Exercise 2.1.2 A Nonhyperbolic Cauchy Problem for the Wave Equation)
3.1 The backward parabolic equation
We are interested in finding a solutionuto the evolution equation
∂tu+Au=0 on×(0,T), (3.1.1)
u=0 on∂×(0,T) (3.1.2)
given the final data
u=uT on× {T}.
(3.1.3)
Method of eigenfunctions
To illustrate the basic ideas and methods, we first consider the simple case of A= −div(a∇)+c, wherea,care measurable bounded functions on,0< ≤ a,0≤c. The results, however, are not trivial, even for the one-dimensional heat equation∂tu−∂x2u =0 when=(0,1).
The operator Awith the Dirichlet boundary conditions (3.1.2) is self-adjoint in the Hilbert space L2() with the domain H(2)∩H˚(1)() when we assume in addition thata ∈C1(), so there is a complete orthonormal system of eigenfunc- tionsek(x) of Awith eigenvaluesλk+1≥λk≥0,k=1,2. . . .For any solution of (3.1.1), (3.1.2) we have
u(t,x)=
uk(t)ek(x).
(3.1.4)
This function solves (3.1.1), (3.1.2) if and only if the coefficientsuk satisfy the following ordinary differential equations:
∂tuk+λkuk=0 when 0<t<T,uk(0)=u0k, whereu0=u(0). Solving them, we obtainuk(t)=u0ke−λkt, so
u(t,x)=
u0ke−λktek(x). (3.1.5)
The backward uniqueness easily follows from this representation, because the coefficientsu0kare uniquely determined from the relationsu0ke−λkT =uT k, which follow from (3.1.3) and (3.1.5). Since theλkbehave likeCkpwith somep>0 for largek(e.g., for the one-dimensional heat equation and=(0,1) we haveek(x)=
√2 sin(kπx) andλk=π2k2). These relations show that for anyuT ∈L2() (and even H(m)()) a solutionu does not exist, and when it does, it is exponentially unstable: foru that is thekth term of the sum (3.1.5) withu0k=εeλkT we have u02()=εeλkT, whileuT2()=ε.
The next natural question is how to computeu(oru0), provided that it exists.
The first answer was obtained in the pioneering paper of Fritz John [Jo1]. We describe a couple of more recent algorithms.
3.1. The backward parabolic equation 43
Quasi-reversibility
The method of quasi-reversibility has been suggested in the book of Lattes and Lions [LL].
We replace (3.1.1) by the regularized “higher”-order equation
∂tu(;α)+Au(;α)−αA2u(;α)=0.
(3.1.6)
Again, by using eigenfunctions we obtain for the coefficientsuk(;α) the ordinary differential equations
∂tuk(;αk)+(λk−αλ2k)uk(;αk)=0,0<t<T,uk(T;α)=uT k. Solving them we get
u(t;α)=
uT ke(λk−αλ2k)(T−t)ek(x). (3.1.7)
Since theλkgo to infinity askincreases, this series is convergent inL2() for any uT in this space and for anyt<T. Moreover, whenαgoes to 0, the regularized solutionsuαare convergent to the solutionu(inL2() for anyt ∈(0,T)), provided thatuexists.
Regularization by pseudo-parabolic equations
Another example of a (weakly) convergent numerical algorithm is due to Gajewski and Zacharias [GZ]. They suggested regularizing equation (3.1.1) by the equation
∂t(uα−αAuα)+Auα=0.
Exercise 3.1.1. Prove the existence of solutions of the pseudo-parabolic regular- ization and their convergence touwhenuexists.
Method of logarithmic convexity
The next natural question is about stability in the backward evolutionary equations and about the rate of convergence of numerical algorithms. It is not a simple question unless one makes use of a priori bounds and logarithmic convexity.
To show that the logarithm F(t) of the squared norm of a solution f(t)= u(t)22() is a convex function, we use again the simple but important example of A= −div(a∇)+cassuming thata,c,−∂ta,−∂tcare measurable, bounded, and nonnegative, and thata≥ε0>0.
Now we will prove that
F≥0. (3.1.8)
We have
F= f/f, F=(f f−(f)2)/f2.
