Chapter 4

Introduction to MathematicalModelling and Partial DifferentialEquations of Mathematical Physics

4.1 Notation

For the notation involving functions, we will sometimes suppress dependent variables:u(x, y, ...) = u. Functions are usually assumed smooth enough to enable us to carry outall necessary differentiations. All functions and derivatives will be assumed continuous, unlessotherwise stated.

The subscript notation will be used for the partial derivatives:

∂

∂xF (x, y) = Fx,

∂

∂yF (x, y) = Fy,

etc.

The sets are denoted by a, b, ...; set operations will be used.

The orthonormal basis of the space R3 is denoted i, j,k or ex, ey, ez as needed in thecontext. For the orthonormal basis of Rn, n ≥ 2, we will use the notation e1, e2, ....

Vectors are denoted by boldface: a. Vectors in a given orthonormal basis will be written asan ordered set of coordinates in that basis, or in terms of unit vectors, for example,

n =

n1

n2

n3

≡ [n1, n2, n3]T = n1ex + n2ey + n3ez.

Here T is the matrix transposition symbol. Matrices will also use boldface symbols: Ax = b,etc.

Linear operators will be denoted by straight Roman symbols, e.g.,

L =∂

∂t− c

∂

∂x.

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The symbol denotes the end of a proof/derivation.

Symbol ∂D will denote the boundary of a bounded domain D.

We will use “math” and “physics” notation for the repeated integrals freely. In “physics”notation, the integrand is at the last place, so that differentials stand by their integrals, andlimits are obvious. For example, the following “math” integral is written in the “physics” form:

P (x) =

∫ 1

0

(∫ b

a

F (x, y, z) dy

)dz ≡

∫ 1

0

dz

∫ b

a

dy F (x, y, z).

4.2 Mathematical Modelling and PDEs

We start from a discussion of mathematical models, in particular, how they arise, and how theyare formulated, and which models lead to partial differential equations.

Definition 4.1. A mathematical model is a relation between parameters of interest in a givenapplication, which allows for some analysis and/or solution relevant to the description of theprocesses of interest.

The above-mentioned parameters of interest are referred to as variables. Variables haveappropriate units of measurement. Examples of common physical variables include:

• position x [meters];

• time t [seconds];

• charge q [coulombs];

• energy density E [joules],

and so on.

Other areas of knowledge need other variables, e.g., in econometrics, one is interested invariables such as prices, revenue, employment rate, and so on. In these notes, physical modelsare emphasized, since they are related to the equations of interest in a more direct way.

There are two main kinds of variables we will deal with on a regular basis:

(a) Independent variables: x, t, ...

(b) Dependent variables: f(x), T (x, t), u(x, y, z), ...

Vector variables will be denoted by boldface. For example, we will freely write things like

• r = [x, y, z]T = xi+ yj+ zk ∈ R3,

• u = u(t, r) = [u1(t, x, y, z), u2(t, x, y, z)]T ∈ R2,

and so on. The number of dimensions for a vector will follow from the context.

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4.2.1 Origins of Mathematical Models

According to the above definition, a mathematical model arises in the form of one or moreequations. What are the possible origins of such equations? On other words, what is thebasis for the formulation of a mathematical model? The model equations are usually derivedfrom the following principles.

(a) Fundamental/philosophical (axiomatic) principles, such as conservation of energy,mass, momentum; other specific assumptions such as symmetry of the problem, and soon. For example, in a nuclear reaction where two particles are transformed into other twoparticles, the conservation of charge would have the form

q1 + q2 = q3 + q4.

(b) Experimental facts and curve fitting, e.g., linear or exponential dependence betweenparameters. Some examples are:

• Newton’s second law (1-dimensional version): Acceleration of a body is proportionalto the force applied; a ∼ F . More specifically, a = F/m, m = const is the object’smass.

• Coulomb’s law: the magnitude of the force between two charges is inversely pro-portional to the squared distance between them, and directly proportional to bothcharges:

Fc = kq1q2r2

, k = const.

(c) Heuristic relations. When a problem is too complicated, for example, in cell biology,some relations or laws are completely unknown, with no experimental data available.Then one has to assume some relationship. For example, in many models of biologicalcells, it is assumed that the diffusivity coefficient of Brownian motion of organic moleculesin the cell is constant. It is clear that it does not have to be constant, but there is noreliable data available to formulate a reasonable alternative.

Another example is constitutive equations (equations of state) in solids, liquids, gases,and plasmas. Equations of state are often given by relations between the internal energyand other problem parameters. The equation of state is necessary to close the set ofequations, but is often unknown, except for simplest cases (ideal gas, incompressiblefluid, etc.). Often linear or polynomial forms of these equations are assumed. Can yousee why?

4.2.2 Equations in Mathematical Models

A mathematical model can involve one or more equations of one or more of the followingcommon types.

(a) Algebraic equations. Examples are the basic equations like F (x) = kx for the springforce, E = mc2 for the relation between the rest mass and rest energy of a particle, andso on.

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(b) Integral equations involving integral relations. An example is the total mass of a bodyoccupying volume V ⊂ R3 expressed in terms of its variable density:

M =

∫∫∫V

ρ(x, y, z) dV.

(c) Ordinary differential equations (ODEs) are relations between function(s) of onevariable, their derivatives, and independent variables. For example, the displacementx = x(t) of a mass in the linear mass-spring system is governed by an autonomous ODE

mx′′(t) = −kx(t),

where m is the mass and k is the spring constant. Primes are used here and below todenote ordinary derivatives.

(d) Partial differential equations (PDEs) are relations involving partial derivatives offunction(s) of several variables. Let q = q(t, x, y) be some dependent variable. Here andin many places below, we will often use the short-hand notation

∂

∂tq(t, x, y) ≡ qt,

∂

∂xq(t, x, y) ≡ qx,

and so on.

Basic examples are the main linear equations of Mathematical Physics that arise in virtu-ally all areas of knowledge where PDEs arise. For instance, we will see that the tempera-ture of a thin rod u(x, t) evolves according to the heat equation (or temperature diffusionequation)

ut = a2uxx + f, (4.1)

where a = const is some material parameter, and f = f(x, t) is the “forcing” functiondefining the density of heat sources/sinks in the rod. In the ‘regular’ notation the aboveequation is written as

∂u

∂t= a2

∂2u

∂x2+ f.

Another important equation of mathematical physics is the wave equation, arising in awide variety of applications where waves occur.For example, we will see that small verticaldeviations u(x, t) of a one-dimensional horizontal string from equilibrium are describedby a one-dimensional wave equation

utt = c2uxx + f, (4.2)

where c is the wave speed, and f = f(x, t) is the acceleration due to external forces.

A third PDE that will be studied in detail in this course is the Poisson equation. This isa time-independent equation, second-order in space, involving a Laplace operator ∆. Intwo spatial dimensions (x, y), it has the form

∆ϕ(x, y) ≡ ϕxx + ϕyy = g(x, y). (4.3)

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Here ϕ(x, y) is the unknown function, and g(x, y) is the given “forcing” term. If g(x, y) =0, the equation becomes

∆ϕ(x, y) = 0, (4.4)

and is called the Laplace equation. The Laplace/Poisson equation is another truly funda-mental one that occurs throughout the realm of mathematical modelling.

An example of a fourth-order linear PDE describing transverse oscillations of an elasticbeam is given by

wtt + k2wxxxx = 0, (4.5)

where w(x, t) is the (small) deviation of the beam at x from its original horizontal equi-librium position.

One more famous PDE is the Korteweg-de Vries (KdV) equation

ht + hhx + hxxx = 0 (4.6)

for the water surface elevation h(x, t). The equation describes the propagation of longwaves over shallow channels. KdV is a nonlinear third-order PDE.

(e) Other types of equations also exist and are widely used in modern-day science. Theyinclude integral equations, integro-differential equations, delay differential equations, dif-ference equations, and so on. Some of these are within the topics of 4th-year and graduateApplied Mathematics curriculum.

Remark 4.2. A well-posed mathematical model is expected to have a solution, preferably aunique solution that reflects, to some extent, the physical reality, or the desired properties ofthe process being modeled, in any area. As it is well-known, and as we will observe again inthis course, due to this uniqueness requirement, models involving ODEs have to also includean appropriate number of initial conditions. Models involving PDEs have to be appended withinitial and/or boundary conditions. Existence and uniqueness theorems and appropriate setupof initial/boundary conditions will be discussed below.

4.2.3 Linear PDEs

For partial differential equations, the same holds. For example, a PDE in 1+1 dimensions (onespace, one time) is linear if it can be represented as

L[u(x, t)] = g(x, t), (4.7)

where L is a linear operator, i.e. involves the sum of terms having u(x, t) and its derivatives inthe ‘total power’ one only.

L[u(x, t)] = a0u(x, t) + a1ut(x, t) + a2ux(x, t) + a3utt(x, t) + a4utx(x, t) + ..... (4.8)

Coefficients ai of a linear PDE are constants or functions of independent variables only (x, t inthis example).

A PDE (4.7) is called linear homogeneous when g(x, t) = 0.

