“If one looks at the different problems of the integral calculus which arise naturally when one wishes to go deep into the different parts of physics, it is impossible not to be struck by the analogies existing. Whether it be electrostatics or electrodynamics, the propogation of heat, optics, elasticity, or hydrodynamics, we are led always to differential equations of the same family.” —Henri Poincar´e
5A
Evolution vs. nonevolution equations
Recommended: §1B,§1C,§2B,§4B.
An evolution equation is a PDE with a distinguished “time” coordinate,
t. In other words, it describes functions of the form u(x;t), and the equation has the form:
Dtu = Dxu
whereDt is some differential operator involving only derivatives in thetvariable
(e.g. ∂t,∂t2, etc.), whileDxis some differential operator involving only derivatives in thex variables (e.g. ∂x,∂y2,4, etc.)
Example 5A.1. The following are evolution equations: (a) The heat equation “∂tu=4u” of§1B.
(b) The wave equation “∂t2u=4u” of§2B.
(c) The telegraph equation “κ2∂2tu + κ1∂tu = −κ0u + 4u” of§2C.
(d) The Schr¨odinger equation “∂tω = i1~Hω” of§3B (hereHis a Hamiltonian operator).
(e) Liouville’s Equation, the Fokker-Plank equation, and Reaction-Diffusion
Equations. ♦
Nonexample 5A.2. The following arenot evolution equations: (a) The Laplace Equation “4u= 0” of §1C.
(b) The Poisson Equation “4u=q” of§1D.
(c) The Helmholtz Equation “4u=λu” (whereλ∈Cis a constant —i.e. an eigenvalue of4).
(d) The Stationary Schr¨odinger equation H ω0 = E ·ω0 (where E ∈ C is a
constant eigenvalue). ♦
In mathematical models of physical phenomena, most PDEs are evolution equa- tions. Nonevolutionary PDEs generally arise asstationary stateequations for evolution PDEs (e.g. Laplace’s equation) or asresonance states (e.g. Sturm- Liouville, Helmholtz).
Order: Theorderof the differential operator∂x2∂3yis 2+3 = 5. More generally, the order of the differential operator ∂k1
1 ∂ k2
2 . . . ∂ kD
D is the sum k1 +. . .+kD.
The order of a general differential operator is the highest order of any of its terms. For example, the Laplacian is second order. Theorder of a PDE is the highest order of the differential operator that appears in it. Thus, the Transport Equation, Liouville’s Equation, and the (nondiffusive) Reaction Equation isfirst order, but all the other equations we have looked at (the heat equation, the wave equation, etc.) are ofsecond order.
5B
Initial value problems
Prerequisites: §5A.
LetX⊂RD be some domain, and letLbe a differential operator onC∞(X;R). Consider evolution equation
∂tu = Lu, (5B.1)
for an unknown functionu :X×R6− −→ R. An initial value problem(IVP) for equation (5B.1) is the following problem:
Given some function f0 :X−→R (theinitial conditions), find a continuous function u : X×R6− −→ R which satisfies (5B.1) and also satisfies u(x,0) =f0(x), for allx∈X.
For example, suppose the domain X is an iron pan being heated on a gas flame stove. You turn off the flame (so there is no further heat entering the system) and then throw some vegetables into the pan. Thus, (5B.1) is the Heat Equation, and f0 describes the initial distribution of heat: cold vegetables in a
hot pan. The initial value problem asks: “How fast do the vegetables cook? How fast does the pan cool?”
Next, consider the second order-evolution equation
∂2tu = Lu, (5B.2)
for a unknown functionu:X×R6−−→R. Aninitial value problem(orIVP, orCauchy problem) for (5B.2) is as follows:
Given a function f0 : X −→ R (the initial position), and/or another function f1 : X −→ R (the initial velocity), find a con- tinuously differentiable function u : X×R6− −→ R which satisfies (5B.2) and also satisfies u(x,0) = f0(x) and ∂tu(x,0) = f1(x), for allx∈X.
For example, suppose (5B.1) is the wave equation on X = [0, L]. Imagine [0, L] as a vibrating string. Thus, f0 describes the initial displacement of the
string, and f1 its initial momentum.
