Lecture 4. The parabolic equations and time dependent Stokes...

Lecture 4. The parabolic equations and timedependent Stokes problem

Ching-hsiao (Arthur) Cheng

Department of MathematicsNational Central University

Taiwan, ROC

The National Center for Theoretic Sciences, Summer 2012

Ching-hsiao Cheng Lecture 4. The parabolic theory and Stokes equations

Parabolic equations

Let Ω ⊆ Rn be a bounded and smooth domain. We consider

ut + Lu = f in Ω× (0,T ),

u = u0 on Ω× t = 0,boundary conditions on ∂Ω× (0,T ),

where Lu is a (time-dependent) uniformly elliptic operator

defined byLu = − ∂

∂xi

(aij ∂u∂xj

)+ bi ∂u

∂xi+ cu.

Here the coefficients a, b, c may depend on t . We recall that L

is called uniformly elliptic if there exists constant λ > 0 such that

aijξiξj ≥ λ|ξ|2 ∀ ξ ∈ Rn.

∂t + L is called uniformly parabolic if L is uniformly elliptic.


Boundary conditions

Two types of boundary conditions are considered.

1 Dirichlet boundary condition:

u = 0 on ∂Ω× (0,T ).

2 Neumann boundary condition:

aij ∂u∂xj

Ni = g on ∂Ω× (0,T ).

A Robin type of boundary condition can also be considered, but

the theory behind that is similar to the Neumann problem, so

we ignore the discussion of such boundary condition.


The parabolic equations

In other words, we consider the Dirichlet problem

ut + Lu = f in Ω× (0,T ),

u = u0 on Ω× t = 0, (D)

u = 0 on ∂Ω× (0,T ),

or the Neumann problem

ut + Lu = f in Ω× (0,T ),

u = u0 on Ω× t = 0, (N)

aij ∂u∂xj

Ni = g on ∂Ω× (0,T ).


The weak formulation and weak solutions

Assume that aij ,bi , c ∈ L∞(Ω× (0,T )), f ∈ L2(0,T ; L2(Ω)),

and u0 ∈ L2(Ω).

Definition (Weak solutions with Dirichlet boundary conditions)

A function u ∈ L2(0,T ; H10 (Ω)) with ut ∈ L2(0,T ; H−1(Ω)) is

said to be a weak solution of (D) provided that

〈ut , ϕ〉+ B(u, ϕ) =(f , ϕ)

L2(Ω)∀ϕ ∈ H1

0 (Ω), a.e. t ∈ (0,T ),

andu(0) = u0 ,

where B(u, ϕ) is defined by

B(u, ϕ) ≡∫

Ω

[aij ∂u∂xj

∂ϕ

∂xi+ bi ∂u

∂xiϕ+ cuϕ

]dx .

The integral equality is called the variational formulation of (D).


The weak formulation and weak solutions

Assume further that g ∈ L2(0,T ; L2(∂Ω)).

Definition (Weak solutions with Neumann boundary conditions)

A function u ∈ L2(0,T ; H1(Ω)) with ut ∈ L2(0,T ; H1(Ω)′) is saidto be a weak solution of (D) provided that

〈ut , ϕ〉+ B(u, ϕ) = (f , ϕ)L2(Ω) + (g, ϕ)L2(∂Ω)

∀ϕ ∈ H1(Ω), a.e. t ∈ (0,T ),

andu(0) = u0 ,

where B(u, ϕ) is defined by

B(u, ϕ) ≡∫

Ω

[aij ∂u∂xj

∂ϕ

∂xi+ bi ∂u

∂xiϕ+ cuϕ

]dx .

The integral equality is called the variational formulation of (N).


The meaning of u(0) = u0

For H = H10 (Ω;Rn) or H = H1(Ω), the initial value of a function

u ∈ L2(0,T ;H) does not make sense. However, since ut is

required to belong to L2(0,T ;H ′) in the definition of the weak

solution, the following “time embedding lemma”

Lemma

Suppose u ∈ L2(0,T ;H), with ut ∈ L2(0,T ;H ′) for some Hilbertspace H so that H →L2(Ω) ⊆ H ′. Then u ∈ C([0,T ]; L2(Ω)),and

maxt∈[0,T ]

‖u(t)‖L2(Ω) ≤(

1 +1T

)[‖u‖L2(0,T ;H) + ‖ut‖L2(0,T ;H ′)

].

suggests that u ∈ C([0,T ]; L2(Ω)); thus u(0) makes sense, and

limt→0+

‖u(t)− u0‖L2(Ω) = 0.