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4.2.4 The main property of linearity

Any linear or nonlinear ODE or PDE for an unknown function u can be written in the form

R[u] = f, (4.9)

where R[u] represents all terms that involve u, and f is a set of terms that may only involvethe independent variable(s) and constants.

The operator R[u] is linear if it has the form (4.8).

Theorem 4.3. A differential operator R[u] is linear if and only if

R[Cu] = CR[u], R[u+ v] = R[u] + R[v] (4.10)

for any two functions u, v from its domain, and any constant C.

Alternatively, one can say this in one formula. If operator R[u] is linear, then

R[Au+Bv] = AR[u] +BR[v] (4.11)

for any two functions u, v from the domain of R[u], and any two constants A,B.

The above theorem is sometimes used as a definition of linearity.

The following important theorem holds, which is a manifestation of the main property oflinear homogeneous differential equations. It is true in any number of dimensions (independentvariables).

Theorem 4.4 (The Superposition Principle). Consider a linear homogeneous PDE

L[u(x)] = 0, (4.12)

where L is a linear differential operator, and x ∈ Rn. If u1(x), ..., uk(x) are solutions of a linearhomogeneous equation (4.12), then their linear combination ( superposition)

u(x) = c1u1(x) + ...+ ckuk(x) (4.13)

(c1, ..., ck ∈ R) is also a solution of the PDE (4.12).

Remark 4.5. For nonlinear equations, special kinds of nonlinear superposition sometimes hold;they are of a significantly more complicated form than (4.13).

Problems 4.2

Problem 4.2a. Give at least two examples of mathematical models derived using each of theprinciples (a), (b), (c) mentioned in Section 4.2.1.

Problem 4.2b. Find at least two examples of mathematical models involving integral equa-tions.

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Problem 4.2c. Find examples of mathematical models outside natural sciences, involving: a)ordinary differential equations, b) partial differential equations; c) integral equations.

Problem 4.2d. When there is only one space variable, then Laplace’s equation is the ordinarydifferential equation u′′ = 0 rather than a partial differential equation. Write down all possiblesolutions.

Problem 4.2e. Find some simple polynomial solutions to the heat, wave and Laplace equations(4.1), (4.2), (4.4).

Problem 4.2f. Can you find any exact solution(s) to the linear PDE (4.5)? A nonlinear PDE(4.6)? (Try it by your own trial and error, without looking for external information!)

Problem 4.2g. Suppose that a problem is given by two equations: E1 = 0 and E2 = 0. Is it“valid” to add or subtract the two equations, considering E1+E2 = 0 (or E1−E2 = 0) insteadof E1 = 0, E2 = 0? Is it “valid” to consider a linear combination aE1 + bE2 = 0, for someconstants a, b, instead of E1 = 0 and E2 = 0? If so, in what sense? give examples In whatsense? Present an answer in terms of equivalence of the set of solutions.

Problem 4.2h. For each of the following PDEs, the unknown function is assumed to dependon (x, t).

• Write the equation in the form R[u(x, t)] = g(x, t), where R[·] is the differential operator,and g(x, t) does not involve u.

• Consider R[αu(x, t)+βv(x, t)], α, β = const, to conclude whether the differential operatorR[·] is linear.

• Conclude whether the equation is linear homogeneous, linear nonhomogeneous, or non-linear.

(a) ut = 2x3uxx + 2u. (b) exut − euuxx = 3. (c) tutt + 2ut = x2ux + 11xt.

Problem 4.2i. Find examples of mathematical models involving nonlinear PDEs, not discussedin class or these notes. Provide equations and necessary details.

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4.3 The Continuity Equation

In the following sections, we will present, sometimes derive, and discuss some properties of thebasic linear equations of mathematical physics.

We start from one of the simplest equations: the continuity equation.

Consider a flow of a fluid (or a gas) as a physical continuum, in the direction of the axisof x. The parameters of interest are the density ρ(x, t) [kg/m3], the fluid velocity (averagevelocity of the particles) v(x, t) [m/s], and perhaps other ones. Note that we assume that thedependent variables only depend on one variable in the flow direction, x [m], and the time t[s]. No flow/change in the transverse (y, z) plane is assumed. Such models are referred to asone-dimensional (due to the fact that there is only one spatial variable x), 1+1-dimensional(one space and one time variable). Such assumptions are valid to model flows in all space withno y, z variation, as well as flows in pipes (in the x-direction) with smooth walls.

The continuity equation is an infinitesimal, i.e., local formula expressing the conservation ofmass. We will derive it for an x-directed pipe of a constant cross-section A [m2]. The value ofA will turn out to be insignificant.

4.3.1 A Naive Derivation

We will now derive the continuity equation using some “physically appealing”, but drastic andunjustified simplifications. Remarkably, we will get the correct equation! The full rigorousderivation will follow.

x x + xA

Q(x,t) Q(x + x,t)

x direction

Figure 4.1: The one-dimensional setup.

Consider an infinite pipe of Figure 4.1. Suppose that x is any given point, and ∆x is “small”.Then within the interval I = (x, x + ∆x), we assume that the fluid density ρ(x, t) does notchange much. The volume of the pipe for the interval I is ∆V = A∆x. Hence the total fluidmass within I at time t is

∆M(t) ≃ ρ(x, t)A∆x [kg]. (4.14)

Consider a “small” time interval from t to t + ∆t. At the “next instant” t + ∆t, the mass in∆V is given by

∆M(t+∆t) ≃ ρ(x, t+∆t)A∆x [kg]. (4.15)

Without loss of generality, suppose that the fluid flow is to the right, i.e.,

v(x, t) ≡ v(x, t) · ex ≥ 0.

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(If it is not so, and somewhere v(x, t) is negative, you will see that formulas will take care ofthemselves.)

Consider a “small” time interval from t to t+∆t. What fluid mass will come in through theleft? The last particle that will make it through the left end within ∆t is the one that is about∆ℓ = v(x, t)∆t left of the left end. So the volume that comes in is given by

∆V in, left ≃ A∆ℓ = Av(x, t)∆t [m3].

So the fluid mass coming in through the left end of the pipe piece ∆V is

∆M in, left ≃ Aρ(x, t)v(x, t)∆t [kg]. (4.16)

We naturally denote the incoming fluid flux (mass entering ∆V per unit time) by the fluid flux

Q(x, t) ≃ Aρ(x, t)v(x, t) [kg/s].

Similarly, for the fluid leaving through the right, the mass that leaves our pipe chunk ∆Vwithin the time ∆t is

∆M out, right ≃ ρ(x+∆x, t)v(x+∆x, t)∆t = Q(x+∆x)∆t [kg], (4.17)

with the flux on the right given by

Q(x+∆x, t) ≃ Aρ(x+∆x, t)v(x+∆x, t) [kg/s].

The conservation of mass within the pipe piece ∆V , for the time between t and t + ∆t isgiven by

∆M(t+∆t)−∆M(t) = ∆M in, left −∆M out, right (4.18)

which says that the change of mass of the fluid in the fixed volume ∆V is due to the massbrought in and out of ∆V within ∆t.

Putting the ingredients (4.14), (4.15), (4.16), (4.17) into (4.18), we get, in detail,

[ρ(x, t+∆t)− ρ(x, t)]A∆x = − [ρ(x+∆x, t)v(x+∆x, t)− ρ(x, t)v(x, t)]A∆t. (4.19)

(Check the signs!) Now divide the equation (4.19) by A∆t∆x, and take simultaneously thelimits

lim∆x → 0∆t → 0

1

∆t[ρ(x, t+∆t)− ρ(x, t)] = − lim

∆x → 0∆t → 0

1

∆x[ρ(x+∆x, t)v(x+∆x, t)− ρ(x, t)v(x, t)] .

These limits are nothing but the definitions of the partial derivatives. Hence we have “derived”the famous one-dimensional continuity equation

∂

∂tρ(x, t) +

∂

∂x(ρ(x, t)v(x, t)) = 0, (4.20)

or

ρt + (ρv)x = 0,

which holds, under the assumptions of flow smoothness, at every point x of the infinite pipe,for all times t.

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4.3.2 A Rigorous Derivation

We now use the same figure to derive the continuity equation (4.20) rigorously, using the samenotation as above, and the illustration of Figure 4.1.

The total fluid mass within ∆V at time t is exactly equal to

∆M(t) =

∫ x+∆x

x

ρ(s, t)Ads [kg]. (4.21)

(Note that if ρ does not depend on x, we get (4.14).)

Now, what is the rate at which fluid mass is coming in through the left end? It is measuredin [kg/s], and is clearly given by

Q(x, t) = Aρ(x, t)v(x, t) [kg/s]. (4.22)

The fluid mass is leaving though the right end at the rate

Q(x+∆x, t) = Aρ(x+∆x, t)v(x+∆x, t) [kg/s]. (4.23)

(The terms “coming in”/“leaving” clearly have to be switched, if the sign of the correspondingvelocity is negative.)

The law of mass rates, i.e., the fact that the rate of change of the total mass ∆M(x, t) (4.30)is due to the fluxes coming in/out, is expressed by

d

dt∆M(t) = Q(x, t)−Q(x+∆x, t), (4.24)

where all terms are measured in [kg/s].

