In this paper, we investigate a final boundary value problem for a class of fractional with parameter β pseudo-parabolic partial differential equations with nonlinear reaction term. For 0<β<1, the solution is regularity-loss, we establish the well-posedness of solutions. In the case that β>1, it has a feature of regularity-gain. Then, the instability of a mild solution is proved. We introduce two methods to regularize the problem. With the help of the modified Lavrentiev regularization method and Fourier truncated regularization method, we propose the regularized solutions in the cases of globally or locally Lipschitzian source term. Moreover, the error estimates is established.
Citation: Vo Van Au, Hossein Jafari, Zakia Hammouch, Nguyen Huy Tuan. On a final value problem for a nonlinear fractional pseudo-parabolic equation[J]. Electronic Research Archive, 2021, 29(1): 1709-1734. doi: 10.3934/era.2020088
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In this paper, we investigate a final boundary value problem for a class of fractional with parameter β pseudo-parabolic partial differential equations with nonlinear reaction term. For 0<β<1, the solution is regularity-loss, we establish the well-posedness of solutions. In the case that β>1, it has a feature of regularity-gain. Then, the instability of a mild solution is proved. We introduce two methods to regularize the problem. With the help of the modified Lavrentiev regularization method and Fourier truncated regularization method, we propose the regularized solutions in the cases of globally or locally Lipschitzian source term. Moreover, the error estimates is established.
We consider the final value problem:
{ut−mΔut+(−Δ)βu=f(t,x;u),in(0,T]×Ω,u(t,x)=0,on(0,T]×∂Ω,u(T,x)=φ(x),inΩ, | (PT) |
where
For
ut−mΔut−Δu=f(t,x;u),in(0,T]×Ω, | (1.1) |
with conditions
For
ut−mΔut+(−Δ)βu=up+1,inR+×Ω, | (1.2) |
supplemented with initial condition
As mention above, initial value problems of nonlinear pseudo-parabolic equations have been considered in many papers see [1,4,5,6,7,8,9,10,11,12,13,14,15,19,21,22,23,24,32,35,36,37]. However, there are not many results devoted to Problem (
The solution of Problem (
The plan of the paper is as follows. Section 2 contains notations and formulation of a solution of Problem (
Let us recall that the spectral problem
{(−Δ)βej(x)=λβjej(x),inΩ,β>0,ej(x)=0,on∂Ω, | (2.1) |
admits a family of eigenvalues
0<λ1≤λ2≤λ3≤...≤λj≤...↗∞. |
The notation
‖w‖Lq(0,T;B)=(∫T0‖w(t)‖qBdt)1q,for1≤q<∞, |
‖w‖L∞(0,T;B)=esssupt∈(0,T)‖w(t)‖B,forq=∞. |
The norm of the function space
‖w‖Ck([0,T];B)=k∑i=0supt∈[0,T]‖w(i)(t)‖B<∞. |
For any
Hν(Ω)={w∈L2(Ω):∞∑j=1λ2νj⟨w,ej⟩2L2(Ω)<∞}, |
where
‖w‖Hν(Ω)=(∞∑j=1λ2νj⟨w,ej⟩2L2(Ω))12. |
The Gevrey of order
Gβ(η1,η2)(Ω):={w∈L2(Ω):∞∑j=1λη1βjexp(η2λβj)⟨w,ej⟩2L2(Ω)<∞}, |
and its norm given by
‖w‖Gβ(η1,η2)(Ω)=(∞∑j=1λη1βjexp(η2λβj)⟨w,ej⟩2L2(Ω))12. |
Next, we give the formulation of solution of the problem (
Now, assume that Problem (
(1+mλj)u′j(t)+λβjuj(t)=fj(u)(t), |
where
uj(t)=exp((T−t)λβj1+mλj)φj−11+mλj∫Ttexp((τ−t)λβj1+mλj)fj(u)(τ)dτ, |
where
Definition 2.1 (Mild solution of Problem (
u(t,x)=∞∑j=1[exp((T−t)λβj1+mλj)φj−11+mλj∫Ttexp((τ−t)λβj1+mλj)fj(u)(τ)dτ]ej(x). | (2.2) |
for all
Now we introduce the main results in this paper.
