methods in applied mechanics iii_daniel m. tartakovsky.pdf

Class notes for MAE 294c:Methods in Applied Mechanics III

Instructor: Daniel M. Tartakovsky

May 30, 2005

Table of Contents

1 Mathematical Modeling of Physical Phenomena 1

1.1 Stages of a modeling process . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1

1.2 A historic perspective . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1

2 Derivation of Equations of Mathematical Physics 3

2.1 Vibrating strings . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3

2.1.1 Physical phenomenon . . . . . . . . . . . . . . . . . . . . . . . . . . . 3

2.1.2 Construction of a physical model . . . . . . . . . . . . . . . . . . . . . 3

2.1.3 The choice of a function describing the process . . . . . . . . . . . . . . 4

2.1.4 Physical laws governing the vibration of a string . . . . . . . . . . . . . 5

2.1.5 Derivation of governing equations for u . . . . . . . . . . . . . . . . . . 5

2.1.6 Initial and boundary conditions . . . . . . . . . . . . . . . . . . . . . . 6

2.2 Vibrating membranes . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7

2.3 Generalizations of the wave equation . . . . . . . . . . . . . . . . . . . . . . . . 8

2.4 Heat conduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9

2.4.1 Physical phenomena . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9


2.4.3 The choice of a function describing the process . . . . . . . . . . . . . . 10

2.4.4 Physical laws governing heat conduction in solids . . . . . . . . . . . . . 10

2.4.5 Derivation of governing equations for u . . . . . . . . . . . . . . . . . . 11

2.4.6 Initial and boundary conditions . . . . . . . . . . . . . . . . . . . . . . 12

2.4.7 Stationary temperature fields . . . . . . . . . . . . . . . . . . . . . . . . 13

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MAE 294c. Methods in Applied Mechanics III Instructor: Daniel M. Tartakovsky

2.5 Continuity equations for fluids . . . . . . . . . . . . . . . . . . . . . . . . . . . 13

2.5.1 Physical phenomena . . . . . . . . . . . . . . . . . . . . . . . . . . . . 13


2.5.3 The choice of functions describing the process . . . . . . . . . . . . . . 14

2.5.4 Derivation of governing equations . . . . . . . . . . . . . . . . . . . . . 14

3 Classification of PDEs 15

3.1 Basic terminology . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 15

3.2 Well-posed problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 16

3.3 Classification of second-order PDEs . . . . . . . . . . . . . . . . . . . . . . . . 16

3.3.1 Transformation of second-order PDEs to their canonical form . . . . . . 17

3.3.2 Classification of second-order PDEs . . . . . . . . . . . . . . . . . . . . 19

3.3.3 A curse of dimensionality . . . . . . . . . . . . . . . . . . . . . . . . . 20

3.3.4 Classification of second-order PDEs in n = 2 dimensions . . . . . . . . 20

4 Method of Characteristics 23

4.1 Linear first-order PDEs . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 23

4.1.1 Constant wave velocity . . . . . . . . . . . . . . . . . . . . . . . . . . . 24

4.1.2 Variable wave velocity . . . . . . . . . . . . . . . . . . . . . . . . . . . 24

4.2 Classification of second-order PDEs in two dimensions (Cntd.) . . . . . . . . . . 254.3 Linear advection with a source . . . . . . . . . . . . . . . . . . . . . . . . . . . 26

4.4 Linear advection and reactions . . . . . . . . . . . . . . . . . . . . . . . . . . . 27

4.5 Quasi-linear first-order PDEs . . . . . . . . . . . . . . . . . . . . . . . . . . . . 284.6 Nonlinear first-order PDEs . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 29

4.6.1 Elementary traffic model . . . . . . . . . . . . . . . . . . . . . . . . . . 31

4.7 Shock waves . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 31

4.7.1 Discontinuous solutions . . . . . . . . . . . . . . . . . . . . . . . . . . 32

4.7.2 Elementary traffic model with shocks . . . . . . . . . . . . . . . . . . . 34

4.7.3 Conditions for shock formation . . . . . . . . . . . . . . . . . . . . . . 36

4.8 Nonlinear first-order PDEs with a source . . . . . . . . . . . . . . . . . . . . . . 37

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5 Laplace Transformation 39

5.1 Motivation and Definition . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 39

5.2 Properties of the Laplace transformation . . . . . . . . . . . . . . . . . . . . . . 40

5.2.1 Linearity . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 40

5.2.2 Applicability . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 40

5.3 Solutions of linear ODEs with constant coefficients . . . . . . . . . . . . . . . . 43

5.3.1 Forced vibrations with damping . . . . . . . . . . . . . . . . . . . . . . 43

5.3.2 A damped absorber . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 43

5.4 Solutions of integral equations with convolution kernels . . . . . . . . . . . . . . 44

5.5 Solutions of linear PDEs . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 45

5.5.1 Wave equation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 45

5.5.2 Diffusion equation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 47

6 Nonhomogeneous Problems and Greens Functions 49

6.1 Motivation and basic ideas . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 49

6.2 The Dirac delta function . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 51

6.3 Properties of the Greens functions . . . . . . . . . . . . . . . . . . . . . . . . . 52

6.3.1 Causality . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 52

6.3.2 Reciprocity . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 52

6.4 Calculation of the Greens functions . . . . . . . . . . . . . . . . . . . . . . . . 53

6.5 Free surface flow . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 56

7 Nonlinear Diffusion 59

7.1 Time-dependent diffusion coefficients1 . . . . . . . . . . . . . . . . . . . . . . . 59

7.2 Boltzmanns transformation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 60

7.3 Kirchhoffs transformation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 61

7.3.1 Steady-state diffusion . . . . . . . . . . . . . . . . . . . . . . . . . . . . 61

7.3.2 Steady-state diffusion with gravity . . . . . . . . . . . . . . . . . . . . . 61

7.4 Method of spatial moments . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 63

7.5 Hopf-Coles transformation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 66

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8 Approximate Solutions of ODEs: Local Analyses 67

8.1 Basic definitions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 67

8.2 Fuchs theory of ODEs . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 68

8.3 Local behavior of linear ODEs near ordinary points . . . . . . . . . . . . . . . . 69

8.4 Local behavior of linear ODEs near regular singular points . . . . . . . . . . . . 70

8.5 Local behavior of linear ODEs near irregular singular points . . . . . . . . . . . 70

8.6 Local behavior of inhomogeneous linear ODEs . . . . . . . . . . . . . . . . . . 71

8.7 Nonlinear ODEs: Spontaneous singularities . . . . . . . . . . . . . . . . . . . . 72

8.8 Early-time behavior of nonlinear ODEs . . . . . . . . . . . . . . . . . . . . . . 73

8.9 Late-time behavior of nonlinear ODEs . . . . . . . . . . . . . . . . . . . . . . . 74

8.10 Local behavior of nonlinear ODEs on bounded domains . . . . . . . . . . . . . . 74

9 Scaling and Self-Similarity 77

9.1 Dimensions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 77

9.2 Dimensional analysis . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 77

9.2.1 Transformation to dimensionless parameters . . . . . . . . . . . . . . . 78

9.2.2 Example . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 79

9.3 Similarity and scaling . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 79

9.4 Self-similar solutions: Linear diffusion . . . . . . . . . . . . . . . . . . . . . . . 80

9.5 Self-similar solutions: Nonlinear diffusion . . . . . . . . . . . . . . . . . . . . . 82

9.6 Anomalous Exponents and Self-Similarity of the Second Kind . . . . . . . . . . 84

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Chapter 1

Mathematical Modeling of PhysicalPhenomena

1.1 Stages of a modeling process

A typical modeling process of a physical phenomenon consists of the stages shown in Figure 1.1.

MathematicalModel Solutions Interpretation

PhysicalModel

Figure 1.1: A schematic representation of a typical modeling process.

Construction of a model consists of the following steps

1. Construction of a physical model, i.e., idealization of a physical process,

2. The choice of (i) a function u(x1, . . . , t) describing the process and (ii) physical laws towhich it obeys,

3. Derivation of governing equations for u,

4. Derivation of complementary conditions

(a) boundary conditions(b) initial conditions.

1.2 A historic perspective

Some of the earliest seminal contributions to the field of mathematical physics include

Isaac Newton, 1687, Philosophiae Naturalis Principia Mathematica.

1


Newtons laws of motion Newtons Law of Gravitation

Brook Taylor, 1715, Methodus Incrementorum. An equation for a vertically vibrating string,2u

t2= D

2u

x2, where u is a vertical displacement.

Joseph-Louis Lagrange, 1788, Mecanique analytique. An equation for electric fields,2u

x2+2u

y2= 0, where u is electric charge.

Jean Baptiste Joseph Fourier, 1822, Theorie analytique de la chaleur. An equation for heatconduction in a rod,

u

t= D

2u

x2, where u is temperature.

James Clerk Maxwell, 1873, Treatise on Electricity and Magnetism. Maxwells equations.

Sofia Kovalevskaya, 1874, PhD dissertation at the University of Gottingen. A prove of the existence and uniqueness of solutions to the Cauchy problem.

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Chapter 2

Derivation of Equations ofMathematical Physics

Below we use the general framework outlined in Section 1.1 to derive a few mathematical models.

2.1 Vibrating strings

2.1.1 Physical phenomenon

vibration of a string

Figure 2.1: A highly stretched string with firmly fixed ends. It is shown in the state of equilibrium,i.e., before a force has been applied.

2.1.2 Construction of a physical model

The process is idealized by assuming that

A string is perfectly flexible, i.e., it offers no resistance to bending; A string is perfectly elastic, i.e., the tension in the string depends only on its local stretching. All points of the string move in the direction, perpendicular to its equilibrium state; The vibrations (deviations from the state of equilibrium) are small.

3


2.1.3 The choice of a function describing the process

The process is described by the vertical displacement of a string u(x, t) from its equilibrium state(Figure 2.2). Note that

u(x0, t) provides the law of motion for a point x0 of the string,

u(x, t0) describes a profile of the string at t0 = 0

x = 0

u(x, t)(x, t)

x = L

Figure 2.2: A profile of the vibrating string (solid line) at time t away from its equilibrium state(dashed line). The function u(x, t) designates the vertical displacement of a string at point x andtime t.