The existence and uniqueness of the weak solution

Note that the general form of the weak formulation above is

〈ut , ϕ〉+ (∇u,∇ϕ)L2(Ω) = F (ϕ) ∀ϕ ∈ H, a.e. t ∈ (0,T ).

Construction of a weak solution - the Galerkin method: Let

ek∞k=1 be an orthogonal basis in H which is orthogonal in Hand orthonormal in L2(Ω). For each k ∈ N, let

uk (x , t) =k∑

`=1

dk` (t)e`(x)

satisfy

(ukt (t), ϕ)L2(Ω) + B(uk (t), ϕ

)= F (ϕ) ∀ ϕ ∈ span(e1, · · · , ek ).

and

dk` (0) = (u0, e`)L2(Ω) ∀ 1 ≤ ` ≤ k .



Since every test function ϕ in the span can be written as a

linear combination of e1, · · · , ek , by the bi-linearity of B we find

that the equality above is equivalent tok∑

`=1

[dk ′` (t)(e`, ej )L2(Ω) + dk

` (t)B(e`, ej

)]= F (ej ) ∀ 1 ≤ j ≤ k .

Then dk (t) = [dk1 (t), · · · ,dk

k (t)]T satisfies the following ODE:

dk ′(t) + M(t)dk (t) = Fk (t),

where Mij = B(ei , ej)T, Fk = [F (e1),F (e2), · · · ,F (ek )]T. The

fundamental theorem of ODE suggests that dk exists in a time

interval [0,Tk ]. The goal is to show that the limit of uk , if exists,

is the weak solution.



Question 1: Is there a positive lower bound of Tk? If not, the

limit of uk means nothing.

Question 2: How do we ensure that uk converges? If uk does

converge, in what space and in what sense?

Answer: We need to look at the so-called energy estimates.

The starting point is that uk satisfies(ukt (t), ej

)L2(Ω)

+ B(uk (t), ej

)= F (ej ) ∀ 1 ≤ j ≤ k ;

thus by the bi-linearity of B and linearity of F ,(ukt (t),uk (t)

)L2(Ω)

+ B(uk (t),uk (t)

)= F

(uk (t)

)⇒ 1

2ddt‖uk (t)‖2

L2(Ω) + B(uk (t),uk (t)

)= F

(uk (t)

)



By the uniform ellipticity of L (or the parabolicity of ∂t + L),

B(uk ,uk )

≥ λ‖∇uk‖2L2(Ω) − ‖b‖L∞(Ω)‖∇uk‖L2(Ω)‖uk‖L2(Ω) − ‖c‖L∞(Ω)‖uk‖2

L2(Ω)

≥ (λ− ε)‖∇uk‖2L2(Ω) −

[ 14ε‖b‖2

L∞(Ω) + ‖c‖L∞(Ω)

]‖uk‖2

L2(Ω).

Moreover,

F (uk ) ≤ ‖F‖H ′‖uk‖H ≤14ε‖F‖2H ′ + ε‖uk‖2H,

where‖F‖H ′ = ‖f‖L2(Ω)

when (D) is considered, or

‖F‖H ′ ≤ C[‖f‖L2(Ω) + ‖g‖L2(∂Ω)

]when (N) is considered.



Therefore, letting ε = λ/4, we find that

12

ddt‖uk (t)‖2

L2(Ω) +λ

2‖∇uk (t)‖2

L2(Ω) ≤λ

4‖uk (t)‖2

L2(Ω) +1λ‖F (t)‖2

H ′ .

Integrating the inequality in time over the time interval (0, t), we

obtain that

‖uk (t)‖2L2(Ω) + λ

∫ t

0‖∇uk (s)‖2

L2(Ω)ds

≤ ‖uk (0)‖2L2(Ω) +

2λ‖F‖2

L2(0,T ;H ′)) +λ

2

∫ t

0‖uk (s)‖2

L2(Ω)ds

≤ ‖u0‖2L2(Ω) +

2λ‖F‖2

L2(0,T ;H ′)︸︷︷︸≡M

+λ

2

∫ t

0‖uk (s)‖2

L2(Ω)ds.