In this section, we prove that the Problem (
We will make the following assumptions:
‖f(t,⋅;w1)−f(t,⋅;w2)‖L2(Ω)≤K‖w1−w2‖L2(Ω), | (3.1) |
with
‖f(t,⋅;w)‖L2(Ω)≤K‖w‖L2(Ω),w∈C([0,T];L2(Ω)). | (3.2) |
Theorem 3.1 (
Proof. We prove the existence of the solution
H(w)(t,x)=∞∑j=1[exp((T−t)λβj1+mλj)φj−11+mλj∫Ttexp((τ−t)λβj1+mλj)fj(w)(τ)dτ]ej(x), | (3.3) |
and we aim to show that
‖Hk(w1)(t,⋅)−Hk(w2)(t,⋅)‖L2(Ω)≤(Kmλ1exp(Tmλ1)(T−t))kk!‖w1−w2‖L∞(0,T;L2(Ω)). | (3.4) |
We will prove (3.4) by induction. For
‖H(w1)(t,⋅)−H(w2)(t,⋅)‖L2(Ω)≤∫Tt‖∞∑j=111+mλjexp((τ−t)λβj1+mλj)(fj(w1)(τ)−fj(w2)(τ))ej‖L2(Ω)dτ≤∫Tt√∞∑j=11(1+mλj)2exp(2(τ−t)λβj1+mλj)|fj(w1)(τ)−fj(w2)(τ)|2dτ≤1mλ1exp(Tmλ1)∫Tt‖f(τ,⋅,w1)−f(τ,⋅,w2)‖L2(Ω)dτ≤Kmλ1exp(Tmλ1)∫Tt‖w1(τ,⋅)−w2(τ,⋅)‖L2(Ω)dτ≤Kmλ1exp(Tmλ1)(T−t)‖w1−w2‖L∞(0,T;L2(Ω)). | (3.5) |
Assume now that (3.4) is satisfied for
‖Hk0+1(w1)(t,⋅)−Hk0+1(w2)(t,⋅)‖L2(Ω)≤Kmλ1exp(Tmλ1)∫Tt‖Hk0(w1)(τ,⋅)−Hk0(w2)(τ,⋅)‖L2(Ω)dτ≤1k0!(Kmλ1exp(Tmλ1))k0+1‖w1−w2‖L∞(0,T;L2(Ω))∫Tt(T−τ)k0dτ≤1(k0+1)!(Kmλ1exp(Tmλ1))k0+1(T−t)k0+1‖w1−w2‖L∞(0,T;L2(Ω))≤1(k0+1)!(KTmλ1exp(Tmλ1))k0+1‖w1−w2‖L∞(0,T;L2(Ω)). |
Therefore, by the induction principle we get (3.4). Since the right hand side of (3.5) is independent of
‖Hk(w1)−Hk(w2)‖L∞(0,T;L2(Ω))≤1k!(KTmλ1exp(Tmλ1))k‖w1−w2‖L∞(0,T;L2(Ω)). |
When
0<1k1!(KTmλ1exp(Tmλ1))k1<1. |
We claim that the mapping
Hk1(H(u))=H(Hk1(u))=H(u). |
Due to the uniqueness of the fixed point of
Theorem 3.2 (Regularity). Let
a) If
‖u‖L∞(0,T;L2(Ω))≤C(T,m,λ1)(‖φ‖L2(Ω)+‖f(0)‖L1(0,T;L2(Ω))mλ1). | (3.6) |
b) If
‖u‖L∞(0,T;Hβ(Ω))≤C(T,m,λ1,β)(‖φ‖Hβ(Ω)+T‖f(0)‖L∞(0,T;L2(Ω))mλ1−β1). | (3.7) |
Here, we recall
Proof. First, we set
M1(φ)(t,x):=∞∑j=1exp((T−t)λβj1+mλj)φjej(x),M2(u)(t,x):=−∞∑j=1[11+mλj∫Ttexp((τ−t)λβj1+mλj)fj(u)(τ)dτ]ej(x). |
a) First, using the Parseval's relation, we infer that
‖M1(φ)(t,⋅)‖L2(Ω)=(∞∑j=1exp(2(T−t)λβj1+mλj)φ2j)1/2≤exp(Tmλ1−β1)‖φ‖L2(Ω), | (3.8) |
where for
‖M2(u)(t,⋅)‖L2(Ω)≤∫Tt‖∞∑j=111+mλjexp((τ−t)λβj1+mλj)fj(u)(τ)ej‖L2(Ω)dτ |
=∫Tt(∞∑j=11(1+mλj)2exp(2(τ−t)λβj1+mλj)|fj(u)(τ)|2)1/2dτ≤1mλ1exp(Tmλ1−β1)∫Tt‖f(τ,⋅;u)‖L2(Ω)dτ, | (3.9) |
where we have used
exp(2(τ−t)λβj1+mλj)≤exp(Tmλ1−β1),0≤t≤τ≤T. |
Using (3.1), we obtain
‖f(t,⋅;u)−f(t,⋅;0)‖L2(Ω)≤K‖u(t,⋅)‖L2(Ω), |
then
‖f(t,⋅;u)‖L2(Ω)≤K‖u(t,⋅)‖L2(Ω)+‖f(t,⋅;0)‖L2(Ω), | (3.10) |
Combining (3.10) with (3.9), leads to
‖M2(u)(t,⋅)‖L2(Ω)≤1mλ1exp(Tmλ1−β1)‖f(0)‖L1(0,T;L2(Ω))+Kmλ1exp(Tmλ1−β1)∫Tt‖u(τ,⋅)‖L2(Ω)dτ. | (3.11) |
From (3.8) and (3.11) yields
‖u(t,⋅)‖L2(Ω)≤exp(Tmλ1−β1)(‖φ‖L2(Ω)+1mλ1‖f(0)‖L1(0,T;L2(Ω)))+Kmλ1exp(Tmλ1−β1)∫Tt‖u(τ,⋅)‖L2(Ω)dτ. |
Thanks to Grönall's inequality
‖u(t,⋅)‖L2(Ω)≤exp(Tmλ1−β1)(‖φ‖L2(Ω)+1mλ1‖f(0)‖L1(0,T;L2(Ω)))×exp(Kmλ1exp(Tmλ1−β1)(T−t)), |
this imlpies (3.6).