Iff0 6≡0, andf1 ≡0, then the string is initially at rest, but is released from
a displaced state —in other words, it is plucked (e.g. in a guitar or a harp). Hence, the initial value problem asks: “How does a guitar string sound when it is plucked?”
On the other hand, if f0≡0, andf1 6≡0, then the string is initially flat, but
is imparted with nonzero momentum —in other words, it isstruck (e.g. by the hammer in the piano). Hence, the initial value problem asks: “How does a piano string sound when it is struck?”
5C
Boundary value problems
Prerequisites: §0D,§1C. Recommended: §5B.
IfX⊂RD is a finite domain, then ∂Xdenotes itsboundary. Theinterior ofX is the setint(X) of all points in Xnoton the boundary.
Example 5C.1.
(a) IfI = [0,1]⊂R is theunit interval, then ∂I ={0,1} is a two-point set, and int(I) = (0,1).
(b) IfX= [0,1]2 ⊂R2 is theunit square, then int(X) = (0,1)2. and
∂X = {(x, y)∈X; x= 0 orx= 1 ory= 0 ory= 1}.
(c) In polar coordinates on R2, let D = {(r, θ) ; r ≤1, θ∈[−π, π)} be the
unit disk. Then ∂D = {(1, θ) ; θ∈[−π, π)} is the unit circle, and
int(D) ={(r, θ) ; r <1, θ ∈[−π, π)}.
(d) In spherical coordinates on R3, let B = x∈R3 ; kxk ≤1 be the 3- dimensional unit ballinR3. Then ∂B = S := {x∈RD ; kxk = 1 is the unit sphere, and int(B) =x∈RD ; kxk <1 .
(e) In cylindrical coordinates onR3, letX={(r, θ, z) ; r≤R, −π≤θ≤π, 0≤z≤L} be thefinite cylinderinR3. Then∂X = {(r, θ, z) ; r =R orz = 0 or z = L}. ♦
A boundary value problem (BVP) is a problem of the following kind:
Find a continuous functionu:X−→Rsuch that
1. u satisfies some PDE at allx in theinterior ofX.
2. u also satisfies some other equation(maybe a differential equa- tion)for all s on the boundaryof X.
The condition u must satisfy on the boundary of X is called a boundary
condition. Note that there is no ‘time variable’ in our formulation of a BVP; thus, typically the PDE in question is an ‘equilibrium’ equation, like the Laplace equation or the Poisson equation.
If we try to solve an evolution equation with specified initial conditionsand specified boundary conditions, then we are confronted with an ‘initial/boundary value problem’. Formally, aninitial/boundary value problem(I/BVP) is a problem of the following kind:
Find a continuous functionu:X×R6−−→Rsuch that
1. u satisfies some (evolution) PDE at all x in the interior of
X×R6−.
2. u satisfies some boundary condition for all (s;t) in(∂X)×R6−.
3. u(x; 0)also satisfies some initial condition(as described in§5B)
for all x∈X.
We will consider four kinds of boundary conditions: Dirichlet, Neumann, Mixed, andPeriodic. Each of these boundary conditions has a particular physical interpretation, and yields particular kinds of solutions for a partial differential equation.
0 0.05 0.1 0.15 0.2 0.25 0.2 0.4 0.6 0.8 1 x
Figure 5C.1: f(x) =x(1−x) satisfies homogeneous Dirichlet boundary condi- tions on the interval [0,1].
5C(i) Dirichlet boundary conditions
Let X be a domain, and let u:X −→ Rbe a function. We say that u satisfies
homogeneous Dirichlet boundary conditions (HDBC)on Xif: For alls∈∂X, u(s)≡0.
Physical interpretation.
Thermodynamic. (Heat equation, Laplace Equation, or Poisson Equation) In this case, u represents a temperature distribution. We imagine that the domain X represents some physical object, whose boundary ∂X is made out of metal or some other material which conducts heat almost perfectly. Hence, we can assume thatthe temperature on the boundary is always equal to the temperature of the surrounding environment.