Let X (t) ≡ ‖uk (t)‖2L2(Ω). Then X (t) satisfies

X (t) ≤ M +λ

2

∫ t

0X (s)ds.

Theorem (The Gronwall inequality)

If a ∈ L1(0,T ) is a non-negative function , and x(t) satisfies

x(t) ≤ M +

∫ t

0a(s)x(s)ds.

Then x(t) ≤ M exp(∫ t

0a(s)ds

)for all t ∈ [0,T ].

Therefore, ‖uk (t)‖2L2(Ω)is uniformly bounded, and

‖uk (t)‖2L2(Ω) + λ

∫ t

0‖∇uk (s)‖2

L2(Ω)ds ≤ Meλt2 ∀ t ∈ [0,T ].



Implications:

1 ‖uk (t)‖2L2(Ω) =

k∑=1|dk

` (t)|2 is bounded uniformly in k ; thus

Tk ≥ T for some fixed T > 0.

2 For ϕ ∈ H, write ϕ = ϕ1 + ϕ2 with ϕ1 ∈ span(e1, · · · , ek )

and ϕ2 ⊥ ϕ1. By the definition of the dual space norm,

‖ukt‖H ′ = sup‖ϕ‖H=1

〈ukt , ϕ〉 = sup‖ϕ‖H=1

(ukt , ϕ1)L2(Ω)

= sup‖ϕ‖H=1

[F (ϕ1)− B(uk , ϕ1)

]≤ C

[‖F‖H ′ + ‖uk‖H

].

Since uk is bounded in L2(0,T ;H) uniformly in k , ukt is

uniformly bounded in L2(0,T ;H ′).



3 Since L2(0,T ;H) and L2(0,T ;H ′) are both Hilbert spaces,

by Banach-Alouglu theorem, there exists a subsequence

ukj of uk such that ukj and ukj t converge weakly in

L2(0,T ;H) and L2(0,T ;H ′), respectively.

4 ukj satisfies

(ukj t , v)L2(Ω) + B(ukj , v) = F (v)

∀ v =k∑

`=1

d`(t)e`(x), a.e. t ∈ (0,T );

thus passing j to the limit, we conclude that∫ T

0

[⟨ut (t), v(t)

⟩+ B

(u(t), v(t)

)]dt =

∫ T

0F(v(t)

)dt

∀ v ∈ L2(0,T ;H).



5 The use of v(x , t) = χ(a,b)(t)ϕ(x) for some 0 ≤ a < b ≤ T

implies that∫ b

a

[⟨ut (t), ϕ

⟩+ B

(u(t), ϕ

)]dt =

∫ b

aF(ϕ)dt ∀ ϕ ∈ H.

Therefore, the Lebesgue differentiation theorem

suggests that u satisfies the variational formulation.

Question: u(0) = u0? This is the same as asking that if

limt→0

limj→∞

ukj (t) = limj→∞

limt→0

ukj (t),

where the limit as j →∞ is taken in the sense of L2, and the

limit as t → 0 is taken in the sense of C0.



Answer: Let ζ ∈ C1([0,T ]) such that ζ(0) = 1 and ζ(T ) = 0.

Then∫ T

0

[(ukj t (t), ζ(t)ϕ

)L2(Ω)

+ B(ukj (t), ζ(t)ϕ

)]dt =

∫ T

0F(ζ(t)ϕ

)dt

∀ ϕ ∈ span(e1, · · · , ekj ),

and ∫ T

0

[⟨ut (t), ζ(t)ϕ

⟩+ B

(u(t), ζ(t)ϕ

)]dt =

∫ T

0F(v(t)

)dt

∀ ϕ ∈ H.

Integrating by parts in time for the first term and then passing j

to the limit, we find that

limj→∞

(ukj (0), ϕ

)L2(Ω)

=(u(0), ϕ

)L2(Ω)

∀ϕ ∈ H

which implies that u(0) = u0; thus u is a weak solution.



Uniqueness: Suppose that u1 and u2 are two weak solutions.

Let u = u1 − u2. Then u ∈ L2(0,T ;H) with ut ∈ L2(0,T ;H ′)satisfies∫ T

0

[〈ut (t), v(t)〉+ B

(u(t), v(t)

)]dt = 0 ∀ v ∈ L2(0,T ;H)

andu(0) = 0.

