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The method of finite differences for nonlinear functional differential equations of the first order

  • Elżbieta Puźniakowska-Gałuch EMAIL logo
Veröffentlicht/Copyright: 10. Mai 2019
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Abstract

Nonlinear functional partial differential equations with initial conditions are considered on the cone. The weak convergence of a sequence of successive approximations is proved. The proof is given by the duality principle.

MSC 2010: 35R10; 35F25; 35A05

A Appendix

In this section the proofs of the used lemmas are presented. First we will present the proof of Lemma 3.3.

Proof of Lemma 3.3.

Take w=zϵ1-zϵ2, ϵ1,ϵ2>0 and

P(t,x,s)=(t,x,sz(t,x)ϵ1+(1-s)z(t,x)ϵ2,sxzϵ1(t,x)+(1-s)xzϵ2(t,x)).

The function w satisfies

tw(t,x)+f(t,x,z(t,x)ϵ1,xzϵ1(t,x))-f(t,x,z(t,x)ϵ2,xzϵ2(t,x))
=tw(t,x)+01uf(P(t,x,s))w(t,x)𝑑s+i=1n01vif(P(t,x,s))𝑑sxiw(t,x)
=(zϵ1)(t,x)-(zϵ2)(t,x).

Let gC be a given function on

𝒦τ={(t,x):0tτ,xρ0-Rt}

such that g(t,x)=0 on the boundary of 𝒦τ. After multiplying the above equation by g(t,x) and integrating by parts in 𝒦τ, we obtain

𝒦τtw(t,x)g(t,x)𝑑t𝑑x+𝒦τg(t,x)01uf(P(t,x,s))w(t,x)𝑑s𝑑t𝑑x
+𝒦τg(t,x)i=1n01vif(P(t,x,s))𝑑sxiw(t,x)𝑑t𝑑x
=Bτw(τ,x)g(τ,x)𝑑x-𝒦τw(t,x)A(g(t,x))𝑑t𝑑x
+t=τi=1ng(t,x)w(t,x)01vif(P(t,x,s))𝑑s𝑑t+𝒦τg(t,x)01uf(P(t,x,s))w(t,x)𝑑s𝑑t𝑑x,

where

A(g(t,x))=tg(t,x)+i=1nxi[g(t,x)01vif(P(t,x,s))𝑑s].

On the other hand,

𝒦τg(t,x)((zϵ1)(t,x)-(zϵ2)(t,x))𝑑t𝑑x
max𝒦τ|g(t,x)|[2ϵ1K2ρ0R+RmesSρ0maxSρ0|xzϵ1(0,x)-xzϵ2(0,x)|
   +R2mesSρ0maxSρ0|zϵ1(0,x)-zϵ2(0,x)|+2ϵ2K2ρ0R]
=max𝒦τ|g(t,x)|C[ϵ1+ϵ2+δ],

where Sρ0 is defined by (2.1),

C=max{2K2ρ0R,RmesSρ0,R2mesSρ0}

and δ is defined by (3.1). Hence, it follows that

Bτw(τ,x)g(τ,x)𝑑x-𝒦τw(t,x)A(g(t,x))𝑑t𝑑x+t=τi=1ng(t,x)w(t,x)01vif(P(t,x,s))𝑑s𝑑t
(A.1)+𝒦τg(t,x)01uf(P(t,x,s))w(t,x)dsdtdxmax𝒦τ|g(t,x)|C[ϵ1+ϵ2+δ].

From the Riesz representation theorem, we have

(A.2)uf(P(t,x,s))w(t,x)=((t,x)+B)𝒦tw(s,y)𝑑Nf(s,y),

where Nf is a measure in the Stieltjes integral. Then, using (A.2), we can modify

𝒦τg(t,x)01uf(P(t,x,s))w(t,x)𝑑s𝑑t𝑑x

in the following way:

𝒦τg(t,x)01uf(P(t,x,s))w(t,x)𝑑s𝑑x𝑑t=𝒦τg(t,x)×nχ((t,x)+B)𝒦t(s,y)w(s,y)𝑑Nf(s,y)𝑑x𝑑t
=×nw(s,y)(𝒦τχ((t,x)+B)𝒦t(s,y)g(t,x)𝑑x𝑑t)𝑑Nf(s,y)
=×nw(t,x)(𝒦τχ((s,y)+B)𝒦t(t,x)g(s,y)𝑑y𝑑s)𝑑Nf(t,x).