According to the mean value theorem for integration, for any continuous function g(x) within[a, b], there exists z ∈ [a, b] such that∫ b

a

g(s)ds = g(z)(b− a).

Moreover, the Leibnitz integral rule states that if f = f(x, t) and its partial derivative fx arecontinuous over the region of the (x, t)-plane including [a, b], then

d

dt

∫ b

a

f(x, t) dx =

∫ b

a

ft dx.

Consider the LHS of (4.24). by using the two above facts, it becomes

d

dt∆M(x, t) =

∫ x+∆x

x

ρt(s, t)Ads = Aρt(q, t)∆x,

where q ∈ [x, x+∆x]. Dividing the formula (4.24) by A∆x and taking the limit, we have.

lim∆x→0

ρt(q, t) = − lim∆x→0

1

∆x[ρ(x+∆x, t)v(x+∆x, t)− ρ(x, t)v(x, t)]. (4.25)

Noting that lim∆x→0 q = x, and the right-hand side of (4.25) yields a partial derivative, oneobtains the desired one-dimensional continuity equation (4.20).

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4.3.3 The Advection Equation

Another famous equation is obtained when the one-dimensional flow along x takes place witha constant speed v(x, t) = c > 0. The corresponding form of the continuity equation (4.20) isgiven by

∂

∂tρ(x, t) + c

∂

∂xρ(x, t) = 0, (4.26)

or

ρt + cρx = 0,

and is called the advection equation.

For the infinite domain x ∈ R, one can in fact construct the general solution of (4.26), i.e.,the solution that holds for all possible initial conditions. It is done by the following observation.Consider a function F (s), with F ′(s) ≡ dF/ds continuous. Now consider F (x − ct). For thelatter, one has, by the chain rule,

∂

∂xF (x− ct) = F ′(x− ct)

∂

∂x(x− ct) = F ′(x− ct),

∂

∂tF (x− ct) = F ′(x− ct)

∂

∂t(x− ct) = −cF ′(x− ct).

Hence

∂

∂tF (x− ct) + c

∂

∂xF (x− ct) = 0,

and for any smooth F (s), F (x − ct) is a solution to the advection equation (4.26). It is alsoa general solution! Assume that the fluid had some given (arbitrary, smooth) initial densitydistribution ρ(x, t = 0) = H(x). Then for all times, the function

ρ(x, t) = H(x− ct) (4.27)

solves the advection equation (4.26). It also satisfies the initial condition (IC): ρ(x, 0) = H(x).Hence we have solved all initial value problems (IVPs) of the form

ρt + cρx = 0, x ∈ R, 0 ≤ t,ρ(x, 0) = H(x), x ∈ R, (4.28)

where the short-hand notation for partial derivatives is used. The solution of the generalproblem (4.28) for any smooth IC is given by (4.27).

Remark 4.6. The physical meaning of the solution (4.27) is that the original density distri-bution is traveling to the right without shape change. See Figure 4.2.

Remark 4.7. The advection equation occurs not only in fluid/gas dynamics, but in manyother areas, including, for example, traffic flow problems. It is indeed the simplest PDE!

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x

H(x) H(x-ct)

Figure 4.2: A right-traveling wave.

4.3.4 The Continuity Equation in Two Dimensions

Now consider a flow within a two-dimensional domain D ⊂ R2. We will derive a continuityequation for such a flow. It will hold in the bulk of D, away from the boundary.

Remark 4.8. On the domain boundary itself, partial differential equations do not hold; theyare replaced by boundary conditions (BCs) that need to be derived separately using similartechniques. We will see many have examples below.

D

xx x + x

y

y

y + y

n

n

n

n

flow streamlines

v

B

Figure 4.3: A two-dimensional setup for the continuity equation derivation.

Let (x, y) ∈ D \ ∂D (Figure 4.3). The fluid density is given by ρ = ρ(t, x, y), and the fluidvelocity is a two-component vector given by

v = [v1(t, x, y), v2(t, x, y)]T ≡ v1(t, x, y)ex + v2(t, x, y)ey. (4.29)

We use Figure 4.3. The streamlines are integral curves of the fluid velocity field v, i.e., curvesto which the velocity is tangent. We consider a small box B = (x, x+∆ x)× (y, y+∆ y) ⊂ D.The outward unit normals on the box boundary ∂B are denoted by n.

The derivation of the continuity equation consists in the formulation of the law relating ratesof change of the fluid mass in the box B, and taking a limit with ∆x,∆ y → 0. It is in manyways parallel to the “rigorous” derivation of the one-dimensional continuity equation, so weskip some technical details here.

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Let L [m] be the thickness of the fluid layer in the z direction. (By the assumptions, nothingis going on in the direction of z, so L is any constant that will eventually cancel.

The mass of fluid in the box B is given by

∆M(t) =

∫∫B

ρ(t, r, s)Ldrds = L

∫ x+∆x

x

dr

∫ y+∆y

x

ds ρ(t, r, s) [kg]. (4.30)

The rate of change of total mass in the box B is given by

d

dt∆M(t) = L

∫ x+∆x

x

dr

∫ y+∆y

x

ds ρt(t, r, s) [kg/s]. (4.31)

This rate of change is due to the flow of fluid in/out of the domain through the boundary. Thefluid flows in through a portion of the boundary dℓ, if at dℓ, v ·n < 0 (i.e., velocity and outwardnormal are counter-directed).

Hence the rate of change (4.31) also equals to the total flux:

Φ(t) =

∫∂B

ρ(v · n)Ldℓ [kg/s].

Indeed, only particles aiming at the boundary ∂B whose distance to the boundary is |(v ·n)|∆twill make it to the boundary in time ∆t.

Writing the expression for Φ(t) in detail, and equating ddt∆M(t) = Φ(t), we get the equation

for the rate of change of mass for the box B:

L∫ x+∆x

xdr

∫ y+∆y

xds ρt(t, r, s) = L

∫ x+∆x

x[v2(t, r, y)− v2(t, r, y +∆y)]dr

+L∫ y+∆y

y[v1(t, x, s)− v1(t, x+∆x, s)]ds.

(4.32)

Using the the mean value theorem for integration in both x and y directions as needed, dividingthe formula (4.32) by L∆x∆y, and taking the limit ∆x,∆y → 0, we eventually get

∂

∂tρ(t, x, y) +

∂

∂x(ρ(t, x, y)v1(t, x, y)) +

∂

∂y(ρ(t, x, y)v2(t, x, y)) = 0, (4.33)

or, in the nabla notation

ρt +∇ · (ρv) ≡ ρt + div(ρv) = 0, (4.34)

where ∇ = [∂/∂x, ∂/∂y]T ≡ ex ∂/∂x+ ey ∂/∂y is the divergence operator.

Remark 4.9. It is clear that in three dimensions, the continuity equation has essentially thesame form. In R3, the velocity vector has three components,

v = [v1(t, x, y, z), v2(t, x, y, z), v3(t, x, y, z)]T , (4.35)

and the continuity equation is given by (4.34) with

∇ = (∂/∂x, ∂/∂y, ∂/∂z).

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Keywords 4.3

Existence and uniqueness theorems for ODEs and PDEs; Initial and boundary conditions;Nolinear advection equation/inviscid Burgers’ equation; Traffic flow models.

Problems 4.3

Problem 4.3a. Derive the continuity equation for the flow in a pipe assuming the (very slow)change of the cross-section: A = A(x). The slowness is needed to assume that the velocity isstill only along the x-direction.

Problem 4.3b. Derive the continuity equation for the flow in a pipe assuming the pipe“breathing” (very slowly), i.e., A = A(t). The slowness is again needed to assume thatv(x, t) = v(x, t)ex.

Problem 4.3c. Show that the advection equation with velocity v(x, t) = c = −a < 0 has aleft-traveling general solution ρ(x, t) = H(x+ at).

Problem 4.3d. Fill in details in the derivation of (4.34).

14

4.4 The Heat Equation

The detailed understanding of heat conduction in solids represents an area that is both sci-entifically interesting and practically important. There are many aspects of the problem, de-pending on temperature ranges, material properties, and geometry, that may have to be takeninto account. A short list includes the material heat exchange law, phase transitions, crystalstructure, chemical reactions, surface phenomena/interaction with the environment, and otherrelated physical phenomena.

The heat conduction equation expresses the conservation of the physical quantity called thethermal energy, a part of the internal body energy related to the temperature.

A closely related set of problems arises when diffusion of one liquid/gas in another similarsubstance is considered. Heat conduction and diffusion equations are in many ways similar.

We will derive the linear heat conduction equation under the simplest assumptions. It willcoincide with the diffusion equation, also derived under the simplest assumptions.

4.4.1 The One-Dimensional Linear Heat Equation

Consider a rod of length L, along the x-axis, made of homogeneous material, with side surfaceperfectly insulated, and with constant cross-section of area A. Assume no heat flow in y- andz-directions, i.e., the temperature of the rod at time t, at any point x, y, z, is assumed to havethe form

T (t, x, y, z) = u(x, t).

