NUMERICAL METHODS FOR PARABOLIC …chenlong/226/ParabolicEquation.pdfNUMERICAL METHODS FOR PARABOLIC...

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NUMERICAL METHODS FOR PARABOLIC EQUATIONS LONG CHEN As a model problem of general parabolic equations, we shall mainly consider the fol- lowing heat equation and study corresponding finite difference methods and finite element methods (1) u t - Δu = f in Ω × (0,T ), u = 0 on Ω × (0,T ), u(·, 0) = u 0 in Ω. Here u = u(x, t) is a function of spatial variable x Ω R n and time variable t (0,T ). The ending time T could be +. The Laplace operator Δ is taking with respect to the spatial variable. For the simplicity of exposition, we consider only homogenous Dirichlet boundary condition and comment on the adaptation to Neumann and other type of boundary conditions. Besides the boundary condition on Ω, we also need to assign the function value at time t =0 which is called initial condition. For parabolic equations, the boundary Ω × (0,T ) Ω ×{t =0} is called the parabolic boundary. Therefore the initial condition can be also thought as a boundary condition. 1. BACKGROUND ON HEAT EQUATION For the homogenous Dirichlet boundary condition without source term, in the steady state, i.e., u t =0, we obtain the Laplace equation Δu =0 in Ω and u| Ω =0. So u =0 no matter what the initial condition is. Indeed the solution will decay to zero exponentially. Let us consider the simplest 1-D problem u t = u xx in R 1 × (0,T ), u(·, 0) = u 0 . Apply Fourier transfer in space ˆ u(k,t)= Z R u(x, t)e -ikx dx. Then c u x =(-iku, d u xx = -k 2 ˆ u, and b u t u t . So we get the following ODE for each Fourier coefficient ˆ u(k,t) ˆ u t = -k 2 ˆ u, ˆ u(·, 0) = ˆ u 0 The solution in the frequency domain is ˆ u(k,t)=ˆ u 0 e -k 2 t . We apply the inverse Fourier transform back to (x, t) coordinate and get u(x, t)= 1 4πt Z R e - (x-y) 2 4t u 0 (y)dy. For a general bounded domain Ω, we cannot apply Fourier transform. Instead we can use the eigenfunctions of A = -Δ 0 (Laplace operator with zero Dirichlet boundary con- dition). Since A is SPD, we know that there exists an orthogonormal basis formed by 1

Transcript of NUMERICAL METHODS FOR PARABOLIC …chenlong/226/ParabolicEquation.pdfNUMERICAL METHODS FOR PARABOLIC...

Page 1: NUMERICAL METHODS FOR PARABOLIC …chenlong/226/ParabolicEquation.pdfNUMERICAL METHODS FOR PARABOLIC EQUATIONS LONG CHEN As a model problem of general parabolic equations, we shall

NUMERICAL METHODS FOR PARABOLIC EQUATIONS

LONG CHEN

As a model problem of general parabolic equations, we shall mainly consider the fol-lowing heat equation and study corresponding finite difference methods and finite elementmethods

(1)

ut −∆u = f in Ω× (0, T ),u = 0 on ∂Ω× (0, T ),

u(·, 0) = u0 in Ω.

Here u = u(x, t) is a function of spatial variable x ∈ Ω ⊂ Rn and time variable t ∈(0, T ). The ending time T could be +∞. The Laplace operator ∆ is taking with respectto the spatial variable. For the simplicity of exposition, we consider only homogenousDirichlet boundary condition and comment on the adaptation to Neumann and other typeof boundary conditions. Besides the boundary condition on ∂Ω, we also need to assign thefunction value at time t = 0 which is called initial condition. For parabolic equations, theboundary ∂Ω× (0, T )∪Ω×t = 0 is called the parabolic boundary. Therefore the initialcondition can be also thought as a boundary condition.

1. BACKGROUND ON HEAT EQUATION

For the homogenous Dirichlet boundary condition without source term, in the steadystate, i.e., ut = 0, we obtain the Laplace equation

∆u = 0 in Ω and u|∂Ω = 0.

So u = 0 no matter what the initial condition is. Indeed the solution will decay to zeroexponentially. Let us consider the simplest 1-D problem

ut = uxx in R1 × (0, T ), u(·, 0) = u0.

Apply Fourier transfer in space

u(k, t) =

∫Ru(x, t)e−ikx dx.

Then ux = (−ik)u, uxx = −k2u, and ut = ut. So we get the following ODE for eachFourier coefficient u(k, t)

ut = −k2u, u(·, 0) = u0

The solution in the frequency domain is u(k, t) = u0e−k2t. We apply the inverse Fourier

transform back to (x, t) coordinate and get

u(x, t) =1√4πt

∫Re−

(x−y)24t u0(y) dy.

For a general bounded domain Ω, we cannot apply Fourier transform. Instead we canuse the eigenfunctions of A = −∆0 (Laplace operator with zero Dirichlet boundary con-dition). Since A is SPD, we know that there exists an orthogonormal basis formed by

1

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eigenfunctions of A, i.e., L2 = spanφ1, φ2, . . . , . We expand the function in such basesu(x, t) =

∑k u

k(t)φk(x). The heat equation then becomes

ukt (t) = λkuk, uk(0) = uk0

and the solution isuk = uk0e

−λkt, for k = 1, 2, . . . .

Again each component will exponentially decay to zero since the eigenvalue λk of A ispositive. And the larger the eigenvalue, the faster the decay rate. This spectral analysisis mainly for theoretical purpose and the numerical application is restricted to special do-mains. In practice, for domains of complex geometry, it is much harder to finding out alleigenvalue and eigenfunctions than solving the heat equation numerically. In the followingsections we will talk about finite difference and finite element methods.

2. FINITE DIFFERENCE METHODS FOR 1-D HEAT EQUATION

In this section, we consider a simple 1-D heat equation

ut = uxx + f in (0, 1)× (0, T ),(2)

u(0) = u(1) = 0, u(x, 0) = u0(x).(3)

to illustrate the main issues in the numerical methods for solving parabolic equations.Let Ω = (0, 1) be decomposed into a uniform grid 0 = x0 < x1 < . . . < xN+1 = 1

with xi = ih, h = 1/N , and time interval (0, T ) be decomposed into 0 = t0 < t1 <. . . < tM = T with tn = nδt, δt = T/M . The tensor product of these two grids give atwo dimensional rectangular grid for the domain Ω× (0, T ). We now introduce three finitedifference methods by discretizing the equation (2) on grid points.

2.1. Forward Euler method. We shall approximate the function value u(xi, tn) by Uniand uxx by second order central difference

uxx(xi, tn) ≈Uni−1 + Uni+1 − 2Uni

h2.

For the time derivative, we use the forward Euler scheme

(4) ut(xi, tn) ≈ Un+1i − Uni

δt.