We further assume that this environment has a constant temperatureTE
(for example, X is immersed in a ‘bath’ of some uniformly mixed fluid), which remains constant during the experiment (for example, the fluid is present in large enough quantities that the heat flowing into/out of X does not measurably change it). We can then assume that the ambient temperature is TE ≡0, by simply subtracting a constant temperature of
TE off the inside and the outside. (This is like changing from measuring
temperature in degrees Kelvin to measuring in degrees Celsius; you’re just adding 273o to both sides, which makes no mathematical difference.)
Electrostatic. (Laplace equation or Poisson Equation) In this case, u repre- sents an electrostatic potential. The domain X represents some compart- ment or region in space, whose boundary∂Xis made out of metal or some other perfect electrical conductor. Thus, the electrostatic potential within the metal boundary is a constant, which we can normalize to be zero.
Acoustic. (Wave equation) In this case,urepresents the vibrations of some vi- brating medium (e.g. a violin string or a drum skin). Homogeneous Dirich-
–1 0 1 x –0.5 0 0.5 1 y 0 0.2 0.4 0.6 0.8 1 z –1 –0.5 0 0.5 1 x –1 –0.5 0 0.5 y 0 0.2 0.4 0.6 0.8 1 z (A) (B)
Figure 5C.2: (A)f(r, θ) = 1−r satisfies homogeneous Dirichlet boundary con- ditions on the disk D = {(r, θ) ; r ≤ 1}, but is not smooth at zero. (B)
f(r, θ) = 1−r2 satisfies homogeneous Dirichlet boundary conditions on the disk D={(r, θ) ; r≤1}, and is smooth everywhere.
let boundary conditions mean that the medium is fixed on the boundary
∂X (e.g. a violin string is clamped at its endpoints; a drumskin is pulled down tightly around the rim of the drum).
The set of infinitely differentiablefunctions from X toR which satisfy homoge- neous Dirichlet Boundary Conditions will be denotedC0∞(X;R) orC0∞(X). Thus, for example
C0∞[0, L] =
n
f : [0, L]−→R; f is smooth, andf(0) = 0 = f(L)
o
The set ofcontinuousfunctions fromXtoRwhich satisfy homogeneous Dirichlet Boundary Conditions will be denotedC0(X;R) or C0(X).
Example 5C.2.
(a) Suppose X= [0,1], and f :X −→R is defined by f(x) = x(1−x). Then
f(0) = 0 =f(1), andf is smooth, so f ∈ C0∞[0,1]. (See Figure 5C.1). (b) Let X= [0, π].
1. For anyn∈N, let Sn(x) = sin (n·x) (see Figure 6D.1 on page 113). Then Sn∈ C0∞[0, π].
2. If f(x) = 5 sin(x)−3 sin(2x) + 7 sin(3x), then f ∈ C0∞[0, π]. More generally, any finite sum
N
X
n=1
BnSn(x) (for some constants Bn) is in
3. Iff(x) = ∞
X
n=1
BnSn(x) is a uniformly convergent Fourier sine series1,
thenf ∈ C∞
0 [0, π].
(c) LetD={(r, θ) ; r≤1} be the unit disk. Letf :D−→Rbe the ‘cone’ in Figure 5C.2(A), defined: f(r, θ) = (1−r). Thenf is continuous, andf ≡0 on the boundary of the disk, so f satisfies Dirichlet boundary conditions. Thus,f ∈ C0(D). However,f is not smooth (it is nondifferentiable at zero), sof 6∈ C∞
0 (D).
(d) Letf :D−→Rbe the ‘dome’ in Figure 5C.2(B), definedf(r, θ) = 1−r2. Thenf ∈ C0∞(D).
(e) LetX= [0, π]×[0, π] be the square of sidelengthπ.
1. For any (n, m) ∈ N2, let Sn,m(x, y) = sin (n·x)·sin (m·y). Then
Sn,m ∈ C0∞(X). (see Figure 9A.2 on page 181).
2. If f(x) = 5 sin(x) sin(2y) −3 sin(2x) sin(7y) + 7 sin(3x) sin(y), then
f ∈ C0∞(X). More generally, any finite sum
N X n=1 M X m=1 Bn,mSn,m(x) is in C0∞(X). 3. If f = ∞ X n,m=1
Bn,mSn,m is a uniformly convergent two dimensional
Fourier sine series2, thenf ∈ C0∞(X).