Equivalently, one can consider a slab, infinite in y- and z-directions, and occupying the space0 ≤ x ≤ L in the x-direction. If the thermal processes only occur in the x-direction, and

∂

∂yT (t, x, y, z) =

∂

∂zT (t, x, y, z) = 0,

then the equation will be the same both for the slab and for the rod. The temperature ismeasured in kelvins [K].

We assume some initial temperature distribution u(x, 0) at time t = 0 and are interested inan equation expressing the dynamics of u(x, t) for all t > 0, x ∈ (0, L). Figure 4.1 will be usedfor the derivation.

The thermal energy E of the rod is measured in joules [J]. We will be interested in thermalenergy density e = e(x, t) . For an infinitesimal volume dV of the solid, its internal energy isgiven by dE = e dV .

The thermal energy density is given by

e(x, t) = Cu(x, t),

where C [J/(m3·K)] is the specific heat capacity of a unit volume of the rod substance, and isassumed constant.

15

Consider an element of the rod located between in the interval I = [x, x + ∆x]. The fullthermal energy of that piece is given by

∆E(t) =

∫ x+∆x

x

CAu(s, t) ds [J]. (4.36)

where A is the cross-section area. (The infinitesimal volume is dV = Ads where ds is thelength of the rod part under question.)

The time rate of change of thermal energy between t and t+∆t is given by

d

dt∆E(t) = Qin −Qout [J/s]. (4.37)

Without loss of generality, assume that the rod is hotter on the left and cooler on the rightof I. Then the heat is coming from the left and leaving through the right end of the volumeenclosed by I. The quantities Qin and Qout are time rates of the heat inflow (Qin, through leftend) and loss (Qout, through right end).

Denote the heat flux (energy per unit time per a unit area) through a vertical cross-sectionat x by q(x, t) [J/(m2· s)]. What is the formula for q(x, t)? A simple one, that however has agood agreement with experiment under normal conditions, is the following.

Definition 4.10. In the Fourier law of heat exchange, the heat flux vector is given by

q = −k gradu,

where k is a constant coefficient.

Projecting on the x axis, in scalar notation, one has

q(x, t) = −kux. (4.38)

Remark 4.11. It is clear that if the temperature is increasing to the right (ux > 0), then theheat flux is to the left (q < 0), and vice versa.

In the energy balance (4.37), we have

Qin = Aq(x, t), Qout = Aq(x+∆x, t) [J/s]. (4.39)

We now take (4.38), substitute (4.36) and (4.39), and exchange integration and differentia-tion. Canceling A, we have

AC

∫ x+∆x

x

ut(s, t) ds = −kA(ux(x, t)− ux(x+∆x, t)). (4.40)

Using the mean value theorem and taking the limit like in Section 4.3.2, one obtains

ut = (k/C)uxx. (4.41)

Since k, C > 0, define a new constant material parameter a2 = k/C. One arrives at the linearhomogeneous heat conduction equation

ut = a2uxx. (4.42)

This concludes the derivation.

16

Remark 4.12. In the case when there are heat sources present (these may depend on x, t),the heat equation takes the form

ut = a2uxx + f, (4.43)

where f = f(x, t) is the “forcing” function defining the density of heat sources/sinks in the rod.

Remark 4.13. The same PDE also describes the diffusion processes, hence equations (4.42),(4.43) are also referred to as diffusion equations. The latter is commonly written as

ut = Duxx + f, (4.44)

where the unknown u(x, t) is the density, or alternatively the concentration, of the diffusingsubstance, and D ≥ 0 is the diffusion coefficient. D is often taken to be constant.

How the heat equation works

Consider the homogeneous heat equation (4.41). How does it “work”? The problem is normallyformulated for a rod of length L, and initial and boundary conditions are given. The timeevolution of the initial condition, i.e., the dynamics of u(x, t) in time, is what one is interestedin.

Let the u(x, t) have some shape at time t. What will it be at the next instant? If the graphof u(x, t) as a function of x is known, can we sketch the graph of u(x, t+∆t) for a “small” ∆t?

In the Figure 4.4, such a situation is schematically shown. If at some point in x, u(x, t) hasa maximum, then ux = 0 and uxx < 0 at that point. The heat equation (4.41) tells us that, atthat point, ut = a2uxx < 0, hence that point is going down as time goes on. This is true for allpoints with uxx < 0. The graph there is concave down.

Similarly, when at a point x, the graph u(x, t) (as a function of x) is concave up, i.e., uxx > 0,one has ut = a2uxx > 0, and such points will raise as time goes from t to t+∆t.

x

u(x,t+ !t)

u

u(x,t)uxx>0,

ut>0

uxx<0,

ut<0

ut =a2 uxx

Figure 4.4: Schematic: “how the heat equation works”.

It is easy to see that any straight line

u(x, t) = αx+ β, (4.45)

17

for any constants α, β, is an equilibrium (time-independent) solution of the homogeneous heatequation (4.41). It is correct to say (though we have not proven it so far) that unless thereis a forcing term in the equation, as in the limit t → ∞, any 1-dimensional temperaturedistribution approaches a linear one, (4.45). The constants α, β will be determined by theboundary conditions.

4.4.2 Initial/Boundary Value Problems (IBVPs) for the Linear HeatEquation

By itself, without initial and/or boundary conditions, any physical ODE or PDE has an infinitenumber of solutions. For example, it is easy to show that the following functions satisfy thehomogeneous heat equation (4.41), for any value of the constants involved:

u(x, t) = αx+ β; u(x, t) = 2ca2t+ cx2, u(x, t) = e−a2t cos x,

etc. A physical problem however is expected to have only one solution.

Definition 4.14. A well-posed problem for a differential equation (DE) is a problem that hasa unique solution u(x, t).

[This is the truncated version of the definition of J. Hadamard.]

In order to formulate a well-posed problem, the PDE has to be appended by an appropriatenumber of initial and boundary conditions. This yields an Initial-Boundary Value Problem(IBVP).

Consider a PDE for u(x, t), 0 ≤ x ≤ L.

Rule 4.15.

• The number of necessary initial conditions (IC) is equal to the highest order of the timederivative in the PDE. [This is a direct analogy with ODEs.] If one IC is required, itis given by u(0, x); if two are required, one has to specify u(0, x) and ut(0, x), the latterbeing the initial rate of change of u. [Note that initial conditions do depend on x, i.e. aredifferent for different x ∈ (0, L).]

• For 2nd-order PDEs with one spatial variable, defined in an interval 0 < x < L, thenumber of boundary conditions is two. Boundary conditions (BC) are given at twoboundary points of the domain 0 < x < L, i.e., at x = 0 and x = L. [BCs are notartificial; they are chosen according to the physics of the problem.]

• Generally, for PDEs that are 2nd-order in space and involve n ≥ 1 spatial dimensions,the rule is that a boundary condition must be given at each point of the boundary of thespatial domain.

In this course, we will be formulating initial and boundary value problems for several equa-tions of mathematical physics. For more details about existence, uniqueness and regularity ofPDE IBVP solutions, see, e.g., [Vladimirov V.S., Equations of Mathematical Physics, 1971].

18

Three main types of linear boundary conditions and their meaning for theheat/diffusion equation

The following three types of linear boundary conditions (BC) are most common.

(a) Dirichlet (Type I) BC.

(b) Neumann (Type II) BC.

(c) Robin, or Mixed (Type III) BC.

For the 1+1-dimensional heat equation problem in the spatial domain x ∈ [0, L], one bound-ary condition has to be given at x = 0 and one at x = L.

The following table contains some explanation. The BCs there are only considered at x = 0.They have the obviously corresponding form at x = L.

Table 4.1: Types of linear BCs

Name Type BC A physical application

Dirichlet Homogeneous u(0, t) = 0 Zero temperature/concentration is kept

(Type I) at x = 0 for all times.

Non-homogeneous u(0, t) = µ(t) Temperature/concentration at x = 0

changes in time, according to the law µ(t).

Neumann Homogeneous ux(0, t) = 0 The left end is insulated (flux q = −kux

(Type II) is zero at x = 0 for all times.)

Non-homogeneous ux(0, t) = ν(t) Heat/substance flux through the end

x = 0 is q(0, t) = −kν(t).

Robin Homogeneous αu(0, t) + βux(0, t) = 0 Heat: Left end exchanges heat with the

(Type III) α, β = const. environment (of zero temperature)

according to the Newton’s law of cooling:

q(x, 0) ≡ −kux(0, t) = h(0− u(0, t)),

h = const.

Non-homogeneous αu(0, t) + βux(0, t) = Heat: left end exchanges heat with the

= θ(t) environment (of temperature θ(t))

according to the Newton’s law of cooling:

q(x, 0) ≡ −kux(0, t) = h(θ(t)− u(0, t)).

Examples of IBVPs

Example 4.16. Describe [dimensionless] heat conduction in a rod made of material witha2 = 11 units, of length 1 m, with initial temperature given by 120 + 200x(1 − x) units,provided that the left end temperature is controlled to equal 140 − 20e−t units at any timet > 0, and the right end is insulated.