b) We observe that for
‖M1(φ)(t,⋅)‖Hβ(Ω)=(∞∑j=1λ2βjexp(2(T−t)λβj1+mλj)φ2j)1/2≤exp(Tmλ1−β1)‖φ‖Hβ(Ω). | (3.12) |
We conclude that
‖M2(u)(t,⋅)‖Hβ(Ω)≤∫Tt‖∞∑j=111+mλjexp((τ−t)λβj1+mλj)fj(u)(τ)ej‖Hβ(Ω)dτ=∫Tt√∞∑j=1λ2βj(1+mλj)2exp(2(τ−t)λβj1+mλj)|fj(u)(τ)|2dτ=∫Tt√∞∑j=11m2λ2−2βjexp(2(τ−t)λβj1+mλj)|fj(u)(τ)|2dτ=1mλ1−β1exp(Tmλ1−β1)∫Tt√∞∑j=1|fj(u)(τ)|2dτ≤1mλ1−β1exp(Tmλ1−β1)∫Tt‖f(τ,⋅;u)‖L2(Ω)dτ. | (3.13) |
Inequality (3.10) associated with (3.13) leads to
‖M2(u)(t,⋅)‖Hβ(Ω)≤∫Tt1mλ1−β1exp(Tmλ1−β1)(K‖u(τ,⋅)‖L2(Ω)+‖f(τ,⋅;0)‖L2(Ω))dτ≤Tmλ1−β1exp(Tmλ1−β1)‖f(0)‖L∞(0,T;L2(Ω))+Kmλ1−β1exp(Tmλ1−β1)∫Tt‖u(τ,⋅)‖L2(Ω)dτ. | (3.14) |
Estimates (3.12) and (3.14) lead to
‖u(t,⋅)‖Hβ(Ω)≤‖∞∑j=1exp((T−t)λβj1+mλj)φjej‖Hβ(Ω)+∫Tt‖∞∑j=111+mλjexp((τ−t)λβj1+mλj)fj(u)(τ)ej‖Hβ(Ω)dτ≤exp(Tmλ1−β1)(‖φ‖Hβ(Ω)+Tmλ1−β1‖f(0)‖L∞(0,T;L2(Ω)))+Kmλ1−2β1exp(Tmλ1−β1)∫Tt‖u(τ,⋅)‖Hβ(Ω)dτ, | (3.15) |
where we have used that
‖u(t,⋅)‖Hβ(Ω)≤(‖φ‖Hβ(Ω)+Tmλ1−β1‖f(0)‖L∞(0,T;L2(Ω))) |
×exp(KTmλ1−2β1exp(Tmλ1−β1)+Tmλ1−β1), |
we implies (3.7). The proof is complete.
Definition 4.1 (Ill-posed). The well-posed problem in the sense of Hadamard is to satisfy the following properties:
ⅰ) a solution exists;
ⅱ) the solution is unique;
ⅲ) the solution's behaviour changes continuously with the initial conditions.
If at least one of the three properties above does not satisfy, the problem is ill-posed.
Next, we give an example which shows that the solution
˜φ(k)(x)=λ−1kek(x),∀k∈N∗, | (4.1) |
˜f(t,x;w)=∞∑j=1λjT−1exp(−Tλβj1+mλj)⟨w(t,⋅),ej⟩L2(Ω)ej(x),m>1, | (4.2) |
for
˜u(k)(t,x)=∞∑j=1[exp((T−t)λβj1+mλj)˜φ(k)j]ej(x)−∞∑j=1[11+mλj∫Ttexp((τ−t)λβj1+mλj)˜fj(u(k))(τ)dτ]ej(x), | (4.3) |
where
˜φ(k)j=⟨˜φ(k),ej⟩L2(Ω), |
and
˜fj(˜u(k))(t)=⟨˜f(t,⋅;˜u(k)),ej⟩L2(Ω). |
Indeed, we consider the function
G(w)(t,x)=∞∑j=1[exp((T−t)λβj1+mλj)˜φ(k)j]ej(x)−∞∑j=1[11+mλj∫Ttexp((τ−t)λβj1+mλj)˜fj(w)(τ)dτ]ej(x). |
Then for any
‖G(w1)(t,⋅)−G(w2)(t,⋅)‖L2(Ω)≤∫Tt‖∞∑j=1[11+mλjexp((τ−t)λβj1+mλj)(˜fj(w1)(τ)−˜fj(w2)(τ))]ej‖L2(Ω)dτ=∫Tt[∞∑j=1(11+mλjexp((τ−t)λβj1+mλj))2λ2jT−2exp(−Tλβj1+mλj)|w1,j(τ)−w2,j(τ)|2]1/2dτ=∫Tt[∞∑j=11m2T2exp((τ−t−T)λβj1+mλj)|w1,j(τ)−w2,j(τ)|2]1/2dτ≤∫Tt1mT‖w1(τ,⋅)−w2(τ,⋅)‖L2(Ω)dτ≤1m‖w1−w2‖C([0,T];L2(Ω)), |
where we denote
This implies that
‖G(w1)−G(w2)‖C([0,T];L2(Ω))≤1m‖w1−w2‖C([0,T];L2(Ω)). |
Hence
‖˜u(k)(t,⋅)‖L2(Ω)≥‖∞∑j=1exp((T−t)λβj1+mλj)˜φ(k)jej‖L2(Ω)⏟=:M3(˜φ(k))(t)−‖∞∑j=1[11+mλj∫Ttexp((τ−t)λβj1+mλj)˜fj(˜u(k))(τ)dτ]ej‖L2(Ω)⏟=:M4(˜u(k))(t). |
It is easy to see that (here, noting that from (4.2), we have
|M4(˜u(k))(t)|=‖G(˜u(k))(t)−G(0)(t)‖L2(Ω)≤1m‖˜u(k)‖C([0,T];L2(Ω)). |
Hence
‖˜u(k)(t,⋅)‖L2(Ω)≥|M3(˜φ(k))(t)|−1m‖˜u(k)‖C([0,T];L2(Ω)). |