Together with the initial condition and the source Fni = f(xi, tn), we then end with the

system

Un+1i − Uni

δt=Uni−1 + Uni+1 − 2Uni

h2+ Fni , 1 ≤ i ≤ N, 1 ≤ n ≤M(5)

U0i = u0(xi), 1 ≤ i ≤ N,n = 0.(6)

To write (5) in a compact form, we introduce the parameter λ = δt/h2 and the vectorUn = (Un1 , U

n2 , . . . , U

nN )t. Then (5) can be written as, for n = 0, · · · ,M

Un+1 = AUn + δtF n,

where

A = I + λ∆h =

1− 2λ λ 0 0λ 1− 2λ λ 0. . . . . . . . . . . .0 λ 1− 2λ λ0 0 λ 1− 2λ

.

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NUMERICAL METHODS FOR PARABOLIC EQUATIONS 3

Starting from t = 0, we can evaluate point values at grid points from the initial conditionand thus obtain U0. After that, the unknown at next time step is computed by one matrix-vector multiplication and vector addition which can be done very efficiently without storingthe matrix. Therefore it is also called time marching.

Remark 2.1. Because of the homogenous Dirichlet boundary condition, the boundaryindex i = 0, N + 1 is not included. For Neumann boundary condition, we do need toimpose equations on these two boundary nodes and introduce ghost points for accuratelydiscretize the Neumann boundary condition; See Chapter: Finite difference methods forelliptic equations.

The first issue is the stability in time. When f = 0, i.e., the heat equation without thesource, in the continuous level, the solution should be exponential decay. In the discretelevel, we have Un+1 = AUn and want to control the magnitude of U in certain norms.

Theorem 2.2. When the time step δt ≤ h2/2, the forward Euler method is stable in themaximum norm in the sense that if Un+1 = AUn then

‖Un‖∞ ≤ ‖U0‖∞ ≤ ‖u0‖∞.

Proof. By the definition of the norm of a matrix

‖A‖∞ = maxi=1,··· ,N

N∑j=1

|aij | = 2λ+ |1− 2λ|.

If δt ≤ h2/2, then ‖A‖∞ = 1 and consequently

‖Un‖∞ ≤ ‖A‖∞‖Un−1‖∞ ≤ ‖Un−1‖∞.

Exercise 2.3. Construct an example to show, numerically or theoretically, that if δt >h2/2, then

‖Un‖∞ > ‖U0‖∞.

Theorem 2.2 and Exercise 2.3 imply that to ensure the stability of the forward Eulermethod, we have to choose the time step δt in the size of h2 which is very restrictive, say,h = 10−3 then t = 10−6/2. Although in one step, it is efficient, it will be very expensiveto reach the solution at the ending time T by moving forward with such tiny time step. Fortime dependent equations, we shall consider not only the computational cost for one singletime step but also the total time to arrive a certain stopping time.

The second issue is the accuracy. We first consider the consistency error. Define unI =(u(x1, tn), u(x2, tn), · · · , u(xN , tn))t. We denote a general differential operator as L andits discritization as Lnh . Note that the action Lu is the operator L acting on a continuousfunction u while Lnh is applied to vectors such as Un or unI . The consistent error orso-called truncation error is to pretend we know the exact function values and see whatthe error from the approximation of the differential operator. More precisely, for the heatequation we define Lu = ut −∆u and LnhV = (V n+1 − V n)/δt −∆h/h

2V n and thetruncation error

τn = LnhunI − (Lu)nI .

The error is denoted byEn = Un − unI .

Estimate of the truncation error is straightforward using the Taylor expansion.

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Lemma 2.4. Suppose u is smooth enough. For the forward Euler method, we have

‖τn‖∞ ≤ C1δt+ C2h2.

The convergence or the estimate of the error En is a consequence of the stability andconsistency. The method can be written as

LnhUn = F n = fnI = (Lu)nI .

We then obtain the error equation

(7) Lnh(Un − unI ) = (Lu)nI − LnhunI ,which can be simply written as LnhE

n = −τn.

Theorem 2.5. For the forward Euler method, when δt ≤ h2/2 and the solution u is smoothenough, we have

‖Un − unI ‖∞ ≤ Ctn (δt+ h2).

Proof. We write out the specific error equation for the forward Euler method

En+1 = AEn − δt τn.Consequently

En = AnE0 − δtn−1∑l=1

An−l−1τ l.

Since E0 = 0, by the stability and consistency, we obtain

‖En‖∞ ≤ δtn−1∑l=1

‖τ l‖∞ ≤ Cnδt(δt+ h2) = Ctn (δt+ h2).

2.2. Backward Euler method. Next we introduce the backward Euler method to removethe strong constraint of the time step-size for the stability. The method is simply usingthe backward difference to approximate the time derivative. We list the resulting linearsystems:

Uni − Un−1i

δt=Uni−1 + Uni+1 − 2Uni

h2+ Fni , 1 ≤ i ≤ N, 1 ≤ n ≤M(8)

U0i = u0(xi), 1 ≤ i ≤ N,n = 0.(9)

In the matrix form (8) reads as

(10) (I − λ∆h)Un = Un−1 + δtF n.

Starting fromU0, to compute the value at the next time step, we need to solve an algebraicequation to obtain

Un = (I − λ∆h)−1(Un−1 + δtF n).

The inverse of the matrix, which involves the stiffness matrix of Laplacian operator, is noteasy in high dimensions. For 1-D problem, the matrix is tri-diagonal and can be solvedvery efficiently by the Thomas algorithm.

The gain is the unconditional stability.

Theorem 2.6. For the backward Euler method without source term, i.e., (I −λ∆h)Un =Un−1, we always have the stability

‖Un‖∞ ≤ ‖Un−1‖∞ ≤ ‖u0‖∞.

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NUMERICAL METHODS FOR PARABOLIC EQUATIONS 5

Proof. We shall rewrite the backward Euler scheme as

(1 + 2λ)Uni = Un−1i + λUni−1 + λUni+1.

Therefore for any 1 ≤ i ≤ N ,

(1 + 2λ)|Uni | ≤ ‖Un−1‖∞ + 2λ‖Un‖∞,

which implies(1 + 2λ)‖Un‖∞ ≤ ‖Un−1‖∞ + 2λ‖Un‖∞,

and the desired result then follows.

The truncation error of the backward Euler method can be obtained similarly.

Lemma 2.7. Suppose u is smooth enough. For the backward Euler method, we have

‖τn‖∞ ≤ C1δt+ C2h2.

We then use stability and consistency to give error analysis of the backward Eulermethod.

Theorem 2.8. For the backward Euler method, when the solution u is smooth enough, wehave

‖Un − unI ‖∞ ≤ Ctn (δt+ h2).

Proof. We write the error equation for the backward Euler method as

(1 + 2λ)Eni = En−1i + λEni−1 + λEni+1 − δt τni .

Similar to the proof of the stability, we obtain

‖En‖∞ ≤ ‖En−1‖∞ + δt‖τn‖∞Consequently, we obtain

‖En‖∞ ≤ δtn−1∑l=1

‖τ l‖∞ ≤ Ctn(δt+ h2).