Let v(t) = χ(0,s)u(t) be a test function, we conclude that

‖u(s)‖2L2(Ω) + λ‖∇u(s)‖2

L2(Ω) ≤λ

2

∫ s

0‖u(t)‖2

L2(Ω)dt ;

thus the Gronwall inequality implies that ‖u(t)‖L2(Ω) = 0 for all

t ∈ [0,T ]. This suggests that the weak solution is unique.



Some remarks:

1 Unlike the case of elliptic equations, even if b or c is not

identical to zero, the weak solution always exists. This is

because the L2-norm of u is controlled by the ut term.

2 The estimates we derive for uk in L∞(0,T ; L2(Ω)) and

L2(0,T ;H) can also be done formally by testing the

equation against u (and so on). The estimates obtained in

this way are called a priori estimates which will suggest

what the solution space should be.

3 Almost all the a priori estimates can be derives rigorously,

so later on we only try to obtain the estimates formally.


The regularity theory of parabolic equations

The regularity theory of parabolic equations is based on elliptic

regularity in the following sense: to improve the regularity of u,

we first try to get better regularity of ut and then convert the

equation toLu = f − ut in Ω,

and use elliptic regularity to obtain better regularity of u.


Compatibility conditions

The compatibility conditions are crucial in the study of regularity

theory of parabolic equations (and hyperbolic equations) when

the domain of interests has boundary. It roughly says that the

time derivatives of boundary condition have to match up the

initial data to certain degrees in order to obtain higher regularity

of the solution. To illustrate the idea, let us look at the following

simple example.

Example

Considerut −∆u = f in Ω× (0,T ),

u = u0 on Ω× t = 0,u = 0 on ∂Ω× (0,T ).



Example (Continued...)

Having established the existence of the unique weak solution

u ∈ L2(0,T ; H10 (Ω)) with ut ∈ L2(0,T ; H−1(Ω)), suppose now

u ∈ L2(0,T ; H2(Ω)). Then ut ∈ L2(0,T ; L2(Ω)) which implies

that u ∈ C([0,T ]; H1(Ω)). Therefore, u ∈ C([0,T ]; H1/2(∂Ω))

which suggest that

limt→0‖u − u0‖H1/2(∂Ω) = 0.

Since u = 0 on boundary, the equality above suggests that

u0 = 0 on ∂Ω. This is the first order compatibility condition that

u0 has to satisfy.


Example (Continued...)

Suppose that we would like to establish u ∈ L2(0,T ; H4(Ω)).

Then ut ∈ L2(0,T ; H2(Ω)) and utt ∈ L2(0,T ; L2(Ω)). The time

embedding lemma implies that u ∈ C([0,T ]; H3(Ω)) and

ut ∈ C([0,T ]; H1(Ω)). Similar to the previous case, that

u ∈ C([0,T ]; H3(Ω)) implies that u0 = 0 on the boundary.

Moreover, ut = f + ∆u in Ω and ut = 0 on ∂Ω, the fact that

ut ∈ C([0,T ]; H1(Ω)) implies that

limt→0‖ut − ut (0)‖H1/2(∂Ω) = 0

which implies the second order compatibility condition

∆u0 = f (0) on ∂Ω.



Define uk =∂k−1

∂tk−1

∣∣∣t=0

(f − Lu) for k ∈ N.

Definition (Compatibility conditions for (D))

For k ∈ N ∪ 0, the (k + 1)th-order compatibility condition forthe parabolic initial-boundary value problem (D) is given by

uk = 0 on ∂Ω.

Definition (Compatibility conditions for (N))

For k ∈ N ∪ 0, the (k + 1)th-order compatibility condition forthe parabolic initial-boundary value problem (N) is given by

aij (0)∂uk

∂xjNi =

∂k g∂tk (0)− Ni

k−1∑`=0

(k`

)∂k−`aij

∂tk−` (0)∂u`

∂xjon ∂Ω ,


The functional framework for parabolic equations

We introduce the function spaces that consist of space-time

dependent functions so that one time derivative scales like two

space derivatives. For k ≥ 1, we define

V kD (T ; Ω)

≡

u ∈ L2(0,T ; Hk (Ω))∣∣∣∂ j

t u ∈ L2(0,T ; Hk−2j (Ω)) if 0≤ j≤[

k+12

],

andV k

N (T ; Ω)

