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Research article Special Issues

Development of a tissue augmented Bayesian model for expression quantitative trait loci analysis

  • Expression quantitative trait loci (eQTL) analyses detect genetic variants (SNPs) associated with RNA expression levels of genes. The conventional eQTL analysis is to perform individual tests for each gene-SNP pair using simple linear regression and to perform the test on each tissue separately ignoring the extensive information known about RNA expression in other tissue(s). Although Bayesian models have been recently developed to improve eQTL prediction on multiple tissues, they are often based on uninformative priors or treat all tissues equally. In this study, we develop a novel tissue augmented Bayesian model for eQTL analysis (TA-eQTL), which takes prior eQTL information from a different tissue into account to better predict eQTL for another tissue. We demonstrate that our modified Bayesian model has comparable performance to several existing methods in terms of sensitivity and specificity using allele-specific expression (ASE) as the gold standard. Furthermore, the tissue augmented Bayesian model improves the power and accuracy for local-eQTL prediction especially when the sample size is small. In summary, TA-eQTL's performance is comparable to existing methods but has additional flexibility to evaluate data from different platforms, can focus prediction on one tissue using only summary statistics from the secondary tissue(s), and provides a closed form solution for estimation.

    Citation: Yonghua Zhuang, Kristen Wade, Laura M. Saba, Katerina Kechris. Development of a tissue augmented Bayesian model for expression quantitative trait loci analysis[J]. Mathematical Biosciences and Engineering, 2020, 17(1): 122-143. doi: 10.3934/mbe.2020007

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  • Expression quantitative trait loci (eQTL) analyses detect genetic variants (SNPs) associated with RNA expression levels of genes. The conventional eQTL analysis is to perform individual tests for each gene-SNP pair using simple linear regression and to perform the test on each tissue separately ignoring the extensive information known about RNA expression in other tissue(s). Although Bayesian models have been recently developed to improve eQTL prediction on multiple tissues, they are often based on uninformative priors or treat all tissues equally. In this study, we develop a novel tissue augmented Bayesian model for eQTL analysis (TA-eQTL), which takes prior eQTL information from a different tissue into account to better predict eQTL for another tissue. We demonstrate that our modified Bayesian model has comparable performance to several existing methods in terms of sensitivity and specificity using allele-specific expression (ASE) as the gold standard. Furthermore, the tissue augmented Bayesian model improves the power and accuracy for local-eQTL prediction especially when the sample size is small. In summary, TA-eQTL's performance is comparable to existing methods but has additional flexibility to evaluate data from different platforms, can focus prediction on one tissue using only summary statistics from the secondary tissue(s), and provides a closed form solution for estimation.


    It is well known that optimal control problems play a very important role in the fields of science and engineering. In the operation of physical and economic processes, optimal control problems have a variety of applications. Therefore, highly effective numerical methods are key to the successful application of the optimal control problem in practice. The finite element method is an important method for solving optimal control problems and has been extensively studied in the literature. Many researchers have made various contributions on this topic. A systematic introduction to the finite element method for partial differential equations (PDEs) and optimal control problems can be found in [1,2]. For example, a priori error estimates of finite element approximation were established for the optimal control problems governed by linear elliptic and parabolic state equations, see [3,4]. Using adaptive finite element method to obtain posterior error estimation; see [5,6]. Furthermore, some superconvergence results have been established by applying recovery techniques, see [7,8].

    The two-grid method based on two finite element spaces on one coarse and one fine grid was first proposed by Xu [9,10,11]. It is combined with other numerical methods to solve many partial differential equations, e.g., nonlinear elliptic problems [12], nonlinear parabolic equations [13], eigenvalue problems [14,15,16] and fractional differential equations [17].

    Many real applications, such as heat conduction control of storage materials, population dynamics control and wave control problems governed by integro-differential equations, need to consider optimal control problems governed by elliptic integral equations and parabolic integro-differential equations. More and more experts and scholars began to pay attention to the numerical simulation of these optimal control problems. In [18], the authors analyzed the finite element method for optimal control problems governed by integral equations and integro-differential equations. In [19], the authors considered the error estimates of expanded mixed methods for optimal control problems governed by hyperbolic integro-differential equations. As far as we know, there is no research on a two-grid finite element method for parabolic integro-differential control problems in the existing literature.

    In this paper, we design a two-grid scheme of fully discrete finite element approximation for optimal control problems governed by parabolic integro-differential equations. It is shown that when the coarse and fine mesh sizes satisfy h=H2, the two-grid method achieves the same convergence property as the finite element method. We are interested in the following optimal control problems:

    minuKU{12T0yyd2+u2dt}, (1.1)
    ytdiv(Ay)+t0div(B(t,s)y(s))ds=f+u,  xΩ, tJ, (1.2)
    y(x,t)=0,  xΩ, tJ, (1.3)
    y(x,0)=y0(x),  xΩ, (1.4)

    where Ω is a bounded domain in R2 and J=(0,T]. Let K be a closed convex set in U=L2(J;L2(Ω)), fL2(J;L2(Ω)), ydH1(J;L2(Ω)) and y0H1(Ω). K is a set defined by

    K={uU:Ωu(x,t)dx0}; (1.5)

    A=A(x)=(aij(x)) is a symmetric matrix function with aij(x)W1,(Ω), which satisfies the ellipticity condition

    a|ξ|22i,j=1ai,j(x)ξiξja|ξ|2, (ξ,x)R2×ˉΩ, 0<a<a.

    Moreover, B(t,s)=B(x,t,s) is also a 2×2 matrix; assume that there exists a positive constant M such that

    B(t,s)0,+Bt(t,s)0,M.