By using the definition of the L2-norm and the differential equation (3.1.1) we obtain
f=2
u∂tu =2
u(div(a∇u)−cu)=2
(−a|∇u|2−cu2), where we have used integration by parts and the boundary condition (3.1.2). Fur- ther,
f=2
(−2a∇uã∂t∇u−∂ta|∇u|2−2cu∂tu−∂tcu2)
≥
(4 div(a∇u)∂tu−4cu∂tu)=4
(∂tu)2 =4∂tu22,
where we integrated by parts again, used the conditions ona,c, and expressed
∂tufrom the differential equation (3.1.1). Sometimes we will drop, for brevity, the symbolin norms and scalar products. Now, we have
(ff −(f)2)≥4∂tu22u22−(2(u, ∂tu)2)2 ≥0 according to the Schwarz inequality.
We have proved that F is convex. Therefore, F(t)≤(1−t/T)F(0)+ t/T F(T), or
k
f(t)≤ f(0)1−t/T f(T)t/T when 0≤t ≤T.
Using the definition of f and the final data, we get u(t)2()≤M1−t/TuTt2/T() (3.1.9)
under the constraintu(t)2()≤M. It is easy to understand that this estimate is sharp. It can be considered a conditional stability estimate (under the a priori boundu(t)2≤M).
Exercise 3.1.2. Prove that under the additional constraint∂tu2()≤ M1 on (0,T) one has
u(0)22()≤ −M M1T(1−ln(−M1T/(Mln))/ln, where= uT2()/M.
{Hint: use Taylor’s formula foru(t)22around t=0; then boundu(0)22by using (3.1.9) and the constraint on∂tu. Minimize the bound with respect tot.}
Now we will prove a quite general result on logarithmic convexity and therefore stability in the backward evolution equations. This result is a weaker version of the theorem of Agmon and Nirenberg [AN].
LetA=A(t) be a linear operator in the (complex) Hilbert spaceHwith domain D(t). By , ( , ) we denote respectively the norm and the scalar product in H.
We consider the following generalization of equation (3.1.1):
∂tu+Au ≤αu on (0,T).
(3.1.10)
3.1. The backward parabolic equation 45 We assume that
A= A++A−, (3.1.11)
where A+ is a linear symmetric operator in H with domain D(t) and A− is a skew-symmetric operator. Moreover, they satisfy the following conditions:
A−u2≤α(A+uu + u2) (3.1.12)
and
∂t(A+u,u)≤2R(A+u, ∂tu)+α(A+uu + u2) (3.1.13)
for some constantα.
Theorem 3.1.3. Let u(t)∈D(t), u∈C1([0,T];H)be a solution of the differen- tial inequality(3.1.10), where A satisfies the conditions(3.1.11)–(3.1.13).
Then
u(t) ≤C1u(0)1−λu(T)λ
with C1≤exp((2α+2)T+2eC T/C) and λ=(1−e−Ct)/(1−e−C T) or (eC(t−T)−e−C T)/(1−e−C T)with C depending onα. In addition, whenα=0 we can take C1=1andλ=t/T .
In the proof we will use two elementary lemmas. We set q= u2, ψ= 2R(f,u)/q, andl=lnq−t
0ψ.
Lemma 3.1.4. There is a constant C ≥0depending onαsuch that∂t2l+C|∂tl| + C >0on(0,T). In addition, C can be taken as0whenα=0.
PROOF. Let f =∂tu+Au.
We have
∂tq =2R(u, ∂tu)= −2R(A+u,u)+2R(f,u)= −2R(A+u,u)+ψq, where ψ=2R(f,u)/q, and we have used that due to skew-symmetry, R(A−u,u)=0. Observe also that for a symmetric operator A+ we have I(A+u,u)=0. So
∂tl =∂tq/q−ψ= −2(A+u,u)/q and
∂t2l = −2∂t(A+u,u)/q+2(A+u,u)(−2(A+u,u)+ψq)/q2
≥ −4R(A+u, ∂tu)/q−2αA+uu/q−2α
−4(A+u,u)2/q2+2(A+u,u)ψ/q
=4(A+u2−(A+u,u)2/q)/q+4R(A+u,A−u)/q
−4R(A+u, f)/q−2αA+u/u −2α +4(A+u,u)R(f,u)/q2.
To obtain the inequality we have used condition (3.1.13) and then condition (3.1.11).
