a\displaystyle a =(2λ1+2λ2λ32λ4)α+(λ1λ3)β4λ12λ2+3λ3+2λ4,\displaystyle=(2\lambda_{1}+2\lambda_{2}-\lambda_{3}-2\lambda_{4})\alpha+(\lambda_{1}-\lambda_{3})\beta-4\lambda_{1}-2\lambda_{2}+3\lambda_{3}+2\lambda_{4},
b\displaystyle b =(λ3+λ4)α2+(λ1+λ2λ3λ4)αβ(λ1+λ2+2λ3+λ4)α\displaystyle=(\lambda_{3}+\lambda_{4})\alpha^{2}+(\lambda_{1}+\lambda_{2}-\lambda_{3}-\lambda_{4})\alpha\beta-(\lambda_{1}+\lambda_{2}+2\lambda_{3}+\lambda_{4})\alpha
+(2λ1λ2+2λ3+λ4)β+2λ1+λ2.\displaystyle\phantom{xx}{}+(-2\lambda_{1}-\lambda_{2}+2\lambda_{3}+\lambda_{4})\beta+2\lambda_{1}+\lambda_{2}.

The general solution of this linear system for (λ1,λ2,λ3,λ4)(\lambda_{1},\lambda_{2},\lambda_{3},\lambda_{4}), with free parameter λ4\lambda_{4}\in{\mathbb{R}}, gives (10).

Next, let α=0\alpha=0. The ansatz for AA and BB becomes

A\displaystyle A =uβ(λ1u2(u1)xxλ2u3(u1)x2),\displaystyle=u^{\beta}\big{(}\lambda_{1}u^{2}(-u^{-1})_{xx}-\lambda_{2}u^{3}(-u^{-1})_{x}^{2}\big{)},
B\displaystyle B =uβ(λ3u3(u1)xx+λ4u4(u1)x2).\displaystyle=u^{\beta}\big{(}-\lambda_{3}u^{3}(-u^{-1})_{xx}+\lambda_{4}u^{4}(-u^{-1})_{x}^{2}\big{)}.

Then

AxvBx\displaystyle A_{x}-vB_{x} =(λ1λ3)uβuxxx+((λ1λ3)β4λ12λ2+3λ3+2λ4)uβ1uxxux\displaystyle=(\lambda_{1}-\lambda_{3})u^{\beta}u_{xxx}+\big{(}(\lambda_{1}-\lambda_{3})\beta-4\lambda_{1}-2\lambda_{2}+3\lambda_{3}+2\lambda_{4}\big{)}u^{\beta-1}u_{xx}u_{x}
+((2λ1λ2+2λ3+λ4)β+2λ1+λ2)uβ2ux3.\displaystyle\phantom{xx}{}+\big{(}(-2\lambda_{1}-\lambda_{2}+2\lambda_{3}+\lambda_{4})\beta+2\lambda_{1}+\lambda_{2}\big{)}u^{\beta-2}u_{x}^{3}.

Identifying the coefficients with those in (1) again gives a linear system for the parameters (λ1,λ2,λ3,λ4)(\lambda_{1},\lambda_{2},\lambda_{3},\lambda_{4}). The general solution reads as

λ1\displaystyle\lambda_{1} =β2aβ+a+2b+3β+32β+3λ4,\displaystyle=\frac{\beta^{2}-a\beta+a+2b+3}{\beta+3}-\frac{2}{\beta+3}\lambda_{4},
(11) λ2\displaystyle\lambda_{2} =2a+b+6β+3+β+4β+3λ4,\displaystyle=-\frac{2a+b+6}{\beta+3}+\frac{\beta+4}{\beta+3}\lambda_{4},
λ3\displaystyle\lambda_{3} =β2aββ+a+2bβ+32β+3λ4,\displaystyle=\frac{\beta^{2}-a\beta-\beta+a+2b}{\beta+3}-\frac{2}{\beta+3}\lambda_{4},

These expressions are the same as (10) with α=0\alpha=0. ∎

For the following theorem, we recall definition (4) of sαs_{\alpha} and set Sα(u)=𝕋sα(u)𝑑xS_{\alpha}(u)=\int_{\mathbb{T}}s_{\alpha}(u)dx.

Theorem 2.

Let uu be a smooth positive solution to (1) and let α0\alpha\geq 0. If K(α,β)0K(\alpha,\beta)\geq 0 (see definition (7)), then SαS_{\alpha} is a Lyapunov functional, i.e. dSα/dt0dS_{\alpha}/dt\leq 0 for all t>0t>0. If K(α,β)>0K(\alpha,\beta)>0 then there exists c(α,β)>0c(\alpha,\beta)>0 such that for all t>0t>0,

dSαdt+c(α,β)𝕋uα+β(u2(1α)(uα1)xx2+u4(1α)(uα1)x4)𝑑x0\frac{dS_{\alpha}}{dt}+c(\alpha,\beta)\int_{\mathbb{T}}u^{\alpha+\beta}\big{(}u^{2(1-\alpha)}(u^{\alpha-1})_{xx}^{2}+u^{4(1-\alpha)}(u^{\alpha-1})_{x}^{4}\big{)}dx\leq 0

and there exists another constant C(α,β)>0C(\alpha,\beta)>0 such that for all t>0t>0,

dSαdt+C(α,β)𝕋uα+β(u2uxx2+u4ux4)𝑑x0.\frac{dS_{\alpha}}{dt}+C(\alpha,\beta)\int_{\mathbb{T}}u^{\alpha+\beta}\big{(}u^{-2}u_{xx}^{2}+u^{-4}u_{x}^{4}\big{)}dx\leq 0.
Proof.

Let first α0\alpha\geq 0 with α1\alpha\neq 1. We calculate the time derivative of the entropy, using integration by parts twice:

dSαdt\displaystyle\frac{dS_{\alpha}}{dt} =𝕋sα(u)tudx=𝕋vJx𝑑x=𝕋Jvx𝑑x\displaystyle=\int_{\mathbb{T}}s_{\alpha}^{\prime}(u)\partial_{t}udx=-\int_{\mathbb{T}}vJ_{x}dx=\int_{\mathbb{T}}Jv_{x}dx
=𝕋(AxvBx)vx𝑑x=𝕋(Avxx(vvxx+vx2)B)𝑑x\displaystyle=\int_{\mathbb{T}}(A_{x}-vB_{x})v_{x}dx=-\int_{\mathbb{T}}\big{(}Av_{xx}-(vv_{xx}+v_{x}^{2})B\big{)}dx
=𝕋uα+β(α1)2[(λ1λ3)(vxxv)2+(λ2λ3λ4)vxxv(vxv)2λ4(vxv)4]𝑑x,\displaystyle=-\int_{\mathbb{T}}\frac{u^{\alpha+\beta}}{(\alpha-1)^{2}}\bigg{[}(\lambda_{1}-\lambda_{3})\bigg{(}\frac{v_{xx}}{v}\bigg{)}^{2}+(\lambda_{2}-\lambda_{3}-\lambda_{4})\frac{v_{xx}}{v}\bigg{(}\frac{v_{x}}{v}\bigg{)}^{2}-\lambda_{4}\bigg{(}\frac{v_{x}}{v}\bigg{)}^{4}\bigg{]}dx,
(12) =𝕋uα+β(α1)2P(vxv,vxxv)𝑑x,\displaystyle=-\int_{\mathbb{T}}\frac{u^{\alpha+\beta}}{(\alpha-1)^{2}}P\bigg{(}\frac{v_{x}}{v},\frac{v_{xx}}{v}\bigg{)}dx,

where

(13) P(ξ1,ξ2)=(λ1λ3)ξ22+(λ2λ3λ4)ξ2ξ1λ4ξ14.P(\xi_{1},\xi_{2})=(\lambda_{1}-\lambda_{3})\xi_{2}^{2}+(\lambda_{2}-\lambda_{3}-\lambda_{4})\xi_{2}\xi_{1}-\lambda_{4}\xi_{1}^{4}.

The right-hand side of (12) is nonpositive if P(ξ1,ξ2)0P(\xi_{1},\xi_{2})\geq 0 for all (ξ1,ξ2)2(\xi_{1},\xi_{2})\in{\mathbb{R}}^{2}. Taking into account that λ1λ3=1\lambda_{1}-\lambda_{3}=1, this is the case if and only if

4λ4(λ2λ3λ4)20.-4\lambda_{4}-(\lambda_{2}-\lambda_{3}-\lambda_{4})^{2}\geq 0.

In view of definition (10) of λ2\lambda_{2} and λ3\lambda_{3}, we may interpret λ4\lambda_{4} as a free parameter und λ2=λ2(λ4)\lambda_{2}=\lambda_{2}(\lambda_{4}) and λ3=λ3(λ4)\lambda_{3}=\lambda_{3}(\lambda_{4}) as affine functions in λ4\lambda_{4}. Therefore, we require that

(14) f(λ4):=4λ4(λ2(λ4)λ3(λ4)λ4)20.f(\lambda_{4}):=-4\lambda_{4}-\big{(}\lambda_{2}(\lambda_{4})-\lambda_{3}(\lambda_{4})-\lambda_{4}\big{)}^{2}\geq 0.