    In this paper, we adopt the standard notation Wm,p(Ω) for Sobolev spaces on Ω with a norm m,p given by vpm,p=|α|mDαvpLp(Ω), as well as a semi-norm ||m,p given by |v|pm,p=|α|=mDαvpLp(Ω). We set Wm,p0(Ω)={vWm,p(Ω):v|Ω=0}. For p=2, we denote Hm(Ω)=Wm,2(Ω), Hm0(Ω)=Wm,20(Ω), and m=m,2, =0,2.

    We denote by Ls(J;Wm,p(Ω)) the Banach space of all Ls integrable functions from J into Wm,p(Ω) with the norm vLs(J;Wm,p(Ω))=(T0||v||sWm,p(Ω)dt)1s for s[1,) and the standard modification for s=. For simplicity of presentation, we denote vLs(J;Wm,p(Ω)) by vLs(Wm,p). Similarly, one can define the spaces H1(J;Wm,p(Ω)) and Ck(J;Wm,p(Ω)). In addition C denotes a general positive constant independent of h and Δt, where h is the spatial mesh size and Δt is a time step.

    The outline of this paper is as follows. In Section 2, we first construct a fully discrete finite element approximation scheme for the optimal control problems (1.1)–(1.4) and give its equivalent optimality conditions. In Section 3, we derive a priori error estimates for all variables, and then analyze the global superconvergence by using the recovery techniques. In Section 4, we present a two-grid scheme and discuss its convergence. In Section 5, we present a numerical example to verify the validity of the two-grid method.

    In this section, we shall construct a fully discrete finite element approximation scheme for the control problems (1.1)–(1.4). For sake of simplicity, we take the state space Q=L2(J;V) and V=H10(Ω).

    We recast (1.1)–(1.4) in the following weak form: find (y,u)Q×K such that

    minuKU{12T0yyd2+u2dt}, (2.1)
    (yt,v)+(Ay,v)=t0(B(t,s)y(s),v)ds+(f+u,v),  vV, tJ, (2.2)
    y(x,0)=y0(x),  xΩ, (2.3)

    where (,) is the inner product of L2(Ω).

    Since the objective functional is convex, it follows from [2] that the optimal control problems (2.1)–(2.3) have a unique solution (y,u), and that (y,u) is the solution of (2.1)–(2.3) if and only if there is a co-state pQ such that (y,p,u) satisfies the following optimality conditions:

    (yt,v)+(Ay,v)=t0(B(t,s)y(s),v)ds+(f+u,v),  vV, tJ, (2.4)
    y(x,0)=y0(x),  xΩ, (2.5)
    (pt,q)+(Ap,q)=Tt(B(s,t)p(s),q)ds+(yyd,q),  qV, tJ, (2.6)
    p(x,T)=0,  xΩ, (2.7)
    (u+p,˜uu)0,  ˜uK, tJ. (2.8)

    As in [20], the inequality (Eq 2.8) can be expressed as

    u=max{0,ˉp}p, (2.9)

    where ˉp=ΩpdxΩdx denotes the integral average on Ω of the function p.

    Let Th denote a regular triangulation of the polygonal domain Ω, hτ denote the diameter of τ and h=maxτThhτ. Let VhV be defined by the following finite element space:

    Vh={vhC0(ˉΩ)V,vh|τP1(τ),  τTh}. (2.10)

    And the approximated space of control is given by

    Uh:={˜uhU: τTh, ˜uh|τ=constant}. (2.11)

    Set Kh=UhK.

    Before the fully discrete finite element scheme is given, we introduce some projection operators. First, we define the Ritz-Volterra projection [21] Rh: VVh, which satisfies the following: for any y,pV

    (A((yRhy),vh)t0(B(t,s)(yRhy),vh)ds=0,   vhVh, (2.12)
    i(yRhy)ti+hi(yRhy)tiCh2im=0mytm2, i=0,1. (2.13)
    (A((pRhp),vh)Tt(B(s,t)(pRhp),vh)ds=0,   vhVh, (2.14)
    i(pRhp)ti+hi(pRhp)tiCh2im=0mptm2, i=0,1. (2.15)

    Next, we define the standard L2-orthogonal projection [22] Qh: L2(Ω)Uh, which satisfies the following: for any ϕL2(Ω)

    (ϕQhϕ,wh)=0,   whUh, (2.16)
    ϕQhϕs,2Ch1+su1, s=0,1,  ϕH1(Ω), (2.17)

    At last, we define the element average operator [7] πh:L2(Ω)Uh by

    πhψ|τ=τψdxτdx,  ψL2(Ω), τTh. (2.18)

    We have the approximation property

    ψπhψs,rCh1+sψ1,r, s=0,1,  ψW1,r(Ω). (2.19)

    We now consider the fully discrete finite element approximation for the control problem. Let Δt>0, N=T/ΔtZ and tn=nΔt, nZ. Also, let

    ψn=ψn(x)=ψ(x,tn),dtψn=ψnψn1Δt,δψn=ψnψn1.

    Like in [23], we define for 1s and s=, the discrete time dependent norms

    |||ψ|||Ls(J;Wm,p(Ω)):=(Nln=1lΔtψnsm,p)1s, |||ψ|||L(J;Wm,p(Ω)):=max1lnNlψnm,p,

    where l=0 for the control variable u and the state variable y, and l=1 for the co-state variable p.