The following useful relations hold. Let (x, t) denote the slope of the string, i.e., the anglebetween the positive x-axis and the positive direction of the tangent, as shown in Figure 2.2. Then

tan(x, t) =u

x. (2.1)

The velocity v and acceleration a of a point x of the string are given by

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

and a(x) =2u(x, t)t2

, (2.2)

respectively.

Since we assumed that the vibrations u are small, the terms similar to

u2, uu

x,

(u

x

)2, etc u (2.3)

can be neglected. This fact has an interesting implications. For example, since the length |M1M2|of a segment M1M2 of the string in Figure 2.3 is

|M1M2| = x+4xx

1 +

(u

x

)2dx

x+4xx

dx = 4x, (2.4)

it does not change with time!

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x x+4xT(x, t)

T(x+4x, t)(x, t) (x+4x, t)M2M1

Figure 2.3: A profile of the vibrating string (solid line) at time t away from its equilibrium state(dashed line). The function u(x, t) designates the vertical displacement of a string at a point x andtime t. A tangential force T is called the tension in the string.

2.1.4 Physical laws governing the vibration of a string

Hookes law (R. Hooke, 1676): A force applied to a body is directly proportional to theresulting deformation of the body;

dAlemberts (J. dAlembert, 1742) principle,i

Fi = 0, (2.5)

which states that all forces Fi acting on a body (including the inertial force) must be inequilibrium.

2.1.5 Derivation of governing equations for u

Consider a small segment of the string M1M2 (Figure 2.3). Its deformation induces the tension inthe string, which is the tangential force T shown in Figure 2.3. Let T = |T| denote the magnitudeof T. According to (2.4), the length of the string does not change with time. Hence neither doesT, i.e., we showed that T = T(x).

According to dAlemberts principle (2.5), the projections x and u of all forces on thehorizontal and vertical coordinates, respectively, must equilibrate. Since

x{T(x+4x)} = T (x+4x) cos(x+4x, t), x{T(x)} = T (x) cos(x, t),u{T(x+4x)} = T (x+4x) sin(x+4x, t), u{T(x)} = T (x) sin(x, t),

(2.6)

dAlemberts principle (2.5) gives two equations

T (x+4x) cos(x+4x, t) T (x) cos(x, t) = 0 (2.7)

and

T (x+4x) sin(x+4x) T (x) sin(x, t) +4xf(x, t) = 4x2u

t2. (2.8)

In (2.8), f denotes the density of an external force, so that 4xf is a total external force actingon the segment M1M2; and is the density of the string, so that 4x is the mass of the segmentM1M2.

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Next we recall that

cos =1

1 + (tan)2. (2.9)

Substituting (2.1) into (2.9), while recalling (2.3), yields

cos(x, t) =1

1 +(ux

)2 1. (2.10)It then follows from (2.7) that

T (x+4x) T (x) = 0. (2.11)In words, under our assumptions, the tension in the string remain constant T (x, t) T0.

Since

sin =tan

1 + (tan)2 tan, (2.12)

recalling (2.1) gives

sin ux. (2.13)

This allows us to rewrite (2.8) as

T0

[u

x(x+4x) u

x(x, t)

]+4xf(x, t) = 4x

2u

t2. (2.14)

Dividing by 4x and taking the limit as 4x 0 yields

T02u

x2+ f(x, t) =

2u

t2(2.15)

or

2u

t2= c2

2u

x2+ g(x, t), (2.16)

where c2 = T0/ and g = f/. Eq. (2.16) is called a wave equation.

2.1.6 Initial and boundary conditions

To complete the description of the problem one needs to specify initial and boundary conditionsfor Eq. (2.16).

Initial conditions describe the behavior of the string at some initial time t = t0. It is commonto set t0 = 0. Since (2.16) contains second time derivative, there must be two initial conditions,which specify the initial shape (x) and velocity (x) of the string, i.e.,

u(x, 0) = (x) andu

t(x, 0) = (x), for 0 x L. (2.17)

Boundary conditions depend on the state of the string at the boundaries x = 0 and x = L.For example,

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If both ends of the string are fixed,

u(0, t) = u(L, t) = 0, t 0; (2.18)

If the ends of the string move according to prescribed laws 1(t) and 2(t),

u(0, t) = 1(t), u(L, t) = 2(t), t 0; (2.19)

If forces v1(t) and v2(t) acting on the ends of the string are known,u

x(0, t) = v1(t),

u

x(L, t) = v2(t); (2.20)

Etc.

2.2 Vibrating membranes

The process is described by u(x, t), the vertical deviation of a point x = (x1, x2)T of a perfectlyflexible, perfectly elastic membrane from its equilibrium position.

x1

u

x2

x = (x1, x2)T

u(x, t)

Figure 2.4: Small vibrations of a membrane.

Following the procedure described in Section 2.1, one can show that u(x, t) satisfies thetwo-dimensional wave equation

2u

t2= a24u+ g(x, t), (2.21)

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where 4 2 denotes the Laplacian.The initial conditions for (2.21) are

u(x, 0) = (x),u

t(x, 0) = (x), x . (2.22)

Here = denotes the union of a domain , on which (2.21) is defined, and its boundary).

If the membranes boundary is fixed, the corresponding boundary condition is

u(x, t) = 0, x . (2.23)

2.3 Generalizations of the wave equation

Consider small vibrations of a uniform isotropic body. Let v = (v1, v2, v3)T denote the vectorof the deviation of a point x = (x1, x2, x3)T in the body at time t from its equilibrium state.Specifically, consider the deviations

v1(x2, x3, t), v2(x1, x3, t), v3(x1, x2, t).

Since each slice of the body vibrates as a membrane, all vi (i = 1, 2, 3) satisfy the two-dimensionalwave equation (2.21),

2v1t2

=a2(2v1x22

+2v1x23

)+ f1(x2, x3, t)

2v2t2

=a2(2v2x21

+2v2x23

)+ f2(x1, x3, t) (2.24)

2v3t2

=a2(2v3x21

+2v3x22

)+ f3(x1, x2, t).

Equations (2.24) can be rewritten in the vector form as2vt2

= a24v + f(x, t). (2.25)

Next we assume that the field is potential, i.e., there exist scalar functions g(x, t and u(x, t)such that

f = g and v = u. (2.26)Then it follows from (2.25) that

{2u

t2 a24u g

}= 0. (2.27)

Therefore we may take

2u

t2= a24u+ g, 4

3i=1

2

x2i, (2.28)

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which, of course, is the three-dimensional wave equation.

In general, the process of vibration can be described by the general wave equation

2u

t2=

ni,j=1

xi

(pij

u

xj

) qu+ f. (2.29)

Here the state variable u(x, t) is a function of independent variables x Rn, and , pij , q, and fare known functions. From physical considerations, > 0, pij > 0, and q 0.

2.4 Heat conduction

2.4.1 Physical phenomena

heat conduction (Fouriers law),

electrical conduction (Ohms law),

diffusive transport (Ficks law)

flow in porous media (Darcys law),

etc.


The process is idealized by assuming that 1) a materials properties do not change with temperatureand 2) thermal energy is transported by conduction only, i.e., that convection can be ignored.

Conduction:

The dominant method of heat conduction in metal is through the movement of electrons.This method of conduction does not operate in non-metals because there are no free elec-trons (other than graphite). When a metal is heated, the electrons closest to the heat sourcevibrate more rapidly. These electrons collide with atoms and gain more kinetic energy(movement energy). This causes the electrons to move around faster and to collide withother free electrons which, in turn, gain more kinetic energy. Kinetic energy is transferredbetween the electrons and through the metal from the point closest to the heat source towardpoints further away. Since the electrons travel very short distances at very large velocities,conduction of heat happens very quickly.

In metals and insulators, there is conduction of heat due to the vibration of atoms. Asatoms closest to the heat source absorb heat/thermal energy, they induce the vibration in theneighboring atoms, which, in turn, make their neighboring atoms vibrate more, etc.

Examples of conduction:

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The wire gauzes used on tripods are metal therefore they are good heat conductors.Gauzes on cookers are also metal so that heat is conducted quickly and food is cookedfast.

Poor thermal conductors (insulators) are used for saucepan handles so that they dontheat up and can still be handled.

Metals are used for the containers which heat liquids e.g. pans and kettles. Air is a poor conductor therefore materials that trap air are used for insulation in lofts

and hot water cylinders.

Convection:

When particles of the cold air are heated by a heat source, they gain kinetic energy andthe air expands. The density of particles in the hot air decreases relative to that of thesurrounding cold air, which causes the hot air to rise and displace the cool air. Since coolparticles are more dense, they fall and move toward the heat source to take the place of thewarm particles. They then heat up and rise while other particles cool down and fall.

Example of convection: Refrigerators are kept cool by convection. Land and sea breezes are due to convection. Atmospheric winds. Hot water systems.

2.4.3 The choice of a function describing the process

The process is described by the temperature u(x, t) of a solid body.

2.4.4 Physical laws governing heat conduction in solids

The relationship between thermal energy density (the amount of thermal energy per unitvolume) e(x, t) and temperature u(x, t) is given by

e = u, (2.30)where (x) is the specific heat (the amount of heat per unit mass required to raise thetemperature by one degree Celsius) of the rod, and (x) its density (mass per unit length).

Fouriers (Joseph Fourier, 1822) law of conduction [see also Newtons (Isaak Newton, 1701)law of cooling]: Heat energy will flow from the region of high temperature to the region oflow temperature

q = ku, (2.31)where q is the heat flux vector and k is thermal conductivity.

Conservation of energy (the first law of thermodynamics) Thales of Miletus (circa 635BC - 543 BC), . . . , Julius Robert von Mayer (1842):

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Heat conducted OUT Heat generated withinHeat conducted IN

Change in energy stored within

Figure 2.5: Conservation of energy.

2.4.5 Derivation of governing equations for u

Inside a heated body , we select a small cube (Figure 2.6), whose dimensions are4x1,4x2 and4x3, and whose volume is 4V = 4x14x24x3. If (x) is the density of this cube, its mass isM = 4V .