We choose the optimal value of λ4\lambda_{4} by computing the critical value of ff:

0=f(λ4)\displaystyle 0=f^{\prime}(\lambda_{4}) =2(2α2+(3a+8β+3)α3aβ+β2+9(a+b)15β)(β2α+3)2\displaystyle=\frac{2(-2\alpha^{2}+(-3a+8\beta+3)\alpha-3a\beta+\beta^{2}+9(a+b)-15\beta)}{(\beta-2\alpha+3)^{2}}
18(α1)2(β2α+3)2λ4.\displaystyle\phantom{xx}{}-\frac{18(\alpha-1)^{2}}{(\beta-2\alpha+3)^{2}}\lambda_{4}.

This yields

(15) λ4=2α2+(3a+8β+3)α3aβ+β2+9(a+b)15β9(α1)2.\lambda_{4}=\frac{-2\alpha^{2}+(-3a+8\beta+3)\alpha-3a\beta+\beta^{2}+9(a+b)-15\beta}{9(\alpha-1)^{2}}.

Inserting λ4\lambda_{4} in (14) leads to

0f(λ4)=49(α1)2(2α2+(3a4β+9)α2β2+(3a+9)β9(a+b+1)),0\leq f(\lambda_{4})=\frac{4}{9(\alpha-1)^{2}}\big{(}-2\alpha^{2}+(3a-4\beta+9)\alpha-2\beta^{2}+(3a+9)\beta-9(a+b+1)\big{)},

which is equivalent to K(α,β)0K(\alpha,\beta)\geq 0, see (7).

If K(α,β)>0K(\alpha,\beta)>0, there exists c0(α,β)>0c_{0}(\alpha,\beta)>0 such that for all (ξ1,ξ2)2(\xi_{1},\xi_{2})\in{\mathbb{R}}^{2}, P(ξ1,ξ2)c0(α,β)(ξ12+ξ14)P(\xi_{1},\xi_{2})\geq c_{0}(\alpha,\beta)(\xi_{1}^{2}+\xi_{1}^{4}). Inserting this information in (12), we infer that

dSαdt\displaystyle\frac{dS_{\alpha}}{dt} c0(α,β)𝕋uα+β(α1)2[(vxxv)2+(vxv)4]𝑑x\displaystyle\leq-c_{0}(\alpha,\beta)\int_{\mathbb{T}}\frac{u^{\alpha+\beta}}{(\alpha-1)^{2}}\bigg{[}\bigg{(}\frac{v_{xx}}{v}\bigg{)}^{2}+\bigg{(}\frac{v_{x}}{v}\bigg{)}^{4}\bigg{]}dx
=c0(α,β)𝕋uα+β2[(uxxu)2+2(α2)uxxu(uxu)2+(2α26α+5)(uxu)4]𝑑x.\displaystyle=-c_{0}(\alpha,\beta)\int_{\mathbb{T}}u^{\alpha+\beta-2}\bigg{[}\bigg{(}\frac{u_{xx}}{u}\bigg{)}^{2}+2(\alpha-2)\frac{u_{xx}}{u}\bigg{(}\frac{u_{x}}{u}\bigg{)}^{2}+(2\alpha^{2}-6\alpha+5)\bigg{(}\frac{u_{x}}{u}\bigg{)}^{4}\bigg{]}dx.

The discriminant equals

1(2α26α+5)(α2)2=(α1)2,1\cdot(2\alpha^{2}-6\alpha+5)-(\alpha-2)^{2}=(\alpha-1)^{2},

and it is positive for all α1\alpha\neq 1. Therefore, there exists k(α)>0k(\alpha)>0 such that

dSαdtc0(α,β)k(α)𝕋uα+β((uxxu)2+(uxu)4)𝑑x,\frac{dS_{\alpha}}{dt}\leq-c_{0}(\alpha,\beta)k(\alpha)\int_{\mathbb{T}}u^{\alpha+\beta}\bigg{(}\bigg{(}\frac{u_{xx}}{u}\bigg{)}^{2}+\bigg{(}\frac{u_{x}}{u}\bigg{)}^{4}\bigg{)}dx,

and this gives the conclusion for c(α,β)=c0(α,β)/(α1)2c(\alpha,\beta)=c_{0}(\alpha,\beta)/(\alpha-1)^{2} and C(α,β)=c0(α,β)k(α)C(\alpha,\beta)=c_{0}(\alpha,\beta)k(\alpha) when α1\alpha\neq 1.

It remains to analyze the case α=1\alpha=1. Here, we cannot formulate the flux as J=AxvBxJ=A_{x}-vB_{x} with v=s1(u)=loguv=s_{1}^{\prime}(u)=\log u, since JJ does not contain logarithmic terms. Our idea is to write J=AxwBxJ=A_{x}-wB_{x} with w=u1w=-u^{-1} and functions AA and BB that depend on ww, wxw_{x}, and wxxw_{xx}. The time derivative of the entropy S1S_{1} can be written in terms of ww and its derivatives only, since the logarithmic term v=loguv=\log u only appears with its derivatives.

The formulation J=AxwBxJ=A_{x}-wB_{x} corresponds to the expression used for α=0\alpha=0. In fact, we have

A\displaystyle A =uβ(λ1u2(u1)xxλ2u3(u1)x2)=uβw(λ1wxxw+λ2(wxw)2),\displaystyle=u^{\beta}\big{(}\lambda_{1}u^{2}(-u^{-1})_{xx}-\lambda_{2}u^{3}(-u^{-1})_{x}^{2}\big{)}=\frac{u^{\beta}}{w}\bigg{(}\lambda_{1}\frac{w_{xx}}{w}+\lambda_{2}\bigg{(}\frac{w_{x}}{w}\bigg{)}^{2}\bigg{)},
B\displaystyle B =uβ(λ3u3(u1)xx+λ4u4(u1)x2)=uβw2(λ3wxxw+λ4(wxw)2),\displaystyle=u^{\beta}\big{(}-\lambda_{3}u^{3}(-u^{-1})_{xx}+\lambda_{4}u^{4}(-u^{-1})_{x}^{2}\big{)}=\frac{u^{\beta}}{w^{2}}\bigg{(}\lambda_{3}\frac{w_{xx}}{w}+\lambda_{4}\bigg{(}\frac{w_{x}}{w}\bigg{)}^{2}\bigg{)},

where λ1,λ2,λ3\lambda_{1},\lambda_{2},\lambda_{3} are given by (11) and λ4\lambda_{4} is a free parameter. As before, the time derivative becomes

dS1dt=𝕋tulogudx=𝕋Jxv𝑑x=𝕋Jvx𝑑x=𝕋(AvxxB(vxw)x)𝑑x.\frac{dS_{1}}{dt}=\int_{\mathbb{T}}\partial_{t}u\log udx=-\int_{\mathbb{T}}J_{x}vdx=\int_{\mathbb{T}}Jv_{x}dx=-\int_{\mathbb{T}}\big{(}Av_{xx}-B(v_{x}w)_{x}\big{)}dx.

Set ξ1=wx/w\xi_{1}=w_{x}/w and ξ2=wxx/w\xi_{2}=w_{xx}/w. Since vxx=wxx/w+(wx/w)2=ξ2+ξ12v_{xx}=-w_{xx}/w+(w_{x}/w)^{2}=-\xi_{2}+\xi_{1}^{2} and (vxw)x=wxx=wξ2(v_{x}w)_{x}=-w_{xx}=-w\xi_{2}, we obtain

dS1dt\displaystyle\frac{dS_{1}}{dt} =𝕋uβw((λ1ξ2+λ2ξ12)(ξ2+ξ12)+(λ3ξ2+λ4ξ12)ξ2)𝑑x\displaystyle=-\int_{\mathbb{T}}\frac{u^{\beta}}{w}\big{(}(\lambda_{1}\xi_{2}+\lambda_{2}\xi_{1}^{2})(-\xi_{2}+\xi_{1}^{2})+(\lambda_{3}\xi_{2}+\lambda_{4}\xi_{1}^{2})\xi_{2}\big{)}dx
=𝕋uβ+1((λ1ξ2+λ2ξ12)(ξ2+ξ12)(λ3ξ2+λ4ξ12)ξ2)𝑑x\displaystyle=-\int_{\mathbb{T}}u^{\beta+1}\big{(}-(\lambda_{1}\xi_{2}+\lambda_{2}\xi_{1}^{2})(-\xi_{2}+\xi_{1}^{2})-(\lambda_{3}\xi_{2}+\lambda_{4}\xi_{1}^{2})\xi_{2}\big{)}dx
=𝕋uβ+1P1(ξ1,ξ2)𝑑x,\displaystyle=-\int_{\mathbb{T}}u^{\beta+1}P_{1}(\xi_{1},\xi_{2})dx,

where

(16) P1(ξ1,ξ2)=(λ1λ3)ξ22+(λ1+λ2λ4)ξ2ξ12λ2ξ14.P_{1}(\xi_{1},\xi_{2})=(\lambda_{1}-\lambda_{3})\xi_{2}^{2}+(-\lambda_{1}+\lambda_{2}-\lambda_{4})\xi_{2}\xi_{1}^{2}-\lambda_{2}\xi_{1}^{4}.