≡

u ∈ L2(0,T ; Hk (Ω))∣∣∣∂ j

t u ∈ L2(0,T ; Hk−2j (Ω)) if 0≤ j≤[

k2

],(

k − 2[

k2

])∂

[ k2 ]+1

t u ∈ L2(0,T ; H1(Ω)′),

equipped with norm

‖u‖VkD(T ;Ω) =

[ k+12 ]∑

j=0

‖∂ jt u‖L2(0,T ;Hk−2j (Ω)),


The functional framework for parabolic equations

and

‖u‖VkN (T ;Ω) =

[ k2 ]∑

j=0

‖∂ jt u‖L2(0,T ;Hk−2j (Ω)) +

(k − 2

[k2

])‖∂[ k

2 ]+1t u‖L2(0,T ;H1(Ω)′) .

For the regularity of the boundary data of the Neumann

problem, for k ∈ N ∪ 0,−1 we define

G k+0.5∂Ω (T )

≡

g∈L2(0,T ; Hk+ 12 (∂Ω))

∣∣∣∂ jt g∈L2(0,T ; Hk+ 1

2−2j (∂Ω)) if 0≤ j≤[

k+12

]∂

k/2+1t g ∈ L2(0,T ; H−

12 (∂Ω))

equipped with norm

‖g‖Gk+0.5∂Ω (T ) =

[ k+12 ]∑

j=0

‖∂ jt g‖L2(0,T ;Hk+0.5−2j (∂Ω))

+(k + 1− 2

[k+1

2

])‖∂[ k+3

2 ]t g‖L2(0,T ;H−0.5(∂Ω)) .


The regularity theory for parabolic equations

Theorem

Assume that ∂Ω is of class Cm+2, ∂t + L is uniformly parabolic,

aij ,bi , c ∈ Cm+1(Ω× [0,T ]), and[aij] is symmetric. Then for

any u0 ∈ Hm+1(Ω) ∩ H10 (Ω), f ∈ Vm

D (T ; Ω), the unique weak

solution u to the parabolic initial-boundary value problem (D) in

fact belongs to Vm+2D (T ; Ω) and satisfies

‖u‖Vm+2D (T ;Ω) ≤ C

[‖u0‖Hm+1(Ω) + ‖f‖Vm

D (T ;Ω)

],

provided that the compatibility conditions are valid up to([m2

]+ 1)-th order.



Theorem

Assume that ∂Ω is of class Cm+2, ∂t + L is uniformly parabolic,

aij ,bi , c ∈ Cm+1(Ω× [0,T ]), and[aij] is symmetric. Then for

any u0 ∈ Hm+1(Ω), f ∈ VmN (T ; Ω), g ∈ G m+0.5

∂Ω (T ), the unique

weak solution u to the parabolic initial-boundary value problem

(N) in fact belongs to Vm+2N (T ; Ω) and satisfies

‖u‖Vm+2N (T ;Ω) ≤ C

[‖u0‖Hm+1(Ω) + ‖f‖Vm

N (T ;Ω) + ‖g‖Gm+0.5∂Ω (T )

],

provided that the compatibility conditions are valid up to[m + 12

]-th order.



Let Ω = S1 × (0,1) so that ∂Ω = x2 = 0 ∪ x2 = 1, and we

consider

ut −∆u = f in Ω× (0,T ),

u = u0 on Ω× t = 0, (U)

u = 0 on ∂Ω× (0,T ).

Suppose that f ∈ L2(0,T ; H2(Ω)) with ft ∈ L2(0,T ; L2(Ω)), and

u0 ∈ H3(Ω) satisfies the first and the second order compatibility

conditions; that is,

u0 ∈ H10 (Ω) and u1 ≡ f (0) + ∆u0 ∈ H1

0 (Ω).

We show that u ∈ L2(0,T ; H4(Ω)) with ut ∈ L2(0,T ; H2(Ω)) and

utt ∈ L2(0,T ; L2(Ω)).



Let w be the weak solution towt −∆w = ft in Ω× (0,T ),

w = u1 on Ω× t = 0, (W)

w = 0 on ∂Ω× (0,T ).

We note that since u1 ∈ H10 (Ω) and ft ∈ L2(0,T ; L2(Ω)), the

existence of the weak solution is guaranteed by the former

result.