Exercise 2.9. Apply the backward Euler method to the example you constructed in Exer-cise 2.3 to show that numerically the scheme is stable.

From the error analysis, to have optimal convergent order, we also need δt ≈ h2 whichis still restrictive. Next we shall give a unconditional stable scheme with second ordertruncation error O(δt2 + h2).

2.3. Crank-Nicolson method. To improve the truncation error, we need to use centraldifference for time discritization. We keep the forward discretization as (Un+1

i − Uni )/δtbut now treat it is an approximation of ut(xi, tn+1/2). That is we discretize the equationat (xi, tn+1/2). The value Un+1/2

i is taken as average of Uni and Un+1i . We then end with

the scheme: for 1 ≤ i ≤ N, 1 ≤ n ≤M

(11)Un+1i − Uni

δt=

1

2

Uni−1 + Uni+1 − 2Unih2

+1

2

Un+1i−1 + Un+1

i+1 − 2Un+1i

h2+ F

n+1/2i ,

and for 1 ≤ i ≤ N, n = 0

U0i = u0(xi).

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In matrix form, (47) can be written as

(I − 1

2λ∆h)Un+1 = (I +

1

2λ∆h)Un + F n+1/2.

It can be easily verify that the truncation error is improved to second order in time

Lemma 2.10. Suppose u is smooth enough. For Crank-Nicolson method, we have

‖τn‖∞ ≤ C1δt2 + C2h

2.

Exercise 2.11. Prove the maximum norm stability of Crank-Nicolson method with as-sumptions on λ. What is the weakest assumption on λ you can impose?

In the next section, we shall prove Crank-Nicolson method is unconditionally stable inthe l2 norm and thus obtain the following second order convergence; See Exercise 13.

Theorem 2.12. For Crank-Nicolson method, we have

‖En‖ ≤ Ctn(C1δt2 + C2h

2).

3. VON NEUMANN ANALYSIS

A popular way to study the L2 stability of the heat equation is through Fourier analysisand its discrete version. When Ω = R, we can use Fourier analysis. For Ω = (0, 1), wecan use eigenfunctions of Laplacian operator. To study the matrix equation, we establishthe discrete counter part.

Lemma 3.1. If A = diag(b, a, b) be a N ×N matrix, then the eigenvalue of A is

λk = a+ 2b cos θk, k = 1, · · · , Nand its corresponding eigenvector is

φk =√

2 (sin θk, sin 2θk, · · · , sinNθk),

where

θk = kθ =kπ

N + 1.

Proof. It is can be easily verified by the direct calculation.

Remark 3.2. The eigenvectors do not depend on the values a and b!

We then define a scaling of l2 inner product of two vectors as

(U ,V )h = hU tV = h

N∑i=1

UiVi,

which is a mimic of L2 inner product of two functions if we thought Ui and Vi representthe values of corresponding functions at grid points xi.

It is straightforward to verify the eigenvectors φkNk=1 forms an orthonormal basis ofRN with respect to (·, ·)h. Indeed the scaling

√2 is introduced for the normalization. For

any vector V ∈ RN , we expand it in this new basis

V =

N∑k=1

vkφk, with vk = (V ,φk)h.

The discrete Parseval identity is‖V ‖h = ‖v‖,

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NUMERICAL METHODS FOR PARABOLIC EQUATIONS 7

where v = (v1, · · · , vN )t is the vector formed by the coefficients. Theˆin the coefficientindicates this is a mimic of Fourier transform in the discrete level.

Suppose

Un+1 = AUn,

with A = diag(b, a, b). Then the stability in ‖ · ‖h is related to the spectrum radius of A.That is

‖Un+1‖h ≤ ‖A‖h‖Un‖h,

and since A is symmetric with respect to (·, ·)h,

‖A‖h = ρ(A) = max1≤k≤N

|λk(A)|.

Note that ‖ · ‖h = h1/2‖ · ‖. The stability in the scaled l2 norm is equivalent to that in ‖ · ‖norm. Here the scaling h is chosen to be consistent with the scaling of discrete Laplaceoperator.

Now we analyze the l2-stability of the three numerical methods we have discussed.Recall that λ = δ/h2.

Forward Euler Method.

Un+1 = AUn, A = diag(λ, 1− 2λ, λ).

Thus

ρ(A) = max1≤k≤N

|1− 2λ+ 2λ cos θk|.

It is easy to verify that

λ ≤ 1/2 =⇒ ρ(A) ≤ 1.

Let us write A = AN , the uniform stability (with respect to N ) implies λ ≤ 1/2, i.e.,

supNρ(AN ) ≤ 1 =⇒ λ ≤ 1/2.

Therefore we obtain the same condition as that for the maximum norm stability.

Backward Euler Method.

Un+1 = A−1Un, A = diag(−λ, 1 + 2λ,−λ).

Thus

ρ(A−1) =1

ρ(A)= max

1≤k≤N

1

|1 + 2λ− 2λ cos θk|,

and

ρ(A−1) ≤ 1 for any λ > 0.

Therefore backward Euler is also unconditionally stable in l2 norm.

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Crank-Nicolson Method.

Un+1 = A−1BUn, where

A =1

2diag(−λ, 2 + 2λ,−λ),

B =1

2diag(λ, 2− 2λ, λ)

SinceA andB have the same eigenvectors, we have

ρ(A−1B) = max1≤k≤N

λk(B)

λk(A)= max

1≤k≤N

|1− λ+ λ cos θk||1 + λ− λ cos θk|

,

andρ(A−1B) ≤ 1 for any λ > 0.

Therefore we proved Crank-Nicolson method is unconditionally stable in l2 norm.

Question 3.3. Is Crank-Nicolson method unconditional stable in the maximum norm? Youcan google “Crank Nicolson method maxnorm stability” to read more.

For general schemes, one can write out the matrix form and plug in the coefficients todo the stability analysis. This methodology is known as von Neumann analysis. Formallywe replace

(12) Unj ← ρneiθj

in the scheme and obtain a formula of the amplification factor ρ = ρ(θ). The uniformstability is obtained by considering inequality

max0≤θ≤π

ρ(θ).

To fascinate the calculation, one can factor out ρneiθj and consider only the difference ofindices.

The eigen-function is changed to eiθ = cos(θ) + i sin(θ) to account for possible dif-ferent boundary conditions. The the frequency of discrete eigenvectors φk = eiθk(1:N)t ischanged to a continuous variable θ in (12). We skip the index k and take the maximum ofθ over [0, π] while the range of the discrete angle θk satisfying hπ ≤ θk ≤ (1− h)π.

As an example, the readers are encouraged to do the following exercise. Use one exer-cise as an example.

Exercise 3.4 (The θ method). For θ ∈ [0, 1], we use the scheme: for 1 ≤ i ≤ N, 1 ≤ n ≤M

(13)Un+1i − Uni

δt= (1− θ)

Uni−1 + Uni+1 − 2Unih2

+ θUn+1i−1 + Un+1

i+1 − 2Un+1i

h2,

and for 1 ≤ i ≤ N, n = 0U0i = u0(xi).