Observe that λ1λ3=1\lambda_{1}-\lambda_{3}=1. Thus, dS1/dt0dS_{1}/dt\geq 0 if the polynomial P1P_{1} is nonnegative for all (ξ1,ξ2)2(\xi_{1},\xi_{2})\in{\mathbb{R}}^{2}, which is the case if and only if

f1(λ4):=4λ2(λ4)(λ1(λ4)+λ2(λ4)λ4)20.f_{1}(\lambda_{4}):=-4\lambda_{2}(\lambda_{4})-\big{(}-\lambda_{1}(\lambda_{4})+\lambda_{2}(\lambda_{4})-\lambda_{4})^{2}\geq 0.

We choose the optimal value λ4\lambda_{4} from f1(λ4)=0f_{1}^{\prime}(\lambda_{4})=0, i.e.

λ4=19(β2(3a+14)β+9(a+b)+3).\lambda_{4}=\frac{1}{9}\big{(}\beta^{2}-(3a+14)\beta+9(a+b)+3\big{)}.

Then f(λ4)0f(\lambda_{4})\geq 0 is equivalent to K(1,β)0K(1,\beta)\geq 0.

Finally, if K(1,β)>0K(1,\beta)>0, we have P1(ξ1,ξ2)c0(1,β)(ξ22+ξ14)P_{1}(\xi_{1},\xi_{2})\geq c_{0}(1,\beta)(\xi_{2}^{2}+\xi_{1}^{4}) and consequently,

dS1dt\displaystyle\frac{dS_{1}}{dt} c0(1,β)𝕋uβ+1(u2(1u)xx2+u4(1u)x4)𝑑x\displaystyle\leq-c_{0}(1,\beta)\int_{\mathbb{T}}u^{\beta+1}\bigg{(}u^{2}\bigg{(}\frac{1}{u}\bigg{)}_{xx}^{2}+u^{4}\bigg{(}\frac{1}{u}\bigg{)}_{x}^{4}\bigg{)}dx
=c0(1,β)𝕋uβ+1(u2uxx24u3uxxux2+5u4ux4)𝑑x.\displaystyle=-c_{0}(1,\beta)\int_{\mathbb{T}}u^{\beta+1}\big{(}u^{-2}u_{xx}^{2}-4u^{-3}u_{xx}u_{x}^{2}+5u^{-4}u_{x}^{4}\big{)}dx.

The discriminant is positive and there exists k(1)>0k(1)>0 such that ξ224ξ2ξ12+5ξ14k(1)(ξ22+ξ14)\xi_{2}^{2}-4\xi_{2}\xi_{1}^{2}+5\xi_{1}^{4}\geq k(1)(\xi_{2}^{2}+\xi_{1}^{4}). We infer that

dS1dtc0(1,β)k(1)𝕋uβ+1(u2uxx2+u4ux4)𝑑x,\frac{dS_{1}}{dt}\leq-c_{0}(1,\beta)k(1)\int_{\mathbb{T}}u^{\beta+1}\big{(}u^{-2}u_{xx}^{2}+u^{-4}u_{x}^{4}\big{)}dx,

which concludes the proof with c(1,β)=c0(1,β)c(1,\beta)=c_{0}(1,\beta) and C(1,β)=c0(1,β)k(1)C(1,\beta)=c_{0}(1,\beta)k(1). ∎

Remark 3 (Examples).

The DLSS equation corresponds to (1) if a=2a=-2, b=0b=0, and β=0\beta=0. Then condition (7) becomes 0<α320<\alpha\leq\frac{3}{2}, which is the optimal interval. Choosing a=b=0a=b=0, we obtain the thin-film equation, and condition (7) is equivalent to 32α+β3\frac{3}{2}\leq\alpha+\beta\leq 3, which again is the optimal parameter range. ∎

Remark 4 (Systematic integration by parts).

The result of Theorem 2 can be also derived by the method of systematic integration by parts of [15]. Our proof is taylored in such a way that it can be directly “translated” to the discrete level. Indeed, the method of [15] needs several chain rules that are not available on the discrete level and our technique needs to be used.

Still, systematic integration by parts and our strategy are strongly related. Systematic integration by parts means that we are adding so-called integration-by-parts formulas whose derivatives vanishe. In this way, we can derive (12),

dSαdt=𝕋(AxvBx)vx𝑑x+c1𝕋(AvxBvvx)x𝑑x=𝕋(AvxxB(vvx)x)𝑑x,\frac{dS_{\alpha}}{dt}=\int_{\mathbb{T}}(A_{x}-vB_{x})v_{x}dx+c_{1}\int_{\mathbb{T}}(Av_{x}-Bvv_{x})_{x}dx=-\int_{\mathbb{T}}(Av_{xx}-B(vv_{x})_{x})dx,

by choosing c1=1c_{1}=-1. In the method of systematic integration by parts, we are adding a term of the type c2𝕋(uα+β(vx/v)3)x𝑑xc_{2}\int_{\mathbb{T}}(u^{\alpha+\beta}(v_{x}/v)^{3})_{x}dx and optimize c2c_{2}. By contrast, the constant c1c_{1} is fixed, but we optimize λ4\lambda_{4} in the formulation of AA and BB. In both cases, just one parameter needs to be optimized. ∎

3. General discretized equation

We “translate” the computations of the previous section to the discrete level. For this, we use the discrete entropy

(17) Sαh(u)=hi𝕋hsα(ui),where α0,α1,S_{\alpha}^{h}(u)=h\sum_{i\in{\mathbb{T}}_{h}}s_{\alpha}(u_{i}),\quad\mbox{where }\alpha\geq 0,\ \alpha\neq 1,

where ui=u(i)u_{i}=u(i) for i𝕋hi\in{\mathbb{T}}_{h} and sαs_{\alpha} is defined in (4). We recall scheme (6):

(18) tui=1h(Ji+1/2Ji1/2),Ji+1/2=1h(Ai+1Ai)12h(vi+1+vi)(Bi+1Bi),\partial_{t}u_{i}=-\frac{1}{h}(J_{i+1/2}-J_{i-1/2}),\quad J_{i+1/2}=\frac{1}{h}(A_{i+1}-A_{i})-\frac{1}{2h}(v_{i+1}+v_{i})(B_{i+1}-B_{i}),

where the functions AiA_{i} and BiB_{i} are given by

(19) Ai=u¯iα+β(α1)2vi(λ1ξ2,i+λ2(α1)ξ1,i2),Bi=u¯iα+β(α1)2vi2(λ3ξ2,i+λ4(α1)ξ1,i2),A_{i}=\frac{\bar{u}_{i}^{\alpha+\beta}}{(\alpha-1)^{2}v_{i}}(\lambda_{1}\xi_{2,i}+\lambda_{2}(\alpha-1)\xi_{1,i}^{2}),\quad B_{i}=\frac{\bar{u}_{i}^{\alpha+\beta}}{(\alpha-1)^{2}v_{i}^{2}}(\lambda_{3}\xi_{2,i}+\lambda_{4}(\alpha-1)\xi_{1,i}^{2}),

u¯i\bar{u}_{i} is an arbitrary average of uu around the point ihih, and ξ2,i\xi_{2,i} and ξ1,i2\xi_{1,i}^{2} are discrete analogs of vxx/vv_{xx}/v and (vx/v)2(v_{x}/v)^{2}:

(20) ξ2,i:=1vih2(vi+12vi+vi1),ξ1,i2:=12vi2h2((vi+1vi)2+(vivi1)2).\xi_{2,i}:=\frac{1}{v_{i}h^{2}}(v_{i+1}-2v_{i}+v_{i-1}),\quad\xi_{1,i}^{2}:=\frac{1}{2v_{i}^{2}h^{2}}\big{(}(v_{i+1}-v_{i})^{2}+(v_{i}-v_{i-1})^{2}\big{)}.

The parameters λi\lambda_{i} for i=1,2,3,4i=1,2,3,4 are given by (10) and (15). For later use, we note that

ξ1,i2\displaystyle\xi_{1,i}^{2} =12vi2h2((vi+122vi2+vi12)2vi(vi+12vi+vi1))\displaystyle=\frac{1}{2v_{i}^{2}h^{2}}\big{(}(v_{i+1}^{2}-2v_{i}^{2}+v_{i-1}^{2})-2v_{i}(v_{i+1}-2v_{i}+v_{i-1})\big{)}
=12vi2h2(vi+122vi2+vi12)ξ2,i\displaystyle=\frac{1}{2v_{i}^{2}h^{2}}(v_{i+1}^{2}-2v_{i}^{2}+v_{i-1}^{2})-\xi_{2,i}

implies that

(21) 12h2(vi+122vi2+vi12)=vi2(ξ2,i+ξ1,i2),1h2(vi+12vi+vi1)=viξ2,i,\frac{1}{2h^{2}}(v_{i+1}^{2}-2v_{i}^{2}+v_{i-1}^{2})=v_{i}^{2}(\xi_{2,i}+\xi_{1,i}^{2}),\quad\frac{1}{h^{2}}(v_{i+1}-2v_{i}+v_{i-1})=v_{i}\xi_{2,i},

Recall definition (17) of SαhS_{\alpha}^{h} and condition (7) for K(α,β)K(\alpha,\beta).