    Then the fully discrete approximation scheme is to find (ynh,unh)Vh×Kh, n=1,2,,N, such that

    minunhKh{12Nn=1Δt(ynhynd2+unh2)}, (2.20)
    (dtynh,vh)+(Aynh,vh)=(ni=1ΔtB(tn,ti1)yih,vh)+(fn+unh,vh),  vhVh, (2.21)
    y0h=Rhy0. (2.22)

    Again, we can see that the above optimal control problem has a unique solution (ynh,unh), and that (ynh,unh)Vh×Kh is the solution of (2.20)–(2.22) if and only if there is a co-state pn1hVh such that (ynh,pn1h,unh) satisfies the following optimality conditions:

    (dtynh,vh)+(Aynh,vh)=(ni=1ΔtB(tn,ti1)yih,vh)+(fn+unh,vh),  vhVh, (2.23)
    y0h=Rhy0, (2.24)
    (dtpnh,qh)+(Apn1h,qh)=(Ni=nΔtB(ti,tn1)pi1h,qh)+(ynhynd,qh),  qhVh, (2.25)
    pNh=0, (2.26)
    (unh+pn1h,˜uhunh)0,  ˜uhKh. (2.27)

    Similarly, employing the projection (2.9), the optimal condition (2.27) can be rewritten as follows:

    unh=max{0,¯pn1h}πhpn1h, (2.28)

    where ¯pn1h=Ωpn1hΩ1.

    In the rest of the paper, we shall use some intermediate variables. For any control function ˜uK satisfies the following:

    (dtynh(˜u),vh)+(Aynh(˜u),vh)=(ni=1ΔtB(tn,ti1)yih(˜u),vh)+(fn+˜un,vh),  vhVh, (2.29)
    y0h(˜u)=Rhy0, (2.30)
    (dtpnh(˜u),qh)+(Apn1h(˜u),qh)=(Ni=nΔtB(ti,tn1)pi1h(˜u),qh)+(ynh(˜u)ynd,qh),  qhVh, (2.31)
    pNh(˜u)=0. (2.32)

    In this section, we will discuss a priori error estimates and superconvergence of the fully discrete case for the state variable, the co-state variable and the control variable. In order to do it, we need the following lemmas.

    Lemma 3.1. Let (ynh(u),pn1h(u)) be the solution of (2.29)–(2.32) with ˜u=u and (y,p) be the solution of (2.4)–(2.8). Assume that the exact solution (y,p) has enough regularities for our purpose. Then, for Δt small enough and 1nN, we have

    |||yyh(u)|||L(L2)+|||pph(u)|||L(L2)C(Δt+h2), (3.1)
    |||(yyh(u))|||L(L2)+|||(pph(u))|||L(L2)C(Δt+h). (3.2)

    Proof. For convenience, let

    ynynh(u)=ynRhyn+Rhynynh(u)=:ηny+ξny,pnpnh(u)=pnRhpn+Rhpnpnh(u)=:ηnp+ξnp.

    Taking t=tn in (2.4), subtracting (2.29) from (2.4) and then using (2.12), we have

    (dtξny,vh)+(Aξny,vh)=(dtynynt,vh)(dtηny,vh)+[tn0(B(tn,s)Rhy(s),vh)ds(ni=1ΔtB(tn,ti1)yih(u),vh)]. (3.3)

    Choosing vh=dtξny in (3.3), we get

    (dtξny,dtξny)+(Aξny,dtξny)=(dtynynt,dtξny)(dtηny,dtξny)+[tn0(B(tn,s)Rhy(s),dtξny)ds(ni=1ΔtB(tn,ti1)yih(u),dtξny)]. (3.4)

    Notice that

    (dtξny,Aξny)12Δt(A12ξny2A12ξn1y2). (3.5)

    Multiplying Δt and summing over n from 1 to l (1lN) on both sides of (3.4), and by using (3.5) and ξ0y=0, we find that

    12A12ξly2+ln=1dtξny2Δtln=1(dtynynt,dtξny)Δtln=1(dtηny,dtξny)Δt+ln=1[tn0(B(tn,s)Rhy(s),dtξny)ds(ni=1ΔtB(tn,ti1)yih(u),dtξny)]Δt=:3i=1Ai. (3.6)

    Now, we estimate the right-hand terms of (3.6). For A1, from the results given in [24], we have

    A1Cln=1(tntn1yttdt)2Δt+ln=1dtξny2ΔtC(Δt)2tl0ytt2dt+14ln=1dtξny2ΔtC(Δt)2ytt2L2(L2)+14ln=1dtξny2Δt. (3.7)

    For A2, using (2.13), the Hölder inequality and the Cauchy inequality, we have

    A2Cln=1ηnyηn1yΔt2Δt+14ln=1dtξny2ΔtCln=11Δttntn1(ηy)tdt2+14ln=1dtξny2ΔtCln=11Δt((tntn1(ηy)t2dt)12(tntn112dt)12)2+14ln=1dtξny2ΔtCh4tl0yt22dt+14ln=1dtξny2ΔtCh4yt2L2(H2)+14ln=1dtξny2Δt. (3.8)

    At last, for A3, it follows from the Cauchy inequality, Cauchy mean value theorem and assumptions on A and B that

    A3=ln=1[tn0(B(tn,s)Rhy(s),dtξny)ds(ni=1ΔtB(tn,ti)yih(u),dtξny)+(ni=1ΔtB(tn,ti)yih(u),dtξny)(ni=1ΔtB(tn,ti1)yih(u),dtξny)]ΔtC(Δt)2(Rhyt2L2(L2)+Rhy2L2(L2))+Cln=1ξny2Δt+Cln=1Δtni=1ξiy2Δt+a4ξly2, (3.9)