We set (A+u,u)= A+uuθ. We represent A+u asβu+u⊥, whereu⊥is orthogonal tou. By scalar multiplication, we findβ=(A+u,u)/u2, and then by the Pythagorean theorem we getu⊥2= A+u2(1−θ2). SinceR(A−u,u)=0 due to skew-symmetry, we conclude that
−4R(A+u,A−u)= −4R(u⊥,A−u)≤4u⊥A−u
≤2(u⊥2+ A−u2)
≤2(A+u2(1−θ2)+αA+uu +αu2) due to the Schwarz inequality, our formula foru⊥, and condition (3.1.12). Summing up and using the Schwarz inequality and the inequalityf ≤αuseveral times, we obtain
∂t2l ≥4A+u2q−1(1−θ2)−2A+u2q−1(1−θ2)−2αA+u/u
−2α−4αA+u/u −2αA+u/u −2α−4αA+u/u
=2σ2(1−θ2)−12ασ −4α when we setσ = A+u/u.
Observe that|∂tl| =2|θ|σ. Ifθ2<14, then
∂t2l≥σ2−12ασ+36α2−36α2−4α≥ −36α2−4α.
If14 ≤θ2≤1, then|∂tl| ≥ |σ|and
∂t2l ≥ −12ασ−4α,so∂t2l+12α|∂tl| +4α≥0.
In the both cases we have the required inequality withC=max{36α2+4α,12α}.
The proof is complete.
Lemma 3.1.5. Under the conditions of Theorem3.1.3we have lnu(t) ≤lnu(0)(e−γt−e−γT)/(1−eγT)
+lnu(T)(1−e−γt)/(1−e−γT)+C2,
whereγis either C or−C, C2 ≤(2α+2)T +2eC T/C, and C2=0whenα=0.
PROOF. LetLbe a solution to the differential equation∂t2L+C|∂tL| +C =0 on (0,T) coinciding withlat the endpoints 0,T. The existence ofL can be proven by using a priori estimates on L, ∂tL that follow from the observation that L satisfies the linear differential equation with the coefficientCsign∂tLof∂tL. By subtracting the equation for L from the inequality forl given by Lemma 3.1.4, we conclude that ∂t2(l−L)+C|∂t(l−L)| ≥0. Then l−L ≤0 by maximum principles, and it suffices to boundL. From extremum principles it follows that the functionL(t) cannot achieve a (local) minimum on (0,T), so∂tL >0 on (0, τ), and∂tL <0 on (τ,T) for some τ between 0 and T. Therefore, L satisfies the linear differential equation∂t2L+Cτ∂tL+C=0, whereCτ isC on (0, τ) and
−Coutside the interval.
3.1. The backward parabolic equation 47 We will first bound the solutionvto the differential equation∂t2v+Cτ∂tv=0 on (0,T) coinciding withlat the endpoints. Thenw=L−vsolves the inhomo- geneous equation and has zero boundary data.
Since∂tvsolves a linear first-order homogeneous ODE, it does not change its sign. Letl(0)≤l(T). Then∂tv≥0. Consider the solutionV to the ODE∂t2V + C∂tV =0 with the same boundary data. We have
∂t2(v−V)+C∂t(v−V)=(C−Cτ)∂tv≥0,
and v−V is zero at the endpoints. By the maximum principles, v≤V. The functionV is easy to calculate by solving the linear ordinary differential equation with constant coefficients and satisfying the boundary conditions. It is
l(0)(e−Ct−e−C T)/(1−e−C T)+l(T)(1−e−Ct)/(1−e−C T).
Whenl(0)>l(T) we obtain a similar bound withCreplaced by−C.
To boundwwe will solve the two ODE forw :∂t2w+C∂tw+C =0 on (0, τ) and∂t2w−C∂tw+C =0 on (τ,T) and we will satisfy zero boundary data at 0,T and the continuity conditions forw, ∂tw att=τ from the left and from the right. After standard calculations we obtain
w(t)=(2τ−T +2C−1−2C−1eCτ)(eCt−eC T)/(eC T−2eCτ+e2Cτ)
−T +twhenτ <t,
w(t)=(T−2τ +2C−1−2C−1eC(T−τ))(1−e−Ct)/(−eC(T−2τ)+2e−Cτ−1)
−t whent< τ.
The numerator and denominator ofw in the first case increase inτ, replacing the first one by its maximal absolute value C2(eC T−1)−T and the second one by its minimal absolute valueeC T −1, we boundwon (τ,T) by 2eC T/C−T +t. Similarly, on (0, τ) we have the bound 2eC T/C−t. Replacing these two functions by their maximal values (att =0 andt=T) we obtain the same bound 2eC T/C).