Testing the equation against wt , we find that

‖wt‖2L2(Ω) +

12

ddt‖∇w‖2

L2(Ω) ≤ ‖ft‖L2(Ω)‖wt‖L2(Ω)

thus Young’s inequality implies that

‖wt‖2L2(Ω) +

ddt‖∇w‖2

L2(Ω) ≤ ‖ft‖2L2(Ω)



Therefore, w ∈ L∞(0,T ; H1(Ω)) and wt ∈ L2(0,T ; L2(Ω))

satisfying that

maxt∈[0,T ]

‖w(t)‖2H1(Ω) +

∫ T

0‖wt (t)‖2

L2(Ω)dt ≤ ‖u1‖2H1(Ω) + ‖ft‖2

L2(0,T ;L2(Ω)).

The elliptic estimates further suggest that ‖w(t)‖2L2(0,T ;H2(Ω))

shares the same upper bound.

In order to show that u ∈ L2(0,T ; H4(Ω)), we need to show that

ut ∈ L2(0,T ; H2(Ω)). To see this, we only need to show that

w = ut ; nevertheless, integrating the weak formulation of (W) in

time over the time interval (0, t),



we find that

(w , ϕ)L2(Ω) − (u1, ϕ)L2(Ω) +(∇∫ t

0w(s)ds,∇ϕ

)L2(Ω)

= (f , ϕ)L2(Ω) −(f (0), ϕ

)L2(Ω)

∀ ϕ ∈ H10 (Ω).

Let v(t) = u0 +

∫ t

0w(s)ds, then v satisfies

(vt , ϕ)L2(Ω) + (∇v ,∇ϕ)L2(Ω) = (f , ϕ)L2(Ω) + (u1 − f (0)−∆u0, ϕ)L2(Ω).

This implies that v is a weak solution to (U). However, we know

that the weak solution is unique, so v = u; thus w = vt = ut .



Remark: If Ω = Tn, then in principle one can test the equation

against ∆3u to obtain that

12

ddt‖∇∆u‖2

L2(Tn) + ‖∆2u‖2L2(Tn) ≤ ‖∆f‖L2(Tn)‖∆2u‖L2(Tn)

which, together with Young’s inequality, further implies that

ddt‖∇∆u‖2

L2(Tn) + ‖∆2u‖2L2(Tn) ≤ ‖f‖

2H2(Tn)

Therefore, if f ∈ L2(0,T ; H2(Tn)) (but without assuming that

ft ∈ L2(0,T ; L2(Tn))), one conclude that u ∈ L∞(0,T ; H3(Tn))

and u ∈ L2(0,T ; H4(Tn)).


The time dependent Stokes problem

Let u : Ω→ Rn and p : Ω→ R denote the fluid velocity and

pressure, respectively. We consider

ut −∆u +∇p = f in Ω× (0,T ),

divu = 0 in Ω× (0,T ),

u = u0 on Ω× t = 0,u = 0 on ∂Ω× (0,T ).

(S)

Recall that

H10,div(Ω) =

u ∈ H1

0 (Ω;Rn)∣∣∣ divu = 0

.

Similar to the weak formulation to the parabolic equations and

the steady Stokes equations, we have the following



Definition

A vector-valued function u ∈ L2(0,T ; H10,div(Ω)) with time

derivative ut ∈ L2(0,T ; H10,div(Ω)′) is said to be a weak solution

to the Stokes equations (S) if

〈ut , ϕ〉+ (∇u,∇ϕ)L2(Ω;Rn2 )= (f , ϕ)L2(Ω;Rn) ∀ ϕ∈H1

0,div(Ω), a.e. t ∈ (0,T )

andu = u0 on Ω× t = 0.

The integral equality is called the variational formulation of the

Stokes equations (S).



Let L2div(Ω) denote the collection of functions u in L2(Ω;Rn)

such that divu = 0.

Theorem

For any u0 ∈ L2div(Ω) and f ∈ L2(0,T ; L2(Ω;Rn)), there exists a

unique weak solution u to the Stokes equations (S).

The proof of this theorem is almost identical to the existence

and uniqueness of the weak solution to the parabolic equations,

except that the Stokes problem is vector-valued, and we need

to use H10,div(Ω) basis in the Galerkin method.

The existence of p is guaranteed by the Lagrange multiplier

lemma.