This leads to
‖˜u(k)‖C([0,T];L2(Ω))≥mm+1sup0≤t≤T|M3(˜φ(k))(t)|. | (4.4) |
We continue to estimate the right hand side of the latter inequality. Indeed, since
{⟨ek,ej⟩L2(Ω)=1,k=j,⟨ek,ej⟩L2(Ω)=0,k≠j, |
we have
|M3(˜φ(k))(t)|=√∞∑j=1exp(2(T−t)λβj1+mλj)⟨˜φ(k)j,ej⟩2L2(Ω)=√∞∑j=1exp(2(T−t)λβj1+mλj)⟨λ−1kek,ej⟩2L2(Ω)=1λkexp((T−t)λβk1+mλk). |
Since the function
sup0≤t≤T|M3(˜φ(k))(t)|≥sup0≤t≤T(1λkexp((T−t)λβk1+mλk))=exp(Tλβk1+mλk)1λk. | (4.5) |
Combining (4.4) and (4.5) yields
‖˜u(k)‖C([0,T];L2(Ω))≥mm+1exp(Tλβk1+mλk)1λk. |
As
limk↗∞‖˜φ(k)‖L2(Ω)=limk↗∞‖ek‖L2(Ω)λk=limk↗∞1λk=0, |
but
limk↗∞‖˜u(k)‖C([0,T];L2(Ω))≥limk↗∞mm+1exp(Tλβk1+mλk)1λk=∞. |
Thus, it is shown that Problem (
In order to obtain the stable numerical solutions, we propose two regularization methods to solve the Problem (
●
●
In this subsection, the functions
Uεα(t,x)=∞∑j=1[Lj(α,T)exp((T−t)λβj1+mλj)φεj]ej(x)−∞∑j=111+mλj[∫TtLj(α,T)exp((τ−t)λβj1+mλj)fj(Uεα)(τ)dτ]ej(x), | (4.6) |
where we set
Lj(α,T)=exp(−Tλβj1+mλj)αλβj+exp(−Tλβj1+mλj), | (4.7) |
and the coefficient
The following technical lemma plays the key role in our analysis.
Lemma 4.2. For
(a)|Lj(α,T)exp((T−t)λβj1+mλj)|≤αtT−1(Tlog(α−1T))1−tT, | (4.8) |
(b)|Lj(α,T)exp((τ−t)λβj1+mλj)|≤αt−τT(Tlog(α−1T))τ−tT. | (4.9) |
Proof.
|Lj(α,T)exp((T−t)λβj1+mλj)|=exp(−tλβj1+mλj)αλβj+exp(−Tλβj1+mλj)=exp(−tλβj1+mλj)(αλβj+exp(−Tλβj1+mλj))tT(αλβj+exp(−Tλβj1+mλj))1−tT≤1(αλβj+exp(−Tλβj1+mλj))1−tT≤1(αλβj+exp(−Tλβj))1−tT. | (4.10) |
Using the inequality
1ν1ζ+exp(−ζν2)≤ν2ν1log(ν2ν1), | (4.11) |
for
1(αλβj+exp(−Tλβj))1−tT≤(α−1Tlog(α−1T))1−tT. |
Whereupon
|Lj(α,T)exp((T−t)λβj1+mλj)|≤(α−1Tlog(α−1T))1−tT. | (4.12) |
Using (4.12) into (4.10), we obtain (4.8).
With the same argument as in the proof of (4.8), we obtain (4.9). This concludes the proof of the lemma.
A. The well-posedness of the regularized solution (4.6). In this subsection, we will obtain the existence and regularity results for the regularized solution (4.6).
Theorem 4.3 (Existence-uniqueness). Suppose that
Proof. For any
F:C([0,T];L2(Ω))→C([0,T];L2(Ω)) |
by
F(ϑ)(t,x):=∞∑j=1[Lj(α,T)exp((T−t)λβj1+mλj)φεj]ej(x)−∞∑j=1[11+mλj∫TtLj(α,T)exp((τ−t)λβj1+mλj)fj(ϑ)(τ)dτ]ej(x). | (4.13) |
We also define
Fk(ϑ)=F⋯F(F(ϑ))⏟k−times. |
We shall prove by induction, for any couple
‖Fk(ϑ1)(t,⋅)−Fk(ϑ2)(t,⋅)‖H2(Ω)≤(Kα−1Tmlog(α−1T)(T−t))kk!‖ϑ1−ϑ2‖C([0,T];H2(Ω)). | (4.14) |
For
αt−τT(Tlog(α−1T))τ−tT≤α−1Tlog(α−1T),for all0≤t≤τ≤T, |
then we have
‖F(ϑ1)(t,⋅)−F(ϑ2)(t,⋅)‖L2(Ω)≤∫Tt‖∞∑j=1[11+mλjLj(α,T)exp((τ−t)λβj1+mλj)(fj(ϑ1)(τ)−fj(ϑ2)(τ))]ej‖L2(Ω)dτ≤∫Tt[∞∑j=11(1+mλj)2|Lj(α,T)exp((τ−t)λβj1+mλj)|2|fj(ϑ1)(τ)−fj(ϑ2)(τ)|2]1/2dτ≤Kα−1Tmλ1log(α−1T)∫Tt[∞∑j=1|ϑ1,j(τ)−ϑ2,j(τ)|2]1/2dτ≤Kα−1Tmλ1log(α−1T)∫Tt‖ϑ1(τ,⋅)−ϑ2(τ,⋅)‖L2(Ω)dτ≤Kα−1T(T−t)mλ1log(α−1T)‖ϑ1−ϑ2‖C([0,T];L2(Ω)),ϑi,j=⟨ϑi,ej⟩L2(Ω),i=1,2. |