Theorem 5.

Let (ui)i𝕋h(u_{i})_{i\in{\mathbb{T}}_{h}} be a positive solution to (18)–(20) and let α0\alpha\geq 0, α1\alpha\neq 1. If K(α,β)0K(\alpha,\beta)\geq 0 then dSαh/dt0dS_{\alpha}^{h}/dt\leq 0 for all t>0t>0. Moreover, if K(α,β)>0K(\alpha,\beta)>0,

dSαhdt+c(α,β)hi𝕋hu¯iα+β(ξ2,i2+ξ1,i4)0\frac{dS_{\alpha}^{h}}{dt}+c(\alpha,\beta)h\sum_{i\in{\mathbb{T}}_{h}}\bar{u}_{i}^{\alpha+\beta}\big{(}\xi_{2,i}^{2}+\xi_{1,i}^{4}\big{)}\leq 0

with the same constant c(α,β)>0c(\alpha,\beta)>0 as in the proof of Theorem 2.

Proof.

Let α>0\alpha>0, α1\alpha\neq 1. We compute the time derivative of the discrete entropy, using summation by parts twice:

dSαhdt\displaystyle\frac{dS_{\alpha}^{h}}{dt} =hi𝕋hsα(u)tui=i𝕋hvi(Ji+1/2Ji1/2)=i𝕋h(vi+1vi)Ji+1/2\displaystyle=h\sum_{i\in{\mathbb{T}}_{h}}s_{\alpha}^{\prime}(u)\partial_{t}u_{i}=-\sum_{i\in{\mathbb{T}}_{h}}v_{i}(J_{i+1/2}-J_{i-1/2})=\sum_{i\in{\mathbb{T}}_{h}}(v_{i+1}-v_{i})J_{i+1/2}
=1hi𝕋h(vi+1vi)((Ai+1Ai)12(vi+1+vi)(Bi+1Bi))\displaystyle=\frac{1}{h}\sum_{i\in{\mathbb{T}}_{h}}(v_{i+1}-v_{i})\bigg{(}(A_{i+1}-A_{i})-\frac{1}{2}(v_{i+1}+v_{i})(B_{i+1}-B_{i})\bigg{)}
=1hi𝕋h((vi+1vi)(Ai+1Ai)12(vi+12vi2)(Bi+1Bi))\displaystyle=\frac{1}{h}\sum_{i\in{\mathbb{T}}_{h}}\bigg{(}(v_{i+1}-v_{i})(A_{i+1}-A_{i})-\frac{1}{2}(v_{i+1}^{2}-v_{i}^{2})(B_{i+1}-B_{i})\bigg{)}
=1hi𝕋h((vi+12vi+vi1)Ai12(vi+122vi2+vi12)Bi).\displaystyle=-\frac{1}{h}\sum_{i\in{\mathbb{T}}_{h}}\bigg{(}(v_{i+1}-2v_{i}+v_{i-1})A_{i}-\frac{1}{2}(v_{i+1}^{2}-2v_{i}^{2}+v_{i-1}^{2})B_{i}\bigg{)}.

In the last step, we recognize the discrete analog of the chain rule (vxv)x=12(v2)xx(v_{x}v)_{x}=\frac{1}{2}(v^{2})_{xx}. Inserting (21), we find that

dSαhdt\displaystyle\frac{dS_{\alpha}^{h}}{dt} =hi𝕋h(ξ2,iviAi(ξ2,i+ξ1,i2)vi2Bi)\displaystyle=-h\sum_{i\in{\mathbb{T}}_{h}}\big{(}\xi_{2,i}v_{i}A_{i}-(\xi_{2,i}+\xi_{1,i}^{2})v_{i}^{2}B_{i}\big{)}
=hi𝕋hu¯iα+β(α1)2((λ1λ3)ξ2,i2+(λ2λ3λ4)ξ2,iξ1,i2λ4ξ1,i4)\displaystyle=-h\sum_{i\in{\mathbb{T}}_{h}}\frac{\bar{u}_{i}^{\alpha+\beta}}{(\alpha-1)^{2}}\big{(}(\lambda_{1}-\lambda_{3})\xi_{2,i}^{2}+(\lambda_{2}-\lambda_{3}-\lambda_{4})\xi_{2,i}\xi_{1,i}^{2}-\lambda_{4}\xi_{1,i}^{4}\big{)}
=hi𝕋hu¯iα+β(α1)2P(ξ1,i,ξ2,i),\displaystyle=-h\sum_{i\in{\mathbb{T}}_{h}}\frac{\bar{u}_{i}^{\alpha+\beta}}{(\alpha-1)^{2}}P(\xi_{1,i},\xi_{2,i}),

where PP is the same polynomial as in (13). The proof of Theorem 2 shows that P(ξ1,i,ξ2,i)0P(\xi_{1,i},\xi_{2,i})\geq 0 if K(α,β)0K(\alpha,\beta)\geq 0. Moreover, if the strict inequality K(α,β)>0K(\alpha,\beta)>0 holds, P(ξ1,i,ξ2,i)c0(α,β)(ξ2,i2+ξ1,i4)P(\xi_{1,i},\xi_{2,i})\geq c_{0}(\alpha,\beta)(\xi_{2,i}^{2}+\xi_{1,i}^{4}) with the same constant as in Theorem 2, which translates into the inequality

dSαdtc0(α,β)i𝕋hu¯iα+β(α1)2(ξ2,i2+ξ1,i4),\frac{dS_{\alpha}}{dt}\leq-c_{0}(\alpha,\beta)\sum_{i\in{\mathbb{T}}_{h}}\frac{\bar{u}_{i}^{\alpha+\beta}}{(\alpha-1)^{2}}\big{(}\xi_{2,i}^{2}+\xi_{1,i}^{4}\big{)},

finishing the proof. ∎

In Theorem 2, the existence of positive solutions is assumed. We show that such solutions exist globally, at least in case α=0\alpha=0.

Proposition 6.

Let α=0\alpha=0 and ui0>0u_{i}^{0}>0 for i𝕋hi\in{\mathbb{T}}_{h}. Then there exists a global solution (ui)i𝕋h(u_{i})_{i\in{\mathbb{T}}_{h}} to scheme (18)–(20) and ui(0)=ui0u_{i}(0)=u_{i}^{0} for i𝕋hi\in{\mathbb{T}}_{h} and constants κ1κ0>0\kappa_{1}\geq\kappa_{0}>0 such that

0<κ0ui(t)κ1for all i𝕋h,t>0.0<\kappa_{0}\leq u_{i}(t)\leq\kappa_{1}\quad\mbox{for all }i\in{\mathbb{T}}_{h},\ t>0.
Proof.

Scheme (18) is a system of ordinary differential equations. According to the Picard–Lindelöf theorem, there exists a unique local positive differentiable solution (ui)i𝕋h(u_{i})_{i\in{\mathbb{T}}_{h}} on the maximal time interval [0,T)[0,T) for some T>0T>0. This solution can be extended to [0,)[0,\infty) if the functions ui(t)u_{i}(t) are uniformly positive and bounded. By Theorem 5,

hi𝕋h(logui(t))hi𝕋h(logui0),h\sum_{i\in{\mathbb{T}}_{h}}(-\log u_{i}(t))\leq h\sum_{i\in{\mathbb{T}}_{h}}(-\log u_{i}^{0}),

Moreover, scheme (18) conserves the mass, hi𝕋hui(t)=hi𝕋hui0h\sum_{i\in{\mathbb{T}}_{h}}u_{i}(t)=h\sum_{i\in{\mathbb{T}}_{h}}u_{i}^{0}. This shows that

hi𝕋h(ui(t)logui(t))hi𝕋h(ui0logui0).h\sum_{i\in{\mathbb{T}}_{h}}(u_{i}(t)-\log u_{i}(t))\leq h\sum_{i\in{\mathbb{T}}_{h}}(u_{i}^{0}-\log u_{i}^{0}).

Since sslogss\mapsto s-\log s diverges for s0s\to 0 and ss\to\infty, there exist constants κ0>0\kappa_{0}>0 and κ1>0\kappa_{1}>0 such that κ0ui(t)κ1\kappa_{0}\leq u_{i}(t)\leq\kappa_{1} for all i𝕋hi\in{\mathbb{T}}_{h} and t>0t>0. Therefore, we can extend the solution globally. ∎

Remark 7 (Noncentral scheme for Ji+1/2J_{i+1/2}).

A more direct discrete analog of vxx/vv_{xx}/v and vx/vv_{x}/v is given by

ξ2,i=1vih2(vi+12vi+vi1),ξ1,i2=1vi2h2(vivi1)2.\xi_{2,i}=\frac{1}{v_{i}h^{2}}(v_{i+1}-2v_{i}+v_{i-1}),\quad\xi_{1,i}^{2}=\frac{1}{v_{i}^{2}h^{2}}(v_{i}-v_{i-1})^{2}.

