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1.4. EXAMPLE 2: THE HEAT EQUATION

9

1.3.4A numerical comparison

We consider the above 1-dimensional eigenvalue problem

(1.20)

−

((1 + x)w′(x))′ + w(x) = λw(x), w(0) = w(1) = 0,

 

 

and solve it with the finite di erence and finite element methods as well as with the global functions method. The results are given in Table 1.1.

Clearly the global function method is the most powerful of them all. With 80 basis functions the eigenvalues all come right. The convergence rate is exponential.

With the finite di erence and finite element methods the eigenvalues exhibit quadratic convergence rates. If the mesh width h is reduced by a factor of q = 2, the error in the eigenvalues is reduced by the factor q2 = 4.

1.4Example 2: The heat equation

The instationary temperature distribution u(x, t) in an insulated container satisfies the equations

∂u(x, t)

− u(x, t) = 0,

x Ω, t > 0,

 

 

 

∂t

(1.21)

 

∂u(x, t)

= 0,

x ∂Ω, t > 0,

 

 

 

∂n

u(x, 0) = u0(x), x Ω.

Here Ω is a 3-dimensional domain2 with boundary ∂Ω. u0(x), x = (x1, x2, x3)T R3, is a given bounded, su ciently smooth function.

(1.22) u =

X ∂2u

∂xi2

is called the Laplace operator and ∂n∂u denotes the derivative of u in direction of the outer normal vector n. To solve the heat equation the method of separation of variables is employed. We write u in the form

(1.23)

 

u(x, t) = v(t)w(x).

 

If a constant λ can be found such that

 

 

 

 

 

 

 

(1.24)

w(x) + λw(x) = 0, w(x) 6= 0, x in Ω,

 

∂w(x, t)

 

 

 

 

 

= 0,

x on ∂Ω,

 

 

 

 

 

∂n

 

then the product u = vw is a solution of (1.21) if and only if

(1.25)

 

 

dv(t)

+ λv(t) = 0,

 

 

 

 

 

 

 

dt

 

the solution of which has the form a·exp(−λt). By separating variables, the problem (1.21) is divided in two subproblems that are hopefully easier to solve. A value λ, for which (1.24)

2In the sequel we understand a domain to be bounded and simply connected.

10

CHAPTER 1. INTRODUCTION

Finite di erence method

k

λk(n = 10)

 

λk(n = 20)

λk(n = 40)

λk(n = 80)

1

15.245

 

15.312

15.331

15.336

2

56.918

 

58.048

58.367

58.451

3

122.489

 

128.181

129.804

130.236

4

206.419

 

224.091

229.211

230.580

5

301.499

 

343.555

355.986

359.327

6

399.367

 

483.791

509.358

516.276

7

492.026

 

641.501

688.398

701.185

8

578.707

 

812.933

892.016

913.767

9

672.960

 

993.925

1118.969

1153.691

10

794.370

 

1179.947

1367.869

1420.585

 

 

 

 

 

 

 

 

 

 

 

 

Finite element method

 

 

 

 

 

 

 

 

 

 

 

 

 

k

λk(n = 10)

 

λk(n = 20)

λk(n = 40)

λk(n = 80)

1

15.447

 

15.367

15.345

15.340

2

60.140

 

58.932

58.599

58.511

3

138.788

 

132.657

130.979

130.537

4

257.814

 

238.236

232.923

231.531

5

426.223

 

378.080

365.047

361.648

6

654.377

 

555.340

528.148

521.091

7

949.544

 

773.918

723.207

710.105

8

1305.720

 

1038.433

951.392

928.983

9

1702.024

 

1354.106

1214.066

1178.064

10

2180.159

 

1726.473

1512.784

1457.733

 

 

 

 

 

 

Global function method

 

 

 

 

 

 

 

 

 

 

 

 

 

k

λk(n = 10)

 

λk(n = 20)

λk(n = 40)

λk(n = 80)

1

15.338

 

15.338

15.338

15.338

2

58.482

 

58.480

58.480

58.480

3

130.389

 

130.386

130.386

130.386

4

231.065

 

231.054

231.053

231.053

5

360.511

 

360.484

360.483

360.483

6

518.804

 

518.676

518.674

518.674

7

706.134

 

705.631

705.628

705.628

8

924.960

 

921.351

921.344

921.344

9

1186.674

 

1165.832

1165.823

1165.822

10

1577.340

 

1439.083

1439.063

1439.063

 

 

 

 

 

 

 

 

 

 

 

 

Table 1.1: Numerical solutions of problem (1.20)

1.5. EXAMPLE 3: THE WAVE EQUATION

11

has a nontrivial (i.e. a nonzero) solution is called an eigenvalue; w then is called a corresponding eigenfunction.