In sum, we have
l ≤L =v+w ≤l(0)(e−γt−e−γT)/(1−e−γT) +l(T)(1−e−γt)/(1−e−γT)+2eC T/C,
whereγisCor−C. From the definitions ofland f we have lnq≤l+2Tαand l≤lnq+2Tα. In addition, the factors ofl(0) and ofl(T) in the above inequality are between 0 and 1. In sum, we conclude that lnq(t)≤l(t)+2Tα≤ ã ã ã +4Tα, where. . .denotes the terms withl(0),l(T) replaced by lnq(0), lnq(T), and the first claim follows.
When α=0, the function l is lnq and it is convex by Lemma 3.1.4. This
completes the proof.
PROOF OFTHEOREM3.1.3. The needed estimate follows from the inequality of Lemma 3.1.5 by taking exponents of both parts. When α=0 this lemma also implies thatC1=expC2is 1. Lettingαgo to 0, we conclude from Lemma 3.1.4 thatCgoes to 0, and calculating the limit ofλ, we conclude the proof.
This theorem can be applied to a second-order parabolic equation with the Dirichlet or Neumann lateral boundary data and even to “parabolic” equation changing direction of time.
EXAMPLE3.1.6 (A PARABOLICEQUATION OFSECONDORDER). Let be a bounded domain inRn,∂∈C2. Let
Au= −
∂k(aj k∂ju)+
bj∂ju+cu (3.1.14)
with the domainD(t)=H˚(1)()∩H(2)(). We assume thataj k, ∂taj k,b,divb∈ C1(×[0,T]),c, ∂tc∈ L∞(×[0,T]). Hereb=(b1, . . . ,bn).
We will check conditions (3.1.11) – (3.1.13) with A+u = −
∂k(aj k∂ju)+
−1
2divb+c
u, A−u =
∂j(bju)−1
2(divb)u.
It is obvious that A= A++A−. Integrating by parts and using the boundary condition u=0 on ∂, we conclude that A+ is symmetric and A− is skew- symmetric.
Let us consider condition (3.1.12). The left side is the sum of integrals of B∂ju∂ku, B∂juu,Bu2 over , where B is the product of eitherbj andbk or of divbandbj. Letting = 2(), we obtain the bound in the first case
B∂ju∂ku = −
∂k(B∂ju)u ≤C(∂k∂juu + ∂juu) by using the Schwarz inequality. HereCdepends on|B|1. Since the operatorA+ is a second-order elliptic operator andusatisfies zero Dirichlet conditions on∂, the Schauder-type a priori estimates (Theorem 4.1) imply that∂k∂ju + ∂ku ≤ C(A+u + u), and we obtain (3.1.12).
To prove (3.1.13) it is sufficient by using the symmetry ofaj k and integration by parts to observe that
∂t
A+uu=
∂t(aj k∂ju∂ku)=
∂taj k∂ju∂ku+2A+u∂tu
and to bound
∂taj k∂j∂kuu, which can be done as before.
Exercise 3.1.7 (The Third Boundary Value Condition). Under the assump- tions of Example 3.1.6 with respect to the coefficients of A, prove uniqueness of a solutionu of the backward initial problem with the lateral boundary condi- tions
aj k∂juνk+bu=0 on∂×(0,T)
3.1. The backward parabolic equation 49 under the regularity assumptions that ∂tu, ∂j∂ku, ∂t2u, ∂j∂k∂tu ∈L2() at any t ∈(0,T) and that the corresponding norms are bounded uniformly with respect tot.
From the proof of Theorem 3.1.3 and from Example 3.1.6 one can see that to satisfy conditions (3.1.11) – (3.1.13), the operator Amust be subordinated to its symmetric part A+. At present it is not clear whether backward uniqueness holds without this assumption whenAdepends ont.
EXAMPLE3.1.8 (FORWARD-BACKWARDPARABOLICEQUATION). Let be the unit interval (0,1) in R, and A= −∂x(a∂xu). Leta, ∂ta∈C1(×[0,T]). We assume that
for anyteither∂ta ≤0 onora≥on (3.1.15)
for some positiveε. Then the conditions of Theorem 3.1.3 are satisfied, and there- fore we have uniqueness and conditional stability of a solutionuto the differential equation ∂tu =∂x(a∂xu) on Q=×(0,T) given zero lateral boundary data u =0 on∂×(0,T) and the final data. We consideruwith∂x2u∈ L2(Q).