(x1, x2, x3)

(x1 +4x1, x2, x3)

(x1, x2, x3)

x2

x3

x1

Figure 2.6: An elementary volume used to derive the heat conduction equation.

Next, we write down the conservation of energy (Figure 2.5) in mathematical terms. Recall-ing (2.30), the amount of heat energy in the cube at times t is

H(x, t) = e(x, t)4V = (x)(x)u(x, t)4V. (2.32)The change in heat energy during the time interval [t, t+4t] is

4H = H(x, t+4t)H(x, t) = (x)(x)4V [u(x, t+4t) u(x, t)] . (2.33)

The amount of heat entering the cube through the side x1 + 4x1 during the time interval[t, t+4t] is Q1(x1 +4x1) = q(x1 +4x1)4x24x24t. According to Fouriers law (2.31) this

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gives

Q1(x1 +4x1) = k(x1 +4x1) ux1

(x1 +4x1)4x24x34t. (2.34)

Likewise, the amount of heat leaving the cube through the side x1 during the time interval [t, t +4t] is

Q1(x1) = k(x1)u

x1(x1)4x24x34t. (2.35)

The amount of heat 4Q1 = Q1(x1 +4x1)Q1(x1) conducted in and out of the cube in the x1direction is

4Q1 =[k(x1 +4x1) u

x1(x1 +4x1) k(x1) u

x1(x1)

]4x24x34t. (2.36)

In a similar manner, one can derive expressions for4Q2 and4Q3, the amounts of heat conductedin and out of the cube in the x2 and x3 directions, respectively.

Let us assume that inside the cube there are sources of heat, whose density if f(x, t). Thenthe total amount of heat generated inside the cube during the time interval [t, t+4t] is

Qg = f4x14x24x34t. (2.37)

The conservation of energy in Figure 2.5 can now be written as 4H = 4Q1 + 4Q2 +4Q3 +Qg, or

u(t+4t) u(t)

4t =14x1

[k(x1 +4x1) u

x1(x1 +4x1) k(x1) u

x1(x1)

]+

14x2

[k(x2 +4x2) u

x2(x2 +4x2) k(x2) u

x2(x2)

]+

14x3

[k(x3 +4x3) u

x3(x3 +4x3) k(x3) u

x3(x3)

]+ f. (2.38)

Taking the limit as 4xi 0 (i = 1, 2, 3) and 4t 0 yields a diffusion equation

u

t=

3i=1

xi

(ku

xi

)+ f, x . (2.39)

If a heated body is homogeneous, i.e., , , k = const, then (2.39) can be rewritten asu

t= c24u+ g, x , (2.40)

where c2 = k/, g = f/, and 4 = 2 is the Laplacian.

2.4.6 Initial and boundary conditions

Since diffusion equation (2.40) contains only first time derivative, initial conditions consist of onlyone equation that specifies the initial temperature distribution (x), i.e,

u(x, 0) = (x), x . (2.41)

- 12 -


Boundary conditions for (2.39) or (2.40) are defined on the surface of a heated body. Forexample,

If temperature is prescribed on the bounding surface,u(x, t) = (x, t), x , t 0; (2.42)

If the surface of a body is insulated, i.e., has a prescribed heat flow v(x, t),ku(x, t) n(x) = v(x, t), x , t 0, (2.43)

where n is the unit outward normal vector for the surface ;

If the surface is perfectly insulated, v = 0.

Conditions (2.42) and (2.43) are called the Dirichlet and Neumann boundary conditions, respec-tively.

2.4.7 Stationary temperature fields

If the boundary conditions and sources of thermal energy do not vary with time, temperaturewill eventually reach a steady-state regime, i.e., the regime in which u = u(x). In steady state,equation (2.39)

3i=1

xi

(ku

xi

)+ f = 0, x . (2.44)

If k = const, this gives Poissons equation

4u+ fk= 0, x . (2.45)

In the absence of sources of heat (f = 0), we arrive at Laplaces equation4u = 0, x . (2.46)

which lies at the foundation of the potential theory.

2.5 Continuity equations for fluids

2.5.1 Physical phenomena

flow of fluids


single-phase flow isothermal conditions, etc.

- 13 -


2.5.3 The choice of functions describing the process

The process is described by

the density of a fluid (x, t) pressure in a fluid p(x, t) the velocity of a fluid v(x, t)

2.5.4 Derivation of governing equations

Consider a volume occupied by a moving fluid. At time t, the mass of a fluid in this volume is

M(t) =(x, t)dx. (2.47)

The change in mass 4M =M(t2)M(t1) from time t1 to t2 is

4M =[(x, t2) (x, t1)] dx =

t2t1

tdtdx =

t2t1

tdxdt. (2.48)

The amount of a fluid leaving the volume through its bounding surface during the timeinterval [t1, t2] is

4Q = t2t1

(v) ndsdt = t2t1

(v)dxdt. (2.49)

Here n is the unit outward normal vector to the surface .

If sources of a fluid with density f(x, t) are present inside the volume , the amount of fluidgenerated during the time interval [t1, t2] is

4Qg = t2t1

f(x, t)dxdt. (2.50)

Since the mass conservation implies that 4M = 4Q+4Qg, we obtain t2t1

[

t+ (v) f

]dxdt = 0. (2.51)

Since (2.51) holds for an arbitrary t1, t2, and , it gives a continuity equation

t+ (v) f = 0. (2.52)

For incompressible fluids ( = const), and in the absence of sources (f = 0), (2.52) gives

v = 0. (2.53)

- 14 -

Chapter 3

Classification of PDEs

3.1 Basic terminology

Consider a function u(x1, . . . , xn). An equation

F

(x1, . . . , xn, u,

u

x1, . . . ,

u

xn,k1+...+knu

xk11 . . . xknn

)= 0 (3.1)

is called a partial differential equation (PDE) if n > 1. Otherwise, it is called an ordinary differ-ential equation (ODE).

A function u?(x1, . . . , xn) that satisfies (3.1), i.e., turns it into identity, is called a solution.

The order of the PDE (3.1) is determined by the order of the highest derivative.

The dimensionality of the PDE (3.1) is determined by the number n of independent variablesx1, . . . , xn. When one of these variables is time t, it is common to define the dimensionality asn 1.

Equation (3.1) is called linear if the corresponding differential operator L(u) is linear, i.e.,has the following property,

L(c1u1 + c2u2) = c1L(u1) + c2L(u2), (3.2)

where u1(x1, . . . , xn) and u2(x1, . . . , xn) are any two functions and c1 and c2 are arbitrary con-stants. For example, a differential equation

u

t=2u

x2+ u+ f(x, t) (3.3)

can be written as

L(u) = f(x, t), L(u) =u

t

2u

x2 u. (3.4)

15


It is easy to check that L in (3.4) satisfies the condition (3.2). Hence, equation (3.3) is a linearPDE. If the condition (3.2) does not hold, the corresponding PDE is called nonlinear. An exampleof nonlinear PDEs is

u

t=2u

x2+ u2 + f(x, t). (3.5)

To distinguish between linear and nonlinear differential operators, it is common to denote theformer by L and the latter by N . For example, (3.5) can be written as

N(u) = f(x, t), N(u) =u

t

2u

x2 u2. (3.6)

Equation (3.1) is called a homogeneous PDE if u = 0 is its solution. Otherwise, it is calleda nonhomogeneous PDE. It is easy to check that both (3.4) and (3.6) are homogeneous linear andnonlinear PDEs, respectively, if and only if f = 0.

3.2 Well-posed problems

Consider a function u = f(z). Let z1, z2 M and u1 = f(z1), u2 = f(z2) N , where M andN are two metric spaces with the corresponding metrics M (z1, z2) and N (u1, u2).

A problem is called stable on the spaces M and N if for any > 0 there exists = () > 0such that M (z1, z2) < implies N (u1, u2) < .

A problem is well-posed [Jacques Solomon Hadamard, 1902] on spaces M and N , if

1. a solution to the problem exists,

2. it is unique, and

3. it is stable.

Otherwise, a problem is ill-posed.

3.3 Classification of second-order PDEs

The significance of the wave (2.16) and diffusion (2.40) equations goes beyond their fundamentalimportance for practical applications. In this section we demonstrate that they represent examplesof canonical forms for a general class of second-order PDEs.

As an example, consider an equation

2u

xy= f. (3.7)

- 16 -


The change of variables = x y and = x+ y, transforms this equation into

2u

2=2u

2+ f, (3.8)

which, of course, is the wave equation (2.16).Let u(x) be a system state that depends on x Rn, where n 2. Consider a differential

equation

ni,j=1

aij(x)2u

xixj+ F (x, u,u) = 0, x . (3.9)

It is called a quasi-linear second-order differential equation, since it is linear with respect to thehighest (second) derivative. The coefficients aij(x) are continuous, i.e., aij(x) C(), on Rn. Without the loss of generality, we further assume that aij = aji.

We wish to simplify (3.9) by transforming it to its canonical form.

3.3.1 Transformation of second-order PDEs to their canonical form

Consider a transformation of coordinates

= (x), or l = l(x), l = 1, . . . , n (3.10)

whose Jacobian

(1, . . . , n)(x1, . . . , xn)

6= 0. (3.11)

The latter condition guarantees the existence of the inverse transformation x = x().