In this situation, the scheme for JJ needs to be noncentral,

Ji+1/2=1h(Ai+1Ai)1hvi(Bi+1Bi),J_{i+1/2}=\frac{1}{h}(A_{i+1}-A_{i})-\frac{1}{h}v_{i}(B_{i+1}-B_{i}),

where AiA_{i} and BiB_{i} are as before. Indeed, it follows from summation by parts for α0\alpha\geq 0, α1\alpha\neq 1 that

dSαdt\displaystyle\frac{dS_{\alpha}}{dt} =1hi𝕋h(vi+1vi)((Ai+1Ai)vi(Bi+1Bi))\displaystyle=\frac{1}{h}\sum_{i\in{\mathbb{T}}_{h}}(v_{i+1}-v_{i})\big{(}(A_{i+1}-A_{i})-v_{i}(B_{i+1}-B_{i})\big{)}
=1hi𝕋h[(vi+12vi+vi1)Ai((vi+1vi)vi(vivi1)vi1)Bi]\displaystyle=-\frac{1}{h}\sum_{i\in{\mathbb{T}}_{h}}\big{[}(v_{i+1}-2v_{i}+v_{i-1})A_{i}-\big{(}(v_{i+1}-v_{i})v_{i}-(v_{i}-v_{i-1})v_{i-1}\big{)}B_{i}\big{]}
=1hi𝕋h[(vi+12vi+vi1)Ai(vi(vi+12vi+vi1)+(vivi1)2)Bi]\displaystyle=-\frac{1}{h}\sum_{i\in{\mathbb{T}}_{h}}\big{[}(v_{i+1}-2v_{i}+v_{i-1})A_{i}-\big{(}v_{i}(v_{i+1}-2v_{i}+v_{i-1})+(v_{i}-v_{i-1})^{2}\big{)}B_{i}\big{]}
=hi𝕋h(ξ2,iviAi(ξ2,i+ξ1,i2)vi2Bi)=hi𝕋hu¯iα+βP(ξ1,i,ξ2,i),\displaystyle=-h\sum_{i\in{\mathbb{T}}_{h}}\big{(}\xi_{2,i}v_{i}A_{i}-(\xi_{2,i}+\xi_{1,i}^{2})v_{i}^{2}B_{i}\big{)}=-h\sum_{i\in{\mathbb{T}}_{h}}\bar{u}_{i}^{\alpha+\beta}P(\xi_{1,i},\xi_{2,i}),

and we can conclude as before. ∎

The case α=1\alpha=1 has to be treated in a slightly different way. Since limα1(α1)vi=limα1uiα1=1\lim_{\alpha\to 1}(\alpha-1)v_{i}=\lim_{\alpha\to 1}u_{i}^{\alpha-1}=1, we consider scheme

(22) tui=1h(Ji+1/2Ji1/2),Ji+1/2=1h(Ai+1Ai)12h(wi+1+wi)(Bi+1Bi),\partial_{t}u_{i}=-\frac{1}{h}(J_{i+1/2}-J_{i-1/2}),\quad J_{i+1/2}=\frac{1}{h}(A_{i+1}-A_{i})-\frac{1}{2h}(w_{i+1}+w_{i})(B_{i+1}-B_{i}),

with AiA_{i} and BiB_{i} defined via

(23) Ai=u¯iβ+1(λ1ξ2,iλ2ξ1,i2),Bi=u¯iβ+1(λ3ξ2,iλ4ξ1,i2),A_{i}=-\bar{u}_{i}^{\beta+1}(\lambda_{1}\xi_{2,i}-\lambda_{2}\xi_{1,i}^{2}),\quad B_{i}=\bar{u}_{i}^{\beta+1}(\lambda_{3}\xi_{2,i}-\lambda_{4}\xi_{1,i}^{2}),

and

ξ2,i\displaystyle\xi_{2,i} =12h2((vi+1vi)(wi+1+wi)(vivi1)(wi+wi1)),\displaystyle=\frac{1}{2h^{2}}\big{(}(v_{i+1}-v_{i})(w_{i+1}+w_{i})-(v_{i}-v_{i-1})(w_{i}+w_{i-1})\big{)},
(24) ξ1,i2\displaystyle\xi_{1,i}^{2} =ξ2,i+1h2(vi+12vi+vi1),\displaystyle=\xi_{2,i}+\frac{1}{h^{2}}(v_{i+1}-2v_{i}+v_{i-1}),

where vi=loguiv_{i}=\log u_{i} and wi=ui1w_{i}=-u_{i}^{-1}. The parameters λ1,,λ4\lambda_{1},\ldots,\lambda_{4} are given by (10) and (15).

Proposition 8.

Let α=1\alpha=1 and let (ui)i𝕋h(u_{i})_{i\in{\mathbb{T}}_{h}} be a positive solution to (22), (23), and (24). If K(1,β)0K(1,\beta)\geq 0 then dS1h/dt0dS_{1}^{h}/dt\leq 0 for all t>0t>0. Moreover, if K(1,β)>0K(1,\beta)>0,

dS1hdt+c(1,β)hi𝕋hu¯iβ+1(ξ2,i2+ξ1,i4)0,\frac{dS_{1}^{h}}{dt}+c(1,\beta)h\sum_{i\in{\mathbb{T}}_{h}}\bar{u}_{i}^{\beta+1}\big{(}\xi_{2,i}^{2}+\xi_{1,i}^{4}\big{)}\leq 0,

with the same constant c(1,β)>0c(1,\beta)>0 as in the proof of Theorem 2.

Proof.

We have with vi=loguiv_{i}=\log u_{i} and wi=ui1w_{i}=-u_{i}^{-1}:

dS1hdt\displaystyle\frac{dS_{1}^{h}}{dt} =i𝕋h(vi+1vi)Ji+1/2\displaystyle=\sum_{i\in{\mathbb{T}}_{h}}(v_{i+1}-v_{i})J_{i+1/2}
=h1i𝕋h(vi+1vi)((Ai+1Ai)12(wi+1+wi)(Bi+1Bi))\displaystyle=h^{-1}\sum_{i\in{\mathbb{T}}_{h}}(v_{i+1}-v_{i})\bigg{(}(A_{i+1}-A_{i})-\frac{1}{2}(w_{i+1}+w_{i})(B_{i+1}-B_{i})\bigg{)}
=h1i𝕋h((vi+12vi+vi1)Ai\displaystyle=-h^{-1}\sum_{i\in{\mathbb{T}}_{h}}\bigg{(}(v_{i+1}-2v_{i}+v_{i-1})A_{i}
12((vi+1vi)(wi+1+wi)(vivi1)(wi+wi1))Bi).\displaystyle\phantom{xx}{}-\frac{1}{2}\big{(}(v_{i+1}-v_{i})(w_{i+1}+w_{i})-(v_{i}-v_{i-1})(w_{i}+w_{i-1})\big{)}B_{i}\bigg{)}.

By definition of ξ1,i\xi_{1,i} and ξ2,i\xi_{2,i},

1h2(vi+12vi+vi1)\displaystyle\frac{1}{h^{2}}(v_{i+1}-2v_{i}+v_{i-1}) =ξ2,i+ξ1,i2,\displaystyle=-\xi_{2,i}+\xi_{1,i}^{2},
12h2((vi+1vi)(wi+1+wi)(vivi1)(wi+wi1))\displaystyle-\frac{1}{2h^{2}}\big{(}(v_{i+1}-v_{i})(w_{i+1}+w_{i})-(v_{i}-v_{i-1})(w_{i}+w_{i-1})\big{)} =ξ2,i.\displaystyle=-\xi_{2,i}.

Therefore,

dS1hdt\displaystyle\frac{dS_{1}^{h}}{dt} =hi𝕋hu¯iβ+1((λ1ξ2,iλ2ξ1,i2)(ξ2,i+ξ1,i2)+(λ3ξ2,iλ4ξ1,i)ξ2,i)\displaystyle=-h\sum_{i\in{\mathbb{T}}_{h}}\bar{u}_{i}^{\beta+1}\big{(}(\lambda_{1}\xi_{2,i}-\lambda_{2}\xi_{1,i}^{2})(-\xi_{2,i}+\xi_{1,i}^{2})+(\lambda_{3}\xi_{2,i}-\lambda_{4}\xi_{1,i})\xi_{2,i}\big{)}
=hi𝕋hu¯iβ+1P1(ξ1,i,ξ2,i),\displaystyle=-h\sum_{i\in{\mathbb{T}}_{h}}\bar{u}_{i}^{\beta+1}P_{1}(\xi_{1,i},\xi_{2,i}),

where P1P_{1} is the same polynomial as in (16). It is nonnegative if and only if K(1,β)0K(1,\beta)\geq 0 holds. ∎

4. Discretized multi-dimensional thin-film equation

The ideas of the previous section cannot be adapted in a straightforward way to the multi-dimensional setting, since there are many possibilities to choose the finite differences and the discrete variables. One idea is to employ only scalar variables like Δu\Delta u, |u|2|\nabla u|^{2}, etc., similarly as for the method of systematic integration by parts of [15]. Still, there does not exist a general approach to define the scalar discrete variables, but we show in this section that the multi-dimensional case can be treated at least in principle. As an example, we consider the thin-film equation

tu=divJ,J=uβΔuin 𝕋d,\partial_{t}u=-\operatorname{div}J,\quad J=u^{\beta}\nabla\Delta u\quad\mbox{in }{\mathbb{T}}^{d},

where 𝕋d{\mathbb{T}}^{d} is the multi-dimensional torus and β>0\beta>0, and the logarithmic entropy

S0(u)=𝕋d(logu)𝑑x.S_{0}(u)=\int_{{\mathbb{T}}^{d}}(-\log u)dx.