If λn is an eigenvalue of problem (1.24) with corresponding eigenfunction wn, then

e−λntwn(x)

is a solution of the first two equations in (1.21). It is known that equation (1.24) has infinitely many real eigenvalues 0 ≤ λ1 ≤ λ2 ≤ · · · , (λnt→∞−→ ∞). Multiple eigenvalues are counted according to their multiplicity. An arbitrary bounded piecewise continuous function can be represented as a linear combination of the eigenfunctions w1, w2, . . .. Therefore, the solution of (1.21) can be written in the form

 

∞

(1.26)

u(x, t) =

cne−λntwn(x),

 

n=1

 

X

where the coe cients cn are determined such that

 

 

∞

(1.27)

u0(x) =

cnwn(x).

 

 

n=1

 

 

X

The smallest eigenvalue of (1.24) is λ1 = 0 with w1 = 1 and λ2 > 0. Therefore we see from (1.26) that

(1.28)

u(x, t)

−→

c1.

t→∞

Thus, in the limit (i.e., as t goes to infinity), the temperature will be constant in the whole container. The convergence rate towards this equilibrium is determined by the smallest positive eigenvalue λ2 of (1.24):

∞

∞

 

X

X

 

ku(x, t) − c1k = k

cne−λntwn(x)k ≤ |e−λnt|kcnwn(x)k

n=2

n=2

∞

 

 

 

≤ e−λ2t

X

 

kcnwn(x)k ≤ e−λ2tku0(x)k.

n=2

Here we have assumed that the value of the constant function w1(x) is set to unity.

1.5Example 3: The wave equation

The air pressure u(x, t) in a volume with acoustically “hard” walls satisfies the equations

(1.29)

∂2u(x, t)

− u(x, t) = 0,

 

x Ω, t > 0,

∂t2

 

(1.30)

 

 

∂u(x, t)

= 0,

 

x ∂Ω, t > 0,

 

 

 

 

 

 

 

 

∂n

 

(1.31)

 

 

 

u(x, 0) = u0

(x),

x Ω,

(1.32)

 

∂u(x, 0)

= u1

(x),

x Ω.

 

 

 

 

 

 

 

∂t

Sound propagates with speed − u, along the (negative) gradient from high to low pressure.

12

CHAPTER 1. INTRODUCTION

To solve the wave equation we proceed as with the heat equation in section 1.4: separation of u according to (1.23) leads again to equation (1.24) but now together with

(1.33)

d2v(t)

+ λv(t) = 0.

 

dt2

We know this equation from the analysis of the vibrating string, see (1.11). From there we know that the general solution of the wave equation has the form

(1.13)

u(x, t) = k=0 wk(x) hak · cos(

λk t) + bk · sin(

 

λk t)i .

 

∞

 

 

 

 

 

 

 

X

p

 

 

p

 

 

where the wk, k = 1, 2, . . ., are the eigenfunctions of the eigenvalue problem (1.24). The coe cients ak and bk are determined by (1.31) and (1.32).

If a harmonic oscillation is forced on the system, an inhomogeneous problem

(1.34)

∂2u(x, t)

− u(x, t) = f(x, t),

∂t2

is obtained. The boundary and initial conditions are taken from (1.29)–(1.32). This problem can be solved by setting

 

 

 

 

 

∞

v˜n(t)wn(x),

 

u(x, t) :=

(1.35)

 

 

 

 

n=1

 

 

 

 

 

 

 

 

 

X

 

 

 

 

 

 

 

 

 

 

∞

 

 

 

 

 

 

 

 

 

 

X

φn(t)wn(x).

 

f(x, t) :=

 

 

 

 

 

n=1

 

 

 

 

 

With this approach, v˜n has to satisfy equation

 

 

 

d2v˜n

 

 

 

 

 

(1.36)

 

 

 

 

+ λnv˜n = φn(t).

 

 

dt2

If φn(t) = a sin ωt, then the solution becomes

 

 

 

 

 

 

 

 

 

 

 

 

 

1

 

 

 

 

 

 

 

 

(1.37)

v˜n = An cos pλnt + Bn sin pλnt +

 

a sin ωt.

λn − ω2

An and Bn are real constants that are determined by the initial conditions. If ω gets close

√ √

to λ1, then the last term can be very large. In the limit, if ω = λn, v˜n gets the form

 

p

 

p

 

 

(1.38)

v˜n = An cos

λnt + Bn sin λnt + at sin ωt.

In this case, v˜n is not bounded in time anymore. This phenomenon is called resonance. Often resonance is not desirable; it may, e.g., mean the blow up of some structure. In order to prevent resonances eigenvalues have to be known. Possible remedies are changing the domain (the structure).

Remark 1.1. Vibrating membranes satisfy the wave equation, too. In general the boundary conditions are di erent from (1.30). If the membrane (of a drum) is fixed at its boundary, the condition

(1.39)

u(x, t) = 0

is imposed. This boundary conditions is called Dirichlet boundary conditions. The boundary conditions in (1.21) and (1.30) are called Neumann boundary conditions. Combinations of these two can occur.

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