    where

    ln=1[tn0(B(tn,s)Rhy(s),dtξny)ds(ni=1ΔtB(tn,ti)yih(u),dtξny)]Δt=(tl0B(tl,s)Rhy(s)dsli=1B(tl,ti)RhyiΔt,ξly)+li=1(ΔtB(tl,ti)ξiy,ξly)+l1n=1(tn0(B(tn,s)B(tn+1,s))Rhyds,ξny)l1n=1(tn+1tnB(tn+1,s)(RhyRhyn+1)ds,ξny)l1n=1(ni=1Δt(B(tn,ti)B(tn+1,ti))Rhyi,ξny)l1n=1(ΔtB(tn+1,tn+1)ξn+1y,ξny)
    +l1n=1(ni=1Δt(B(tn,ti)B(tn+1,ti))ξiy,ξny)=(tl0B(tl,s)Rhy(s)dsli=1B(tl,ti)RhyiΔt,ξly)+(li=1ΔtB(tl,ti)ξiy,ξly))+l1n=1(tn0Bt(tn+1,s)ΔtRhyds,ξny)l1n=1(tn+1tnΔtB(tn+1,s)Rhyn+1tds,ξny)l1n=1(ni=1(Δt)2Bt(tn+1,ti)Rhyids,ξny)l1n=1(ΔtB(tn+1,tn+1)ξn+1y,ξny)+l1n=1(ni=1(Δt)2Bt(tn+1,ti)ξiy,ξny)C(Δt)2(Rhyt2L2(L2)+Rhy2L2(L2))+Cln=1ξny2Δt+Cln=1Δtni=1ξiy2Δt+a8ξly2

    and

    ln=1[(ni=1ΔtB(tn,ti)yih(u),dtξny)(ni=1ΔtB(tn,ti1)yih(u),dtξny)]Δt=(li=1Δt(B(tl,ti)B(tl,ti1))Rhyi,ξly)(li=1Δt(B(tl,ti)B(tl,ti1))ξiy,ξly)+l1n=1(ni=1Δt(B(tn,ti)B(tn,ti1))Rhyi,ξny)l1n=1(ni=1Δt(B(tn,ti)B(tn,ti1))ξiy,ξny)l1n=1(n+1i=1Δt(B(tn+1,ti)B(tn+1,ti1))Rhyi,ξny)+l1n=1(n+1i=1Δt(B(tn+1,ti)B(tn+1,ti1))ξiy,ξny)=(li=1(Δt)2Bt(tl,ti)Rhyi,ξly)(li=1(Δt)2Bt(tl,ti)ξiy,ξly)
    +l1n=1(ni=1(Δt)2Bt(tn,ti)Rhyi,ξny)l1n=1(ni=1(Δt)2Bt(tn,ti)ξiy,ξny)l1n=1(n+1i=1(Δt)2Bt(tn+1,ti)Rhyi,ξny)+l1n=1(n+1i=1(Δt)2Bt(tn+1,ti)ξiy,ξny)C(Δt)2Rhy2L2(L2)+Cln=1ξny2Δt+Cln=1Δtni=1ξiy2Δt+a8ξly2,

    where ti is located between ti1 and ti, and we also used

    tn0B(tn,s)Rhy(s)dsni=1B(tn,ti)RhyiΔtCΔt(RhytL2(L2)+RhyL2(L2)).

    From (3.7)–(3.9), we have

    12A12ξly2+12ln=1dtξny2ΔtCh4yt2L2(H2)+C(Δt)2(ytt2L2(L2)+Rhyt2L2(L2)+Rhy2L2(L2))+Cln=1ξny2Δt+Cln=1Δtni=1ξiy2Δt+a4ξly2. (3.10)

    Adding ln=1ξny2Δt to both sides of (3.10), by use of the assumption on A and discrete Gronwall's inequality, we have

    |||(Rhyyh(u))|||L(L2)C(Δt+h2). (3.11)

    Using (2.13), the Poincare inequality and the triangle inequality, we get

    |||yyh(u)|||L(L2)C(Δt+h2), |||(yyh(u))|||L(L2)C(Δt+h). (3.12)

    Taking t=tn1 in (2.6), subtracting (2.31) from (2.6) and then using (2.14), we have

    (dtξnp,qh)+(Aξn1p,qh)=(dtpnpn1t,qh)+(dtηnp,qh)+Ttn1(B(s,tn1)Rhp(s),qh)ds(Ni=nΔtB(ti,tn1)pi1h(u),qh)+(δyndδyn+ynynh(u),qh). (3.13)

    Choosing qh=dtξnp in (3.13), multiplying by Δt and summing over n from l+1 to N (0lN1) on both sides of (3.13), since ξNp=0, we find that

    12A12ξlp2+Nn=l+1dtξnp2ΔtNn=l+1(dtpnpn1t,dtξnp)ΔtNn=l+1(dtηnp,dtξnp)ΔtNn=l+1[Ttn1(B(s,tn1)Rhp(s),dtξnp)ds(Ni=nΔtB(ti,tn1)pi1h(u),dtξnp)]ΔtNn=l+1(δyndδyn+ynynh(u),dtξnp)Δt=:4i=1Bi. (3.14)

    Notice that

    (Aξn1p,dtξnp)12Δt(A12ξn1p2A12ξnp2). (3.15)

    Now, we estimate the right-hand terms of (3.14). Similar to (3.7), we have

    B1C(Δt)2ptt2L2(L2)+14Nn=l+1dtξnp2Δt. (3.16)

    For B2, using (2.15) and the Cauchy inequality, we have

    B2Ch4pt2L2(H2)+14Nn=l+1dtξnp2Δt. (3.17)

    For B3, applying the same estimates as A3, we conclude that

    B3=Nn=l+1[Ttn1(B(s,tn1)Rhp(s),dtξnp)ds(Ni=nΔtB(ti1,tn1)pi1h(u),dtξnp)+(Ni=nΔtB(ti1,tn1)pi1h(u),dtξnp)(Ni=nΔtB(ti,tn1)pi1h(u),dtξnp)]ΔtC(Δt)2(Rhpt2L2(L2)+Rhp2L2(L2))+CNn=l+1ξnp2Δt+CNn=l+1ΔtNi=nξip2Δt+a4ξlp2, (3.18)

    where

    RhptL2(L2)+RhpL2(L2)(ptRhpt)L2(L2)+ptL2(L2)+(pRhp)L2(L2)+pL2(L2).