Under this transformation, u(x) = u[x()] u(). Hence

u

xi=

nk=1

u

k

kxi

(3.12)

and

2u

xixj=

xj

(u

xi

)=

nk=1

xj

(u

k

)kxi

+n

k=1

u

k

xj

(kxi

)

=n

k,l=1

2u

kl

kxi

lxj

+n

k=1

u

k

2kxixj

. (3.13)

Substituting (3.12) and (3.13) into (3.9) yieldsn

k,l=1

akl()2u

kl+ F1 (, u,u) = 0, (3.14)

- 17 -


where

akl() =n

i,j=1

aij(x)kxi

lxj

(3.15)

and

F1 (, u,u) = F (, u,u) +n

k=1

u

k

2kxixj

. (3.16)

We wish to find transformations (3.10), for which

akl = 0,1. (3.17)

In general such coordinate transformations exist not over the whole domain , but only at a fixedpoint x0, i.e., the are local rather than global. This point is transformed into 0 = (x0), and(3.15) gives

akl(0) =n

i,j=1

aij(x0)kxi

(x0)lxj

(x0). (3.18)

Denote

a?ij = aij(x0), ki =

kxi

(x0), and lj =lxj

(x0), (3.19)

so that (3.18) can be rewritten as a quadratic form

akl(0) =n

i,j=1

a?ijkilj . (3.20)

We can now make use of the existing theory1 for quadratic forms. In particular, it is provedthat for any real quadratic form in n variables,

Q(z) =n

i,j=1

bijzizj , (3.21)

there exists an orthogonal point-transformation which reduces it to the diagonal form,

Q(z) =r

i=1

z2i m

i=r+1

z2i , m n, (3.22)

i.e., that the coefficient a?ij become

a?ij = 0, i 6= j (3.23a)a?ii = 1, 1 i r (3.23b)a?ii = 1, r < i m. (3.23c)

1see, e.g., http://mathworld.wolfram.com/QuadraticForm.html and references therein

- 18 -


Moreover, according to Sylvesters Inertia Law2, when a quadratic form (3.21) in n variables isreduced by a nonsingular linear transformation to the form (3.22), the number r of positive squaresappearing in the reduction is an invariant of the quadratic form (3.21) and does not depend on themethod of reduction.

Since the numbers r and m are solely properties of the quadratic form (3.18), and thisquadratic form derives directly from a quasi-linear PDE (3.9), they can be used to classify thisequation. Moreover, a transformation that recasts (3.20) as (3.23), also transforms the generalquasi-linear PDE (3.9) into its canonical form,

ri=1

2u

2i

mi=r+1

2u

2i+ F3(, u,u) = 0, m n. (3.24)

3.3.2 Classification of second-order PDEs

Equation (3.9) is called hyperbolic at a point x0, if in (3.24)

m = n and one of its terms has the sign opposite to the others, i.e., if r = 1 or r = n 1.

Equation (3.9) is called elliptic at a point x0, if in (3.24)

m = n and all its terms have the same sign, i.e., if r = 0 or r = n.

Equation (3.9) is called parabolic at a point x0, if in (3.24)

m < n.

Equation (3.9) is called ultrahyperbolic at a point x0, if in (3.24)

m = n and a few terms have opposite signs, i.e., if 1 r n 1.

Examples:

A wave equation for u(x1, . . . , xN , t),

2u

t2= a2

Ni=1

2u

x2i, a2 = 1.

It has n = N + 1 variables, and m = N + 1 = n second derivatives, one of which has theopposite sign. Hence the wave equation is hyperbolic.

2http://mathworld.wolfram.com/SylvestersInertiaLaw.html

- 19 -


A diffusion equation for u(x1, . . . , xN , t),

u

t= a2

Ni=1

2u

x2i, a2 = 1.

It has n = N + 1 variables, and m = N < n second derivatives. Hence the diffusionequation is parabolic.

Poissons equation for u(x1, . . . , xN ),

a2Ni=1

2u

x2i+ f = 0, a2 = 1.

It has n = N variables, and m = N = n second derivatives, all of which have the samesign. Hence the diffusion equation is elliptic.

3.3.3 A curse of dimensionality

The transformation of coordinates (3.10) consists of n equations for n independent variables l(l = 1, . . . , n). Let us count the number of conditions in (3.23), which these variables have tosatisfy. Equation (3.23a),

a?ij = 0, i 6= j, i, j = 1, . . . , n

containsn(n 1)

2!such conditions. Equations (3.23b) and (3.23c) can be combined into one equation

a?ii = ia?11, i = 1, i = 2, . . . , n

and consist of (n 1) conditions. Thus the total number of conditions contained in (3.23) isn(n 1)

2!+ (n 1), (3.25)

which, for n 3, is always larger than the number n of independent variables l. In other words,the problem does not have a solution in n 3 dimensions!

It does have a solution in n = 2 dimensions, which we explore below.

3.3.4 Classification of second-order PDEs in n = 2 dimensions

In n = 2 dimensions, equation (3.9) becomes

a112u

x21+ 2a12

2u

x1x2+ a22

2u

x22+ F (x, u,u) = 0, x = (x1, x2)T , (3.26)

- 20 -


the transformation of variables (3.10) takes the form = (x), or l = l(x), l = 1, 2 (3.27)

with the Jacobian

(1, 2)(x1, x2)

=

1x1

1x2

2x1

2x2

= 1x1 2x2 1x2 2x1 6= 0. (3.28)Without the loss of generality, we assume that a11(x) 6= 0 in the region of interest.

Transformation of variables (3.27) recasts (3.26) as (3.14) and (3.15) with n = 2,

a11()2u

21+ 2a12()

2u

12+ a22()

2u

22+ F1 (, u,u) = 0, (3.29)

where

a11 = a11

(1x1

)2+ 2a12

1x1

1x2

+ a22

(1x2

)2, (3.30a)

a12 = a111x1

2x1

+ a12

(1x1

2x2

+1x2

2x1

)+ a22

1x2

2x2

, (3.30b)

a22 = a11

(2x1

)2+ 2a12

2x1

2x2

+ a22

(2x2

)2. (3.30c)

We note that we can make a11 = a22 = 0 by defining the transformation 1(x) and 2(x) assolutions of

a11

(v

x1

)2+ 2a12

v

x1

v

x2+ a22

(v

x2

)2= 0. (3.31)

The quadratic form in the left hand side of (3.31) factors as

a11

(v

x1 a12 +

a212 a11a22a11

v

x2

)(v

x1 a12

a212 a11a22a11

v

x2

)= 0.

(3.32)

A solution of this equation and its behavior are determined by the term a212a11a22. The followingcases can occur.

Case 1: a212 a11a22 > 0. The quasi-linear second-order PDE (3.26) is hyperbolic.Since a11 6= 0, in this case (3.32) gives a system of first-order PDEs

v

x1 a12 +

a212 a11a22a11

v

x2= 0 (3.33a)

and

v

x1 a12

a212 a11a22a11

v

x2= 0. (3.33b)

- 21 -


Indeed, if a transformation of coordinates 1(x) and 2(x) is defined as solutions of (3.33),equation (3.29) becomes

2a12()2u

12+ F1 (, u,u) = 0, (3.34)

It is easy to show that

a12 = 2a11a22 a212

a11

1x2

2x2

6= 0, (3.35)

provided that 1/x2, 2/x2 6= 0. Hence,2u

12+ F2 = 0, F2 =

F12a12

, (3.36)

which, as we showed earlier, can be transformed into a wave equation.

Case 2: a212 a11a22 = 0. The quasi-linear second-order PDE (3.26) is parabolic.In this case (3.32) becomes (

v

x1+a12a11

v

x2

)2= 0, (3.37)

which gives rise to one first-order PDE,

v

x1+a12a11

v

x2= 0. (3.38)

If the transformation 1 = 1(x) is defined as its solution, it follows from (3.30) that a11 = 0.Then one can show that, regardless of the choice of a transformation 2 = 2(x), the conditiona212 a11a22 = 0 leads to a12 = 0. Hence, (3.29) reduces to

2u

2+ F2 = 0, F2 =

F1 (, u,u)a22

. (3.39)

Case 3: a212 a11a22 < 0. The quasi-linear second-order PDE (3.26) is elliptic.This case requires complex analyses to bring (3.26) to its canonical form. We only note that

if the coefficients aij in (3.26) are analytical functions, then it is possible to reduce (3.26) to

2u

12= G (, u,u) . (3.40)

To eliminate the complex variables, we now use the transformation of coordinates = 1 + i2and = 1 i2 to obtain the canonical form for elliptic equations

2u

2+2u

2= G1

(, , u,

u

,u

). (3.41)

- 22 -

Chapter 4

Method of Characteristics

In the previous Chapter we demonstrated that any second-order quasi-linear PDE with variablecoefficients (3.9) can be transformed into much simpler canonical forms (3.24). In two dimen-sions (n = 2), we also identified a set of first-order PDEs, either (3.33) or (3.38), which thecorresponding transformations of coordinates must satisfy. In this Chapter, we use the method ofcharacteristics to solve these and similar first-order PDEs.1

4.1 Linear first-order PDEs

Both (3.33) and (3.38) can be rewritten as a first-order wave equationv

t+ c

v

x= 0, (4.1)

where we relabeled the coordinates, t x1 and x x2. This equation describes the behavior ofa function v(x, t) in the Eulerian coordinate system (x, t).2

The method of characteristics is based on recasting the process described by (4.1) in theLagrangian (material) framework, in which a process v is measured by a moving observer, x =x(t). While in an Eulerian coordinate system v = v(x, t), in a Lagrangian coordinate systemv = v[x(t), t]. In the Lagrangian framework, the rate of change of v = v[x(t), t] is described bythe total (also known as substantial or convective) derivative,

dvdt

=v

t+

dxdtv

x. (4.2)

Here dx/dt is the velocity of a moving observer.

The comparison of (4.1) and (4.2) shows that if the observer moves with velocity c, i.e., ifdxdt

= c, (4.3)1Much of our presentation of the method of characteristics borrows heavily from R. Haberman, Applied Differential

Equations, 4th Ed., Prentice Hall, 2004.2The Eulerian (field) description of a process v relies on a fixed coordinate system (frame of reference), in which an

immobile observer takes measurements of v(x, t).

23


then the PDE (4.1) can be replaced with an ODEdvdt

= 0. (4.4)

In other words, (4.3) defines a family of curves x(t), along which an original PDE reduces to anODE. Such curves are called characteristics.

4.1.1 Constant wave velocity

For a constant c, (4.3) gives a family of linear characteristics

x = ct+ , (4.5)

where is a constant of integration, such that x = at t = 0. It follows from (4.4) that v(x, t) isconstant along characteristics, i.e., v propagates as a wave with velocity c along the characteristics(4.5). The shape of this wave is determined by the initial condition.