We show first that we can write J=AvBJ=\nabla A-v\nabla B, where v=u1v=-u^{-1} and AA, BB are functions depending on Δv/v\Delta v/v and |v|2/v2|\nabla v|^{2}/v^{2}.

Lemma 9.

It holds that J=AvBJ=\nabla A-v\nabla B, where

A\displaystyle A =uβ(λ1u2Δ(u1)λ2u3|(u1)|2)=uβv(λ1Δvv+λ2|vv|2),\displaystyle=u^{\beta}\big{(}\lambda_{1}u^{2}\Delta(-u^{-1})-\lambda_{2}u^{3}|\nabla(-u^{-1})|^{2}\big{)}=\frac{u^{\beta}}{v}\bigg{(}\lambda_{1}\frac{\Delta v}{v}+\lambda_{2}\bigg{|}\frac{\nabla v}{v}\bigg{|}^{2}\bigg{)},
B\displaystyle B =uβ(λ3u3Δ(u1)+λ4u4|(u1)|2)=uβv2(λ3Δvv+λ4|vv|2)\displaystyle=u^{\beta}\big{(}-\lambda_{3}u^{3}\Delta(-u^{-1})+\lambda_{4}u^{4}|\nabla(-u^{-1})|^{2}\big{)}=\frac{u^{\beta}}{v^{2}}\bigg{(}\lambda_{3}\frac{\Delta v}{v}+\lambda_{4}\bigg{|}\frac{\nabla v}{v}\bigg{|}^{2}\bigg{)}

and the parameters λi\lambda_{i} are defined by

(25) λ1=β+1,λ2=2(β+1),λ3=β,λ4=2β.\lambda_{1}=\beta+1,\quad\lambda_{2}=-2(\beta+1),\quad\lambda_{3}=\beta,\quad\lambda_{4}=-2\beta.
Proof.

We compute

A+u1B\displaystyle\nabla A+u^{-1}\nabla B =(λ1λ3)uβΔu+(βλ1(β+1)λ3)uβ1Δuu\displaystyle=(\lambda_{1}-\lambda_{3})u^{\beta}\nabla\Delta u+\big{(}\beta\lambda_{1}-(\beta+1)\lambda_{3}\big{)}u^{\beta-1}\Delta u\nabla u
+2((2λ1+λ2)+2λ3+λ4)uβ12uu\displaystyle\phantom{xx}{}+2\big{(}-(2\lambda_{1}+\lambda_{2})+2\lambda_{3}+\lambda_{4}\big{)}u^{\beta-1}\nabla^{2}u\nabla u
+((1β)(2λ1+λ2)+β(2λ3+λ4))|u|2u.\displaystyle\phantom{xx}{}+\big{(}(1-\beta)(2\lambda_{1}+\lambda_{2})+\beta(2\lambda_{3}+\lambda_{4})\big{)}|\nabla u|^{2}\nabla u.

Here, 2u\nabla^{2}u denotes the Hessian matrix of uu. Identifying the coefficients of A+u1B\nabla A+u^{-1}\nabla B and J=uβΔuJ=u^{\beta}\nabla\Delta u gives the linear system

λ1λ3\displaystyle\lambda_{1}-\lambda_{3} =1,\displaystyle=1, βλ1(β+1)λ3\displaystyle\quad\beta\lambda_{1}-(\beta+1)\lambda_{3} =0,\displaystyle=0,
(2λ1+λ2)+2λ3+λ4\displaystyle-(2\lambda_{1}+\lambda_{2})+2\lambda_{3}+\lambda_{4} =0,\displaystyle=0, (1β)(2λ1+λ2)+β(2λ3+λ4)\displaystyle\quad(1-\beta)(2\lambda_{1}+\lambda_{2})+\beta(2\lambda_{3}+\lambda_{4}) =0.\displaystyle=0.

The unique solution is given by (25). ∎

Proposition 10.

Let β=2\beta=2. Then dS0/dt0dS_{0}/dt\leq 0 for all t>0t>0.

Proof.

The time derivative of S0S_{0} becomes

dS0dt\displaystyle\frac{dS_{0}}{dt} =𝕋dJvdx=𝕋d(AvB)vdx\displaystyle=\int_{{\mathbb{T}}^{d}}J\cdot\nabla vdx=\int_{{\mathbb{T}}^{d}}(\nabla A-v\nabla B)\cdot\nabla vdx
=𝕋d(ΔvA(vΔv+|v|2)B)𝑑x\displaystyle=-\int_{{\mathbb{T}}^{d}}\big{(}\Delta vA-(v\Delta v+|\nabla v|^{2})B\big{)}dx
=𝕋duβ((λ1λ3)(Δvv)2+(λ2λ3λ4)Δvv|vv|2λ4|vv|4)𝑑x.\displaystyle=-\int_{{\mathbb{T}}^{d}}u^{\beta}\bigg{(}(\lambda_{1}-\lambda_{3})\bigg{(}\frac{\Delta v}{v}\bigg{)}^{2}+(\lambda_{2}-\lambda_{3}-\lambda_{4})\frac{\Delta v}{v}\bigg{|}\frac{\nabla v}{v}\bigg{|}^{2}-\lambda_{4}\bigg{|}\frac{\nabla v}{v}\bigg{|}^{4}\bigg{)}dx.

The polynomial

(26) P0(ξ1,ξ2)=(λ1λ3)ξ22+(λ2λ3λ4)ξ2ξ12λ4ξ14P_{0}(\xi_{1},\xi_{2})=(\lambda_{1}-\lambda_{3})\xi_{2}^{2}+(\lambda_{2}-\lambda_{3}-\lambda_{4})\xi_{2}\xi_{1}^{2}-\lambda_{4}\xi_{1}^{4}

is nonnegative in 2{\mathbb{R}}^{2} if and only if 4λ4(λ2λ3λ4)2=(β2)20-4\lambda_{4}-(\lambda_{2}-\lambda_{3}-\lambda_{4})^{2}=-(\beta-2)^{2}\geq 0. Hence, we need to assume that β=2\beta=2. ∎

Remark 11 (Discussion).

The restriction β=2\beta=2 in the previous lemma comes from the fact that we do not have a free parameter to optimize the inequalities. One may overcome this issue by allowing AA and BB to depend on more variables or by assuming that AA and BB are matrix-valued with variables like 2u\nabla^{2}u and uu\nabla u\otimes\nabla u and to formulate J=divAvdivBJ=\operatorname{div}A-v\operatorname{div}B. However, this leads to several parameters that need to be determined and eventually to complicated numerical schemes which seem to be less interesting in practice.

Another way to understand the restriction β=2\beta=2 is from systematic integration by parts. Indeed, computing

dS0dt\displaystyle\frac{dS_{0}}{dt} =𝕋duβΔu(u1)dx𝕋ddiv(uβ2Δuu)𝑑x\displaystyle=\int_{{\mathbb{T}}^{d}}u^{\beta}\nabla\Delta u\cdot\nabla(-u^{-1})dx-\int_{{\mathbb{T}}^{d}}\operatorname{div}(u^{\beta-2}\Delta u\nabla u)dx
=𝕋duβ2(Δu)2𝑑x(β2)𝕋duβ3Δu|u|2𝑑x,\displaystyle=-\int_{{\mathbb{T}}^{d}}u^{\beta-2}(\Delta u)^{2}dx-(\beta-2)\int_{{\mathbb{T}}^{d}}u^{\beta-3}\Delta u|\nabla u|^{2}dx,

we see that S0S_{0} is a Lyapunov functional if the last term vanishes, which is the case if β=2\beta=2. This computation suggests the following generalization: Let β(0,2)\beta\in(0,2) and consider the Rényi entropy SαS_{\alpha}. Then

dSαdt=𝕋duα+β2(Δu)2𝑑x(α+β2)𝕋duα+β3Δu|u|2𝑑x,\frac{dS_{\alpha}}{dt}=-\int_{{\mathbb{T}}^{d}}u^{\alpha+\beta-2}(\Delta u)^{2}dx-(\alpha+\beta-2)\int_{{\mathbb{T}}^{d}}u^{\alpha+\beta-3}\Delta u|\nabla u|^{2}dx,

and the last term vanishes if α=2β>0\alpha=2-\beta>0. As for the case β=2\beta=2 and α=0\alpha=0, which is discussed below, we expect that the generalization β(0,2)\beta\in(0,2) and α=2β\alpha=2-\beta can be “translated” to the discrete case, but we leave the details to the interested reader. ∎

We turn to the discrete setting. Let 𝕋hd{\mathbb{T}}_{h}^{d} be the discrete multi-dimensional torus, u:𝕋hdu:{\mathbb{T}}_{h}^{d}\to{\mathbb{R}} be a function, and eμe_{\mu} be the μ\mu-th unit vector of d{\mathbb{R}}^{d}. We write ui=u(i)u_{i}=u(i) for i𝕋hdi\in{\mathbb{T}}_{h}^{d} and we introduce the finite differences