19

Solution: the IBVP is given byut = 11uxx, 0 < x < 1, 0 < t;u(x, 0) = 120 + 200x(1− x), (IC)u(0, t) = 140− 20e−t, (BC 1)ux(L, t) = 0. (BC 2)

Example 4.17. Describe heat conduction in a metal rod of length L, with initial temperature100 K, with insulated left end, and the right end exchanging heat with the environment whosetemperature is θ = 200 K.

Solution: the IBVP is given byut = a2uxx, 0 < x < L, 0 < t;u(x, 0) = 100 (IC)ux(0, t) = 0 (BC 1)kux(L, t) = (200− u(L, t))h (BC 2)

Here k, h, a are constant material parameters that can be found in tables for each specificmaterial.

4.4.3 The Linear Heat Equation in 2+1 and 3+1 Dimensions

(A) The Linear Heat Equation in 2+1 Dimensions

The (2+1)-dimensional heat equation arises when the temperature of an object is the same inone direction (e.g., z), but is varying the the other ones. We will denote temperature of everypoint of such object by T = u(t, x, y). A physical example is given by objects that are cylinders(not necessarily circular) whose generatrix is parallel to the z−axis, and the (insulated) facesare parallel to the (x, y) plane. Due to the insulation, there is no heat flow in z-direction. Thefollowing Figure 4.5 is borrowed from Dr. Froese’s notes. Here u = u(t, x, y), and (x, y) arepoints in the spatial domain Ω.

The heat equation in 2+1 dimensions has the form

ut = a2∆u.

It is derived the same way the 2D continuity equation was derived in Section 4.3.4. Here ∆ isthe Laplace operator. In two dimensions in Cartesian coordinates (x, y),

∆ =∂2

∂x2+

∂2

∂y2,

and the heat equation for u = u(t, x, y) is

∂

∂tu = a2

[∂2u

∂x2+

∂2u

∂y2

]. (4.46)

For 2D domains that involve disks or their parts, it is convenient to use polar coordinates. Intwo dimensions in polar coordinates (r, ϕ), the heat equation is formulated for u = u(t, x, y) =v(t, r, ϕ), and has the form

∂

∂tv = a2

[1

r

∂

∂r

(r∂v

∂r

)+

1

r2∂2v

∂ϕ2

]. (4.47)

20

!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!insulation

x

y

ΩΩ

Figure 4.5: The heat equation in two space dimensions: the setup.

In order to formulate an IBVP for the heat equation in 2+1 dimensions, according to theRule 4.15, one has to prescribe an initial condition u(0, x, y) = u0(x, y) for all (x, y) ∈ Ω, andone boundary condition at each boundary point (x, y) ∈ ∂Ω. Examples are now considered.

x

y

L

L0

u=2L-y

u=28

v = 100

v =100 cos "

R

(a) (b)

Figure 4.6: Examples.

Example 4.18. Describe the [dimensionless] heat conduction in a square with a given a2 value,given side length L, two sides insulated, and two sides having temperature prescribed by theFigure 4.6(a). Initially the part is at the temperature of 100 units.

21

Solution: the IBVP for u = u(t, x, y) is given by

ut = a2(uxx + uyy), 0 < x, y < L, 0 < t;u(0, x, y) = 100, 0 < x, y < L, (IC)u(t, x, 0) = 28, 0 < y < L, 0 < t, (BC)u(t, 0, y) = 2L− y, 0 < x < L, 0 < t, (BC)ux(t, L, y) = 0, 0 < y < L, 0 < t, (BC)uy(t, x, L) = 0, 0 < x < L, 0 < t. (BC)

We have specified BCs at every point of the boundary. Note that Neumann, insulation boundaryconditions (zero heat flux) have the appropriate partial derivative in them, corresponding tothe component of the heat flux q = −k∇u that is normal to the corresponding boundary.

Example 4.19. Describe the heat conduction in a disk of a given radius R, with a given a2

value. Initially the disk is at zero temperature. Starting at t = 0, heaters turn on, keeping thedisk boundary parts at constant temperatures as shown in the Figure 4.6(b).

Solution: this purely Dirichlet IBVP for v = v(t, r, ϕ) in polar coordinates is given by

vt = a2(1

r(rvr)r +

1

r2vϕϕ

), 0 ≤ r < R, 0 ≤ ϕ < 2π, 0 < t;

v(0, r, ϕ) = 0, 0 ≤ r < R, 0 ≤ ϕ < 2π, (IC)

v(t, R, ϕ) = 100, 0 ≤ ϕ < π, 0 < t. (BC)

v(t, R, ϕ) = 100 cosϕ, π ≤ ϕ < 2π, 0 < t. (BC)

Note that the BCs are given indeed at the disk boundary: r = R.

(B) The Linear Heat Equation in 3+1 Dimensions

The (3+1)-dimensional heat equation arises when one needs to describe the temperature of a 3Dobject, occupying the spatial domain Ω ⊂ R3. The temperature is then given by u = u(t, x, y, z),and the heat equation takes the form

∂

∂tu = a2

[∂2u

∂x2+

∂2u

∂y2+

∂2u

∂z2

], (4.48)

or simply

ut = ∆u = a2(uxx + uyy + uzz).

The IBVPs are formulated in a similar fashion, with one BC given at each point of the boundary(x, y, z) ∈ ∂Ω.

Remark 4.20. When the problem has a spherical or cylindrical symmetry, it is convenient touse the corresponding coordinate system. This is done by rewriting the Laplacian differentialoperator accordingly. For example, in cylindrical coordinates, the temperature has the formu = u(t, x, y, z) = v(t, r, ϕ, z), and the equation (4.48) becomes

∂

∂tv = a2

[1

r

∂

∂r

(r∂v

∂r

)+

1

r2∂2v

∂ϕ2+

∂2v

∂z2

]. (4.49)

22

Similarly, in spherical coordinates, u = u(t, x, y, z) = w(t, ρ, θ, ϕ), and the heat equationbecomes

∂

∂tw = a2

[1

ρ2∂

∂ρ

(ρ2

∂w

∂ρ

)+

1

r2 sin θ

∂

∂θ

(sin θ

∂w

∂θ

)+

1

r2 sin2 θ

∂2w

∂ϕ2

]. (4.50)

Here ρ is the spherical radius, θ is the polar (zenith) angle measured from the positive directionof the z-axis, and ϕ is the azimuthal angle.

Keywords 4.4

Dirichlet boundary condition; Neumann boundary condition; Robin boundary condition; Spher-ical coordinates; Nabla operator; Laplacian; Existence, uniqueness and regularity of PDE solu-tions; Ill-posed problems.

Problems 4.4

Problem 4.4a. Consider the homogeneous heat equation (4.41) in the domain 0 ≤ x ≤ π,0 ≤ t.

• Show that for any integer n, the separated solutions

un(x, t) = e−a2n2t sin(nx), vn(x, t) = e−a2n2t cos(nx)

do indeed satisfy the PDE (4.41).

• What initial conditions, and what homogeneous boundary conditions do un(x, t) andvn(x, t) satisfy?

Problem 4.4b. Give an example of a physical model that could be described by the followingIBVP.

ut = a2uxx, u = u(x, t), 0 < x < 1, 0 < t;u(x, 0) = sin 2πx;u(0, t) = f(t);ux(1, t) = 0.

Instructions for Problems 4.4c-4.4g.

• Provide an illustration.

• Specify the dependent and independent variables. Specify physical units as needed.

• Formulate the initial/boundary value problem.

Problem 4.4c. Describe temperature dynamics during heat conduction in a silver rod of lengthL meters, with insulated side surface. Suppose that initially the rod has room temperature(22oC), the right end of the rod is insulated, and the left end is kept at constant temperature0oC for all time.

23

Problem 4.4d. A thin tube of length l is filled with helium. At t = 0, through the right endof the tube, oxygen starts to flow in, in such a way that its concentration at the right end ismaintained at 0.5 g/l at all times. The left end is sealed. Write down an intitial/boundaryvalue problem for the diffusion equation, describing the concentration u(x, t) of oxygen in thetube for all times.

Problem 4.4e. Set up an IVBP describing heat conduction in a cubical iron block of sidelength ℓ, if initially it has the temperature T0, and its sides are maintained at temperature T1

for all t > 0.

Problem 4.4f. Repeat problem 4.4c, if instead of insulation of the side surface, startingat t = 0, the rod’s side surface is being heated by an electrical winding at the rate off(x, t) = M t sin(x/L) [J/(m3·s)]. Assume that the rod is thin (i.e., thermal energy is dis-tributed throughout any cross-section instantly). Revisit the derivation of the heat equation,and obtain a new mathematical model. [It will involve the linear heat equation with a forcingterm.] Make sure that the physical units agree.

Problem 4.4g. Set up an IVBP describing heat conduction in a circular cylinder C: 0 ≤x2 + y2 ≤ R2, 0 ≤ z ≤ L, if the face z = 0 is insulated, whereas the face z = L and thecylindrical surface x2 + y2 = R2 are kept at 300 K. Initially the cylinder is at 100 K.

Instructions for Problems 4.4h–4.4j. For a homogeneous heat equation (4.41), if the boundaryconditions are time-independent, it is correct to say that the solution u(x, t) will tend to somelimiting equilibrium solution U(x) as t → ∞. The function U(x) can be found as a time-independent solution of the same IBVP, where one simply ignores the IC.