Now we define a function g(t,x) satisfying the above assumptions. Let us take ηαC(+,[0,1]) such that ηα(r)=0 for rρ0+Rτ and ηα(r)=1 for 0rρ0-Rτ-α. In formula (A.1), we put g(t,x) as a solution to the problem

(A.3)A(g(t,x))=𝒦τχ((s,y)+B)𝒦t(t,x)g(s,y)𝑑y𝑑sMf(t,x)

on 𝒦τ{t=τ}, where Mf(t,x) is the density of Nf with respect to the Lebesgue measure, and

(A.4)g(τ,x)=ηα(x)zm(x)on Bτ,

where zmC(Bτ), |zm(x)|1, and

limmzm(x)=sgn(zϵ1(τ,x)-zϵ2(τ,x))

in L1(Bt). At the upper and bottom boundary of 𝒦τ, there exists a unique zero solution to problem (A.3), (A.4), a.e. g(t,x)=0 for x=ρ0-Rt and t<τ.

Since the matrix vivjf is positive definite, we can find an orthogonal system of coordinates y1,,yn such that

H(t,x,s)=i,j=1nvivjf(P(t,x,s))(sxixjzϵ1(t,x)+(1-s)xixjzϵ2(t,x))
+i=1nviuf(P(t,x,s))(sxjz(t,x)ϵ1+(1-s)xjz(t,x)ϵ2)+i,j=1nvixjf(P(t,x,s))
=i=1nλi(syiyizϵ1(t,x)+(1-s)yiyizϵ2(t,x))
+i=1nviuf(P(t,x,s))(sxjz(t,x)ϵ1+(1-s)xjz(t,x)ϵ2)+i,j=1nvixjf(P(t,x,s))

for 0<λi<μ where μ is the upper boundary of eigenvalues of the matrix of the partial derivative fvivj. From the assumptions on the Δ2zϵ, we have that yiyizϵ1(t,x)K1 and yiyizϵ2(t,x)K1 on 𝒦τ. Then

01H(t,x,s)𝑑snμK1+R1+R2K¯1n,

where the constants are defined by Assumption 3.2. For the function g(t,x) defined as a solution to problem (A.3), (A.4), we get |g(t,x)|C, where C is the constant independent on m and α. For m in inequality (A.1) and α0, we have that g(t,x)1 on Bτ and

A(g(t,x))-𝒦τχ((s,y)+B)𝒦t(t,x)g(s,y)𝑑y𝑑sMf(t,x)0.

The proof of Lemma 3.3 is complete. ∎

Proof of Lemma 4.3.

This lemma is proved by induction. For k=0, the Lipschitz condition is given by Assumption 3.1. Assume that the thesis of Lemma 4.3 is satisfied for k>0. Then

z(k)(x+he)-z(k)(x-he)2hK0n.

The sequence of successive approximations (4.1) satisfies

z(k+1)(x)-z(k+1)(y)=12ni=1n[z(k)(x+hei)-z(k)(y+hei)]+12ni=1n[z(k)(x-hei)-z(k)(y-hei)]
(A.5)-τ[f(k)(z(k)(x+he)-z(k)(x-he)2h)-f(k)(z(k)(y+he)-z(k)(y-he)2h)].

From the mean value theorem, there exists an intermediate point (kτ,x~,u~,v~) such that

f(k)(z(k)(x+he)-z(k)(x-he)2h)-f(k)(z(k)(y+he)-z(k)(y-he)2h)
=i=1nxif(kτ,x~,u~,v~)(x-y)+uf(kτ,x~,u~,v~)(z(kτ,x)(k)-z(kτ,y)(k))
   +i=1nvif(kτ,x~,u~,v~)(z(k)(x+hei)-z(k)(x-hei)2h-z(k)(y+hei)-z(k)(y-hei)2h).

Now we calculate the absolute value of the (k+1)-th difference of the given functions. We assume that |z(k)(x)-z(k)(y)|Kk. Then from (A.5) we have

|z(k+1)(x)-z(k+1)(y)|12ni=1n[Kkx-y+Kkx-y]+(τR2Kk+τR1)x-y
=(Kk(1+τR2)+τR1)x-y=Kk+1x-y,

where Kk+1 is a solution to the equation yk+1=(1+τR2)yk+τR1, k>0 and y0=K0. Then the solution to the problem is

Kk=K0(1+τR2)k+τR1(1+τR2)k-11+τR2-1.

It is obvious that for every 0kN, KkKN. Moreover, the following estimate holds:

KNK0eNτR2+R1R2(eNτR2-1)K0eρ0RR2+R1R2eρ0RR2=K~0.