Hence, (4.14) holds for
Assume that (4.14) holds for
‖FN+1(ϑ1)(t,⋅)−FN+1(ϑ2)(t,⋅)‖Hν(Ω)≤Kα−1Tmlog(α−1T)∫Tt‖FN(ϑ1)(τ,⋅)−FN(ϑ2)(τ,⋅)‖L2(Ω)dτ≤(Kα−1Tmlog(α−1T))N+1∫Tt(T−τ)NN!dτ‖ϑ1−ϑ2‖C([0,T];L2(Ω))≤1(N+1)!(Kα−1Tmlog(α−1T)(T−t))N+1‖ϑ1−ϑ2‖C([0,T];L2(Ω)). |
By the induction principle, we deduce that (4.14) holds for all
1k0!(Kα−1Tmlog(α−1T)(T−t))k0<1. |
It means that
Given a constant
‖w‖α,∞=sup0≤t≤T(α−tT‖w(t,⋅)‖L2(Ω)). | (4.15) |
Theorem 4.4 (Regularity). Assume that
a) If
‖Uεα‖α,∞≤α−1Tlog(T)exp(KT2mλ1log(T))‖φε‖L2(Ω). | (4.16) |
b) If
‖Uεα‖L∞(0,T;H1(Ω))≤α−1Tlog(T)exp(KT2α−1mλ1log(T))‖φε‖H1(Ω). | (4.17) |
Proof. We rewrite (4.6) as
Uεα(t,x)=∞∑j=1[Lj(α,T)exp((T−t)λβj1+mλj)φεj]ej(x)⏟=:M5(φε)(t,x)−∞∑j=1[11+mλj∫TtLj(α,T)exp((τ−t)λβj1+mλj)fj(Uεα)(τ)dτ]ej(x)⏟=:M6(Uεα)(t,x). | (4.18) |
a) Using (3.10) (noting that
‖M6(Uεα)(t,⋅)‖L2(Ω)≤∫Tt√∞∑j=1[11+mλjLj(α,T)exp((τ−t)λβj1+mλj)fj(Uεα)(τ)]2dτ≤1mλ1∫Ttαt−τT(Tlog(α−1T))τ−tT‖f(τ,⋅;Uεα)‖L2(Ω)dτ≤KTmλ1log(T)∫Ttαt−τT‖Uεα(τ,⋅)‖L2(Ω)dτ, | (4.19) |
where we have used
The next observation, from Lemma 4.2a), is that
‖M5(φε)(t,⋅)‖L2(Ω)≤∞∑j=1[Lj(α,T)exp((T−t)λβj1+mλj)φεj]2≤αtT−1(Tlog(α−1T))1−tT‖φε‖L2(Ω)≤αtT−1Tlog(T)‖φε‖L2(Ω), | (4.20) |
noting that
Combining (4.18)-(4.20) and multiplying the two sides of of the result obtained by
α−tT‖Uεα(t,⋅)‖L2(Ω)≤‖M5(φε)(t,⋅)‖L2(Ω)+‖M6(Uεα)(t,⋅)‖L2(Ω)≤α−1Tlog(T)‖φε‖L2(Ω)+KTmλ1log(T)∫Ttα−τT‖Uεα(τ,⋅)‖L2(Ω)dτ. |
Using Grönwall's inequality, we get
α−tT‖Uεα(t,⋅)‖L2(Ω)≤α−1Tlog(T)‖φε‖L2(Ω)exp(KT2mλ1log(T)). | (4.21) |
Since the right side of (4.21) does not depend on
‖Uεα‖α,∞≤α−1Tlog(T)exp(KT2mλ1log(T))‖φε‖L2(Ω). |
This leads to (4.16).
b) From Lemma 4.2a), we first observe that
‖M5(φε)(t,⋅)‖H1(Ω)≤∞∑j=1[λjLj(α,T)exp((T−t)λβj1+mλj)φεj]2≤αtT−1(Tlog(α−1T))1−tT‖φε‖H1(Ω)≤α−1Tlog(T)‖φε‖H1(Ω). | (4.22) |
The next observation, using Lemma 4.2b), is that
‖M6(Uεα)(t,⋅)‖H1(Ω)≤∫Tt√∞∑j=1[λj1+mλjLj(α,T)exp((τ−t)λβj1+mλj)fj(Uεα)(τ)]2dτ≤1m∫Ttαt−τT(Tlog(α−1T))τ−tT‖f(τ,⋅;Uεα)‖L2(Ω)dτ≤KTα−1mlog(T)∫Tt‖Uεα(τ,⋅)‖L2(Ω)dτ, | (4.23) |
where we have used
From (4.22) and (4.23), we deduce that
‖Uεα(t,⋅)‖H1(Ω)≤α−1Tlog(T)‖φε‖H1(Ω)+KTα−1mλ1log(T)∫Tt‖Uεα(τ,⋅)‖H1(Ω)dτ, |
where
‖Uεα(t,⋅)‖H1(Ω)≤α−1Tlog(T)‖φε‖H1(Ω)exp(KTα−1mλ1log(T)(T−t)), |
which implies (4.17). This concludes the proof.
B. Error estimate In this subsection, by using MLR method, the error between the exact solution and the regularized solution is obtained. Now, we can formulate the main theorem.