We check conditions (3.1.11) – (3.1.13) with the obvious choiceA+=A. Then condition (3.1.12) is satisfied with anyα. We will check condition (3.1.13). Let us consider
∂t(A+u,u)=∂t
(−∂x(a∂xu))u=
∂t(a∂xu∂xu)
=
(∂ta(∂xu)2+2a∂xu∂t∂xu)
≤
((∂ta(∂xu)2−2(∂x(a∂xu))∂tu),
where we integrated by parts with respect to x using the zero lateral boundary conditions. The term with∂ta is nonpositive together with∂ta. If∂tais positive at a point offor somet, we integrate by parts in the first integral again and we obtain that this integral is
−
∂x(∂ta∂xu)u ≤Cu(2)()u ≤αAuu
due to the Cauchy-Schwarz inequality and condition (3.1.15), because thena≥ε, the operatorAis uniformly elliptic, andu(2)()≤αAuis the standard elliptic estimate. The second term is 2(Au, ∂tu).
So we have condition (3.1.13)
It is quite interesting that this equation can change type from a forward parabolic equation to a backward parabolic equation at any point of. For simplicity, we considered the one-dimensional case, but nothing will change for equation (3.1.14) withb=0 if in condition (3.1.15) we request nonpositivity of the matrix (∂taj k).
Use of semigroups
The idea of using semigroups was systematically developed by Krein and Prozorovskaya [Kre], pp. 73, 81, and we present some results of their theory.
Let us consider the differential equation
∂tu+ζAu=0 on (0,T), (3.1.16)
whereAis a linear closed operator in a Banach SpaceX with domainD(A) that is dense inX, and X denotes the norm inX. A solutionu(t;ζ) of this equation satisfying the initial datau(0)=u0is a function continuous from [0,T] into X that is differentiable on (0,T),u(t;ζ)∈ D(A), and such that (3.1.16) holds. The basic assumption is that there exists a sector S= {|argζ|< φ}of the complex plane such that for anyζ ∈Sand for any initial datau0∈ Xthere is a solution to this initial problem that is complex-analytic with respect toζ ∈Sand that satisfies the following estimate:
u(1;ζ)X ≤Cφeσtu0X,t =Rζ.
(3.1.17)
This estimate follows from the following bound on the operator norm of the re- solvent of the operatorA:(A−ζI)−1 ≤C/|ζ −σ|whenRζ > σfor someC andσ. The last condition is possible to check whenAcorresponds to several elliptic boundary value problems, including the first boundary value problem for higher- order elliptic equations and the elliptic oblique derivative problem for second-order equations. This is not easy analytical work. The most recent reference is to the paper of Colombo and Vespri [CoV], where they considered operators A corre- sponding to general elliptic boundary value problems for higher-order operators satisfying the complementing and normality conditions under the assumptions that the coefficients of the differential operators are merely continuous and of those the boundary conditions are accordinglyCk()-smooth with natural choice ofk.
Under condition (3.1.17) and under the a priori constraintu(t; 1)X ≤ M, the following stability estimate is valid:
u(t)X ≤C M1−à(t)eσtuTàX(t), (3.1.18)
whereà(ζ) is the harmonic measure of the cut [T,∞) in S with respect to the pointζ. This function is defined as a bounded harmonic function ofζ in ST = S\[T,+∞) with the boundary values 0 on∂Sand 1 on the cut. By using conformal mappings onto standard domains and Giraud’s extremum principle as in Section 9.3, it is possible to show that à exists, is unique, and satisfies the following estimates:t2φ/π/C < à(t) when 0<t < τ, and 1−(T −t)1/2/C < à(t) when T −τ <t <T for someτ.
To obtain (3.1.18) we observe that the functions(ζ)=lne−σζu(;ζ)X is sub- harmonic inSTas the norm of an analytic function with values inX. From (3.1.17) it follows thats(t)≤lnCφuTX whenT <t<+∞. So we have lim infs(ζ)≤ (1−à(ζ)) lnCφM+à(ζ) lnuTX when ζ ∈∂ST. By the maximum principle