Consider the initial condition

v(x, 0) = V (x). (4.6)

Along the characteristics (4.5), x = at t = 0 so that (4.6) gives v(, 0) = V (). Since along thecharacteristics (4.5) v is constant,

v(x, t) = v(x, 0) = V (). (4.7)

At any point (x, t), the parameter can be determined from (4.5),

= x ct. (4.8)

Substituting (4.8) into (4.7) yields the solution of the PDE (4.1) subject to the initial condition(4.8)3,

v(x, t) = V (x ct). (4.9)

4.1.2 Variable wave velocity

If the wave velocity is c = c(x, t), it follows from (4.3) that the characteristics are no longer linear.The procedure outlined above is generalized by using the following recipe:

1. define a characteristic coordinate = (x, t) by

dxdt

= c(x, t), x(0) = ; (4.10)

2. find v(, t) by integrating the ODE (4.4) subject to the initial condition (4.6);3Note that any function F (x ct) satisfies the PDE (4.1). Hence v(x, t) = F (x ct) is the general solution of

(4.1).

- 24 -


3. eliminate in favor of x and t.

Consider, for example,

c(x, t) = (x+ t)2. (4.11)

The ODE (4.3) becomesdxdt

= (x+ t)2, x(0) = . (4.12)

The substitution w = x+ t gives

dwdt

= w2 + 1, (4.13)

whose solution defines a family of characteristics

arctan(x+ t) = t+ arctan . (4.14)

Writing the initial condition (4.6) along the characteristics (4.14) gives v(x, 0) = v(, 0)= V (). Since along the characteristics (4.14) v is constant,

v(x, t) = v(, 0) = V (). (4.15)

Substituting from (4.14) into (4.15) yields the final solution

v(x, t) = V {tan[arctan(x+ t) t]}. (4.16)

4.2 Classification of second-order PDEs in two dimensions (Cntd.)

We are now ready to revisit the problem of classification of second-order quasi-linear PDEs in twodimensions. As an example, consider a PDE

x22u

x21 x1

2u

x22= 0, x1 > 0, x2 > 0. (4.17)

Using the notation of Section 3.3, we have a11 = x2, a12 = 0, a22 = x1. Hence a212 a11a22 =x1x2 > 0, which identifies (4.17) as hyperbolic.

To determine a coordinate transformation i = i(x1, x2) (i = 1, 2) that transforms (4.17)into its canonical form, we have to solve (3.33), which now become

v

x1x1x2

v

x2= 0 and

v

x1+x1x2

v

x2= 0. (4.18)

Applying the method of characteristics to the first of these equations gives

dx2dx1

= x1x2, x2(0) = , (4.19)

- 25 -


whose solution defines a family of characteristics

=(x3/21 + x

3/22

)2/3. (4.20)

Since along these characteristics v is constant, any function F of solves the first PDE in (4.18),i.e., v = F (x3/21 + x

3/22 ) is its general solution. Any choice of F will provide an appropriate

transformation of coordinates 1 = 1(x1, x2). Let us select

1(x1, x2) = x3/21 + x

3/22 . (4.21)

Similarly, applying the method of characteristics to the second equation in (4.18) gives a transfor-mation of coordinates 2 = 2(x1, x2),

2(x1, x2) = x3/22 x3/21 . (4.22)

The inverse transformation is

x1 =(1 2

2

)2/3, x1 =

(1 + 2

2

)2/3. (4.23)

Note that the inverse transformation is invariant with respect to the choice of F . Substituting(4.23) into (4.17) transforms the latter into

2u

12= 1

2(3/21 3/22 )

(22u

1 21

u

2

). (4.24)

As we showed in the beginning of Section 3.3, this equation is readily transformed into a second-order wave equation under the transformation of coordinates x = 1 2 and t = 1 + 2.

The importance of first-order PDEs and the method of characteristics for solving them goesway beyond their use in classification of second-order quasi-linear PDEs. It stems from a plethoraof physical phenomena described by first-order PDEs, which include the immiscible displacementof one fluid by another in a porous medium, traffic flow, shock waves, and contaminant transportin rivers. Some of these applications are discussed and analyzed below.

4.3 Linear advection with a source

Consider the following problem. A contaminant has been released into a river, whose flow velocityis c. Determine the concentration u(x, t) at a location x downstream from the place of release. Weassume that the river flows fast enough to disregard the diffusive effects on the contaminants con-centration, i.e., that the contaminant transport is due to advection only. This problem is describedby a first-order PDE

u

t+ c

u

x= f(x, t), u(x, 0) = 0. (4.25)

The initial condition implies that the river was free of the contaminant before the release. Thesource function f(x, t) specifies the strength, location, time, and duration of the release.

- 26 -


The presence of a source function f necessitates only a slight modification to the method ofcharacteristics recipe described in Section 4.1. Specifically, in Step 2 of the recipe the ODE (4.4)should be replaced with an ODE

dudt

= f [x(, t), t]. (4.26)

Following this modified recipe we obtain an ODE for the characteristics,

dxdt

= c = x = ct+ . (4.27)

Hence (4.26) becomesdudt

= f(ct+ , t). (4.28)

Recalling that is a constant on a characteristic, and using the initial condition (4.25), gives

u = t0f(ct + , t)dt. (4.29)

The final step of the recipe is to eliminate = x ct in favor of x and t,

u(x, t) = t0f [x+ c(t t), t]dt. (4.30)

4.4 Linear advection and reactions

Here we generalize the contaminant transport problem in Section 4.3 in two important ways. First,we allow the rivers velocity c to vary in space x and time t, i.e., consider c = c(x, t). Then thetransport equation (4.25) takes the form

u

t+cu

x= f(x, t), u(x, 0) = uin(x). (4.31)

This equation can be rewritten as

u

t+ c(x, t)

u

x= a(x, t)u+ b(x, t), u(x, 0) = uin(x). (4.32)

where a cx and b f .Second, we allow the contaminant to undergo linear chemical reactions, which adds an extra

term to (4.31),u

t+cu

x= (u ueq) + f(x, t), u(x, 0) = uin(x), (4.33)

where ueq is the equilibrium concentration and is the reaction rate constant. This, of course, isthe same as (4.32) with a = cx and b = f + ueq.

- 27 -


Since the right-hand side of the PDE in (4.32) is now a function of u, the method of charac-teristics recipe described in Section 4.1 once again needs a slight modification. This time, Step 2of the recipe consists of solving an ODE

dudt

= a[x(t), t]u+ b[x(t), t], u(0) = uin(). (4.34)

As an example, we set c(x, t) ex and uin ex2 in (4.31). This results in (4.32) witha = ex, b = 0, and c = ex. To solve the resulting PDE, we follow the recipe. Step 1 is to solvean ODE for the characteristics,

dxdt

= ex, x(0) = . (4.35)

Integrating this equation gives a family of characteristics

ex e = t = = ln (ex t) . (4.36)

Step 2 is to solve the ODE (4.34), which now becomesdudt

= exu, u(0) = uin(). (4.37)

Substituting an expression for ex from (4.36) and integrating along the characteristics yields uuin()

du

u= t0

dt

t + e= u(, t) = uin() t+ e

e. (4.38)

Step 3 is to eliminate in favor of x and t. Substituting from (4.36) into (4.38) yields

u(x, t) = uin [ln(ex t)] ex

ex t . (4.39)

For uin() = e2, this gives a solution for the contaminant distribution in a river,

u(x, t) = exp[ ln2(ex t)] ex

ex t . (4.40)

4.5 Quasi-linear first-order PDEs

The approach we used in the previous section can be readily generalized to account for nonlinearchemical reactions. Such a generalization replaces the reactive term in (4.33) with a reaction termr(u)

u

t+cu

x= r(u) + f(x, t), u(x, 0) = uin(x). (4.41)

To be specific, we consider chemical reactions described by r(u) = (uueq) and set = 2.

- 28 -


Equation (4.41) can be recast in the form of a general quasi-linear first-order PDE,u

t+ c(x, t)

u

x= g(x, t, u), u(x, 0) = uin(x), (4.42)

where g = cx(x, t)u(x, t) + r(u) + f(x, t). We use the method of characteristics to solve thequasi-linear PDE (4.42) with constant velocity c and without sources f = 0.

Step 1 of the recipe is to find the characteristics by solving

dxdt

= c, x(0) = = = x ct. (4.43)

Step 2 of the recipe is modified to solve an ODE

dudt

= r(u), u(0) = uin[x(, t)]. (4.44)

Recalling that r(u) = 2(u2 u2eq) and integrating along the characteristics gives uuin

duu2 u2eq

= 2t = u uequ+ ueq

=uin() uequin() + ueq

e4t. (4.45)

Finally, we use Step 3 to eliminate by substituting (4.43) into (4.45),u uequ+ ueq

=uin(x ct) uequin(x ct) + ueq e

4t. (4.46)

4.6 Nonlinear first-order PDEs

Consider a nonlinear (often referred to as quasi-linear) first-order PDE,u

t+ c(u)

u

x= 0, u(x, 0) = uin(x), (4.47)

which describes, for example, traffic flow under an assumption that the car velocity v depends onlyon the density u(x, t) of cars on the road. The coefficient c(u) in (4.47) is called the characteristicvelocity of traffic flow.

The presence of nonlinearity c(u) does not affect the fundamental idea of the method ofcharacteristics, which is to reduce first-order PDEs to first-order ODEs by writing the former inthe Lagrangian framework. Thus, as before, we have

dxdt

= c(u), x(0) = (4.48)

anddudt

= 0, u(0) = uin(). (4.49)

However, one cannot obtain a family of characteristics from (4.48) alone, since its right-hand sidedepends on the unknown u. In other words, now ODEs (4.48) and (4.49) are fully coupled.

- 29 -


The fact that the right-hand side of the ODE in (4.49) is zero significantly simplifies theanalysis, since it implies that u is constant along characteristics,

u = uin(). (4.50)

Hence we can integrate (4.49) along characteristics to obtain

x = c(u)t. (4.51)

Thus, as in Section 4.1, the characteristics are straight lines. The difference is that now thesestraight lines are not parallel to each other.

Finally, using (4.51) to eliminate in favor of x and t in (4.50) we obtain

u(x, t) = uin[x c(u)t]. (4.52)

Fanlike characterisitics. Consider the following initial value problem,

u

t+ 2u

u

x= 0, u(x, 0) =

{1 x < 02 x > 0.

