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 termf(u)= u(1 u), which can exhibit traveling wave solutions that switch between equilibrium states given byf(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 (x), (t), (t) are all given and (0)= (0), (L)= (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.Letu(x,t)be the solution of the problem (1.1
) (1.3
) with (t) 0, (t) 0. DenoteE(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.
ThenE(t)=E(0),F(t)=F(0),0<t T.
Proof.
(I) Multiplying both the right- and left-hand sides of (1.1) byu(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 tox on the interval[0,L] and noticing (1.3) with (t)= (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.
ThenE(t)=E(0) is obtained.
(II) Multiplying both the right- and left-hand sides of (1.1) byut(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 tox on the interval[0,L] and noticing (1.3) with (t)= (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 integersm,n. Divide[0,L] intom equal subintervals, and[0,T] inton subintervals. Denoteh=L/m, =T/n;xi=ih,0 i m;tk=k,0 k n; h={xi 0 i m}, ={tk 0 k n}; h = h×. We call all of the nodes{(xi,tk) 0 i m} on the linet=tk thek-th time-level nodes. In addition, denotexi+12=12(xi+xi+1),tk+12=12(tk+tk+1),r= h2.
Denote
Uh={u u=(u0,u1, ,um)is the grid function defined on h},U h={u u Uh,u0=um=0}.
For any grid functionu Uh, introduce the following notation:
xui+12=1h(ui+1 ui), x2ui=1h2(ui 1 2ui+ui+1), xui=12h(ui+1 ui 1).
It follows easily that
x2ui=1h( xui+12 xui 12), xui=12( xui 12+ xui+12).
Supposeu,v Uh. Introduce the inner products, norms and seminorms as
(u,v)=h(12u0v0+ i=1m 1uivi+12umvm), xu, xv =h i=1m( xui 12)( xvi 12), u =max0 i m|ui|, u =(u,u), xu =max1 i m| xui 12|,|u|1= xu, xu , u 1= u 2+|u|12,|u|2=h i=1m 1( x2ui)2, u 2= u 2+|u|12+|u|22.
IfUh is a complex space, then the corresponding inner product is defined by
(u,v)=h(12u0v¯0+ i=1m 1uiv¯i+12umv¯m),
withv¯i the conjugate ofvi.
Denote
S ={w w=(w0,w1, ,wn)is the grid function defined on }.
For anyw S, introduce the following notation:
wk+12=12(wk+wk+1),wk¯=12(wk+1+wk 1),Dtwk=1 (wk+1 wk),Dt wk=1 (wk wk 1), twk+12=1 (wk+1 wk), twk=12 (wk+1 wk 1).
It is easy to know that
twk=12( twk 12+ twk+12).
Supposeu={uik 0 i m,0 k n} is a grid function defined on h, thenv={uik 0 i m} is a grid function defined on h,w={uik 0 k n} is a grid function defined on.
Lemma 1.1 ([25], [35]).
(a)Supposeu,v Uh, then
h i=1m 1( x2ui)vi=h i=1m( xui 12)( xvi 12)+( xu12)v0 ( xum 12)vm.
(b)Supposeu U h, then
h i=1m 1( x2ui)ui=|u|12,|u|12 u ·|u|2, u L2|u|1, u L6|u|1.
(c)Supposeu U h, then
u 2 u ·|u|1,
and for arbitrary>0, it holds that
u |u|1+14 u , u 2 |u|12+14 u 2.
(d)Supposeu Uh, then
|u|12 4h2 u 2.
(e)Supposeu Uh, then
u 2 2 u ·|u|1+1L u 2,
and for arbitrary>0, it holds that
u 2 |u|12+(1 +1L) u 2.
(f)Supposeu Uh, then for arbitrary>0, it holds that
xu 2 |u|22+(1 +1L)|u|12.
Proof.
We only prove (c) and (e).
(c) Noticing that...