μ+ui=1h(u(i+heμ)u(i)),μui=1h(u(i)u(iheμ)),\partial_{\mu}^{+}u_{i}=\frac{1}{h}(u(i+he_{\mu})-u(i)),\quad\partial_{\mu}^{-}u_{i}=\frac{1}{h}(u(i)-u(i-he_{\mu})),

where i𝕋hdi\in{\mathbb{T}}_{h}^{d} and μ=1,,d\mu=1,\ldots,d. The discrete divergence of F=(F1,,Fd):𝕋hddF=(F_{1},\ldots,F_{d}):{\mathbb{T}}_{h}^{d}\to{\mathbb{R}}^{d} and the discrete gradient and Laplacian of u:𝕋hdu:{\mathbb{T}}_{h}^{d}\to{\mathbb{R}} are defined by, respectively,

divh+F=μ=1dμ+Fμ,(h±u)μ=μ±u,Δhu=divh+hu.\operatorname{div}_{h}^{+}F=\sum_{\mu=1}^{d}\partial_{\mu}^{+}F_{\mu},\quad(\nabla_{h}^{\pm}u)_{\mu}=\partial_{\mu}^{\pm}u,\quad\Delta_{h}u=\operatorname{div}_{h}^{+}\nabla_{h}^{-}u.

where μ=1,,d\mu=1,\ldots,d. The discrete analogs of vxx/vv_{xx}/v and (vx/v)2(v_{x}/v)^{2} are

ξ2,i=Δhvivi,ξ1,i2=|h+vivi|2,\xi_{2,i}=\frac{\Delta_{h}v_{i}}{v_{i}},\quad\xi_{1,i}^{2}=\bigg{|}\frac{\nabla_{h}^{+}v_{i}}{v_{i}}\bigg{|}^{2},

where vi=ui1v_{i}=-u_{i}^{-1}. The numerical scheme reads as

tui=divh+Ji,Ji=hAivihBi,\displaystyle\partial_{t}u_{i}=-\operatorname{div}_{h}^{+}J_{i},\quad J_{i}=\nabla_{h}^{-}A_{i}-v_{i}\nabla_{h}^{-}B_{i},
(27) Ai=u¯iβvi(λ1ξ2,i+λ2ξ1,i2),Bi=u¯iβvi2(λ3ξ2,i+λ4ξ1,i2),\displaystyle A_{i}=\frac{\bar{u}_{i}^{\beta}}{v_{i}}(\lambda_{1}\xi_{2,i}+\lambda_{2}\xi_{1,i}^{2}),\quad B_{i}=\frac{\bar{u}_{i}^{\beta}}{v_{i}^{2}}(\lambda_{3}\xi_{2,i}+\lambda_{4}\xi_{1,i}^{2}),

and the coefficients are as in (25).

Lemma 12.

Let β=2\beta=2 and ui0>0u_{i}^{0}>0 for i𝕋hdi\in{\mathbb{T}}_{h}^{d}. Then there exists a positive solution (ui)i𝕋hd(u_{i})_{i\in{\mathbb{T}}_{h}^{d}} to (25) and (27), and it holds that dS0h/dt0dS_{0}^{h}/dt\leq 0 for all t>0t>0.

Proof.

By the Picard–Lindelöf theorem, there exists a local smooth positive solution (ui)i𝕋hd(u_{i})_{i\in{\mathbb{T}}_{h}^{d}}. We use the summation-by-parts formula

i𝕋hdvidivh+Fi=i𝕋hd(hvi)Fi\sum_{i\in{\mathbb{T}}_{h}^{d}}v_{i}\operatorname{div}_{h}^{+}F_{i}=-\sum_{i\in{\mathbb{T}}_{h}^{d}}(\nabla_{h}^{-}v_{i})\cdot F_{i}

to compute the time derivative of the entropy:

dS0hdt\displaystyle\frac{dS_{0}^{h}}{dt} =hi𝕋hdtuivi=hi𝕋hddivh+Jivi=hi𝕋hdJihvi\displaystyle=h\sum_{i\in{\mathbb{T}}_{h}^{d}}\partial_{t}u_{i}v_{i}=-h\sum_{i\in{\mathbb{T}}_{h}^{d}}\operatorname{div}_{h}^{+}J_{i}v_{i}=h\sum_{i\in{\mathbb{T}}_{h}^{d}}J_{i}\cdot\nabla_{h}^{-}v_{i}
=hi𝕋hd(hAivihBi)hvi=hi𝕋hd(ΔhviAidivh+(vihvi)Bi).\displaystyle=h\sum_{i\in{\mathbb{T}}_{h}^{d}}(\nabla_{h}^{-}A_{i}-v_{i}\nabla_{h}^{-}B_{i})\cdot\nabla_{h}^{-}v_{i}=-h\sum_{i\in{\mathbb{T}}_{h}^{d}}\big{(}\Delta_{h}v_{i}A_{i}-\operatorname{div}_{h}^{+}(v_{i}\nabla_{h}^{-}v_{i})B_{i}\big{)}.

The product rule reads as

divh+\displaystyle\operatorname{div}_{h}^{+} (vihvi)=μ=1dμ+(viμvi)=1hμ=1dμ+(v(i)(v(i)v(iheμ)))\displaystyle(v_{i}\nabla_{h}^{-}v_{i})=\sum_{\mu=1}^{d}\partial_{\mu}^{+}(v_{i}\partial_{\mu}^{-}v_{i})=\frac{1}{h}\sum_{\mu=1}^{d}\partial_{\mu}^{+}\big{(}v(i)(v(i)-v(i-he_{\mu}))\big{)}
=1h2μ=1d[v(i+heμ)(v(i+heμ)v(i))v(i)(v(i)v(iheμ))]\displaystyle=\frac{1}{h^{2}}\sum_{\mu=1}^{d}\big{[}v(i+he_{\mu})\big{(}v(i+he_{\mu})-v(i)\big{)}-v(i)\big{(}v(i)-v(i-he_{\mu})\big{)}\big{]}
=1h2μ=1d[v(i)(v(i+heμ)2v(i)+v(iheμ))+(v(i+heμ)v(i))2]\displaystyle=\frac{1}{h^{2}}\sum_{\mu=1}^{d}\big{[}v(i)\big{(}v(i+he_{\mu})-2v(i)+v(i-he_{\mu})\big{)}+\big{(}v(i+he_{\mu})-v(i)\big{)}^{2}\big{]}
=viΔhvi+|h+vi|2.\displaystyle=v_{i}\Delta_{h}v_{i}+|\nabla_{h}^{+}v_{i}|^{2}.

Therefore,

dS0hdt\displaystyle\frac{dS_{0}^{h}}{dt} =hi𝕋hd(ΔhviAi(viΔhvi+|h+vi|2)Bi)\displaystyle=-h\sum_{i\in{\mathbb{T}}_{h}^{d}}\big{(}\Delta_{h}v_{i}A_{i}-(v_{i}\Delta_{h}v_{i}+|\nabla_{h}^{+}v_{i}|^{2})B_{i}\big{)}
=hi𝕋hd(ξ2,iviAi(ξ2,i+ξ1,i2)vi2Bi)\displaystyle=-h\sum_{i\in{\mathbb{T}}_{h}^{d}}\big{(}\xi_{2,i}v_{i}A_{i}-(\xi_{2,i}+\xi_{1,i}^{2})v_{i}^{2}B_{i}\big{)}
=hi𝕋hd((λ1λ3)ξ2,i2+(λ2λ3λ4)ξ2,iξ1,i2λ4ξ1,i4)\displaystyle=-h\sum_{i\in{\mathbb{T}}_{h}^{d}}\big{(}(\lambda_{1}-\lambda_{3})\xi_{2,i}^{2}+(\lambda_{2}-\lambda_{3}-\lambda_{4})\xi_{2,i}\xi_{1,i}^{2}-\lambda_{4}\xi_{1,i}^{4}\big{)}
=hi𝕋hd(ξ2,i2(β+2)ξ2,iξ1,i2+2βξ1,i4).\displaystyle=-h\sum_{i\in{\mathbb{T}}_{h}^{d}}\big{(}\xi_{2,i}^{2}-(\beta+2)\xi_{2,i}\xi_{1,i}^{2}+2\beta\xi_{1,i}^{4}\big{)}.

This gives the polynomial (26) which is nonnegative in 2{\mathbb{R}}^{2} if and only if β=2\beta=2. Then S0h(t)=S0h(0)S_{0}^{h}(t)=S_{0}^{h}(0) for all t>0t>0, and we conclude as in the proof of Proposition 6 that 0<c0ui(t)c10<c_{0}\leq u_{i}(t)\leq c_{1} for all i𝕋hdi\in{\mathbb{T}}_{h}^{d} and t>0t>0, where c1c0>0c_{1}\geq c_{0}>0 are some constants. Thus, the solution ui(t)u_{i}(t) can be extended to a global one. ∎

5. Numerical tests

We apply our scheme to the thin-film and DLSS equations on the torus in one and two space dimensions. The system of ordinary differential equations is solved by the command

 scipy.integrate.solve_ivp 

from the SciPy library, which uses the Backward Differentiation Formula (BDF) method of variable order or the implicit Runge–Kutta method of the Radau IIA family of order 5. We used the default values tol = 1e-3} for the bsolute tolerance and tol = 1e-6} fo the relative tolerance. The local erroris computed according to tol + rtol * bs(u).