♦
Exercise 5C.1. (i) Verify examples (b) to (e) above E
(ii) Show thatC∞
0 (X) is a vector space.
(iii) Show thatC0(X) is a vector space.
Arbitrary nonhomogeneous Dirichlet boundary conditions are im- posed by fixing some function b:∂X−→R, and then requiring:
u(s) =b(s), for alls∈∂X. (5C.3) For example, theclassical Dirichlet Problemis to find a continuous function
u :X −→ R satisfying the Dirichlet condition (5C.3), such that u also satisfies Laplace’s Equation: 4u(x) = 0 for all x∈int(X).
1See§7B on page 144. 2
Physical interpretations.
Thermodynamic. u describes a stationary temperature distribution on X, where the temperature is fixed on the boundary. Different parts of the boundary may have different temperatures, so heat may be flowingthrough the region X from warmer boundary regions to cooler boundary regions. But the actual temperature distribution withinXis in equilibrium.
Electrostatic. u describes an electrostatic potential field within the region X. The voltage level on the boundaries is fixed (e.g. boundaries ofXare wired up to batteries which maintain a constant voltage). However different parts of the boundary may have different voltages (the boundary is not a perfect conductor).
Minimal surface. u describes a minimal-energy surface (e.g. a soap film). The boundary of the surface is clamped in some position (e.g. the wire frame around the soap film); the interior of the surface must adapt to find the minimal energy configuration compatible with these boundary conditions. Minimal surfaces of low curvature are well-approximated by harmonic functions.
For example, if X = [0, L], and b(0) and b(L) are two constants, then the
Dirichlet Problem is to findu: [0, L]−→Rsuch that
u(0) =b(0), u(L) =b(L), and ∂x2u(x) = 0, for 0< x < L. (5C.4) That is, the temperature at the left-hand endpoint is fixed at b(0), and at the right-hand endpoint is fixed atb(L). The unique solution to this problem is the functionu(x) =b(L)−b(0)x/L+b(0). (Exercise 5C.2).
E
5C(ii) Neumann boundary conditions
Suppose X is a domain with boundary ∂X, and u :X −→ R is some function. Then for any boundary point s∈∂X, we use “∂⊥u(s)” to denote the outward
normalderivative3ofuon the boundary. Physically,∂⊥u(s) is therate of change in u as you leave X by passing through ∂X in a perpendicular direction.
Example 5C.3.
(a) IfX= [0,1], then∂⊥u(0) =−∂xu(0) and∂⊥u(1) =∂xu(1).
(b) Suppose X= [0,1]2 ⊂R2 is the unit square, and (x, y) ∈ ∂X. There are four cases:
3
This is sometimes indicated as ∂u
∂n or
∂u
• Ifx= 0 (left edge), then∂⊥u(0, y) =−∂xu(0, y). • Ifx= 1 (right edge), then∂⊥u(1, y) =∂xu(1, y). • Ify= 0 (top edge), then ∂⊥u(x,0) =−∂yu(x,0). • Ify= 1 (bottom edge), then ∂⊥u(x,1) =∂yu(x,1).
(If more than one of these conditions is true —for example, at (0,0) —then (x, y) is a corner, and ∂⊥u(x, y) is not well-defined).
(c) Let D={(r, θ) ; r <1} be the unit disk in the plane. Then∂Dis the set {(1, θ) ; θ∈[−π, π)}, and for any (1, θ)∈∂D, ∂⊥u(1, θ) = ∂ru(1, θ).
(d) LetD={(r, θ) ; r < R}be the disk of radiusR. Then∂D={(R, θ) ; θ∈[−π, π)}, and for any (R, θ)∈∂D, ∂
⊥u(R, θ) = ∂ru(R, θ).
(e) LetB={(r, φ, θ) ; r <1}be the unit ball inR3. Then∂B={(r, φ, θ) ; r= 1} is the unit sphere. If u(r, φ, θ) is a function in polar coordinates, then for any boundary point s= (1, φ, θ), ∂⊥u(s) = ∂ru(s).
(f) SupposeX={(r, θ, z) ; r≤R, 0≤z≤L, −π ≤θ < π}, is thefinite cylin-
der, and (r, θ, z)∈∂X. There are three cases:
• If r=R (sides), then ∂⊥u(R, θ, z) = ∂ru(R, θ, z).