    For B4, using the Cauchy inequality and the smoothness of y and yd, we have

    B4=Nn=l+1(δyndδyn+ynynh(u),dtξnp)ΔtC(Δt)2(yt2L2(L2)+(yd)t2L2(L2))+Cynynh(u)2L2(L2)+14Nn=l+1dtξnp2Δt. (3.19)

    Combining (3.16)–(3.19), we have

    12A12ξlp2+14Nn=l+1dtξnp2ΔtC(Δt)2(ptt2L2(L2)+Rhpt2L2(L2)+Rhp2L2(L2)+yt2L2(L2)+(yd)t2L2(L2))+Ch4pt2L2(H2)+Cynynh(u)2L2(L2)+a4ξlp2+CNn=l+1ξnp2Δt+CNn=l+1ΔtNi=nξip2Δt. (3.20)

    By adding Nn=l+1ξnp2Δt to both sides of (3.20) and applying the assumption on A, discrete Gronwall's inequality and (3.12), we conclude that

    |||(Rhpph(u))|||L(L2)C(Δt+h2). (3.21)

    Using (2.15) and the triangle inequality, we get

    |||pph(u)|||L(L2)C(Δt+h2), |||(pph(u))|||L(L2)C(Δt+h); (3.22)

    we have completed the proof of the Lemma 3.1.

    Lemma 3.2. Choose ˜un=Qhun and ˜un=un in (2.29)–(2.32) respectively. Then, for Δt small enough and 1nN, we have

    |||(yh(u)yh(Qhu))|||L(L2)+|||(ph(u)ph(Qhu))|||L(L2)Ch2. (3.23)

    Proof. For convenience, let

    λny=ynh(u)ynh(Qhu), λnp=pnh(u)pnh(Qhu).

    Taking ˜un=un and ˜un=Qhun in (2.29), we easily get

    (dtλny,vh)+(Aλny,vh)=ni=1Δt(B(tn,ti1)λiy,vh)+(unQhun,vh). (3.24)

    By choosing vh=dtλny in (3.24), multiplying by Δt and summing over n from 1 to l (1lN) on both sides of (3.24), we find that

    12A12λly2+ln=1dtλny2Δtln=1(ni=1Δt(B(tn,ti1)λiy,dtλny)Δt+ln=1(unQhun,λnyλn1y)=(li=1ΔtB(tl,ti1)λiy,λly)+l1n=1(ni=1ΔtB(tn,ti1)λiyn+1i=1ΔtB(tn+1,ti1)λiy,λny)+(ulQhul,λly)l1n=1(un+1Qhun+1(unQhun),λny)=(li=1ΔtB(tl,ti1)λiy,λly)+l1n=1(ni=1(Δt)2Bt(tn+1,ti1)λiy,λny)l1n=1(ΔtB(tn+1,tn)λn+1y,λny)+CulQhul1λly+l1n=1(uQhu)t(θn)1λnyΔtCln=1λny2Δt+Cln=1Δtni=1λiy2Δt+a4λly2+Ch4(ul21+ut2L2(H1)), (3.25)

    where we use (2.17) and the assumption on B; additionally, θn is located between tn and tn+1.

    Add ln=1λny2Δt to both sides of (3.25); then for sufficiently small Δt, combining (3.25) and the discrete Gronwall inequality, we have

    |(yh(u)yh(Qhu))|L(L2)Ch2. (3.26)

    Similar to (3.24), we have

    (dtλnp,qh)+(Aλn1p,qh)=(Ni=nΔtB(ti,tn1)λi1p,qh)+(λny,qh),  qhVh. (3.27)

    By choosing qh=dtλnp in (3.27), multiplying by Δt and summing over n from l+1 to N (0lN1) on both sides of (3.27), combining (3.26) and Poincare inequality gives

    12A12λlp2+ln=1dtλnp2ΔtNn=l+1(Ni=nΔtB(ti,tn1)λi1p,dtλnp)ΔtNn=l+1(λny,dtλnp)Δt=(Ni=l+1ΔtB(ti,tl)λi1p,λlp)N1n=l+1(Ni=nΔtB(ti,tn1)λi1p,λnp)+N1n=l+1(Ni=n+1ΔtB(ti,tn)λi1p,λnp)Nn=l+1(λny,dtλnp)Δt=(Ni=l+1ΔtB(ti,tl)λi1p,λlp)N1n=l+1(Ni=n(Δt)2Bt(ti,tn)λi1p,λnp)N1n=l+1(ΔtB(tn,tn)λn1p,λnp)Nn=l+1(λny,dtλnp)ΔtCh4+a4λlp2+CNn=l+1λn1p2Δt+CN1n=l+1ΔtNi=nλip2Δt+12Nn=l+1dtλnp2Δt. (3.28)

    Add Nn=l+1λn1p2Δt to both sides of (3.28); then for sufficiently small Δt, applying the discrete Gronwall inequality and the assumptions on A and B, we have

    |||(ph(u)ph(Qhu))|||L(L2)Ch2. (3.29)

    Using the stability analysis as in Lemma 3.2 yields Lemma 3.3.

    Lemma 3.3. Let (ynh,pnh) and (ynh(Qhu),pnh(Qhu)) be the discrete solutions of (2.29)(2.32) with ˜un=unh and ˜un=Qhun, respectively. Then, for Δt small enough and 1nN, we have

    |||(yh(Qhu)yh)|||L(L2)+|||(ph(Qhu)ph)|||L(L2)C|||Qhuuh|||L2(L2). (3.30)

    Next, we derive the following inequality.