Theorem 4.5 (Error estimate). Let
{limε→0+α=0,limε→0+εα−1=M0<∞. | (4.24) |
Assume that
u∈L∞(0,T;Gβ(η1,η2)(Ω))and‖u‖L∞(0,T;Gβ(η1,η2)(Ω))≤P0, | (4.25) |
for some known positive constants
‖Uεα(t,⋅)−u(t,⋅)‖L2(Ω)≤√2(M0+P0)exp(K2Tm2λ21(T−t))αtT(Tlog(α−1T))1−tT. | (4.26) |
Remark 1. The error estimate in (4.26) is of order
Remark 2. Choose
εrtT[T/log(T/ε−r)]t−TT. |
Proof. For all
‖Uεα(t,⋅)−u(t,⋅)‖L2(Ω)≤‖Uεα(t,⋅)−Uα(t,⋅)‖L2(Ω)⏟=:Mα,ε7(t,x)+‖Uα(t,⋅)−u(t,⋅)‖L2(Ω)⏟=:Mα,ε8(t,x), | (4.27) |
where
Step 1. Estimate
Thanks to Parseval's relation and basic inequality
‖Uεα(t,⋅)−Uα(t,⋅)‖2L2(Ω)≤2∞∑j=1[Lj(α,T)exp((T−t)λβj1+mλj)(φεj−φj)]2⏟=:Mα,ε7a(t)+2∞∑j=1[11+mλj∫TtLj(α,T)exp((τ−t)λβj1+mλj)(fj(Uεα)(τ)−fj(Uα)(τ))dτ]2⏟=:Mα,ε7b(t). | (4.28) |
To estimate
|Mα,ε7a(t)|≤2∞∑j=1α2tT−2(Tlog(α−1T))2−2tT|φεj−φj|2≤2α2tT−2(Tlog(α−1T))2−2tT‖φε−φ‖2L2(Ω)≤2α2tT−2(Tlog(α−1T))2−2tTε2. | (4.29) |
Next, we estimate
|Mα,ε7b(t)|=2∞∑j=1[11+mλj∫TtLj(α,T)exp((τ−t)λβj1+mλj)(fj(Uεα)(τ)−fj(Uα)(τ))dτ]2≤2T∞∑j=11(1+mλj)2∫Tt|Lj(α,T)exp((τ−t)λβj1+mλj)|2|fj(Uεα)(τ)−fj(Uα)(τ)|2dτ≤2Tm2λ21∫Ttα2t−2τT(Tlog(α−1T))2τ−2tT∞∑j=1|fj(Uεα)(τ)−fj(Uα)(τ)|2dτ≤2Tm2λ21∫Ttα2t−2τT(Tlog(α−1T))2τ−2tT‖f(Uεα)(τ)−f(Uα)(τ)‖2L2(Ω)dτ |
≤2K2Tm2λ21∫Ttα2t−2τT(Tlog(α−1T))2τ−2tT‖Uεα(τ,⋅)−Uα(τ,⋅)‖2L2(Ω)dτ. | (4.30) |
Combining the results in (4.28)-(4.30), we get
‖Uεα(t,⋅)−Uα(t,⋅)‖2L2(Ω)≤2α2tT−2(Tlog(α−1T))2−2tTε2+2K2Tm2λ21∫Ttα2t−2τT(Tlog(α−1T))2τ−2tT‖Uεα(τ,⋅)−Uα(τ,⋅)‖2L2(Ω)dτ. |
Multiplying by
α−2tT(Tlog(α−1T))2tT‖Uεα(t,⋅)−Uα(t,⋅)‖2L2(Ω)≤2α−2(Tlog(α−1T))2ε2+2K2Tm2λ21∫Ttα−2τT(Tlog(α−1T))2τT‖Uεα(τ,⋅)−Uα(τ,⋅)‖2L2(Ω)dτ. |
Grönwall's inequality allows to obtain
α−2tT(Tlog(α−1T))2tT‖Uεα(t,⋅)−Uα(t,⋅)‖2L2(Ω)≤2α−2(Tlog(α−1T))2ε2exp(2K2Tm2λ21(T−t)). |
Similar calculations yield
‖Uεα(t,⋅)−Uα(t,⋅)‖L2(Ω)≤√2α−1εexp(K2Tm2λ21(T−t))αtT(Tlog(α−1T))1−tT. | (4.31) |
Step 2. Estimate
Θw(t,x)=∞∑j=1[Lj(α,T)⟨w(t,⋅),ej⟩L2(Ω)]ej(x),forw∈C([0,T];L2(Ω)). |
It clearly follows that
Θu(t,x)=∞∑j=1[Lj(α,T)exp((T−t)λβj1+mλj)φj]ej(x)−∞∑j=1[11+mλj∫TtLj(α,T)exp((τ−t)λβj1+mλj)fj(u)(τ)dτ]ej(x). |
The triangle inequality allows to write
‖Uα(t,⋅)−u(t,⋅)‖2L2(Ω)≤2‖Uα(t,⋅)−Θu(t,⋅)‖2L2(Ω)+2‖Θu(t,⋅)−u(t,⋅)‖2L2(Ω)=:Mα8a(t)+Mα8b(t). |
We estimate for
|Mα8a(t)|=2∞∑j=1[∫Tt11+mλjLj(α,T)exp((τ−t)λβj1+mλj)(fj(Uα)(τ)−fj(u)(τ))dτ]2≤2K2Tm2λ21∫Ttα2t−2τT(Tlog(α−1T))2τ−2tT‖Uα(τ,⋅)−u(τ,⋅)‖2L2(Ω)dτ. |
The term
|Mα8b(t)|=∞∑j=1(Lj(α,T)−1)2|uj(t)|2=2∞∑j=1(exp(−Tλβj1+mλj)αλβj+exp(−Tλβj1+mλj)−1)2|uj(t)|2=2∞∑j=1(αλβjαλβj+exp(−Tλβj1+mλj))2|uj(t)|2≤2α2∞∑j=1(exp(−tλβj1+mλj)αλβj+exp(−Tλβj1+mλj))2λ2βjexp(2Tλβj1+mλj)|uj(t)|2≤2α2∞∑j=1α2tT−2(Tlog(α−1T))2−2tTλ2βjexp(2Tmλ1λβj)|uj(t)|2≤2α2tT(Tlog(α−1T))2−2tT‖u(t,⋅)‖2Gβ(η1,η2)(Ω), |
where we have used
Combining all these inequalities, we deduce
‖Uα(t,⋅)−u(t,⋅)‖2L2(Ω)≤2α2tT(Tlog(α−1T))2−2tT‖u(t,⋅)‖2Gβ(η1,η2)(Ω)+2K2Tm2λ21∫Ttα2t−2τT(Tlog(α−1T))2τ−2tT‖Uα(τ,⋅)−u(τ,⋅)‖2L2(Ω)dτ. |
Multiplying by
α−2tT(Tlog(α−1T))2tT‖Uα(t,⋅)−u(t,⋅)‖2L2(Ω) |
≤2(Tlog(α−1T))2‖u(t,⋅)‖2Gβ(η1,η2)(Ω)+2K2Tm2λ21∫Ttα−2τT(Tlog(α−1T))2τT‖Uα(τ,⋅)−u(τ,⋅)‖2L2(Ω)dτ. |
Using Grönwall's inequality, we thus obtain
‖Uα(t,⋅)−u(t,⋅)‖L2(Ω)≤√2αtT(Tlog(α−1T))1−tT‖u(t,⋅)‖Gβ(η1,η2)(Ω)exp(K2Tm2λ21(T−t)). | (4.32) |
Combining (4.27), (4.31) and (4.32), we complete the proof of Theorem 4.5.