The regularity theory of the Stokes equations

Since the time dependent Stokes equations looks very similar

to the parabolic equation, we would expect similar regularity

theory for the time dependent Stokes equations, as well as the

need of the compatibility conditions. To fully describe the

compatibility conditions, the initial state of the time derivatives

of u have to be computed, while this involves the computation

of the initial state of the time derivatives of p. For example,

ut (0) must satisfy

u1 ≡ ut (0) = ∆u0 −∇p(0) + f (0) in Ω;

thus we need to know how p(0) is obtained in order to proceed.



Nevertheless, taking the divergence and the normal trace of the

Stokes equations, we find that p(0) solves the elliptic equation

−∆p(0) = divf (0) in Ω,

∂p(0)

∂N= f (0) · N + ∆u0 · N on ∂Ω.

When u0 ∈ Hm(Ω;Rn) ∩ H10,div(Ω) for some m ≥ 2, the solution

p0 ∈ L2(Ω)/R is uniquely determined by the equation.

In general, we have the following



Definition

For k ∈ N, with pk−1 defined by

−∆pk−1 = div(∂k−1t f )(0) in Ω,

∂pk−1

∂N=[(∂k−1

t f )(0) + ∆uk−1]· N on ∂Ω,

the initial state of the k -th time derivative of u is given by

uk ≡ ∂kt |t=0u = ∆uk−1 −∇pk−1 + (∂k

t f )(0).

By elliptic regularity, for 1 ≤ k ≤[m+1

2

],

[ m+12 ]∑

k=1

[‖uk‖Hm+1−2k (Ω) + ‖pk−1‖Hm+2−2k (Ω)

]≤ C

[‖u0‖Hm+1(Ω) +

[ m+12 ]∑

k=0

‖∂kt f‖L2(0,T ;Hm−2k (Ω))

].



Definition

For k ∈ N ∪ 0, the (k + 1)-th order compatibility condition forthe time dependent Stokes equations (S) is uk = 0 on ∂Ω.

Theorem

Assume that ∂Ω is of class Cm+2 for some m ∈ N ∪ 0. Thenfor any u0 ∈ Hm+1(Ω;Rn) ∩ H1

0,div(Ω) and f ∈ VmD (T ; Ω), the

unique weak solution u to the Stokes equations (S) in factbelongs to Vm+2

D (T ; Ω) and satisfies

‖u‖Vm+2D (T ;Ω) ≤ C

[‖u0‖Hm+1(Ω) + ‖f‖Vm

D (T ;Ω)

],

provided that the compatibility conditions are valid up to([m2

]+ 1)-th order.


The decay of the solution to the Stokes equations

Before proceeding, we introduce the space Vs defined by

Vs =

u ∈ Hs(Tn;Rn)∣∣∣ ∫

Tnu(x) dx = 0

.

For any u0 ∈ Vs, we denote etAu0 as the solution u to

ut −∆u +∇p = 0 in Tn × (0,T ),

divu = 0 in Tn × (0,T ),

u = u0 on Tn × t = 0.

Lemma

‖etA‖B(Vs,Vs+1) ≤ Ct−12 for t ∈ (0,1]; that is,

‖etAu0‖Vs+1 ≤ Ct−12 ‖u0‖Vs ∀ u0 ∈ Vs.



Proof: Since Tn is a domain without boundary, the regularity

theory of the Stokes equation suggests that u = etAu0 ∈ Vs+1.

Moreover,12

ddt‖Dk u‖2

L2(Tn) + ‖∇Dk u‖2L2(Tn) = 0;

thus12

ddt‖u‖2

Hs(Tn) + ‖u‖2Hs+1(Tn) = 0.

Therefore,

2∫ t

0‖u(t)‖2

Hs+1(Tn)dt = ‖u0‖2Hs(Tn) − ‖u(t)‖2

Hs(Tn) ≤ ‖u0‖2Hs(Tn).



On the other hand, the Poincare inequality further implies that

ddt‖u‖2

Hs(Tn) ≤ −C‖u‖2Hs(Tn)

for some constant C > 0. As a consequence,

‖u(t)‖2Hs+1(Tn) ≤ e−C(t−r)‖u(r)‖2

Hs+1(Tn).

Integrating the inequality above in r over the time interval (0, t),

we conclude that

t‖u(t)‖2Hs+1(Tn) ≤

∫ t

0‖u(r)‖2

Hs+1(Tn) ≤12‖u0‖2

Hs(Tn).


Lecture 4. The parabolic equations and time dependent Stokes...

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Transcript of Lecture 4. The parabolic equations and time dependent Stokes...