(4.53)

Using the general forms of its characteristics (4.51) and its solution (4.52), we obtain

x = 2u(, 0)t (4.54)

and

u(x, t) =

{1 x 2ut < 02 x 2ut > 0, (4.55)

respectively. The latter can now be solved analytically,

u(x, t) =

{1 x < 2t2 x > 4t,

(4.56)

Since the distance between u = 1 and u = 2 increases, this solution is called an expansion wave.

It remains to determine the behavior of u(x, t) in the region 4t < x < 2t. The reason for thisgap in our solution is the discontinuity of the initial condition in (4.53). We imagine that all valuesof u between 1 and 2 are present initially at x = 0. Straight line characteristics corresponding toeach of these values start at the point x = 0 and t = 0. Since at this point = 0, it follows from(4.54) that the equation for these characteristics is

x = 2ut, 1 < u < 2. (4.57)

This equation gives u(x, t) for the wedge-shaped region

u =x

2t, 2t < x < 4t. (4.58)

- 30 -


4.6.1 Elementary traffic model

As an example, we consider an elementary traffic model, which replaces the function c(u) in (4.47)with its linear counterpart. This is accomplished as follows. We postulate that cars move with themaximum velocity v = vmax at u = 0 and practically stop moving after the density of cars reachessome u = umax. A linear relationship that satisfies these conditions is

v(u) = vmax

(1 u

umax

). (4.59)

Since c(u) (uv)u, this gives

c(u) = vmax

(1 2u

umax

), (4.60)

which, when combined with (4.47), gives an elementary traffic equationu

t+ vmax

(1 2u

umax

)u

x= 0. (4.61)

The initial condition u(x, 0) = uin(x) is determined by the traffic scenario we are interested in.Consider a situation when the traffic light switches from red to green. Behind a red light (x = 0),the traffic density is maximum u = umax, while ahead of the light it is u = 0, so that at themoment t = 0 when the light turns green

u(x, 0) = uin(x) =

{umax, x < 00, x > 0.

(4.62)

For c(u) and uin given by (4.60) and (4.62), respectively, the general solution (4.52) becomes

u(x, t) =

umax, x vmax

(1 2uumax

)t < 0

0, x vmax(1 2uumax

)t > 0.

(4.63)

This gives a final expression for the density of cars in traffic,

u(x, t) =

{umax, x < vmaxt0, x > vmaxt.

(4.64)

The comparison of (4.62) and (4.64) shows that there is a delay between the time the light switchesto green and the time cars start moving. It is because the information propagates backward at thespeed umax.

4.7 Shock waves

For nonlinear (quasilinear) first-order wave equation (4.47),u

t+ c(u)

u

x= 0, u(x, 0) = uin(x), (4.65)

- 31 -


1 2

x

t

c(u1) < c(u2)

1 2

x

t

c(u1) > c(u2)

Figure 4.1: Linear characteristics (4.51) for nonlinear (quasilinear) first-order wave equation(4.65).

the method of characteristics will not alway work in a way described in the previous section.Specifically, it fails when the characteristics intersect. Figure 4.1 shows two possible scenariosfor the behavior of characteristics (4.51). Two characteristicsone starting at x = 1 with u1 =uin(1) and the other starting at x = 2 with u2 = uin(2)intersect if the faster one catches upwith the slower one, i.e., if c(u1) > c(u2). In this case, which is know as a compression wave, thedistance between the densities u1 and u2 decreases with time, and eventually they intersect, i.e.,at some point x two densities coexist. In other words, density becomes a multi-valued function ofspace, which in many cases is not physically realizable.

4.7.1 Discontinuous solutions

What is the reason for arriving at this nonphysical behavior (e.g., multi-valued density)? Since ourmathematical treatment of (4.65) did not use any approximations, the source of error must be inthe mathematical formulation of a physical process, i.e., in the formulation of (4.65).

All our analysis so far implicitly assumed that u(x, t) is a continuous function. This as-sumption will now be overturned by postulating that at some point xs the function u(x, t) has ajump discontinuity, also called a shock wave, that propagates in time, i.e., xs(t), with the velocitydxs(t)/dt. Schematic representation of a jump discontinuity of u(x, t) at point xs is shown inFigure 4.2, where xs and x+s indicate that the point xs is being approached from the left and theright, respectively. At the moment, the point of discontinuity xs, its trajectory xs(t), and velocitydxs(t)/dt are all unknown.

We start by reformulating our mathematical model. Consider the conservation law for asegment 0 < x < b of the road, which now includes a jump discontinuity 0 < xs(t) < b,

ddt

bau(x, t)dx = Mass INMass OUT (4.66)

where Mass IN = q(a, t), Mass OUT = q(b, t), q = vu, and v(u) is the velocity of a substancewhose density is u. Accounting for the shock, this equation can be rewritten as

ddt

[ xs(t)a

u(x, t)dx+ bxs(t)

u(x, t)dx

]= q(a, t) q(b, t) (4.67)

- 32 -


x

u(xs , t)

u(x, t)

u(x+s , t)

xs(t)

Figure 4.2: Jump discontinuity of u(x, t) at a point xs(t).

Using Leibnitzs rule of differentiation, we get

dxsdt[u(xs , t) u(x+s , t)

]+ xs(t)a

u

tdx+

bxs(t)

u

tdx = q(a, t) q(b, t). (4.68)

Since the continuity equation

u

t+q

x= 0 (4.69)

holds on both sides of the shock x = xs(t), the integrals become xs(t)a

u

tdx+

bxs(t)

u

tdx =

xs(t)a

q

xdx

bxs(t)

q

xdx

= q(xs , t) + q(a, t) q(b, t) + q(x+s , t) (4.70)

and (4.68) yields an expression for the shock velocity

dxsdt

=q(xs , t) q(x+s , t)u(xs , t) u(x+s , t)

={q}{u} , (4.71)

where {A} denotes the magnitude of the jump of a function A. In gas dynamics, (4.71) is calledthe Rankine-Hugoniot condition.

A useful property of the shock velocity is provided by the entropy condition,

c[u(xs , t)] >dxsdt

> c[u(x+s , t)]. (4.72)

This general property is tested below.

A shock and a compression wave. Consider the following initial value problem

u

t+ 2u

u

x= 0, u(x, 0) =

{2 x < 01 x > 0.

(4.73)

- 33 -


Following the recipe, we write an equation for characteristics,

dxdt

= 2u. (4.74)

Since u(x, t) is constant along characteristics, (4.74) gives a family of linear characteristics

x = 2u(, 0)t+ . (4.75)

Taking into account the initial condition in (4.73), one can see that the characteristics < 0 traveltwice as fast as the characteristics > 0. Hence the two sub-families will intersect, leading tonon-physical multi-valued solutions.

The problem is fixed by introducing a jump discontinuity that travels with the shock velocity(4.71). It follows from the initial condition in (4.73) that u(x, t) = 2 on the left-hand side of theshock and u(x, t) = 1 on the right-hand side of the shock. It remains to find the trajectory of theshock xs(t) from (4.71).

To make use of (4.71), we note that q in the continuity equation (4.69) is related to c(u) in(4.65), or to c(u) = 2u in (4.73), by

c(u)u

x=q

u

u

x. (4.76)

Hence for u(x, t) in (4.73) to be a conservative quantity,q

u= c(u), = q =

c(u)du = 2

udu = u2. (4.77)

Equation (4.71) becomes

dxsdt

=u2(xs , t) u2(x+s , t)u(xs , t) u(x+s , t)

= u(xs , t) + u(x+s , t) = 2 + 1 = 3, (4.78)

i.e., the velocity of the shock is constant. Since the discontinuity in u(x, 0) occurs at x = 0, wefind that the trajectory of the shock is

dxsdt

= 3, xs(0) = 0 = xs(t) = 3t. (4.79)

Since c(u) = 2u, we find that c[u(xs )] = 4 and c[u(x+s )] = 2. Hence the entropy condition(4.72) holds.

4.7.2 Elementary traffic model with shocks

Let us return to the task of modeling traffic flow. Following our approach in Section 4.6.1, weemploy the elementary traffic model which results in (4.61),

u

t+ vmax

(1 2u

umax

)u

x= 0. (4.80)

- 34 -


However, now we are interested in the opposite traffic scenario, i.e., a situation when the trafficlight switches from green to red. Before the light turns red, the traffic density is uniform, u = u0.The traffic light is located at x = 0 and turns red at time t = 0. At this time, the traffic densitybehind the light is uniform,

u(x, 0) = u0, x < 0. (4.81)Since the cars stop at the light, the traffic density is maximum at x = 0,

u(0, t) = umax, t > 0. (4.82)This is the first time we have introduced a boundary condition for the first-order wave equation.

An equation for characteristics is

dxdt

= vmax

(1 2u

umax

), x(0) = . (4.83)

Thus characteristics are given by

x = vmax(1 2u

umax

)t. (4.84)

They form two families of parallel curves

x = vmax(1 2u0

umax

)t and x = vmaxt (4.85)

that propagate the traffic densities u0 and umax. The characteristic velocity c1 of the first familyof characteristics (originating at points x < 0) is

c1 = vmax

(1 2u0

umax

), (4.86)

while that of the second family (originating at the point x = 0) isc2 = vmax. (4.87)

The characteristics from x = 0 move backward, since c2 < 0. The characteristics originating atpoints x < 0 can move either backward or forward depending on the ratio 2u0/umax. Regardless,the latter move faster the the former, since c1 > c2.

Thus a shock separating u0 and umax will form. Since it follows from (4.77) that the trafficflow is

q(u) =c(u)du = vmax

(1 2u

umax

)du = vmaxu

(1 u

umax

), (4.88)

the shock velocity is determined from (4.71) asdxsdt

=q(xs , t) q(x+s , t)u(xs , t) u(x+s , t)

=q(u0) q(umax)u0 umax =

q(u0)u0 umax . (4.89)

Since u0 < umax and q(u0) > 0, the shock velocity is negative, and the shock propagates back-ward. The corresponding shock trajectory is

xs =q(u0)

u0 umax t. (4.90)

- 35 -


4.7.3 Conditions for shock formation

We have established that shocks form when faster waves start behind slower waves, i.e., when thecharacteristic velocity c(u) is a decreasing function of x,

c(u)x

=dcdu

u

x< 0. (4.91)

Thus for equations with dc/du < 0 shocks form when u(x, t) is an increasing function of x,so that u must increase with x at a shock. Likewise, for equations with dc/du > 0, shocksform when u decreases with x at a shock. If these conditions are not met, characteristics do notintersect, shocks do not form, and discontinuous initial conditions correspond to expansion waves(see Section 4.6.1).