In Subsection 4.2.1 has addressed the Problem (
We begin by establishing the locally Lipschitz properties of
|f(t,x;u)−f(t,x;v)|≤Kϱ|u−v|,ifmax{|u|,|v|}≤ϱ, | (4.33) |
and
Kϱ=sup{|f(t,x;u)−f(t,x;v)u−v|,|u|,|v|≤ϱ,u≠v,(t,x)∈[0,T)×Ω}, |
Example 1. Let
|f1(u)−f1(v)|=|u|u|2−v|v|2|=|(u−v)|u|2+v(|u|2−|v|2)|=(|u|2+|v||u|+|v|2)|u−v|. |
Clearly,
Example 2. Let
fϱ(t,x;u):=f(t,x;˜u),where˜u:={−ϱ,ifu∈(−∞,−ϱ),u,ifu∈[−ϱ,ϱ],ϱ,ifu∈(ϱ,∞). | (4.34) |
With this definition, we claim that
Lemma 4.6. Let
|fϱε(t,x;u)−fϱε(t,x;v)|≤Kϱε|u−v|,∀(t,x)∈[0,T)×Ω,u,v∈R, | (4.35) |
where
Proof. The proof can be found in [2].
Remark 3. For
˜u≡u,almost everywhere in[0,T)×Ω,fϱε(t,x;u)≡f(t,x;˜u),almost everywhere in[0,T)×Ω. |
Based on the above analysis, we propose the regularized solution by using FTR method as follows
VεΛε(t,x)=[Λε]∑j=1[exp((T−t)λβj1+mλj)φεj−11+mλj∫Ttexp((τ−t)λβj1+mλj)fϱε,j(VεΛε)(τ)dτ]ej(x), | (4.36) |
where
The regularity estimates of the solution
Theorem 4.7 (Error estimate). Suppose that we can choose
limε→0+exp((Λε)βmλ1T)ε=N0<∞, | (4.37) |
and let us choose
Kϱε=|log(log(ε−r))mλ1T|1/2⟶∞,asεtends to0+. |
Let
u∈L∞(0,T;Gβ(η1,η2)(Ω)),withη1≥2+2δ,η2≥2Tmλ1, |
and
‖u‖L∞(0,T;Gβ(η1,η2)(Ω))≤~P0,for some known constant~P0≥0. |
Then the following stability estimate holds for any
‖VεΛε(t,⋅)−u(t,⋅)‖L2(Ω)≤√2(~P0(Λε)β(1+δ)+N0)(log(1ε−r))T−tTexp(−t(Λε)βmλ1). | (4.38) |
Remark 4. We can choose
εrtT(log(ε−r))1−tT→0asε→0+,for allt∈[0,T). |
Remark 5. Obviously, from (4.26), one can raise the question about
Proof. We divide the proof into two parts. In Part a, we prove that the integral equation (4.36) has a unique solution
Part a. The existence and uniqueness of a solution of the integral equation (4.36).