    Lemma 3.4. Choose ˜un=Qhun and ˜un=unh in (2.29)(2.32) respectively. Then, we have

    Nn=1(Qhununh,pn1h(Qhu)pn1h)Δt0. (3.31)

    Proof. For n=0,1,,N, let

    rnp=pnh(Qhu)pnh, rny=ynh(Qhu)ynh.

    From (2.29)–(2.32), we have

    (dtrny,vh)+(Arny,vh)ni=1Δt(B(tn,ti1)riy,vh)=(Qhununh,vh),  vh Vh, (3.32)
    (dtrnp,qh)+(Arn1p,qh)Ni=nΔt(B(ti,tn1)ri1p,qh)=(rny,qh),  qh Vh. (3.33)

    Notice that

    Nn=1(Δtni=1B(tn,ti1)riy,rn1p)=Nn=1(ΔtNi=nB(ti,tn1)ri1p,rny)

    and

    Nn=1(dtrny,rn1p)Δt+Nn=1(dtrnp,rny)Δt=0.

    By choosing vh=rn1p in (3.32), qh=rny in (3.33), and then multiplying the two resulting equations by Δt and summing it over n from 1 to N, we have

    Nn=1(Qhununh,pn1h(Qhu)pn1h)Δt=Nn=1rny2Δt, (3.34)

    which completes the proof of the lemma.

    Lemma 3.5. Let u be the solution of (2.4)–(2.8) and unh be the solution of (2.23)–(2.27). Assume that all of the conditions in Lemmas 3.1–3.4 are valid. Then, for Δt small enough and 1nN, we have

    |||Qhuuh|||L2(L2)C(h2+Δt). (3.35)

    Proof. Take ˜u=unh in (2.8) and ˜uh=Qhun in (2.27) to get the following two inequalities:

    (un+pn,unhun)0 (3.36)

    and

    (unh+pn1h,Qhununh)0. (3.37)

    Note that unhun=unhQhun+Qhunun. Adding the two inequalities (3.36) and (3.37), we have

    (unh+pn1hunpn,Qhununh)+(un+pn,Qhunun)0. (3.38)

    Thus, by (3.38), (2.16), (2.8) and Lemma 3.4, we find that

    |||Qhuuh|||2L2(L2)=Nn=1(Qhununh,Qhununh)ΔtNn=1(Qhunun,Qhununh)Δt+Nn=1(pn1hpn,Qhununh)Δt+Nn=1(un+pn,Qhunun)Δt=Nn=1(pn1hpn1h(Qhu),Qhununh)Δt+Nn=1(pn1pn,Qhununh)Δt+Nn=1(pn1h(u)pn1,Qhununh)Δt+Nn=1(un+pn,Qhunun)Δt+Nn=1(pn1h(Qhu)pn1h(u),Qhununh)ΔtNn=1(pn1pn,Qhununh)Δt+Nn=1(pn1h(u)pn1,Qhununh)Δt+Nn=1(pn1h(Qhu)pn1h(u),Qhununh)Δt=:3i=1Fi. (3.39)

    It follows from the Cauchy inequality, Lemma 3.1, Lemma 3.2 and Poincare's inequality that

    F1C(Δt)2pt2L2(L2)+14Nn=1Qhununh2Δt, (3.40)
    F2C(h4+(Δt)2)+14Nn=1Qhununh2Δt, (3.41)
    F3Ch4+14Nn=1Qhununh2Δt. (3.42)

    Substituting the estimates for F1F3 into (3.39), we derive (3.35).

    Using (3.11), (3.21), Lemmas 3.2–3.5 and the triangle inequality, we derive the following superconvergence for the state variable.

    Lemma 3.6. Let u be the solution of (2.4)–(2.8) and unh be the solution of (2.23)–(2.27). Assume that all of the conditions in Lemmas 3.1–3.5 are valid. Then, for Δt small enough and 1nN, we have

    |||(Rhyyh)|||L(L2)+|||(Rhpph)|||L(L2)C(h2+Δt). (3.43)

    Now, the main result of this section is given in the following theorem.

    Theorem 3.1. Let (y,p,u) and (ynh,pn1h,unh) be the solutions of (2.4)–(2.8) and (2.23)–(2.27), respectively. Assume that y, p and u have enough regularities for our purpose; then, for Δt small enough and 1nN, we have

    |||yyh|||L(L2)+|||pph|||L(L2)C(h2+Δt), (3.44)
    |||(yyh)|||L(L2)+|||(pph)|||L(L2)C(h+Δt), (3.45)
    |||uuh|||L2(L2)C(h+Δt). (3.46)

    Proof. The proof of the theorem can be completed by using Lemmas 3.1–3.5, (2.17) and the triangle inequality.

    To provide the global superconvergence for the control and state, we use the recovery techniques on uniform meshes. Let us construct the recovery operators Ph and Gh. Let Phv be a continuous piecewise linear function (without the zero boundary constraint). The value of Phv on the nodes are defined by a least squares argument on element patches surrounding the nodes; the details can be found in [25,26].

    We construct the gradient recovery operator Ghv=(Phvx,Phvy) for the gradients of y and p. In the piecewise linear case, it is noted to be the same as the Z-Z gradient recovery (see [25,26]). We construct the discrete co-state with the admissible set

    ˆunh=max{0,¯pn1h}pn1h. (3.47)

    Now, we can derive the global superconvergence result for the control variable and state variable.