If dc/du does not change signs, then discontinuous initial conditions result in either a shockor an expansion wave.

If dc/du does change signs at least ones, then discontinuous initial conditions may result inboth a shock and an expansion wave.

A shock and an expansion wave. Consider the following initial value problem

u

t+ u2

u

x= 0, u(x, 0) =

{1 x < 02 x > 0.

(4.92)

The initial conditions indicate that u will increase with x, i.e., u/x > 0. On the other hand,

dcdu

= 2u (4.93)

can be negative (leading to a shock) and positive (leading an expansion wave).An equation for the characteristics is

dxdt

= u2, x(0) = , (4.94)

and an equation for u gives

dudt

= 0, u(, 0) =

{1 < 02 > 0

= u(, t) ={1 < 02 > 0.

(4.95)

Since along the characteristics u is constant, (4.94) gives

x = u2(, 0)t. (4.96)

Eliminating in favor of x and t gives

u(x, t) =

{1 x < t2 x > 4t.

(4.97)

- 36 -


The fan-like characteristics for the region 4t < x < t are obtained as before (Section 4.6). Thesecharacteristics correspond to the discontinuity in the initial condition in (4.92) and originate at thepoint x = 0. Hence, (4.96) gives an equation for these characteristics

x = u2t, 4t < x < t, (4.98)

which leads to a fan-like family of characteristics

ufan = x

t, 1 < ufan < 2. (4.99)

In accordance with (4.91), the part of this solution,

ufan =x

t, 0 ufan < 2, (4.100)

combined with the positive part of the solution (4.97) represents the expansion wave.Consider now the remaining characteristics < 0, along which u = 1. It follows from

(4.94) that their characteristic velocity is 1. They will intersect the slower moving characteristicsin the expansion wave, which correspond to 1 < u < 0. This leads to the formation of a shock,which separates the region with u = 1 and the region with u =x/t.

Since

q(u) =c(u)du =

u3

3, (4.101)

the shock velocity is obtained from (4.71) as

dxsdt

=131 + (xs/t)3/2

1 + (xs/t)1/2. (4.102)

This ODE has to be solved numerically.

4.8 Nonlinear first-order PDEs with a source

Here we generalize the nonlinear (quasi-linear) first-order PDE in (4.47) by adding a state-dependentright-hand side,

u

t+ c(u)

u

x= 2u, u(x, 0) = uin(x), (4.103)

and set c(u) u. This example loosely corresponds to traffic flow, in which the cars on ahighway are not conserved, but instead leave it at a rate proportional to the car density u.

Along the characteristics

dxdt

= u, x(0) = , (4.104)

- 37 -


the density of cars is no longer constant, but is determined instead as a solution of an ODE

dudt

= 2u, u(, 0) = uin(). (4.105)

The solution of (4.105) is

u = uin()e2t. (4.106)

Substituting it into (4.104) gives

x = uin()2

[e2t 1] . (4.107)

In general, the last step of eliminating in favor of x and t has to be done numerically.

Consider a simple initial condition uin(x) = x. Now (4.107) can be solved analytically togive

=2x

1 + e2t. (4.108)

Eliminating leads to the car density distribution

u =2x

1 + e2te2t =

2xe2t + 1

. (4.109)

- 38 -

Chapter 5

Laplace Transformation

In this Chapter we introduce the Laplace transformation as another tool to solve PDEs by reducingthem to ODEs.

5.1 Motivation and Definition

Consider one-dimensional diffusion equation for u(x, t),u

t= D

2u

x2, a < x < b, t > 0. (5.1)

It is subject to the initial condition u(x, 0) = uin(x) and some appropriate boundary conditions.Multiplying this equation by est and integrating from t = 0 to t = gives

0

u

testdt = D

2

x2

0

uestdt. (5.2)

The left-hand side of this equation can be integrated by parts to yield, after accounting for theinitial condition,

uin(x) + s 0

uestdt = D2

x2

0

uestdt. (5.3)

Denoting

u(x, s) = 0

u(x, t)estdt, (5.4)

we obtain an ODE

uin(x) + su = Dd2udx2

, (5.5)where (real or imaginary) transformation constant s acts a parameter.

The operation on a function u in (5.4) is called the Laplace transformation, the new func-tion u is called the Laplace transform. Often it is convenient to use a shorthand notation, whichreplaces the integral in (5.4), by writing

u(x, s) = L{u(x, t)}. (5.6)

39


5.2 Properties of the Laplace transformation

Some of the basic properties of the Laplace transformation are discussed below.

5.2.1 Linearity

The Laplace transformation L{u} is linear, i.e., for a linear combination of two functions u1 andu2,

L{c1u1 + c2u2} = c1L{u1}+ c2L{u2}. (5.7)

5.2.2 Applicability

Piecewise continuous functions. A function u(t) is called piecewise continuous on an intervala t b, if the interval can be subdivided into a finite number of sub-intervals, in each ofwhich u(t) is continuous and has finite limits at the ends of these sub-intervals. Such functions areintegrable on the interval a t b. As an example of piecewise continuous functions, considera unit step function

Sp(t) =

{0 0 < t < p1 t > p.

(5.8)

Its Laplace transform exists and is given by 0

Sp(t)estdt = p

estdt =1seps, (5.9)

provided s > 0.

However, not all piecewise continuous functions have Laplace transforms. For instance, afunction

fp(t) =

{et 0 < t < pet2

t > p(5.10)

does not have the Laplace transform because the integral 0

fp(t)estdt = p0

e(1s)tdt+ p

et2stdt (5.11)

does not exist.

Functions of exponential order. A function u(t) is of exponential order as t, or O (et),if there exists such that

|u(t)|et


for all t larger than some finite number T . Clearly e2t satisfies this condition, while the functionfp(t) defined above does not.

Any function u(t) that is piecewise continuous and of exponential order as t has theLaplace transform. This condition can be easily proved by noting that |estu(t)| < Me(s)t,where M is some constant. Since the upper bound of |estu(t)| is integrable provided s > , it isintegrable as well.

One has to keep in mind that this condition is sufficient but not necessary. For example, thefunction u(t) = 1/

t has an infinite discontinuity at t = 0, but still has the Laplace transform,

L

{1t

}= 0

1testdt =

2s

0

ex2dx =

pi

s, s > 0. (5.13)

Transforms of derivatives. The passage from (5.2) to (5.3),

L

{du

dt

}= sL{u} u(0), (5.14)

implicitly assumed that a solution of the diffusion equation u is of exponential order as t .Other requirements on u, i.e., that u be continuous with a piecewise continuous derivative, aretypically implied by the fact that u is a solution of a differential equation.

For example,

L

{ t0u(t)dt

}=

1sL{u}. (5.15)

The inverse Laplace transform. The inverse Laplace transformation is denoted by L1. If

L{u(t)} = u(s), (5.16)then

u(t) = L1{u(s)}. (5.17)

The inverse Laplace transformation is a linear operation. Indeed, since

L{c1u1 + c2u2} = c1L{u1}+ c2L{u2} (5.18)we have

L1{c1u1 + c2u2} = c1u1 + c2u2 = c1L1{u1}+ c2L1{u2}. (5.19)This property will be used again and again to compute the inverse Laplace transforms of solutionsof differential equations. A number of the pairs of direct and inverse transforms are tabulated.

A theorem of substitution. Let u(t) be such that its Laplace transform u(s) exists when s > .Then

u(s a) = L{eatu(t)}, s > + a. (5.20)

- 41 -


A theorem of scaling. If u(s) = L{u(t)} when s > , then1au(sa

)= L{u(at)}, s > a, a > 0. (5.21)

A theorem of translation. Consider a function u(t), and let ub(t) denote its translation (seeFig. 5.1),

ub(t) =

{0 0 < t < bu(t b) t > b. (5.22)

Then

ebsu(s) = L{ub(t)}. (5.23)

0 t0 t

u(t)

b

ub(t)

Figure 5.1: The translation of a function u(t).

This theorem is critical for understanding convolutions.

Convolution. The convolution u v of the functions u(t) and v(t) is defined as the function

u(t) v(t) = t0u()v(t )d, 0 t


5.3 Solutions of linear ODEs with constant coefficients

The application of the Laplace transformation to the solution of linear ODEs with constant coeffi-cients is relatively straightforward.

5.3.1 Forced vibrations with damping

As an example, we find the solution of the differential equation

d2udt2

dudt 6u = 2 (5.29)

subject to the initial conditions

u(0) = 1,dudt

= 0. (5.30)

This problem describes forced vibrations with viscous damping.

Applying the Laplace transformation to the both sides of this equation gives an algebraicequation

s2u s su+ 1 6u = 2s. (5.31)

Hence

u =2 + s2 ss(s2 s 6) . (5.32)

By writing this as

u =2 + s2 s

s(s 3)(s+ 2) = 131s+

815

1s 3 +

45

1s+ 2

(5.33)

and consulting a table of Laplace transforms, we obtain the solution

u(t) = 13+

815

e3t +45e2t. (5.34)

5.3.2 A damped absorber

If a damped absorber is added to forced vibrations, the process is described by a coupled systemof ODEs, such as

d2udt2

d2v

dt2+

dvdt u = et 2, (5.35)

2d2udt2

d2v

dt2 2du

dt+ v = t. (5.36)

- 43 -


We assume the homogeneous initial conditions

u(0) =dudt

(0) = v(0) =dvdt

(0). (5.37)

Applying the Laplace transformation to these ODEs gives a system of algebraic equations

s2u s2v + sv u = 1s 1

2s, (5.38)

2s2u s2v 2su+ v = 1s2. (5.39)

These equations can be rewritten as

(s+ 1)u sv = s 2s(s 1)2 , (5.40)

2su (s+ 1)v = 1s2(s 1) . (5.41)

The solutions of this system are

u =1

s(s 1)2 =1s 1s 1 +

1(s 1)2 (5.42)

and

v =2s 1

s2(s 1)2 = 1s2

+1

(s 1)2 . (5.43)

Consulting a table of Laplace transforms, we arrive at the solution of the problem,

u(t) = 1 et + tet and v(t) = t+ tet. (5.44)

5.4 Solutions of integral equations with convolution kernels

Consider a Fredholm integral equation of the second kind,

u(t) = at+ t0u(t) sin(t )d. (5.45)

Since the integral on the right-hand side is a convolution, taking the Laplace transformation of thisequation gives an algebraic equation

u(s) =a

s2+ u(s)

1s2 + 1

, (5.46)

whose solution is

u(s) = a(

1s2

+1s4

). (5.47)

Taking the inverse Laplace transform, we obtain the solution

u(t) = a(1t+

6t3

)(5.48)

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5.5 Solutions of linear PDEs

In this section we use the Laplace transformation to solve linear PDEs, whose coefficients do notvary with time.