• If z= 0 (bottom disk), then∂⊥u(r, θ,0) = −∂zu(r, θ,0).
• If z=L(top disk), then ∂⊥u(r, θ, L) = ∂zu(r, θ, L).
♦ We say that u satisfies homogeneous Neumann boundary conditions
if
∂⊥u(s) = 0 for alls∈∂X. (5C.5)
Physical Interpretations.
Thermodynamic. (Heat, Laplace, or Poisson equation) Supposeu represents a temperature distribution. Recall that Fourier’s Law of Heat Flow (§ 1A on page 3) says that ∇u(s) is the speed and direction in which heat is flowing at s. Recall that ∂⊥u(s) is the component of ∇u(s) which is per- pendicular to ∂X. Thus, homogeneous Neumann BC means that∇u(s) is parallel to the boundary for alls∈∂X. In other words no heat is crossing the boundary. This means that the boundary is a perfect insulator.
Ifurepresents the concentration of a diffusing substance, then∇u(s) is the flux of this substance at s. Homogeneous Neumann Boundary conditions mean that the boundary is animpermeable barrier to this substance.
Electrostatic. (Laplace or Poisson equation) Supposeu represents an electric potential. Thus ∇u(s) is the electric field at s. Homogeneous Neumann BC means that ∇u(s) is parallel to the boundary for all s ∈ ∂X; i.e. no field lines penetrate the boundary.
The set ofcontinuousfunctions fromXtoRwhich satisfy homogeneous Neumann boundary conditions will be denoted C⊥(X). The set ofinfinitely differentiable functions fromXtoRwhich satisfy homogeneous Neumann boundary conditions will be denotedC⊥∞(X). Thus, for example
C⊥∞[0, L] = nf : [0, L]−→R; f is smooth, andf0(0) = 0 = f0(L) o 0 0.02 0.04 0.06 0.08 0.1 0.12 0.14 0.16 0.2 0.4 0.6 0.8 1 x –1 0 1 x –0.5 0 0.5 1 y 0 0.2 0.4 0.6 0.8 1 z (A) (B) Figure 5C.3: (A) f(x) = 12x2−1 3x 3
satsfies homogeneous Neumann boundary conditions
on the interval [0,1]. (B) f(r, θ) = (1−r)2 satisfies homogeneous Neumann boundary
conditions on the diskD={(r, θ) ; r≤1}, but is not differentiable at zero.
Example 5C.4.
(a) LetX= [0,1], and letf : [0,1]−→Rbe defined byf(x) = 12x2−13x3 (See Figure 5C.3(A)). Thenf0(0) = 0 =f0(1), andf is smooth, sof ∈ C∞
⊥[0,1].
(b) LetX= [0, π].
1. For anyn∈N, letCn(x) = cos (n·x) (see Figure 6D.1 on page 113). ThenCn∈ C⊥∞[0, π].
2. If f(x) = 5 cos(x)−3 cos(2x) + 7 cos(3x), then f ∈ C∞
⊥ [0, π]. More
generally, any finite sum
N
X
n=1
AnCn(x) (for some constants An) is in
3. If f(x) = ∞
X
n=1
AnCn(x) is a uniformly convergent Fourier cosine se-
ries4, and the derivative series f0(x) = − ∞
X
n=1
nAnSn(x) is also uni-
formly convergent, thenf ∈ C∞
⊥[0, π].
(c) LetD={(r, θ) ; r ≤1} be the unit disk.
1. Let f : D −→ R be the “witch’s hat” of Figure 5C.3(B), defined:
f(r, θ) := (1−r)2. Then ∂⊥f ≡0 on the boundary of the disk, so f
satisfies Neumann boundary conditions. Also, f is continuous on D; hencef ∈ C⊥(D). However,f is not smooth (it is nondifferentiable at zero), sof 6∈ C⊥∞(D). –1 –0.5 0 0.5 1 x –1 0 y 0 0.2 0.4 0.6 0.8 1 z –1 –0.5 0 0.5 1 –1 0 0 0.5 1 1.5 2 (A) (B)
Figure 5C.4: (A) f(r, θ) = (1−r2)2 satisfies homogeneous Neumann boundary conditions
on the disk, and is smooth everywhere. (B)f(r, θ) = (1 + cos(θ)2)·(1−(1−r2)4) does
notsatisfy homogeneous Neumann boundary conditions on the disk, and is not constant on the
boundary.