• Find the limiting (equilibrium) temperature distributions in the following IBVPs.

• In one plot, sketch initial data together with the equilibrium solution. Indicate on theplot that the equilibrium solution indeed satisfies the boundary conditions.

Problem 4.4h.ut = uxx, 0 < x < 1, 0 < t;u(x, 0) = x;u(0, t) = 12;u(1, t) = 1.

Problem 4.4i.ut = a2uxx, 0 < x < 1, 0 < t;u(x, 0) = x;ux(0, t) = 0;u(1, t) = T.

Problem 4.4j.ut = uxx + kx, 0 < x < L, 0 < t;u(x, 0) = 10;u(0, t) = 0;u(L, t) = 0.

24

Problem 4.4k. Consider the homogeneous Neumann problem describing heat conduction ina totally insulated rod having initial temperature distribution u(x, 0) = u0(x):

ut = a2uxx, 0 < x < L, 0 < t;u(x, 0) = u0(x);ux(0, t) = ux(L, t) = 0.

• Provide a sketch.

• Prove mathematically, using the above IBVP, that the total thermal energy of the rod

E(t) =

∫ L

0

Cu(x, t)dx

is conserved by the process, i.e., does not depend on time: dE/dt = 0.

• Find the limiting equilibrium solution U(x) of the above problem as t → ∞.

Problem 4.4l. For the heat equation ut = a2uxx on the real line, i.e., for x ∈ R, the IBVP isreplaced by an initial value problem (IVP)

ut = a2uxx, x ∈ R, 0 < t;u(x, 0) = u0(x), x ∈ R (4.51)

with a restriction that u(x, t) → 0 as x → ±∞ sufficiently quickly.

• Show that the heat kernel

u(x, t) =U0√4πa2t

e−(x−x0)2/(4a2t) (4.52)

is an exact solution of the heat equation, for any constant amplitude U0 and any shift x0.

• Sketch several curves of (4.52) as t grows.

• Compute∫∞−∞ u(x, t) dx.

• Express u(x, 0) in terms of the Dirac delta-function.

Problem 4.4m. Consider an IBVPut = a2uxx, u = u(x, t), 0 < x < 1, 0 < t;u(x, 0) = x;ut(0, t) = f(t);u(1, t) = 0.

What is the type of the boundary condition at x = 0? Do you think this IBVP is well-posed?Provide a clear mathematical or physical explanation.

25

4.5 The Wave Equation

The one-dimensional linear wave equation (4.2), or its non-forced (homogeneous) version

utt = c2uxx, (4.53)

is an equation describing basic oscillatory phenomena in a wide variety of applications. Here cis the wave speed, and u = u(x, t) is the “displacement field” to be determined.

The wave equation (4.53) is a linear PDE, second-order in space, and second-order in time.

We start from a simplistic “naive” derivation of the wave equation for small oscillations ofa horizontal string. The setup is shown in Figure 4.7. It is assumed that u(x, t) is sufficientlysmooth, and

maxx,t

|u(x, t)| ≪ L, maxx,t

|ux(x, t)| ≪ 1.

u

xL

u(x,t) > 0

x0

Figure 4.7: The setup.

4.5.1 A Naive Derivation

Consider a homogeneous, constant-density string along the axis of x, attached at points (x, y) =(0, 0) and (x, y) = (L, 0). If a string is slightly perturbed in the y−direction, it will undergosmall oscillations.

Let there be no forces acting on the string, except for its own tension. Assume the linearmass density of the string to be λ = const > 0 [kg/m].

The fundamental principle here will be the Newton’s Second Law. We will again be using“physically appealing”, but unjustified simplifications. One of them is that the small oscillationsare up/down, i.e., no string element moves to the left or to the right. We also assume that theslopes of the string, i.e., values ux(x, t), are “small”, as indicated above.

We wish to write the Newton’s Second Law for a string element of length ∆x, where 0 <x < L (Figure 4.8). The mass of the element is

∆M ≃ λ∆x√1 + u2

x(x, t) ≃ λ∆x. (4.54)

26

u

xx x+ x

T(x,t)

T(x+ x,t)

Figure 4.8: Tension forces acting on an infinitesimal string element.

One may consider projecting the Newton’s Second law on the x-axis. It is possible to showthat the x-projection of the total force is given by

F = |T(x+∆x, t)| cos θ(x+∆x, t)− |T(x, t)| cos θ(x, t). (4.55)

Note that since θ = arctanux, then up to our necessary precision,

cos θ(x+∆x, t) =1√

1 + u2x(x+∆x, t)

≃ 1, cos θ(x, t) =1√

1 + u2x(x, t)

≃ 1.

Since there is no motion assumed in the x-direction, the force (4.55) should vanish. UsingF = 0 in (4.55), we get

|T(x+∆x, t)| ≃ |T(x, t)|.

So in our approximation, the absolute value of the tension force is approximately constantthroughout the string. We denote that constant by T > 0.

The Newton’s Second law is now projected on the vertical axis. There will be motion andacceleration. The vertical displacement of the string element is approximately u(x, t), so theacceleration is approximately a = utt(x, t). The force balance law reads

(∆M)a ≃ (λ∆x)utt(x, t) ≃ T sin θ(x+∆x, t)− T sin θ(x, t). (4.56)

Using θ = arctanux, one has sin θ = ux/√

1 + u2x. We keep the leading terms in ux in the

balance law (4.56), and it becomes

(λ∆x)utt(x, t) ≃ T [ux(x+∆x, t)− ux(x, t)]. (4.57)

Division of (4.57) by ∆x and the limit of ∆x → 0 yields the wave equation

λutt = Tuxx.

It is common to denote

c2 = T/λ,

27

which has dimension units of [(m/s)2]. The constant c is referred to as the wave speed. We willsee that it indeed is so! The wave equation finally takes the form (4.2). This completes thederivation. Remark 4.21. If external forces act on the string in the vertical direction, such as the force ofgravity, the linear homogeneous wave equation (4.2) will change to the linear non-homogeneousone:

utt = c2uxx + f, (4.58)

where f = f(x, t) is the vertical acceleration produced by such external forces.

4.5.2 Initial/Boundary Value Problems for the Linear Wave Equa-tion

The linear wave equation (4.58) in 1+1 dimensions is second order in time and second orderin space. According to the Rule 4.15, the well-posed problem for the wave equation in a finiteinterval x ∈ [0, L] should consist of

• the wave equation in the semi-strip 0 < x < 1, 0 < t;

• a boundary condition at x = 0 and another one at x = L,

• two initial conditions at t = 0.

[The latter is typical for problems formulated in terms of acceleration, arising from the Newton’sSecond law.]

The same three main types of linear boundary conditions discussed in Section 4.4.2 are usedfor the wave equation.

For small oscillations of a horizontal string, it is natural to use homogeneous Dirichlet BCs,which mean that each end of the string is attached and does not move for all times. Such aDirichlet problem will have the form

utt = c2uxx, 0 < x < L, 0 < t;u(x, 0) = u0(x), (IC 1)ut(x, 0) = v0(x), (IC 2)u(0, t) = 0, (BC 1)u(L, t) = 0. (BC 2)

The initial conditions indicate the initial displacements of the points of the string, u0(x), andtheir initial velocities v0(x). Depending on positive/negative sign, they are directed up or down.

Example 4.22. Describe small oscillations of a string of length L, with fixed ends, no dis-placement, and initial velocity given by v0(x) = x(L− x)2.

Solution: the IBVP is given byutt = c2uxx, 0 < x < L, 0 < t;u(x, 0) = 0, (IC 1)ut(x, 0) = x(L− x)2, (IC 2)u(0, t) = 0, (BC 1)u(L, t) = 0. (BC 2)

(4.59)

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Another Physical Model – Longitudinal Oscillations of an Elastic Rod

An additional physical model that uses exactly the same linear wave equation (4.58) is a modelof linear longitudinal oscillations of an elastic (e.g., rubber) rod. A setup is given in Figure4.9. Here each physical point of the rod, marked by x ∈ [0, L], undergoes a displacement to the

0 L

Elasic rod at rest/equilibrium.

Reference configuration

x

0x

x1 x2 x3

u(x1, t) u(x2, t) u(x3, t)

u(L, t)

Actual (moving) configuration

Figure 4.9: Longitudinal Oscillations of an Elastic Rod.

right (or to the left), which is given by u(x, t) > 0 (or respectively u(x, t) < 0).

For example, small oscillations of a rod with ends firmly fixed at their initial positions couldbe described by the IBVP (4.59). The constant c2 now represents some material parameterinvolving the density and elastic properties of the rod material.

The following table contains a summary of BCs for the linear wave equation, both basic ap-plications. The BCs there are only considered at x = 0. They have the obviously correspondingform at x = L.

In the table, the application is labeled as follows. Transverse oscillations of a string = STR.Longitudinal oscillations of an elastic rod/spring = LO.

29

Name Type BC A physical application

Dirichlet Homogeneous u(0, t) = 0 STR, LO: The left end of the string/rod

(Type I) is attached at x = 0 for all times.