Give a complete error analysis (stability, consistency, and convergence) of the θ method.Note that θ = 0 is forward Euler, θ = 1 is backward Euler, and θ = 1/2 is Crank-Nicolsonmethod.

Exercise 3.5 (Leap-frog method). The scheme is an explicit version of Crank-Nicolson.For 1 ≤ i ≤ N, 1 ≤ n ≤M

(14)Un+1i − Un−1

i

2δt=Uni−1 + Uni+1 − 2Uni

h2.

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NUMERICAL METHODS FOR PARABOLIC EQUATIONS 9

The computation of Un+1 can be formulated as linear combination of Un and Un−1.Namely it is an explicit scheme. To start the computation, we need two ‘initial’ values.One is the real initial condition

U0i = u0(xi), 1 ≤ i ≤ N.

The other U1, so-called computational initial condition, can be computed using one stepforward or backward Euler method.

Give a complete error analysis (stability, consistency, and convergence) of the leap-frogmethod.

4. VARIATIONAL FORMULATION AND ENERGY ESTIMATE

We multiply a test function v ∈ H10 (Ω) and apply the integration by part to obtain a

variational formulation of the heat equation (1): given an f ∈ L2(Ω) × (0, T ], for anyt > 0, find u(·, t) ∈ H1

0 (Ω), ut ∈ L2(Ω) such that

(15) (ut, v) + a(u, v) = (f, v), for all v ∈ H10 (Ω).

where a(u, v) = (∇u,∇v) and (·, ·) denotes the L2-inner product.We then refine the weak formulation (15). The right hand side could be generalized to

f ∈ H−1(Ω). Since ∆ map H10 (Ω) to H−1(Ω), we can treat ut(·, t) ∈ H−1(Ω) for a

fixed t. We introduce the Sobolev space for the time dependent functions

Lq(0, T ;W k,p(Ω)) := u(x, t) | ‖u‖Lq(0,T ;Wk,p(Ω)) :=

(∫ T

0

‖u(·, t)‖qk,p dt

)1/q

<∞.

Our refined weak formulation will be: given f ∈ L2(0, T ;H−1(Ω)) and u0 ∈ H10 (Ω),

find u ∈ L2(0, T ;H10 (Ω)) and ut ∈ L2(0, T ;H−1(Ω)) such that

(16)〈ut, v〉+ a(u, v) = 〈f, v〉, ∀v ∈ H1

0 (Ω), and t ∈ (0, T )a.e.u(·, 0) = u0

where 〈·, ·〉 is the duality pair of H−1 and H10 . We assume the equation (16) is well posed.

For the existence and uniqueness of the solution [u, ut], we refer to [1]. In most places, weshall still use the formulation (15) and assume f, ut ∈ L2.

Remark 4.1. The topology for the time variable should be also treat in L2 sense. Butin (15) and (16) we still pose the equation point-wise (almost everywhere) in time. Inparticular, one has to justify the point value u(·, 0) does make sense for a L2 type functionwhich can be proved by the regularity theory of the heat equation.

To easy the stability analysis, we treat t as a parameter and the function u = u(x, t) asa mapping

u : [0, T ]→ H10 (Ω),

defined asu(t)(x) := u(x, t) (x ∈ Ω, 0 ≤ t ≤ T ).

With a slight abuse of the notation, here we still use u(t) to denote the map. The norm‖u(t)‖ or ‖u(t)‖1 is taken with respect to the spatial variable.

We then introduce the operator

L : L2(0, T ;H10 (Ω))→ L2(0, T ;H−1(Ω))× L2(Ω)

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as

(Lu)(·, t) = ∂tu−∆u in H−1(Ω), for t ∈ (0, T ] a.e.

(Lu)(·, 0) = u(·, 0).

Then the equation (16) can be written as

Lu = [f, u0].

Here we explicitly include the initial condition. The spatial boundary condition is buildinto the space H1

0 (Ω). In most places, when it is clear from the context, we also use L forthe differential operator only.

We shall prove several stability results of L which are known as energy estimates in [3].

Theorem 4.2. Suppose [u, ut] is the solution of (15) and ut ∈ L2(0, T ;L2(Ω)), then fort ∈ (0, T ] a.e.

(17) ‖u(t)‖ ≤ ‖u0‖+

∫ t

0

‖f(s)‖ds

(18) ‖u(t)‖2 +

∫ t

0

|u(s)|21 ds ≤ ‖u0‖2 +

∫ t

0

‖f(s)‖2−1 ds,

(19) |u(t)|21 +

∫ t

0

‖ut(s)‖2 ds ≤ |u0|21 +

∫ t

0

‖f(s)‖2 ds.

Proof. The solution is defined via the action of all test functions. The art of the energyestimate is to choose an appropriate test function to extract desirable information.

We first choose v = u to obtain

(ut, u) + a(u, u) = (f, u).

We manipulate these three terms as:

• (ut, u) =

∫Ω

1

2(u2)t =

1

2

d

dt‖u‖2 = ‖u‖ d

dt‖u‖;

• a(u, u) = |u|21;

• |(f, u)| ≤ ‖f‖‖u‖ or |(f, u)| ≤ ‖f‖−1|u|1 ≤1

2(‖f‖2−1 + |u|21).

The inequality (17) is an easy consequence of the following inequality

‖u‖ ddt‖u‖ ≤ ‖f‖‖u‖.

From1

2

d

dt‖u‖2 + |u|21 ≤

1

2(‖f‖2−1 + |u|21),

we getd

dt‖u‖2 + |u|21 ≤ ‖f‖2−1.

Integrating over (0, t), we obtain (18).The inequality (19) can be proved similarly by choosing v = ut and left as an exercise.

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NUMERICAL METHODS FOR PARABOLIC EQUATIONS 11

From (18), we can obtain a better stability for the operator

L : L2(0, T ;H10 (Ω))→ L2(0, T ;H−1(Ω))× L2(Ω)

as‖u‖L2(0,T ;H1

0 (Ω)) ≤ ‖u‖0 + ‖f‖L2(0,T ;H−1(Ω)).

Since the equation is posed a.e for t, we could also obtain maximum-norm estimate intime. For example, (17) can be formulated as

‖u‖L∞(0,T ;L2(Ω)) ≤ ‖u0‖+ ‖f‖L1(0,T ;L2(Ω)),

and (19) implies

‖u‖L∞(0,T ;H10 (Ω)) ≤ |u0|1 + ‖f‖L2(0,T ;L2(Ω)).

Exercise 4.3. Prove the stability result (19).

Exercise 4.4. Prove the energy estimate

(20) ‖u‖2 ≤ e−λt‖u0‖2 +

∫ t

0

e−λ(t−s))‖f‖2−1 ds,

where λ = λmin(−∆) > 0. The estimate (20) shows that the effect of the initial data isexponential decay.

