1 Difference methods for the Fisher equation

1.1 Introduction

The Fisher equation belongs to the class of reaction-diffusion equations. In fact, it is one of the simplest semilinear reaction-diffusion equations, the one which has the inhomogeneous term f(u)=?u(1-u), which can exhibit traveling wave solutions that switch between equilibrium states given by f(u)=0. Such an equation occurs, e.?g., in ecology, physiology, combustion, crystallization, plasma physics and in general, phase transition problems. Fisher proposed this equation in 1937 to describe the spatial spread of an advantageous allele and explored its traveling wave solutions [12]. In the same year (1937) as Fisher, Kolmogorov, Petrovskii and Piskunov introduced a more general reaction-diffusion equation [18]. In this chapter, we consider the following initial and boundary value problem of a one-dimensional Fisher equation:

where ? is a positive constant, functions f(x), a(t), ß(t) are all given and f(0)=a(0), f(L)=ß(0). Suppose that the problem (1.1)-(1.3) has a smooth solution.

Before introducing the difference scheme, a priori estimate on the solution of the problem (1.1)-(1.3) is given.

Theorem 1.1. Let u(x,t) be the solution of the problem (1.1)-(1.3) with a(t)=0, ß(t)=0. Denote E(t)=?0Lu2(x,t)dx+2?0t[?0Lux2(x,s)dx+??0L(u3(x,s)-u2(x,s))dx]ds,F(t)=?0Lux2(x,t)dx+??0L[23u3(x,t)-u2(x,t)]dx+2?0t[?0Lus2(x,s)dx]ds. Then E(t)=E(0),F(t)=F(0),0<t?T.

Proof.

(I) Multiplying both the right- and left-hand sides of (1.1) by u(x,t) gives

u(x,t)ut(x,t)-u(x,t)uxx(x,t)+?[u3(x,t)-u2(x,t)]=0,

i.?e.,

12ddt[u2(x,t)]-(u(x,t)ux(x,t))x+ux2(x,t)+?[u3(x,t)-u2(x,t)]=0.

Integrating both the right- and left-hand sides with respect to x on the interval [0,L] and noticing (1.3) with a(t)=ß(t)=0, we have

12ddt?0Lu2(x,t)dx+?0Lux2(x,t)dx+??0L[u3(x,t)-u2(x,t)]dx=0,

which can be rewritten as

ddt{?0Lu2(x,t)dx+2?0t[?0Lux2(x,s)dx+??0L(u3(x,s)-u2(x,s))dx]ds}=0.

Then E(t)=E(0) is obtained.

(II) Multiplying both the right- and left-hand sides of (1.1) by ut(x,t) yields

ut2(x,t)-ut(x,t)uxx(x,t)-?[u(x,t)-u2(x,t)]ut(x,t)=0,

i.?e.,

ut2(x,t)-(ut(x,t)ux(x,t))x+(12ux2(x,t))t+?[13u3(x,t)-12u2(x,t)]t=0.

Integrating both the right- and left-hand sides with respect to x on the interval [0,L] and noticing (1.3) with a(t)=ß(t)=0, we have

12ddt?0Lux2(x,t)dx+?ddt?0L[13u3(x,t)-12u2(x,t)]dx+?0Lut2(x,t)dx=0,

which can be rewritten as

ddt[?0Lux2(x,t)dx+??0L(23u3(x,t)-u2(x,t))dx+2?0t(?0Lus2(x,s)dx)ds]=0,

i.?e.,

dF(t)dt=0,0<t?T.

Thus, F(t)=F(0) is followed. ?

1.2 Notation and lemmas

In order to derive the difference scheme, we first divide the domain [0,L]×[0,T]. Take two positive integers m, n. Divide [0,L] into m equal subintervals, and [0,T] into n subintervals. Denote h=L/m, t=T/n; xi=ih, 0?i?m; tk=kt, 0?k?n; Oh={xi│0?i?m}, Ot={tk│0?k?n};Oht=Oh×Ot. We call all of the nodes {(xi,tk)│0?i?m} on the line t=tk the k-th time-level nodes. In addition, denote xi+12=12(xi+xi+1), tk+12=12(tk+tk+1), r=th2.

Denote

Uh={u│u=(u0,u1,.,um)is the grid function defined onOh},U°h={u│u?Uh,u0=um=0}.

For any grid function u?Uh, introduce the following notation:

dxui+12=1h(ui+1-ui),dx2ui=1h2(ui-1-2ui+ui+1),?xui=12h(ui+1-ui-1).

It follows easily that

dx2ui=1h(dxui+12-dxui-12),?xui=12(dxui-12+dxui+12).

Suppose u,v?Uh. Introduce the inner products, norms and seminorms as

(u,v)=h(12u0v0+Si=1m-1uivi+12umvm),?dxu,dxv?=hSi=1m(dxui-12)(dxvi-12),?u?8=max0?i?m|ui|,?u?=(u,u),?dxu?8=max1?i?m|dxui-12|,|u|1=?dxu,dxu?,?u?1=?u?2+|u|12,|u|2=hSi=1m-1(dx2ui)2,?u?2=?u?2+|u|12+|u|22.

If Uh is a complex space, then the corresponding inner product is defined by

(u,v)=h(12u0v¯0+Si=1m-1uiv¯i+12umv¯m),

with v¯i the conjugate of vi.

Denote

St={w│w=(w0,w1,.,wn)is the grid function defined onOt}.

For any w?St, introduce the following notation:

wk+12=12(wk+wk+1),wk¯=12(wk+1+wk-1),Dtwk=1t(wk+1-wk),Dt?wk=1t(wk-wk-1),dtwk+12=1t(wk+1-wk),?twk=12t(wk+1-wk-1).

It is easy to know that

?twk=12(dtwk-12+dtwk+12).

Suppose u={uik│0?i?m,0?k?n} is a grid function defined on Oht, then v={uik│0?i?m} is a grid function defined on Oh, w={uik│0?k?n} is a grid function defined on Ot.

Lemma 1.1 ([25], [35]).

(a) Suppose u,v?Uh, then

-hSi=1m-1(dx2ui)vi=hSi=1m(dxui-12)(dxvi-12)+(dxu12)v0-(dxum-12)vm.

(b) Suppose u?U°h, then

-hSi=1m-1(dx2ui)ui=|u|12,|u|12??u?·|u|2,?u?8?L2|u|1,?u??L6|u|1.

(c) Suppose u?U°h, then

?u?82??u?·|u|1,

and for arbitrary e>0, it holds that

?u?8?e|u|1+14e?u?,?u?82?e|u|12+14e?u?2.

(d) Suppose u?Uh, then

|u|12?4h2?u?2.

(e) Suppose u?Uh, then

?u?82?2?u?·|u|1+1L?u?2,

and for arbitrary e>0, it holds that

?u?82?e|u|12+(1e+1L)?u?2.

(f) Suppose u?Uh, then for arbitrary e>0, it holds that

?dxu?82?e|u|22+(1e+1L)|u|12.

Proof.

We only prove (c) and (e).

(c) Noticing that...