Non-homogeneous u(0, t) = µ(t) STR, LO: The left end of the string (rod) is

moved up/down (left/right) according to µ(t).

Neumann Homogeneous ux(0, t) = 0 LO: the left end of the elastic rod

(Type II) is free for all times.

Non-homogeneous ux(0, t) = ν(t) LO: to the left end of the

rod, a force F (t) = kν(t) is applied.

Robin Homogeneous αu(0, t) + βux(0, t) = 0 LO: the left end of oscillating rod

(Type III) α, β = const. or spring is attached elastically.

Example 4.23. Describe small longitudinal oscillations of an elastic rod of length L, initiallyundisturbed and at rest, with left end externally driven (position given by µ(t) = A sin 2t) anda free right end.

Solution: the IBVP is given byutt = c2uxx, 0 < x < L, 0 < t;u(x, 0) = 0, (IC 1)ut(x, 0) = 0, (IC 2)u(0, t) = A sin 2t, (BC 1)ux(L, t) = 0. (BC 2)

4.5.3 D’Alembert Solution and the Traveling Wave

Consider the initial value problem for the wave equation (4.53), describing oscillations of a verylong (infinite) string:

utt = c2uxx, −∞ < x < ∞, 0 < t;u(x, 0) = u0(x);ut(x, 0) = v0(x).

(4.60)

General solution of the wave equation in the form of two traveling waves

First we look at the wave equation by itself. We will take care of the initial conditions in (4.60)later.

The linear homogeneous wave equation (4.53) can be written in an interesting way:

L[u] =

(∂2

∂t2− c2

∂2

∂x2

)u(x, t) = 0,

or even further,

L[u] = L1[L2[u]] =

(∂

∂t− c

∂

∂x

)(∂

∂t+ c

∂

∂x

)u(x, t) = 0

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(check!). Similarly, one can use L[u] = L2[L1[u]], because the operators L1 and L2 commute.Conversion of a PDE to such form is referred to as operator splitting.

So the wave equation utt = c2uxx is equivalent to both L1[L2[u]] = 0 and to L2[L1[u]] = 0.In particular, if we want to seel solutions to the wave equation, then solutions to just L1[u] = 0or L2[u] = 0 will clearly satisfy the wave equation.

Now, from Section 4.3.3, we know that the general solution to the advection equation

L1[u] =

(∂

∂t− c

∂

∂x

)= 0

is given by the right-traveling wave (4.27), which we write here again:

u(x, t) = F (x− ct). (4.61)

Similarly, the general solution to the left-advection equation

L2[u] =

(∂

∂t+ c

∂

∂x

)= 0

is given by the left-traveling wave

u(x, t) = G(x+ ct). (4.62)

The wave equation (4.53) considered here is linear homogeneous. Hence we can add solutionsto obtain the general d’Alembert solution of the wave equation

ugen(x, t) = F (x− ct) +G(x+ ct), (4.63)

for any pair of sufficiently smooth functions F,G.

Solution of the IVP (4.60) for the infinite string

Now we will use the general solution form (4.63) to try to satisfy the initial conditions of theIVP (4.60).

(A) First, consider a simplified version of the IVP (4.60) with no initial velocity:utt = c2uxx, −∞ < x < ∞, 0 < t;u(x, 0) = u0(x);ut(x, 0) = 0.

(4.64)

Let u(x, t) = F (x− ct) +G(x+ ct). In order to satisfy the IC ut(x, 0) = 0, we need

ut(x, 0) = (−cF ′(x− ct) + cG′(x+ ct))|t=0 = −cF ′(x) + cG′(x).

hence F,G must be related by F ′(x) = G′(x), or, F (x) = G(x) + a for some a = const.

Now we use the remaining freedom to satisfy the remaining IC u(x, 0) = u0(x). We have

u(x, 0) = (F (x− ct) +G(x+ ct))|t=0 = F (x) +G(x) = 2G(x) + a.

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Hence we find

G(x) =1

2u0(x)−

a

2, F (x) =

1

2u0(x) +

a

2. (4.65)

We have determined F,G such that the general d’Alembert solution (4.63) of the wave equationsatisfies both ICs of (4.64). Note that upon the substitution of (4.65) into (4.63), a cancels.The unique solution of the initial value problem (4.64) is thus given by

udisplacement(x, t) = F (x− ct) +G(x+ ct) =u0(x− ct) + u0(x+ ct)

2. (4.66)

The physical meaning of the above solution is that the initial string shape u0(x) will split intotwo identical shapes of half the height, moving to the left and to the right (Figure 4.10).

u0(x)

x

Figure 4.10: D’Alembert solution for the problem (4.64) with no initial velocity.

(B) Second, consider another simplified version of the IVP (4.60) with no initial displacement:utt = c2uxx, −∞ < x < ∞, 0 < t;u(x, 0) = 0;ut(x, 0) = v0(x).

(4.67)

Let again u(x, t) = F (x− ct) +G(x+ ct). In order to satisfy the IC u(x, 0) = 0, we need

u(x, 0) = F (x) +G(x) = 0,

hence F (x) = −G(x). Now again use the remaining freedom to satisfy the remaining ICut(x, 0) = v0(x). We have

ut(x, 0) = (−cF ′(x− ct) + cG′(x+ ct))|t=0 = 2cG′(x) = v0(x).

Hence we find

G(x) =1

2c

∫ x

q

v0(x) dx; F (x) = − 1

2c

∫ x

q

v0(x) dx, (4.68)

where q = const. The solution of the initial velocity problem (4.67) is then given by

uvelocity(x, t) = F (x− ct) +G(x+ ct) =1

2c

∫ x+ct

x−ct

v0(x) dx. (4.69)

(C) Now proceed with the solution of the general problem (4.60) with nonzero initial displace-ment/velocity.

32

It is clear that the problem (4.60) is linear, both in the PDE and the ICs. It is a direct“sum” of problems (4.64) and (4.67). Hence the solution of the general problem (4.60) is givenby

u = udisplacement + uvelocity.

(Check that if udisplacement is any solution of (4.64), and uvelocity is any solution of (4.67), thensuch u is a solution of (4.60), satisfying the PDE and ICs!)

We write this general d’Almbert solution of the IVP in one formula. It is given by

u(x, t) =u0(x− ct) + u0(x+ ct)

2+ 1

2c

∫ x+ct

x−ct

v0(x) dx, (4.70)

and involves both ICs.

Remark 4.24. Nonlinear PDEs can also have traveling wave solutions, which, however, cannotbe added together. The existence of traveling wave solutions is related to space- and time-translation symmetry that a PDE must have. For example, the KdV equation on u(x, t)written as

ut + 6uux + uxxx = 0 (4.71)

has a traveling wave solution (a solitary wave, soliton)

u(x, t) =c

2

1

cosh2

(√c

2(x− ct)

) (4.72)

for any wave speed c. Note that contrary to what one has for the wave equation, the amplitudeof the soliton solution (4.72) depends on the wave speed.

4.5.4 Wave Equation in Two Dimensions

The wave equation in 2+1 dimensions describes, for example, small elastic vertical oscillationsof a membrane. Let D denote the projection of the membrane on the (x, y) plane. Then theelevation is given by u(t, x, y). In a well-posed IBVP, the wave equation is appended by twoinitial conditions in D, and a boundary condition at each point of the boundary ∂D. In themembrane model, the latter prescribes the elevation of the membrane boundary from the flat(zero) position.

As a first example, consider a problem where a rectangular elastic membrane is fixed on theboundary at zero level, has an initial displacement, and no initial velocity. The IBVP is givenby (cf. Figure a)

utt = c2∆u = c2(uxx + uyy), 0 < x < a, 0 < y < b, 0 < t;u(0, x, y) = u0(x, y), (IC 1)ut(0, x, y) = 0, (IC 2)u(t, 0, y) = u(t, a, y) = 0, 0 < y < b, 0 < t; (BC)u(t, x, 0) = u(t, x, a) = 0, 0 < x < a, 0 < t, (BC)

33

Figure 4.11: Rectangular and circular membrane

Note that in principle, the BCs could be space- and time-dependent.

As a second example, consider a dimensionless problem for the oscillations of a circularmembrane. Here we use polar coordinates, and denote u(t, x, y) = v(t, r, ϕ), 0 ≤ r ≤ 1,0 ≤ ϕ < 2π. Suppose the membrane boundary is fixed (by a solid “wire”) at the radius R = 1to have the shape v(t, 1, ϕ) = sinϕ + sin 2ϕ + sin 3ϕ, and has some initial displacement andinitial velocity. The IBVP is given by (cf. Figure b)

vtt = c2∆v = c2(vrr +

1

rvr +

1

r2vϕϕ

), 0 ≤ r < 1, 0 ≤ ϕ < 2π, 0 < t;

v(0, r, ϕ) = f(r, ϕ), (IC 1)vt(0, r, ϕ) = g(r, ϕ), (IC 2)v(t, 1, ϕ) = sinϕ+ sin 2ϕ+ sin 3ϕ, 0 ≤ ϕ < 2π, 0 < t. (BC)

Keywords 4.5

Linear elasticity; Guitar string; Musical notes; Traveling waves; KdV; Soliton; Multi-Solitonsolution.