5.1. DLSS equation

The DLSS equation is solved by scheme (18) using the logarithmic entropy:

tui=1h(Ji+1/2Ji1/2),Ji+1/2=1h(Ai+1Ai)+12h(ui+11+ui1)(Bi+1Bi),\displaystyle\partial_{t}u_{i}=-\frac{1}{h}(J_{i+1/2}-J_{i-1/2}),\quad J_{i+1/2}=\frac{1}{h}(A_{i+1}-A_{i})+\frac{1}{2h}(u^{-1}_{i+1}+u^{-1}_{i})(B_{i+1}-B_{i}),
Ai=u¯i(53ξ2,i73ξ1,i2),Bi=u¯iui(23ξ2,i+ξ1,i2),\displaystyle A_{i}=\bar{u}_{i}\bigg{(}\frac{5}{3}\xi_{2,i}-\frac{7}{3}\xi_{1,i}^{2}\bigg{)},\quad B_{i}=\bar{u}_{i}u_{i}\bigg{(}-\frac{2}{3}\xi_{2,i}+\xi_{1,i}^{2}\bigg{)},
ξ1,i2=u¯i22h2((ui+11ui1)2+(ui1ui11)2),ξ2,i=u¯i2uih2(ui+112ui1+ui11),\displaystyle\xi_{1,i}^{2}=\frac{\bar{u}_{i}^{2}}{2h^{2}}\big{(}(u^{-1}_{i+1}-u^{-1}_{i})^{2}+(u^{-1}_{i}-u^{-1}_{i-1})^{2}\big{)},\quad\xi_{2,i}=\frac{\bar{u}_{i}^{2}}{u_{i}h^{2}}(u^{-1}_{i+1}-2u^{-1}_{i}+u^{-1}_{i-1}),

where i𝕋hi\in{\mathbb{T}}_{h}. Figure 1 shows the solution to the DLSS equation at various time steps using the initial datum u0(x)=max{1010,cos(πx)16}u^{0}(x)=\max\{10^{-10},\cos(\pi x)^{16}\} and the space grid size h=1/100h=1/100. We see that the solution is not monotone, since it possesses at x=0.5x=0.5 and t=108t=10^{-8} a local maximum. After some time, it approaches the constant steady state given by 01u0(x)𝑑x\int_{0}^{1}u^{0}(x)dx.

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Figure 1. Evolution of the DLSS equation in a semi-logarithmic scale, using the initial datum u0(x)=max{1010,cos(πx)16}u^{0}(x)=\max\{10^{-10},\cos(\pi x)^{16}\}.

The entropy decay for α=0\alpha=0 is illustrated in Figure 2 (left). We used the initial datum u0(x)=2106u^{0}(x)=2-10^{-6} for x(0,0.5)x\in(0,0.5) and u0(x)=106u^{0}(x)=10^{-6} for x(0.5,1)x\in(0.5,1). We observe in the semi-logarithmic plot that the decay is exponential, as expected. The rate degrades for larger times when the 2\ell^{2} error dominates, i.e., when the grid is rather coarse.

The 2\ell^{2} error (in space and time) at time t=0.001t=0.001 is shown in Figure 2 (right), using the initial datum u0(x)=1+0.5sin(2πx)u^{0}(x)=1+0.5\sin(2\pi x) for x(0,1)x\in(0,1). As an explicit solution is not known, we use a numerical solution with h=1/2048h=1/2048 as the reference solution. As expected, the convergence rate is roughly of second order.

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Figure 2. Left: Decay of the logarithmic entropy s0(u(t))s_{0}(u(t)) for two different space grid sizes h=1/20h=1/20 and h=1/200h=1/200. Right: Convergence of the 2\ell^{2} error. The dots are the values from the numerical solution, the solid line is the regression curve.

5.2. Thin-film equation

The thin-film equation is solved by scheme (18) using the logarithmic entropy:

tui=1h(Ji+1/2Ji1/2),Ji+1/2=1h(Ai+1Ai)+12h(ui+11+ui1)(Bi+1Bi),\displaystyle\partial_{t}u_{i}=-\frac{1}{h}(J_{i+1/2}-J_{i-1/2}),\quad J_{i+1/2}=\frac{1}{h}(A_{i+1}-A_{i})+\frac{1}{2h}(u^{-1}_{i+1}+u^{-1}_{i})(B_{i+1}-B_{i}),
Ai=uiβ+1(7β+99ξ2,i+β214β189ξ1,i2),\displaystyle A_{i}=u_{i}^{\beta+1}\bigg{(}\frac{7\beta+9}{9}\xi_{2,i}+\frac{\beta^{2}-14\beta-18}{9}\xi_{1,i}^{2}\bigg{)},
Bi=uiβ+2(7β9ξ2,i+15ββ29ξ1,i2),\displaystyle B_{i}=u_{i}^{\beta+2}\bigg{(}-\frac{7\beta}{9}\xi_{2,i}+\frac{15\beta-\beta^{2}}{9}\xi_{1,i}^{2}\bigg{)},
ξ1,i2=ui22h2((ui+11ui1)2+(ui1ui11)2),ξ2,i=uih2(ui+112ui1+ui11),\displaystyle\xi_{1,i}^{2}=\frac{u_{i}^{2}}{2h^{2}}\big{(}(u^{-1}_{i+1}-u^{-1}_{i})^{2}+(u^{-1}_{i}-u^{-1}_{i-1})^{2}\big{)},\quad\xi_{2,i}=\frac{u_{i}}{h^{2}}(u^{-1}_{i+1}-2u^{-1}_{i}+u^{-1}_{i-1}),

where i𝕋hi\in{\mathbb{T}}_{h}. The solutions at different times, emanating from the initial datum u0(x)=1+0.5sin(2πx)u^{0}(x)=1+0.5\sin(2\pi x), are shown in Figure 3, where we have chosen β=2\beta=2. Again, the solutions converge to the constant steady state. The decay of the logarithmic entropy is illustrated in Figure 4, using β=2\beta=2 and the initial datum u0(x)=1+(11016)sin(2πx)u^{0}(x)=1+(1-10^{-16})\sin(2\pi x) for x(0,1)x\in(0,1). The decay rate is exponential over a large time interval.

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Figure 3. Evolution of the solution to the thin-film equation at times t=0t=0 (densely dotted), t=2104t=2\cdot 10^{-4} (dotted), t=5104t=5\cdot 10^{-4} (dash-dotted), t=1103t=1\cdot 10^{-3} (dashed), t=2103t=2\cdot 10^{-3} (densely dashed), and t=5103t=5\cdot 10^{-3} (solid) and grid sizes h=1/10h=1/10 (left), h=1/200h=1/200 (right).
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Figure 4. Decay of the logarithmic entropy S0(u(t))S_{0}(u(t)) for various space grid sizes.

Finally, we present a numerical example in two space dimensions. As the initial datum, we choose a lantern picture with 77×10077\times 100 pixels in gray scale; see Figure 5 (top left). The evolution of the discrete solution is shown in the remaining panels of Figure 5 for various times. The values u=0u=0 and u=1u=1 correspond to black and white, respectively. Because of the periodic boundary conditions, we observe a small gray band at the lower right boundary. Interestingly, the solution shows a denoising effect, especially for t=3108t=3\cdot 10^{-8}. For larger times, the diffusion drives the solution to the constant steady state. These results are not surprising, as fourth-order parabolic equations have been suggested in the literature for image denoising. For instance, Bertozzi and Greer [2] analyzed

tu=div(g((Δu)2)Δu),\partial_{t}u=-\operatorname{div}\big{(}g((\Delta u)^{2})\nabla\Delta u\big{)},

where gg is a diffusivity function, while Wei [21] considered

tu=div(g(|u|2)Δu).\partial_{t}u=-\operatorname{div}\big{(}g(|\nabla u|^{2})\nabla\Delta u\big{)}.

This model was generalized to fractional derivatives; see, e.g., [12]. An example is the equation

tu=div(g((Δ)1εu)Δu),ε>0,\partial_{t}u=-\operatorname{div}\big{(}g(-(\Delta)^{1-\varepsilon}u)\nabla\Delta u\big{)},\quad\varepsilon>0,

which formally reduces to a general thin-film equation in the limiting case ε=1\varepsilon=1. We do not claim that the thin-film equation is a good image denoising model; our numerical example is just a nice illustration.

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Figure 5. Evolution of the solution to the two-dimensional thin-film equation with β=2\beta=2, t=0t=0 (top left), t=3109t=3\cdot 10^{-9} (top right), t=108t=10^{-8} (bottom left), t=106t=10^{-6} (bottom right).

Finally, we show the entropy decay of the two-dimensional example in Figure 6. The decay rate is exponential until approximately t=102t=10^{-2}. For later times, the numerical error dominates. Observe, however, that we obtain denoising for very small times, like t=109108t=10^{-9}\ldots 10^{-8}, where the decay rate is still exponential.

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Figure 6. Decay of the logarithmic entropy S0(u(t))S_{0}(u(t)) for various space grid sizes.

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