This completes the proof of Lemma 4.3. ∎

The next proof is also carried out by induction.

Proof of Lemma 4.4.

For k=0, the proof follows from the assumptions of the lemma. Assume that the inequality is fulfilled for k. Then, from (4.1), for k+1 we have

z(k+1)(x)=12ni=1n[z(k)(x+hei)+z(k)(x-hei)]
-τ2ni=1nvif(kτ,x,(Thz(k))(kτ,x),v)[z(k)(x+hei)-z(k)(x-hei)]+τf(kτ,x,(Thz(k))(kτ,x),0).

Making some estimations on the given functions we have that

|z(k+1)(x)|12ni=1n|z(k)(x+hei)(1-τnhvif(kτ,x,(Thz(k))(kτ,x),v))
+z(k)(x-hei)(1+τnhvif(kτ,x,(Thz(k))(kτ,x),v))|+τ|f(kτ,x,(Thz(k))(kτ,x),0)|
Mk+τ|f(kτ,x,(Thz(k))(kτ,x),0)|.

We will need the following remark for Lemma 4.5.

Remark A.1.

The function f() is convex, so we have

f(t,x+,u+,v+)-2f(t,x,u,v)+f(t,x-,u-,v-)
=xf(t,x~,u~,v~)(x+-2x+x-)+uf(t,x~,u~,v~)(u+-2u+u-)+vf(t,x~,u~,v~)(v+-2v+v-)
xf(t,x~,u~,v~)(x+-2x+x-)+vf(t,x~,u~,v~)(v+-2v+v-)-R2|u+-2u+u-|.

Proof of Lemma 4.5.

From Assumption 3.2 on the given functions, N0<K1. From Remark A.1, we have

Δ2z(k+1)(x,Δx)=z(k+1)(x+Δx)-2z(k+1)(x)+z(k+1)(x-Δx)
=12ni=1n[Δ2z(k)(x+hei)+Δ2z(k)(x-hei)]
-τ[f(τk,x+Δx,(Thz(k))(kτ,x+Δx),z(k)(x+Δx+he)-z(k)(x+Δx-he)2h)
-2f(τk,x,(Thz(k))(kτ,x),z(k)(x+he)-z(k)(x-he)2h)
+f(τk,x-Δx,(Thz(k))(kτ,x-Δx),z(k)(x-Δx+he)-z(k)(x-Δx-he)2h)]
12ni=1n[Δ2z(k)(x-hei)(1-nτhvf(t,x,u,v))+Δ2z(k)(x+hei)(1+nτhvf(t,x,u,v))]
+τR2|Δ2z(k)(x)|Nk(1+τR2)Δx2.

This gives

Nk+1|Δx|2Δ2z(k+1)(x,Δx)Nk(1+τR2)Δx2.

The assertion follows. ∎

Proof of Lemma 4.7.

From (4.1) we conclude that

tzϵ(t,x)+f(z(k)(x+hei)-z(k)(x-hei)2h)=h22nτi=1nΔ2z(k)(x,hei)h2=ϵωk(x)

for kτ<t<(k+1)τ. Take the cube Q={x:|xi|ρ0}. Using inequality (4.4) and taking (4.2) into account, we find that if kτt(k+1)τ, then

Bt|Δ2z(k)(x,hei)|h2𝑑xQ|Δ2z(k)(x,hei)|h2𝑑xidxdxiC,

whence ϵ0ωk(x)ϵC. Since for almost all xn,

|f(z(k)(x+hei)-z(k)(x-hei)2h)-f(t,x,(Thz(k))(kτ,x),xz(k)(x))|
Ci=1n|z(k)(x+hei)-z(k)(x-hei)2h-xiz(k)(x)|

and for kτt(k+1)τ,

|f(t,x,(Thz(k))(kτ,x),xz(k)(x))-f(t,x,z(kτ,x)ϵ,xzϵ(t,x))|Cτ,

from (4.3) and (4.2), we obtain

f(z(k)(x+hei)-z(k)(x-hei)2h)-f(t,x,z(kτ,x)ϵ,xzϵ(t,x))L1(Bt)C(h+τ).

Applying the above estimate to (4.4) and taking into account that h+τϵC with the condition on the steps of the method (nAhτ), we see that

(zϵ)L1(Bt)ϵC

for t[0,ρ0R], tkτ. ∎

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Received: 2016-12-12
Revised: 2018-04-26
Accepted: 2018-10-18
Published Online: 2019-05-10
Published in Print: 2020-12-01

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