2. Let f :D−→ R be the “bell” of Figure 5C.4(A), defined: f(r, θ) := (1−r2)2. Then∂⊥f ≡0 on the boundary of the disk, andf is smooth everywhere on D, sof ∈ C∞
⊥(D).
3. Let f : D −→ R be the “flower vase” of Figure 5C.4(B), defined
f(r, θ) := (1 + cos(θ)2)·(1−(1−r2)4). Then ∂⊥f ≡0 on the bound- ary of the disk, and f is smooth everywhere on D, so f ∈ C∞
⊥(D).
Note that, in this case, the angular derivative is nonzero, so f is not constant on the boundary of the disk.
(d) LetX= [0, π]×[0, π] be the square of sidelengthπ.
1. For any (n, m) ∈ N2, let Cn,m(x, y) = cos(nx)·cos(my) (see Fig- ure 9A.2 on page 181). ThenCn,m ∈ C⊥∞(X).
4
2. If f(x) = 5 cos(x) cos(2y)−3 cos(2x) cos(7y) + 7 cos(3x) cos(y), then
f ∈ C∞
⊥(X). More generally, any finite sum
N X n=1 M X m=1 An,mCn,m(x) (for some constantsAn,m) is inC⊥∞(X). 3. More generally, iff = ∞ X n,m=0
An,mCn,m is a uniformly convergenttwo
dimensional Fourier cosine series5, and the derivative series
∂xf(x, y) = − ∞ X n,m=0 nAn,msin(nx)·cos(my) ∂yf(x, y) = − ∞ X n,m=0 mAn,mcos(nx)·sin(my)
arealsouniformly convergent, thenf ∈ C⊥∞(X).
Exercise 5C.3 Verify examples (b) to (d) ♦
E
Arbitrary nonhomogeneous Neumann Boundary conditionsare imposed by fixing a functionb:∂X−→R, and then requiring
∂⊥u(s) = b(s) for alls∈∂X. (5C.6) For example, the classical Neumann Problem is to find a continuously dif- ferentiable functionu :X−→ Rsatisfying the Neumann condition (5C.6), such thatu also satisfies Laplace’s Equation: 4u(x) = 0 for allx∈int(X).
Physical Interpretations.
Thermodynamic. Here u represents a temperature distribution, or the con- centration of some diffusing material. Recall that Fourier’s Law (§ 1A on page 3) says that ∇u(s) is the flux of heat (or material) at s. Thus, for anys ∈ ∂X, the derivative ∂
⊥u(s) is the flux of heat/material across
the boundary at s. The nonhomogeneous Neumann Boundary condition
∂⊥u(s) =b(s) means that heat (or material) is being ‘pumped’ across the boundary at a constant rate described by the functionb(s).
Electrostatic. Here,urepresents an electric potential. Thus ∇u(s) is the elec- tric fieldats. Nonhomogeneous Neumann boundary conditions mean that the field vector perpendicular to the boundary is determined by the func- tionb(s).
5
5C(iii) Mixed (or Robin) boundary conditions
These are a combination of Dirichlet and Neumann-type conditions obtained as follows: Fix functions b : ∂X −→ R, and h, h⊥ : ∂X −→ R. Then (h, h⊥, b)-
mixed boundary conditionsare given:
h(s)·u(s) + h⊥(s)·∂⊥u(s) = b(x) for alls∈∂X. (5C.7) For example:
• Dirichlet Conditions corresponds toh≡1 andh⊥ ≡0. • Neumann Conditionscorresponds to h≡0 andh⊥ ≡1. • Noboundary conditions corresponds to h≡h⊥ ≡0. • Newton’s Law of Coolingreads:
∂⊥u = c·(u−TE) (5C.8)
This describes a situation where the boundary is an imperfect conductor (with conductivity constant c), and is immersed in a bath with ambient temperature TE. Thus, heat leaks in or out of the boundary at a rate