Problems 4.5

Problem 4.5a. Derive the wave equation for a transversely oscillating string rigorously fromthe first principles.

Problem 4.5b. Find in the literature and study the derivation of the wave equation for lon-gitudinal oscillations of an elastic rod in the linear elasticity framework.

Problem 4.5c. Formulate the initial/boundary value problem for a given model (PDE, BCs,ICs). Do not solve. Do not provide much detail. Provide an illustration.

(a) Describe small oscillations of a horizontal elastic string of length L [m], whose left end ismoving up/down with elevation given by µ(t) = e−kt sin(ωt) [m], the right end fixed atheight 0 [m], horizontal initial position, and zero initial velocity.

34

(b) Describe small oscillations of a horizontal elastic membrane occupying the square region0 ≤ x, y ≤ L [m], if the sides of the membrane are fixed in a square frame that is keptat zero elevation for all times. Initially, the membrane is flat. At t = 0, the membraneis hit upwards with a round hammer of radius r < L/2 applied around the center ofthe membrane. The hammer provides the part of a membrane it contacts with an initialvelocity of 1 m/s.

(c) Describe small longitudinal oscillations of an elastic bar of natural length L [m], lying ona smooth table, if the left end of the bar is fixed at x = 0, and right end is free. The barwas stretched (to the right) to the length 1.05L, and at time t = 0, released from rest (noinitial velocity).

Problem 4.5d. Produce an IBVP describing small oscillations of a horizontal string of length4 m, with ends at level u = 0, no initial velocity, and with initial displacement given by

u0(x) =

x/100, 0 ≤ x ≤ 2,0.04− x/100, 2 < x ≤ 4,

taking into account the force of gravity. [Hint: the equation will change, due to the forcing.Re-do the derivation of the wave equation to get the correct one. Keep the physical dimensionsin mind.]

Problem 4.5e. Re-do Problem 4.5d, if in addition to the effect of gravity, the string is chargedwith electric charge density r(x) [coulombs/meter] and is immersed in the upward electric fieldE(t) [volts/meter]. Suppose that the string has constant linear density ρ [kg/meter]. [Hint:electric force is given by F = qE newtons, if q is the charge in coulombs].

Problem 4.5f. An elastic rod lies on a perfectly smooth table. Its [dimensionless] motion isdescribed by a homogeneous Neumann problem

utt = 4uxx, 0 < x < 1, 0 < t;u(x, 0) = 0,ut(x, 0) = 5,ux(0, t) = ux(L, t) = 0.

Describe the physical meaning of the initial and boundary conditions. Describe the motion ofthe rod; give a formula for the displacement of the center of mass of the rod at all times.

Problem 4.5g. (Modes)

• Show that the homogeneous wave equation (4.53) has separated solutions (modes)

un(x, t) = A sinπnct

Lsin

πnx

L, vn(x, t) = B sin

πnct

Lcos

πnx

L

that hold for all A,B,L = const, and integer n = 1, 2, 3....

• Sketch the essential shape of some modes as functions of x, 0 ≤ x ≤ L, for severalconsecutive values of t. Sketch the un modes in one plot, and the vn modes in the other.[You can choose A = B = 1.]

35

• Modes un satisfy certain homogeneous boundary conditions at x = 0 and x = L. Specifythose BC.

• Modes vn satisfy a different homogeneous BC at x = 0 and x = L. Specify that BC.

Problem 4.5h. (Plane waves) Show that the complex plane wave solution

w(x, t) = Aei(kx−ωt)

can be a solution to the linear homogeneous wave equation (4.53) for any constant A.

In order to prove that the complex-valued function w(x, t) is a solution of the real PDE(4.53), one has to prove that both real and imaginary parts of w(x, t) solve the PDE, and thenuse the linearity.

[The constants k and ω are the wave number and the angular frequency. What is thenecessary relation between them, in terms of the wave speed c?]

Problem 4.5i. Let us tune a guitar. One starts from the 1st (thinnest) string. Let its lengthbe 65 cm. The string is made of steel wire, circular cross-section. The string diameter is 0.25mm, the density of steel is µ = 0.00785 grams/mm3.

(a) Find the mass per unit length λ [kg/m].

(b) Suppose that the tension force applied to the string is T = 69 N. Compute the coefficientc2 = T/λ of the corresponding wave equation.

(c) Write down the separated “standing wave”-type solutions (modes) un of the wave equation(Problem 4.5g). In particular, write down the fundamental mode: n = 1.

(d) For the fundamental mode, find the period of oscillations τ (in seconds) and the frequencyof oscillations f = 1/τ (in Hz). A properly tuned first guitar string should correspond toE of the 1st octave (329.6 Hz).

(e) How should the tension be changed in order to correct the frequency? What is the correcttension value?

Problem 4.5j. Verify by differentiation that (4.61) and (4.63) are indeed exact solutions ofthe wave equation utt = c2uxx, for any sufficiently smooth functions F,G.

Problem 4.5k. Suppose an infinite string is hit with a hammer, so that the initial conditionsare given by u0(x) = 0 and

v0(x) =

1, − 1 ≤ x ≤ 1,0 otherwise.

(a) Use the d’Alembert’s solution to find the shape of the string for all t > 0.

(b) In one graph, sketch the solution u(x, t) as a function of x, for t = 1, 2, 3, assuming thatc = 1/2. [Hint: for each given time, consider the overlap of the integration interval withinterval where v0 = 0.]

36

Problem 4.5l. A horizontal rubber bar of unperturbed length L lies on a table (assume nofriction). Its left end is attached at x = 0. The right end is connected elastically (through aspring with spring constant k and an unperturbed length ℓ) to a wall located at x = B > L(see picture).

Come up with a correct boundary condition at the right end. It will be a Robin-type BC.

Write down an IBVP for a wave equation, describing small compression waves in this elasticbar, if its initial displacement field is given by u0(x), and initial velocity is zero.

x

0 B

Figure 4.12: Longitudinal oscillations of an elastic rod with an elastically attached end.

Problem 4.5m. Verify by hand that indeed, the soliton solution (4.72) satisfies the KdVequation (4.71) for all c > 0.

Choose a value of c and generate some plots u(x, t) as functions of x for a set of timest = 0, t1, .... Repeat the same plots for a higher and a lower value of c, for the same timeinstants. Make animations if desired. Observe the soliton behaviour.

37

4.6 The Laplace Equation

In two dimensions, the Laplace equation is a linear homogeneous PDE

∆ϕ ≡ ϕxx + ϕyy = 0, (4.73)

where ϕ = ϕ(x, y), and the equation (4.73) holds in an open domain (x, y) ∈ D ⊂ R2.

In three dimensions, one has ϕ = ϕ(x, y, z), (x, y, z) ∈ D ⊂ R3, and the Laplace operator isgiven by

∆ϕ ≡ ϕxx + ϕyy + ϕzz.

A non-homogeneous version of the Laplace equation is the time-independent Poisson PDE

∆ϕ = g, (4.74)

where g is a “forcing” term; g = g(x, y) in 2D, and g = g(x, y, z) in 3D.

The Laplace and Poisson equations arise in many areas, in particular, as time-independent(equilibrium) solutions of heat and wave equations, and in further models, including electro-magnetism, fluid and plasma dynamics, and narrow escape problems. An interesting relatedtime-independent, linear homogeneous equation is the Helmholtz equation.

A well-posed problem for Laplace and Poisson equations includes a boundary condition atevery point of the domain boundary.

Example 4.25. Formulate a problem describing an equilibrium shape of a circular membraneof radius R in the gravity field, if the boundary of the membrane is a wire frame whose elevationabove the x, y plane is given by F (ϕ).

Solution: the BVP for the Poisson on the membrane elevation equation v(t, r, ϕ) in polarcoordinates has the form vrr +

1

rvr +

1

r2vϕϕ =

g

c2, 0 ≤ r < R, 0 ≤ ϕ < 2π, 0 < t;

v(t, R, ϕ) = F (ϕ), 0 ≤ ϕ < 2π, 0 < t. (BC)

Here g is the acceleration of the free fall, and c is the membrane material parameter (wavespeed).

Keywords 4.6

Maxwell’s equations; Narrow escape problems; Helmholtz equation.

38

Problems 4.6

Problem 4.6a. Produce a BVP for the Laplace equation, describing the equilibrium heatdistribution in a thin square block 0 ≤ x, y ≤ L [m], with isolated square sides. Denote thetemperature of every point of the block by u(x, y) [oC]. The sides of the block are maintainedat the following temperatures: the sides x = L, y = 0, and y = L, at u = 0oC. The side x = 0,

at temperature u =

x/100, 0 ≤ x ≤ L/2,L/100− x/100, L/2 < x ≤ L.

[oC].

Provide a sketch.

Problem 4.6b. Consider the Laplace equation in two dimensions: ϕ = ϕ(x, y),

ϕxx + ϕyy = 0

(we do not worry about the domain shape/boundary conditions in this problem).

Seek separated solutions of the Laplace equation, in the form

ϕ(x, y) = A(x)B(y).

See also Problems 4.4a and 4.5g above.

39