5. FINITE ELEMENT METHODS: SEMI-DISCRETIZATION IN SPACE

Let Th, h→ 0 be a quasi-uniform family of triangulations of Ω. The semi-discretizedfinite element method is: given f ∈ V′h × (0, T ], u0,h ∈ Vh ⊂ H1

0 (Ω), find uh ∈L2(0, T ;Vh) such that

(21)

(∂tuh, vh) + a(uh, vh) = 〈f, vh〉, ∀vh ∈ Vh, t ∈ R+.uh(·, 0) = u0,h

The scheme (21) is called semi-discretization since uh is still a continuous (indeeddifferential) function of t. The initial condition u0 is approximated by u0,h ∈ Vh andchoices of u0,h is not unique.

We can expand uh =∑Ni=1 ui(t)ϕi(x), where ϕi is the standard hat basis at the vertex

xi for i = 1, · · · , N , the number of interior nodes, and the corresponding coefficient ui(t)now is a function of time t. The solution uh can be computed by solving an ODE system

(22) Mu+Au = f ,

where u = (u1, · · · , uN )t is the coefficient vector, M ,A are the mass matrix and thestiffness matrix, respectively, and f = (f1, · · · , fN )t. For the linear finite element, one canchose a numerical quadrature to make M diagonal, which is known as the mass lumping,so that the ODE system (22) can be solved efficiently by mature ODE solvers.

We shall apply our abstract error analysis developed in Chapter: Unified Error Analysisto estimate the error u− uh in certain norms. The setting is

• X = L2(0, T ;H10 (Ω)), and ‖u‖X =

(∫ T0|u(t)|21 dt

)1/2

• Y = L2(0, T ;H−1(Ω), and ‖f‖Y =(∫ T

0‖f(t)‖2−1 dt

)1/2

• Xh = L2(0, T ;Vh), and ‖uh‖Xh=(∫ T

0|uh(t)|21 dt

)1/2

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12 LONG CHEN

• Yh = L2(0, T ;V′h), and ‖fh‖Yh =(∫ T

0‖fh(t)‖2−1,h dt

)1/2

. Recall the dualnorm

‖fh‖−1,h = supvh∈Vh

〈fh, vh〉|vh|1

, for fh ∈ V′h.

• Ih = Rh(t) : H10 (Ω) → Vh is the Ritz-Galerkin projection, i.e., Rhu ∈ Vh such

thata(Rhu, vh) = a(u, vh), ∀vh ∈ Vh.

• Πh = Qh(t) : H−1(Ω)→ V′h is the projection

〈Qhf, vh〉 = 〈f, vh〉, ∀vh ∈ Vh.• Ph : Xh →X is the natural inclusion• L : X → Y ×H1

0 (Ω) is Lu = ∂tu−∆u,Lu(·, 0) = u(·, 0), andLh = L|Xh

: Xh → Yh × Vh.The equation we are solving is

Lhuh = Qhf, in V′h,∀t ∈ (0, T ] a.e.

uh(·, 0) = u0,h.

5.1. Stability. Adapt the proof of the energy estimate forL, we can obtain similar stabilityresults for Lh. The proof is almost identical and thus skipped here.

Theorem 5.1. Suppose uh satisfy Lhuh = fh, uh(·, 0) = u0,h, then

‖uh(t)‖ ≤ ‖u0,h‖+

∫ t

0

‖fh(s)‖ ds(23)

‖uh(t)‖2 +

∫ t

0

|uh(s)|21 ds ≤ ‖u0,h‖2 +

∫ t

0

‖fh(s)‖2−1,h ds,(24)

|uh(t)|21 +

∫ t

0

‖∂tuh(s)‖2 ds ≤ |u0,h|21 +

∫ t

0

‖fh(s)‖2 ds.(25)

Note that in the energy estimate (24), the ‖ · ‖−1 is replaced by a weaker norm ‖ · ‖−1,h

since we can apply the inequality

〈fh, uh〉 ≤ ‖fh‖−1,h|uh|1.The weaker norm ‖ · ‖−1,h can be estimated by

‖f‖−1,h ≤ ‖f‖−1 ≤ C‖f‖.

5.2. Consistency. Recall that the consistency error is defined as

LhIhu− Lhuh = LhIhu−ΠhLu.The choice of Ih = Rh simplifies the consistency error analysis.

Lemma 5.2. For the semi-discretization, we have the error equation

Lh(Rhu− uh) = ∂t(Rhu−Qhu), t > 0, in V′h,(26)

(Rhu− uh)(·, 0) = Rhu0 − u0,h.(27)

Proof. Let A = −∆. By our definition of consistency, the error equation is: for t > 0

Lh(Rhu− uh) = LhRhu−QhLu = ∂t(Rhu−Qhu) + (ARhu−QhAu).

The desired result then follows by noting that ARh = QhA in V′h. One can also verify(26) by writing out the variational formulation.

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NUMERICAL METHODS FOR PARABOLIC EQUATIONS 13

The error equation (26) holds in V′h which is a weak topology. The motivation to chooseIh = Rh is that in this weak topology

〈ARhu, vh〉 = (∇Rhu,∇vh) = (∇u,∇v) = 〈Au, vh〉 = 〈QhAu, vh〉,

or simply in operator formARh = QhA.

This technique is firstly proposed by Wheeler [4].Apply the stability to the error equation, we obtain the following estimate on the discrete

error Rhu− uh.

Theorem 5.3. The solution uh of (21) satisfy the following error estimate

(28) ‖Rhu− uh‖ ≤ ‖u0,h −Rhu0‖+

∫ t

0

‖∂tQhu− ∂tRhu‖ ds

(29)

‖Rhu−uh‖2 +

∫ t

0

|(uh−Rhu)|21 ds ≤ ‖u0,h−Rhu0‖2 +

∫ t

0

‖∂tQhu−∂tRhu‖2−1,h ds,

(30)

|Rhu− uh|21 +

∫ t

0

‖∂t(Rhu− uh)‖2 ds ≤ |u0,h −Rhu0|21 +

∫ t

0

‖∂tQhu− ∂tRhu‖2 ds.

We then estimate the two terms involved in these error estimates. The first issue is onthe choice of u0,h. An optimal one is obviously u0,h = Rhu0 so that no error comingfrom approximation of the initial condition. However, this choice requires the inversion ofa stiffness matrix which is not cheap. A simple choice would be the nodal interpolation,i.e. u0,h(xi) = u0(xi) or any other choice with optimal approximation property

(31) ‖u0,h −Rhu0‖ ≤ ‖u0 − u0,h‖+ ‖u0 −Rhu0‖ . h2‖u0‖2,

and similarly|u0,h −Rhu0|1 . h‖u0‖2.

According to (20), the affect of the initial boundary error will be exponentially decay tozero as t goes to infinity. So in practice, we can choose the simple nodal interpolation.

On the estimate of the second term, we assume ut ∈ H2(Ω) and assume H2-regularityresult hold for Poisson equation (for example, the domain is smooth or convex), then

(32) ‖∂tQhu− ∂tRhu‖ = ‖Qh(I −Rh)ut‖ ≤ ‖(I −Rh)ut‖ . h2‖ut‖2.