5.5.1 Wave equation

Consider the wave equation,

2u

t2= c2

2u

x2, x > 0, t > 0 (5.49)

subject to the initial conditions

u(x, 0) = (x),u

t(x, 0) = 0 (5.50)

and the boundary conditions

u(0, t) = 0, u(x, t) = 0. (5.51)

This problem describes vertical displacements in a long string that has been initially displaced.

Accounting for the initial conditions, 0

2u

t2estdt = s

0

u

testdt = s2u s(x), (5.52)

so that the Laplace transformation of the wave equation leads to the ODE

s2u s(x) = c2d2u

dx2, x > 0 (5.53)

subject to the boundary conditionsu(0, s) = 0, u(x, s) = 0. (5.54)

We shall solve this ODE by using the Laplace transformation with respect to x. Let

u(p, s) = 0

u(x, s)epxdx. (5.55)

Then taking the Laplace transformation of the ODE (5.53), while accounting for the boundaryconditions, yields the algebraic equation

s2u s(p) = c2 [p2u C(s)] , C(s) dudx

(0, s), (5.56)

whose solution is

u =c2C sc2p2 s2 . (5.57)

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Taking the inverse Laplace transform with respect to p gives

u(x, s) =cC

ssinh

sx

c sc2L1

{

p2 s2/c2}. (5.58)

Using the convolution to invert the last term in this expression gives

u(x, s) =cC

ssinh

sx

c 1c

x0(y) sinh

s

c(x y)dy. (5.59)

Now we can determine C(s) from the boundary condition at infinity,

0 = expsx

c

[cC

s 1c

0

(y)esy/cdy], x. (5.60)

Hence,

cC

s=

1c

0

(y)esy/cdy. (5.61)

Thus the Laplace transform of the solution is

cu(x, s) = sinhsx

c

0

(y)esy/cdy x0(y) sinh

s

c(x y)dy. (5.62)

To compute the inverse Laplace transform of this expression, we rewrite it as

2cu(x, s) = 0

(y)es(x+y)/cdy + x

(y)es(xy)/cdy + x0(y)es(xy)/cdy. (5.63)

The change of variables

t =x+ yc

, t =y xc

, t =x yc

, (5.64)

in the first, second and third integrals, respectively, gives

2u(x, s) = x/c

(ct x)estdt+ 0

(ct+ x)estdt+ x/c0

(x ct)estdt. (5.65)

The first and third integrals can be combined by introducing a function

1(z) =

{(z) z 0(z) z 0, (5.66)

which is called the odd extension of (z). This gives

2u(x, s) = 0

(x+ ct)estdt+ 0

1(x ct)estdt, (5.67)

whose inverse Laplace transform is

2u(x, t) = (x+ ct) + 1(x ct). (5.68)Finally, since in the domain of interest x ct, (x ct) = 1(x ct) so that

u(x, t) =(x+ ct) + (x ct)

2. (5.69)

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5.5.2 Diffusion equation

Consider the diffusion equation,

u

t= D

2u

x2, x > 0, t > 0 (5.70)

subject to the initial conditionu(x, 0) = 0 (5.71)


kux

(0, t) = q, u(x, t) = 0. (5.72)

This problem describes the distribution of temperature in a long rod when a constant flux of heatq is maintained at the boundary x = 0.

The Laplace transform of this problem is

su = Dd2udx2

, x > 0 (5.73)

subject to the boundary conditions

kdudx

(0, s) =q

s, u(x, s) = 0. (5.74)

The solution of this problem is

u(x, s) =qD

kssex

s/D. (5.75)

It can be inverted by means of a table of Laplace transforms.

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- 48 -

Chapter 6

Nonhomogeneous Problems andGreens Functions

The Greens functions are an ideal tool for solving linear nonhomogeneous PDEs with nonhomo-geneous initial and boundary conditions.

6.1 Motivation and basic ideas

Consider a linear differential equation

c

t= (Dc) vc+ qc+ f x t > 0 (6.1)

which describes, among many other processes, contaminant transport in porous media. All coeffi-cients in this equation are functions of the N -dimensional space x and time t. The coefficientD is a positive semi-definite tensor whose components are Dij(x, t) (i, j = 1, . . . , N ), and v is avector with components vi(x, t) (i = 1, . . . , N ). Equation (6.1) is subject to the initial condition

c(x, 0) = cin(x) x (6.2)


c(x, t) = (x, t) x D t > 0, (6.3)

n Dc = (x, t) x N t > 0, (6.4)

n (Dc vc) = r(x, t) x R t > 0, (6.5)

where D, N, and R are the Dirichlet, Neumann, and Robin segments of the boundary =D N R of . The boundary value problem (6.1) (6.5) is nonhomogeneous due to thepresence of the driving forces f , , , and r.

49


We start by multiplying (6.1) with as yet unspecified function G(x, t;y, ) and integrating itin space and time, t

0

[c

(Dc) + vc qc

]Gdyd =

t0

fGdyd. (6.6)

Since t0

c

Gdyd =

[cG]t=0 dy

t0

cG

dyd, (6.7)

t0

(Dc)Gdyd =

t0

n [GDc cDG] dyd +

t0

c (DG)dyd,

(6.8)and t

0

G vc =

t0

Gn vc

t0

G vcdyd (6.9)

equation (6.6) can be written as

t0

c

[G

+ (DG) +G v + qG

]dyd +

[cG]t=0 dy

t0

n [GDc cDGGvc] dyd =

t0

fGdyd. (6.10)

Due to the initial and boundary conditions (6.2) (6.5), this gives t0

c(y, )(x y)(t )dyd =

t0

fGdyd

+cinG(x,y, t, 0)dy

t0

D

n DGdyd + t0

N

Gdyd + t0

R

rGdyd.

(6.11)where we chose the function G(x,y, t, ) to be the solution of the differential equation

G

= y (DyG) + v yG+ qG+ (x y)(t ) (6.12)

subject to the homogeneous initial conditionG(x,y, t, t) = 0 (6.13)

and the homogeneous boundary conditions

G(x,y, t, ) = 0 y D, (6.14)

n [DG vG] = 0 y N, (6.15)and

n DG = 0 y R. (6.16)

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Thus defined G(x, t;y, ) is called the Greens function for the boundary value problem (6.1) (6.5).

You might recognize the function in (6.12) as the Dirac delta function. We will explore itsproperties below. At this point, it suffices to say that it allows us to obtain from (6.11) the solutionof (6.1) (6.5),

c(x, t) = t0

fGdyd +

cinG(x,y, t, 0)dy

t0

D

n DGdyd + t0

N

Gdyd + t0

R

rGdyd. (6.17)

Note that the Greens function is the solution of the boundary value problem (6.12) (6.12), whichis similar to the original problem (6.1) (6.5). What have we gained? First, homogeneous prob-lems are usually easier to solve than their nonhomogeneous counterparts. Second, the boundaryvalue problem for the Greens function is independent from external forces and, thus, describesthe internal structure of a system. Once determined (analytically or numerically), it can be used toanalyze the response of the system to applied external forces in accordance with (6.17).

6.2 The Dirac delta function

The one-dimensional Dirac delta function is defined as

(x x) ={0 x 6= x x = x. (6.18)

It is not a function in the usual sense of the word. Instead, it is an example of the so-calledgeneralized functions that are defined in relationship to other functions. Specifically, the definitionof the Dirac delta function (and its main property) is that for any continuous function f(x),

f(x) =

f(x)(x x)dx. (6.19)

An immediate consequence of this definition is that it has unit area,

(x x)dx = 1. (6.20)

Other important consequences are that the Dirac delta function is even,

(x x) = (x x), (6.21)that its derivative is given by

df

dx=

f(x)d(x x)

dxdx, (6.22)

that it is the derivative of the Heaviside step function,

(x x) = dH(x x)

dx, H(x x) =

{0 x x1 x > x,

(6.23)

- 51 -


and, finally, that it has the following scaling property:

[c(x x)] = 1|c|(x x). (6.24)

The best way to think of the Dirac delta function (x x) is in terms of a source of infinitestrength concentrated at a point.

The properties of the one-dimensional Dirac delta function are easy to generalize to higherdimensions by using its definition

f(x) =

f(y)(x y)dy. (6.25)

This gives

(x y) = (x1 y1) (xN yN ). (6.26)

6.3 Properties of the Greens functions

6.3.1 Causality

The analysis of the boundary value problem (6.12) (6.12) shows that the Greens functionG(x, t;y, ) represents the response of the system at (x, t) to a perturbation introduced by theinfinite strength instanteneous point source located at (y, ). It becomes obvious from this inter-pretation that

G(x, t;y, ) = 0 for > t, (6.27)since the system does not know about perturbations that occur in the future. This property is calleda causality principle.

6.3.2 Reciprocity

Equations (6.1) and (6.12) can be written in the operator form as

L(c) = f(y, ) L =

y Dy +y v q (6.28)

and

L(G) = (x y)(t ) L = y Dy v y q, (6.29)

respectively. The operator L is called the adjoint of th

methods in applied mechanics iii_daniel m. tartakovsky.pdf

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