    Theorem 3.2. Let u and unh be the solutions of (2.4)–(2.8) and (2.29)–(2.32), respectively. Assume that all of the conditions in Lemmas 3.1–3.5 are valid. Then we have

    |||uˆuh|||L2(L2)C(h2+Δt). (3.48)

    Proof. Using (2.9), (3.47) and Theorem 3.1, we have

    |||uˆuh|||2L2(L2)=Nn=1unˆunh2ΔtCNn=1max{0,¯pn}max{0,¯pn1h}2Δt+CNn=1pnpn1h2ΔtCNn=1¯pn¯pn1h2Δt+CNn=1pnpn1h2ΔtCNn=1pnpn1h2ΔtCNn=1pnpn12Δt+CNn=1pn1pn1h2ΔtC(h4+(Δt)2). (3.49)

    Theorem 3.3. Let (y,p) and (ynh,pn1h) be the solutions of (2.4)–(2.8) and (2.29)–(2.32), respectively. Assume that all of the conditions in Lemmas 3.1–3.5 are valid. Then we have

    |||Ghyhy|||L(L2)+|||Ghphp|||L(L2)C(h2+Δt). (3.50)

    Proof. Notice that

    |||Ghyhy|||L(L2)|||GhyhGhRhy|||L(L2)+|||GhRhyy|||L(L2). (3.51)

    It follows from Lemma 3.6 that

    |||GhyhGhRhy|||L(L2)C|||(yhRhy)|||L(L2)C(h2+Δt). (3.52)

    It can be proved by the standard interpolation error estimate technique (see [1]) that

    |||GhRhyy|||L(L2)Ch2. (3.53)

    Therefore, it follows from (3.52) and (3.53) that

    |||Ghyhy|||L(L2)C(h2+Δt). (3.54)

    Similarly, it can be proved that

    |||Ghphp|||L(L2)C(h2+Δt). (3.55)

    Therefore, we complete the proof.

    In this section, we will present a two-grid scheme and analyze a priori error estimates. Now, we present our two-grid algorithm which has the following two steps:

    Step 1. On the coarse grid TH, find (ynH,pn1H,unH)V2H×KH that satisfies the following optimality conditions:

    (dtynH,vH)+(AynH,vH)=(ni=1ΔtB(tn,ti1)yiH,vH)+(fn+unH,vH),  vH VH, (4.1)
    y0H=RHy0, (4.2)
    (dtpnH,qH)+(Apn1H,qH)=(Ni=nΔtB(ti,tn1)pi1H,qH)+(ynHynd,qH),  qH VH, (4.3)
    pNH=0, (4.4)
    (unH+pn1H,uHunH)0,  uHKH. (4.5)

    Step 2. On the fine grid Th, find (¯˜ynh,¯˜pn1h,¯˜unh)V2h×Kh such that

    (dt¯˜ynh,vh)+(A¯˜ynh,vh)=(ni=1ΔtB(tn,ti1)¯˜yih,vh)+(fn+ˆunH,vh),  vh Vh, (4.6)
    ¯˜y0h=Rhy0, (4.7)
    (dt¯˜pnh,qh)+(A¯˜pn1h,qh)=(Ni=nΔtB(ti,tn1)¯˜pi1h,qh)+(¯˜ynhynd,qh),  qh Vh, (4.8)
    ¯˜pNh=0, (4.9)
    (¯˜unh+¯˜pn1h,uh¯˜unh)0,  uhKh. (4.10)

    Combining Theorem 3.1 and the stability estimates, we easily get the following results.

    Theorem 4.1. Let (y,p,u) and (¯˜ynh,¯˜pnh,¯˜unh) be the solutions of (2.4)–(2.8) and (4.1)–(4.10), respectively. Assume that y, yd, p, pd and u have enough regularities for our purpose; then, for Δt small enough and 1nN, we have

    |||(y¯˜yh)|||L(L2)+|||(p¯˜ph)|||L(L2)C(h+H2+Δt), (4.11)
    |||u¯˜uh|||L2(L2)C(h+H2+Δt). (4.12)

    Proof. For convenience, let

    yn¯˜ynh=ynRhyn+Rhyn¯˜ynh=:ηny+eny,pn¯˜pnh=pnRhpn+Rhpn¯˜pnh=:ηnp+enp.

    Taking t=tn in (2.4), subtracting (4.6) from (2.4) and then using (2.12), we have

    (dteny,vh)+(Aeny,vh)=(tn0B(tn,s)Rhy(s)dsni=1ΔtB(tn,ti1)¯˜yih,vh)+(dtynynt,vh)(dtηny,vh)+(unˆunH,vh),  vhVh. (4.13)

    Selecting vh=dteny in (4.13), multiplying by Δt and summing over n from 1 to l (1lN) on both sides of (4.13), we find that

    12A12ely2+ln=1dteny2Δtln=1(dtηny,dteny)Δt+ln=1(dtynynt,dteny)Δt+ln=1(tn0B(tn,s)Rhy(s)dsni=1ΔtB(tn,ti1)¯˜yih,dteny)Δt+ln=1(unˆunH,dteny)Δt=:4i=1Ii. (4.14)

    Similar to Lemma 3.1, it is easy to show that

    I1+I2Ch4yt2L2(H2)+C(Δt)2ytt2L2(L2)+12ln=1dteny2Δt. (4.15)

    Similar to A3, we find that

    I3C(Δt)2(Rhyt2L2(L2)+Rhy2L2(L2))+Cln=1eny2Δt+Cln=1Δtni=1eiy2Δt+a4ely2. (4.16)

    For I4, using Theorem 3.2, we have

    I4C(H4+(Δt)2)+14ln=1dteny2Δt. (4.17)

    Combining (4.15)–(4.17), the discrete Gronwall inequality, the triangle inequality and (2.13), we get

    |||(y¯˜yh)|||L(L2)C(h+H2+Δt). (4.18)

    By taking t=tn1 in (2.6), subtracting (4.8) from (2.6) and using (2.12), we have

    (dtenp,qh)+(Aen1p,qh)=(Ttn1B(s,tn1)Rhp(s)dsNi=nB(ti,tn1)¯˜pi1hΔt,qh)(dtpnpn1t,qh)+(dtηnp,qh)+(δyndδyn,qh)+(yn¯˜ynh,qh),  qhVh. (4.19)

    By selecting qh=dtenp in (4.19), multiplying by Δt and summing over n from l+1 to N (0lN1) on both sides of (4.19), we find that using (2.15), (4.18) and the triangle inequality, similar to (3.14), gives

    |||(p¯˜ph)|||L(L2)C(h+H2+Δt). (4.20)

    Note that

    ¯˜unh=max{0,¯¯˜pn1h}πh¯˜pn1h,un=max{0,¯pn}pn.