The negative norm can be bounded by the L2-norm as

‖∂tQhu− ∂tRhu‖−1,h ≤ ‖∂tQhu− ∂tRhu‖−1 ≤ C‖∂tQhu− ∂tRhu‖ . h2‖ut‖2.

When using quadratic and above polynomial, we can prove a stronger estimate for thenegative norm

(33) ‖u−Rhu‖−1 ≤ Chr+1‖u‖r

where r is the order of the element, then we could obtain a superconvergence in L2-norm

‖Rhu− uh‖ ≤ Chr+1

(∫ t

0

‖ut‖2r ds)1/2

.

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14 LONG CHEN

5.3. Convergence. The convergence of the discrete error comes from the stability andconsistency. For functions in finite element space, the H1 semi-norm can be estimate by asimple inverse inequality

|Rhu− uh|1 ≤ Ch−1‖Rhu− uh‖.

Theorem 5.4. Suppose the solution u to (16) satisfying ut ∈ L2(0, T ;H2(Ω)) and theH2

regularity holds. Let uh be the solution of (21) with u0,h satisfying (31). We then have

(34) h−1|Rhu− uh|1 + ‖Rhu− uh‖ ≤ Ch2

(‖u0‖2 +

∫ t

0

‖ut‖2 ds).

To estimate the true error u − uh, we need the approximation error estimate of theprojection Rh

h−1‖u−Rhu‖+ |u−Rhu|1 ≤ Ch‖u‖2.

Theorem 5.5. Suppose the solution u to (16) satisfying ut ∈ L2(0, T ;H2(Ω)). Thenthe solution uh of (21) with u0,h having optimal approximation property (31) satisfy thefollowing optimal order error estimate:

(35) h−1|u− uh|1 + ‖u− uh‖ ≤ Ch2

(‖u0‖2 +

∫ t

0

‖ut‖2 ds).

5.4. *Superconvergence and maximum norm estimate. The error eh = Rhu− uh sat-isfies the evolution equation (26) with eh(0) = 0 with the chose u0,h = Rhu0, i.e.,

(36) ∂teh +Aheh = Qh∂t(Rhu− u).

Therefore

eh(t) =

∫ t

0

exp(−Ah(t− s))τhds, with τh = Qh∂t(Rhu− u).

Due to the smoothing effect of the semi-group e−Aht, we have the following estimate.Here we follow the work by Garcia-Archilla and Titi [2].

Lemma 5.6 (Smoothing property of the heat kernel). For τh ∈ Vh, we have

(37) max0≤t≤T

∥∥∥∥∫ t

0

exp(−Ah(t− s))Ahτhds∥∥∥∥ ≤ C| log h| max

0≤t≤T‖τh‖.

Proof. Let λm and λM be the minimal and maximal eigenvalue of Ah. Then it is easy tocheck ∥∥∥e−Ah(t−s)Ah

∥∥∥ ≤λMe

−λM (t−s) if (t− s) ≤ λ−1M ,

(t− s)−1 if λ−1M ≤ (t− s) ≤ λ−1

m ,

λme−λm(t−s) if (t− s) ≥ λ−1

m .

Note that λm = O(1) and λM ≤ Ch−2. We get

max0≤t≤T

∥∥∥∥∫ t

0

e−Ah(t−s)Ahτhds

∥∥∥∥ ≤ C| log h| max0≤t≤T

‖τh‖.

Theorem 5.7 (Superconvergence in H1-norm). Suppose the solution u to (16) satisfyingut ∈ L∞(0, T ;H2(Ω)). Let uh be the solution of (21) with u0,h = Rhu0. Then

(38) max0≤t≤T

|Rhu− uh|1 ≤ C| log h|h2 max0≤t≤T

‖ut‖2.

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NUMERICAL METHODS FOR PARABOLIC EQUATIONS 15

When ut ∈ L2(0, T ;H2(Ω)), then

(39) |Rhu− uh|1 ≤ Ch2

(∫ t

0

‖ut‖22 ds)1/2

.

Proof. We multiply A1/2 to (36) and apply Lemma 5.6 to get

|eh(T )|1 = ‖A1/2h eh(T )‖ =

∥∥∥∥∥∫ T

0

e−Ah(T−s)A1/2h τhds

∥∥∥∥∥≤

∥∥∥∥∥∫ T

0

e−Ah(T−s)Ah ds

∥∥∥∥∥ max0≤t≤T

‖A−1/2h τh‖

≤ C| log h|h2 max0≤t≤T

‖ut‖2.

In the last step, we have used the fact ‖A−1/2h τh‖ = ‖τh‖−1 ≤ ‖τh‖.

To get (39), we use the energy estimate (30).

Since the optimal convergent rate for |u − Rhu|1 or |u − uh|1 is only first order, theestimate (38) and (39)) are called superconvergence.

To control the maximum norm, we use the discrete embedding result (for 2-D only)

‖Rhu− uh‖∞ ≤ C| log h||Rhu− uh|1,

and the error estimate of Rh in the maximum norm

‖u−Rhu‖∞ ≤ C| log h|h2‖u‖2,∞,

to obtain the following result.

Theorem 5.8 (Maximum norm estimate for linear element in two dimensions). Supposethe solution u to (16) satisfying u ∈ L∞(0, T ;W 2,∞) and ut ∈ L2(0, T ;W 2,∞) or ut ∈L∞(0, T ;W 2,∞(Ω)). Let uh be the solution of (21) with u0,h = Rhu0. Then in twodimensions

‖(u− uh)(t)‖∞ ≤ C| log h|h2

[‖u‖2,∞ +

(∫ t

0

‖ut‖22 ds)1/2

],(40)

max0≤t≤T

‖(u− uh)(t)‖∞ ≤ C| log h|2h2 max0≤t≤T

[‖u(t)‖2,∞ + ‖ut(t)‖2] .(41)

For high order elements, we could get superconvergence in L2 norm. Let us define theorder of the polynomial as the degree plus 1, which is the optimal order when measuringthe approximation property in Lp norm. For example, the order of the linear polynomial is2. When the order of polynomial r is bigger than 3 (,i.e., quadratic and above polynomial),we can prove a stronger estimate for the negative norm

(42) ‖u−Rhu‖−1 ≤ Chr+1‖u‖r.

Using the technique in Lemma 5.6 and Theorem 5.7, we have the following estimate onthe L2 norm.

Theorem 5.9 (Superconvergence in L2-norm). Suppose the solution u to (16) satisfyingut ∈ L∞(0, T ;Hr(Ω)). Let uh be the solution of (21) with u0,h = Rhu0. Then

(43) max0≤t≤T

‖(Rhu− uh)(t)‖ ≤ C| log h|hr+1 max0≤t≤T

‖(ut)(t)‖2r.