For
L(ζ)(t,x)=[Λε]∑j=1[exp((T−t)λβj1+mλj)φεj−11+mλj∫Ttexp((τ−t)λβj1+mλj)fϱ,j(ζ)(τ)dτ]ej(x). | (4.39) |
Let us define
Lk(ζ)=L⋯L(L(ζ))⏟k−times. |
To explore the proof, we only need to prove that there exists
‖Lk(ζ1)−Lk(ζ2)‖C([0,T];L2(Ω)) |
≤[Kϱεexp(T(Λε)βmλ1)(T−t)]kk!‖ζ1−ζ2‖C([0,T];L2(Ω)), |
for
Part b. The error estimate between the exact solution
‖VεΛε(t,⋅)−u(t,⋅)‖[L2(Ω)]2≤‖VεΛε(t,⋅)−WΛε(t,⋅)‖L2(Ω)⏟=:MΛε,ε9(t)+‖WΛε(t,⋅)−u(t,⋅)‖L2(Ω)⏟=:MΛε10(t), | (4.40) |
where
WΛε(t,x)=[Λε]∑j=1[exp((T−t)λβj1+mλj)φj−11+mλj∫Ttexp((τ−t)λβj1+mλj)fϱε,j(WΛε)(τ)dτ]ej(x). |
To estimate (4.40), we divide the right-hand side of (4.40) into two steps
VεΛε(t,x)−WΛε(t,x)=[Λε]∑j=1[exp((T−t)λβj1+mλj)(φεj−φj)]ej(x) |
+[Λε]∑j=1[11+mλj∫Ttexp((τ−t)λβj1+mλj)(fϱε,j(WΛε)(τ)−fϱε,j(VεΛε)(τ))dτ]ej(x). | (4.41) |
Using the Hölder's inequality with Parseval's relation, we obtain
|MΛε,ε9(t)|2=2[Λε]∑j=1[exp((T−t)λβj1+mλj)(φεj−φj)]2+2[Λε]∑j=1[11+mλj∫Ttexp((τ−t)λβj1+mλj)(fϱε,j(WΛε)(τ)−fϱε,j(VεΛε)(τ))dτ]2≤2exp(2(T−t)(Λε)βmλ1)[Λε]∑j=1|φεj−φj|2+2Tm2λ21∫Ttexp(2(τ−t)(Λε)βmλ1)[Λε]∑j=1|fϱε,j(WΛε)(τ)−fϱε,j(VεΛε)(τ)|2dτ≤2exp(2(T−t)(Λε)βmλ1)ε2+2TK2ϱεm2λ21∫Ttexp(2(τ−t)(Λε)βmλ1)‖WΛε(τ,⋅)−VεΛε(τ,⋅)‖2L2(Ω)dτ. | (4.42) |
Multiplying both sides (4.42) by
exp(2t(Λε)βmλ1)‖VεΛε(t,⋅)−WΛε(t,⋅)‖2L2(Ω)≤2exp(2T(Λε)βmλ1)ε2+2TK2ϱεm2λ21∫Ttexp(2τ(Λε)βmλ1)‖WΛε(τ,⋅)−VεΛε(τ,⋅)‖2L2(Ω)dτ. |
Applying Grönwall's inequality yields
exp(2t(Λε)βmλ1)‖VεΛε(t,⋅)−WΛε(t,⋅)‖2L2(Ω)≤2exp(2(Λε)βmλ1T)ε2exp(2TK2ϱεm2λ21(T−t)). |
Consequently,
‖VεΛε(t,⋅)−WΛε(t,⋅)‖L2(Ω)≤√2exp((Λε)βmλ1(T−t))εexp(2TK2ϱεm2λ21(T−t)). | (4.43) |
|MΛε10(t)|2=2∞∑j=[Λε]+1[exp((T−t)λβj1+mλj)φj−11+mλj∫Ttexp((τ−t)λβj1+mλj)fϱε,j(WΛε)(τ)dτ]2 |
+2[Λε]∑j=1[11+mλj∫Ttexp((τ−t)λβj1+mλj)(fj(u)(τ)−fϱε,j(WΛε)(τ))dτ]2=:MΛε10a(t)+MΛε10b(t). | (4.44) |
To estimate
|MΛε10a(t)|=2∞∑j=[Λε]+1exp(−2tλβj1+mλj)[exp(Tλβj1+mλj)φj−11+mλj∫Ttexp(−τλβj1+mλj)fϱε,j(WΛε)(τ)dτ]2≤2(Λε)−2β−2βδexp(−2t(Λε)βmλ1)∞∑j=[Λε]+1λ2(β+βδ)jexp(2Tλβjmλ1)⟨u(t,⋅),ej(⋅)⟩2L2(Ω)≤2(Λε)−2β−2βδexp(−2t(Λε)βmλ1)‖u(t,⋅)‖2Gβ(η1,η2)(Ω),for2≤η1,η2≥2Tmλ1. | (4.45) |
Since
|MΛε10b(t)|=2[Λε]∑j=1[11+mλj∫Ttexp((τ−t)λβj1+mλj)(fϱε,j(u)(τ)−fϱε,j(WΛε)(τ))dτ]2≤2Tm2λ21∫Ttexp(2(τ−t)(Λε)βmλ1)[Λε]∑j=1|fϱε,j(u)(τ)−fϱε,j(WΛε)(τ)|2dτ≤2TK2ϱεm2λ21∫Ttexp(2(τ−t)(Λε)βmλ1)‖WΛε(τ,⋅)−u(τ,⋅)‖2L2(Ω)dτ. | (4.46) |
Combining (4.45) and (4.46), we get
‖WΛε(t,⋅)−u(t,⋅)‖2L2(Ω)≤2(Λε)−2β−2βδexp(−2t(Λε)βmλ1)‖u(t,⋅)‖2Gβ(η1,η2)(Ω)+2TK2ϱεm2λ21∫Ttexp(2(τ−t)(Λε)βmλ1)‖WΛε(τ,⋅)−u(τ,⋅)‖2L2(Ω)dτ. | (4.47) |
Multiplying by
‖WΛε(t,⋅)−u(t,⋅)‖L2(Ω)≤√2exp(−t(Λε)βmλ1)(Λε)β+βδ‖u(t,⋅)‖Gβ(η1,η2)(Ω)exp(TK2ϱεm2λ21(T−t)). | (4.48) |
Combining (4.40), (4.43) and (4.48), we obtain (4.38). This completes the proof of Theorem 4.7.
The paper investigate a final boundary value problem for a class of pseudo-parabolic partial differential equations with nonlinear reaction term. For
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