    Using (2.19), (4.20) and the mean value theorem, we have

    |||u¯˜uh|||2L2(L2)=Nn=1un¯˜unh2ΔtCNn=1max{0,¯pn}max{0,¯¯˜pn1h}2Δt+CNn=1pnπh¯˜pn1h2ΔtCNn=1¯pn¯¯˜pn1h2Δt+CNn=1pnpn12Δt+CNn=1pn1πhpn12Δt+CNn=1πhpn1πh¯˜pn1h2ΔtCNn=1pn¯˜pn1h2Δt+CNn=1pnpn12Δt+CNn=1pn1πhpn12Δt+CNn=1πhpn1πh¯˜pn1h2ΔtCNn=1pnpn12Δt+CNn=1pn1πhpn12Δt+CNn=1pn1¯˜pn1h2ΔtC(h2+H4+(Δt)2), (4.21)

    which completes the proof.

    In this section, we present the following numerical experiment to verify the theoretical results. We consider the following two-dimensional parabolic integro-differential optimal control problems

    minuK{1210(yyd2+u2)dt}

    subject to

    (yt,v)+(y,v)=t0(y(s),v)ds+(f+u,v),  vV,y(x,0)=y0(x),  xΩ,

    where Ω=(0,1)2.

    We applied a piecewise linear finite element method for the state variable y and co-state variable p. The stopping criterion of the finite element method was chosen to be the abstract error of control variable u between two adjacent iterates less than a prescribed tolerance, i.e.,

    ul+1hulhϵ,

    where ϵ=105 was used in our numerical tests. For the linear system of equations, we used the algebraic multigrid method with tolerance 109.

    The numerical experiments were conducted on a desktop computer with a 2.6 GHz 4-core Intel i7-6700HQ CPU and 8 GB 2133 MHz DDR4 memory. The MATLAB finite element package iFEM was used for the implementation [27].

    Example: We chose the following source function f and the desired state yd as

    f(x,t)=(2e2t+4π2e2t+4π2+sin(πt))sin(πx)sin(πy)4π2sin(πt),yd(x,t)=(πcos(πt)8π2sinπt+8π2(cos(πt)π)cosπTπ+e2t)sin(πx)sin(πy)

    such that the exact solutions for y, p, u are respectively,

    y=e2tsin(πx)sin(πy),p=sin(πt)sin(πx)sin(πy),u=sin(πt)(4π2sin(πx)sin(πy)).

    In order to see the convergence order with respect to time step size t and mesh size h, we choose t=h or t=h2 with h=14,116,164. To see the convergence order of the two-grid method, we choose the coarse and fine mesh size pairs (12,14),(14,116),(18,164). Let us use yh,ph and uh as two-grid solutions in the following tables. In Tables 1 and 2, we let t=h2 and present the errors of the finite element method and two-grid method for y and p in the L2-norm. Next, in Tables 3 and 4, we set t=h and show the errors of the two methods for y and p in the H1-norm and u in the L2-norm. We can see that the two-grid method maintains the same convergence order as the finite element method. Moreover, we also display the computing times of the finite element method and the two-grid method in these tables. By comparison, we find that the two-grid method is more effective for solving the optimal control problems (1.1)–(1.4).

    Table 1.  Errors of finite element method with t=h2 at t=0.5.
    h yyh pph CPU time (s)
    14 0.1095 0.0856 0.7031
    116 0.0079 0.0045 8.8702
    164 0.0005 0.0002 2253.6396

     | Show Table
    DownLoad: CSV
    Table 2.  Errors of two-grid method with t=h2 at t=0.5.
    (h,H) yyh pph CPU time (s)
    (14,12) 0.1059 0.0853 0.4335
    (116,14) 0.0056 0.0043 5.0842
    (164,18) 0.0006 0.0002 1027.9740

     | Show Table
    DownLoad: CSV
    Table 3.  Errors of finite element method with t=h at t=0.5.
    h yyh1 pph1 uuh CPU time (s)
    14 1.6604 1.1385 0.1358 0.4720
    116 0.6187 0.2143 0.0367 0.6320
    164 0.1687 0.0578 0.0090 24.0800

     | Show Table
    DownLoad: CSV
    Table 4.  Errors of two-grid method with t=h at t=0.5.
    (h,H) yyh1 pph1 uuh CPU time (s)
    (14,12) 1.6755 1.1375 0.0988 0.2880
    (116,14) 0.6288 0.2142 0.0346 0.3870
    (164,18) 0.1716 0.0579 0.0089 7.3120

     | Show Table
    DownLoad: CSV

    In this paper, we presented a two-grid finite element scheme for linear parabolic integro-differential control problems (1.1)–(1.4). A priori error estimates for the two-grid method and finite element method have been derived. We have used recovery operators to prove the superconvergence results. These results seem to be new in the literature. In our future work, we will investigate a posteriori error estimates. Furthermore, we shall consider a priori error estimates and a posteriori error estimates for optimal control problems governed by hyperbolic integro-differential equations.

    The authors declare they have not used Artificial Intelligence (AI) tools in the creation of this article.

    The authors declare there is no conflict of interest.



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