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16 LONG CHEN

When ut ∈ L2(0, T ;Hr(Ω)), then

(44) ‖Rhu− uh‖ ≤ Chr+1

(∫ t

0

‖ut‖2r ds)1/2

.

Proof. From (36), we have

‖eh(T )‖ =

∥∥∥∥∥∫ T

0

e−Ah(T−s)τhds

∥∥∥∥∥≤

∥∥∥∥∥∫ T

0

e−Ah(T−s)A−1/2h ds

∥∥∥∥∥ max0≤t≤T

‖A−1/2h τh‖

≤ C| log h|‖τh‖−1.

In the second step, we have used the estimate∥∥∥∥∥∫ T

0

e−Ah(T−s)A−1/2h ds

∥∥∥∥∥ ≤ C| log h|,

which can be proved by the estimate ‖e−Ah(T−s)A−1/2h ‖ ≤ (t− s)−1

To prove (44), we simply use the energy estimate (29) and (42).

In 1-D, using the inverse inequality ‖Rhu − uh‖∞ ≤ h−1/2‖Rhu − uh‖, we couldobtain the superconvergence in the maximum norm

(45) ‖Rhu− uh‖∞ ≤ Chr+1/2

(∫ t

0

‖ut‖2r ds)1/2

.

Again using the inverse inequality ‖uh −Rhu‖∞ ≤ Ch−1‖uh −Rhu‖, the supercon-vergence of L2 norm, and the maximum norm estimate of Ritz-Galerkin projection, forr ≥ 3, ‖u−Rhu‖∞ ≤ Chr‖u‖r,∞, we can improve the maximum norm error estimate.

Theorem 5.10 (Maximum norm estimate for high order elements in two dimensions). Sup-pose the solution u to (16) satisfying u ∈ L∞(0, T ;W 2,∞) and ut ∈ L2(0, T ;W 2,∞) orut ∈ L∞(0, T ;W 2,∞(Ω)). Let uh be the solution of (21) with u0,h = Rhu0. Then in twodimensions and for r ≥ 3

‖(u− uh)(t)‖∞ ≤ Chr[‖u‖r,∞ +

(∫ t

0

‖ut‖2r ds)1/2

].(46)

6. FINITE ELEMENT METHODS: SEMI-DISCRETIZATION IN TIME

In this section, we consider the semi-discretization in time. We first discretize the timeinterval (0, T ) into a uniform grid with size δt = T/N and denoted by tn = nδt for n =0, . . . N . A continuous function in time will be interpolated into a vector by (Inf)(·, tn) =f(·, tn). Recall that A = −∆ : H1

0 → H−1. We list three schemes in operator form.• Forward Euler Method. u0 = u0

un − un−1

δt+Aun−1 = fn−1.

• Backward Euler Method. u0 = u0

un − un−1

δt+Aun = fn.

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NUMERICAL METHODS FOR PARABOLIC EQUATIONS 17

• Crank-Nicolson Method. u0 = u0

un − un−1

δt+A(un + un−1)/2 = fn−1/2.

Note that these equations hold in H−1 sense. Taking Crank-Nicolson as an example, theequation reads as

(47)1

δt(un − un−1, v) +

1

2(∇un +∇un−1,∇v) = (fn−1/2, v) for all v ∈ H1

0 .

We now study the stability of these schemes. We rewrite the Backward Euler method as

(I + δtA)un = un−1 + δtfn.

Since A is SPD, λmin(I + δtA) ≥ 1 and consequently, λmax((I + δtA)−1) ≤ 1. Thisimplies the L2 stability

(48) ‖un‖ ≤ ‖un−1‖+ δt‖fn‖ ≤ ‖u0‖+

n∑k=1

δt‖fk‖.

The stability (49) is the discrete counter part of (17): discretize the integral∫ tn

0‖f‖ds by

a Riemann sum.Similarly one can derive the L2 stability for the C-N scheme

(49) ‖un‖ ≤ ‖u0‖+

n∑k=1

δt‖fk−1/2‖.

The integral∫ tn

0‖f‖ ds is approximated by the middle point rule.

Remark 6.1. For C-N method, the right hand side can be also chosen as (fn + fn−1)/2.It corresponds to the trapezoid rule of the integral. For nonlinear problem A(u), it can beA((un + un−1)/2) or (A(un) +A(un−1))/2. Which one to chose is problem dependent.

The energy method can be adapted to the semi-discretization in time easily. For exam-ple, we chose v = (un + un−1)/2 in (47) to get

1

2‖un‖2 − 1

2‖un−1‖2 + δt|un−1/2|21 = δt(fn−1/2, un−1/2),

which implies the counter part of (18)

(50) ‖un‖2 +

n∑k=1

δt|uk−1/2|21 ≤ ‖u0‖2 +

n∑k=1

δt‖fk−1/2‖2−1.

Exercise 6.2. Study the stability of the forward and backward Euler method.

We then study the convergence. We use C-N as a typical example. We apply the discreteoperator Ln to the error Inu− un

Ln(Inu− un) = LnInu− InLu =u(·, tn)− u(·, tn−1)

δt− ∂tu(·, tn−1/2).

The consistency error is

(51)∣∣∣∣u(·, tn)− u(·, tn−1)

δt− ∂tu(·, tn−1/2)

∣∣∣∣ ≤ C(δt)2.

By the stability result, we then get

‖Inu− un‖ ≤ Ctnδt2.

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18 LONG CHEN

From the consistency error estimate (51), one can easily see the backward and forwardEuler methods are only first order in time. High order schemes, e.g., Runge-Kutta method,can be constructed by using a high order approximation of the integral. We first rewrite thedifferential equation (in time) as an integral equation:

1

δt

∫ tn

tn−1

ut(·, t) dt+1

δt

∫ tn

tn−1

Au(·, t) dt =1

δt

∫ tn

tn−1

f(·, t) dt.

The first term is simply (u(·, tn)−u(·, tn−1)/δt. The right hand side can be approximatedas accurately as we want since f is known. The subtle part is the middle one. We can applyhigh order quadrature using more points but u is only known at tn−1. So approximation atquadrature points should be computed first. For details, we refer to [?].

Add more on high order (explicit and implicit) discretization in time.

7. FINITE ELEMENT METHOD OF PARABOLIC EQUATION: FULL DISCRETIZATION

The full discretization is simply a combination of discretization in space and time. Weconsider the following three schemes:

Forward Euler Method. u0h = u0,h

(52)1

δt(unh − un−1

h , vh) + a(un−1h , vh) = 〈fn−1

h , vh〉, ∀vh ∈ Vh, 1 ≤ n ≤ N

Backward Euler Method. u0h = u0,h

(53)1

δt(unh − un−1

h , vh) + a(unh, vh) = 〈fnh , vh〉, ∀vh ∈ Vh, 1 ≤ n ≤ N

Crank-Nicolson Method. u0h = u0,h

(54)1

δt(unh − un−1

h , vh) + a((unh + un−1h )/2, vh) = 〈fn−1/2

h , vh〉, ∀vh ∈ Vh, 1 ≤ n ≤ N

We write the semi-discretization and the full discretization as

L → Lh → Lnh.

The discrete error is Rhu(tn)− unh . The setting is

• X = L2(0, T ;H10 (Ω)), and ‖u‖X =

(∫ T0|u(t)|21 dt

)1/2

• Y = L2(0, T ;H−1(Ω), and ‖f‖Y =(∫ T

0‖f(t)‖2−1 dt

)1/2

• X δth = Vh × Vh × · · · × Vh, and ‖uh‖X δt

h=(∑N

k=0 δt|ukh|21)1/2

• Y δth = V′h × V′h × · · · × V′h, and ‖fh‖Y δt

h=(∑N

k=0 δt‖fkh‖2−1

)1/2

.

• InRh : L2(0, T ;H10 (Ω))→ Vh ×Vh × · · · ×Vh is the composition of the nodal

interpolation in time and Ritz projection in space

InRhu = RhInu = (Rhu)(tn).

• InQh : L2(0, T ;H−1(Ω))→ Vh×V′h×· · ·×V′h is the composition of the nodalinterpolation in time and L2 projection in space

InQhf = QhInf = (Qhf)(tn).

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NUMERICAL METHODS FOR PARABOLIC EQUATIONS 19

• Lnh : X δth → Y δt

h × Vh is u0h = u0,h and for n ≥ 1

Lnhunh =

D+t u

nh −∆unh Forward Euler,

D−t unh −∆unh Backward Euler,

D−t unh −∆(unh + un−1

h )/2 Crank-Nicolson.

where D+t is the forward difference and D−t the backward difference in time. The

solution unh solves the full discrete equation:

Lnhunh = In+sQhf, in V′h, for n ≥ 1, u0h = u0,h,

where s = 0 for backward Euler and forward Euler, and s = −1/2 C-N method.

7.1. Stability. We write the discretization using operator form and present the stability inL2 norm. Again denoted by Ah = −∆|Vh .

Forward Euler Method.

(55) unh = (I − δtAh)un−1h + δtfn−1

h .

Backward Euler Method.

(56) unh = (I + δtAh)−1(un−1h + δtfnh ).

Crank-Nicolson Method.

(57) unh = (I +1

2δtAh)−1

[(I − 1

2δtAh)un−1

h + δtfn−1/2h

].

Since A = −∆ is symmetric in the L2 inner product, to obtain the stability in L2 norm,we only need to study the spectral radius of these operators.

Forward Euler Method.

ρ(I − δtAh) = |1− δtλmax(Ah)| ≤ 1,

provided

δt ≤ 2

λmax(Ah).

Note that λmax(Ah) = O(h−2). We need the time step is in the size of h2 to make theforward Euler stable.

Backward Euler Method. For any δt > 0, since Ah is SPD, i.e., λmin(Ah) > 0

ρ((I + δtAh)−1) = (1 + δtλmin(Ah))−1 ≤ 1.

Crank-Nicolson Method. For any δt > 0,

ρ((I +1

2δtAh)−1(I − 1

2δtAh)) = max

1≤k≤N

|1− 12δt λk(Ah)|

1 + 12δt λk(Ah)

≤ 1,

We summarize the stability as the following theorem which is a discrete version of (17).

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20 LONG CHEN

Theorem 7.1. For forward Euler method, when δt < 1/λmax(Ah)

(58) ‖unh‖ ≤ ‖u0h‖+

n−1∑k=0

δt‖fkh‖.

For backward Euler

(59) ‖unh‖ ≤ ‖u0h‖+

n−1∑k=0

δt‖fk+1h ‖.

For Crank-Nicolson method

(60) ‖unh‖ ≤ ‖u0h‖+

n−1∑k=0

δt‖fk+1/2h ‖.

It is interesting to note that the last term in the stability result (58), (59), and (60) aredifferent Riemann sums of the integral in the continuous level.

Exercise 7.2. Derive the discrete version of (18) and (19).

7.2. Consistency. The error equation will be

LnhInRhu− InQhLu = Dnt I

nRhu− In∂tu = (Dnt I

nu− In∂tu) +Dnt I

n(Rhu− u).

Forward Euler.

‖(Dnt I

nu− In∂tu)‖ = ‖u(tn+1)− u(tn)

δt− ∂tu(tn)‖ ≤ Cδt‖∂ttu‖,

and

‖Dnt I

n(Rhu− u)‖ = ‖δt−1

∫ tn+1

tn

(Rhut − ut) ds‖ ≤ δt−1h2

∫ tn+1

tn

‖ut‖2 ds.

Backward Euler.

‖(Dnt I

nu− In∂tu)‖ = ‖u(tn)− u(tn−1)

δt− ∂tu(tn)‖ ≤ Cδt‖∂ttu‖

and

‖Dnt I

n(Rhu− u)‖ = ‖δt−1

∫ tn

tn−1

(Rhut − ut) ds‖ ≤ δt−1h2

∫ tn

tn−1

‖ut‖2 ds.

Crank-Nicolson. For Crank-Nicolson, it is important to note that we are solving

Lnhunh = In−1/2Qhf.

So the error equation is

LnhInRhu− In−1/2QhLu =Rhu(tn)−Rhu(tn−1)

δt− ∂tu(tn−1/2)

+ARh1

2(u(tn) + u(tn−1))−QhAu(tn−1/2)

= DtIn(Rhu− u)

+DtInu− ∂tu(tn−1/2)

+QhA

[1

2(u(tn) + u(tn−1))− u(tn−1/2)

]= I1 + I2 + I3.

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NUMERICAL METHODS FOR PARABOLIC EQUATIONS 21

We then estimate each term as

‖I1‖ = δt−1h2

∫ tn

tn−1

‖ut‖2 ds,

‖I2‖ ≤ C‖∂tttu‖δt2,‖I3‖ ≤ C‖∆∂ttu‖δt2.

7.3. Convergence. The convergence is a consequence of stability, consistency, and theapproximation. For simplicity, we list the result for Crank-Nicolson and assume u0,h =Rhu0.

Theorem 7.3. Let u, unh be the solution of (16) and (54), respectively with u0,h = Rhu0.Then for n ≥ 0,

‖u(tn)− unh‖ ≤ Ch2

∫ tn

0

‖ut‖2 ds+ Cδt2∫ tn

0

(‖∂tttu‖+ ‖∂tt∆u‖).

REFERENCES

[1] L. C. Evans. Partial Differential Equations. American Mathematical Society, 1998.[2] B. GarcIa-Archilla and E. S. Titi. Postprocessing the galerkin method: The finite-element case. SIAM J.

Numer. Anal., 37(2):470–499, 2000.[3] V. Thomee. Galerkin finite element methods for parabolic problems, volume 25 of Springer Series in Com-

putational Mathematics. Springer-Verlag, Berlin, second edition, 2006.[4] M. F. Wheeler. A priori L2 error estimates for Galerkin approximations to parabolic partial differential equa-

tions. SIAM J. Numer. Anal., 10:723–759, 1973.