Math 3I03 – Fall 2009
Lecture Notes
September 10
thName: Bernardo Galv˜ao-Sousa Office: Hamilton Hall 204 Phone: x23419
Email: [email protected]
Office Hours: Tuesday 13:30-15:00 or by appointment
Course Website: http://www.math.mcmaster.ca/bsousa/math3i3
I’ll post the announcements on the website.
No Homework, but some Practice Problems. To pass the course you’ll have to practice a lot, there’s no real trick to pass the course other than practice.
Practice Problems are only a suggestion, you can try to do all the problems in the sections we cover. If you have questions about the lectures or the problems, contact me.
2.2
Review
Definition. An equation relating an unknown function of several variables with some of its partial derivatives
is called a partial differential equation (PDE).
Example (Heat Equation). u=u(t, x)
∂u ∂t =k
∂2u ∂x2
Proposition 2.1. A linear operator Lsatisfies
L(c1u1+c2u2) =c1L(u1) +c2L(u2),
for any two functionsu1, u2 and constants c1, c2.
Example.
Definition. A linear partial differential equation is an equation of the form:
L(u) =f,
whereLis a linear operator and f a known function.
Example.
• The heat equation is linear: ∂u∂t =k∂x∂2u2 +f(x, t) withL=
∂ ∂t−k
∂2
∂x2;
• ∂u ∂t =k
∂2u
∂x2 +α(x, t)u+f(x, t) is linear;
• ∂∂x2u2 +
∂2u
∂y2 = 0 is linear;
• ∂u∂t =k ∂2u
∂x2 +α(x, t)u4 is nonlinear;
• ∂u ∂t +u
∂u ∂x =
∂3u
∂x3 is nonlinear.
Definition. A PDE is called linear homogeneous iff = 0, that is of the formL(u) = 0.
A simpler definition of a linear homogeneous PDE, is a linear PDE which admitsu≡0 as a solution.
A very important property of linear homogeneous PDEs is:
Proposition 2.2 (Principle of Superposition). If u1 and u2 solve a linear homogeneous PDE, then a linear combination of themc1u1+c2u2 for arbitrary constantsc1, c2, also solves the same PDE.
Proof. Let the linear homogeneous PDE be written asL(u) = 0 for some linear operatorL. Then we know that
L(u1) = 0 andL(u2) = 0. By the linearity ofL, we then have
L(c1u1+c2u2) =c1L(u1) | {z }
=0
+c2L(u2) | {z }
=0
= 0.
A PDE may have many solutions: the PDE
∂u ∂t +
∂u
∂x = 0, t >0, x∈R, (Transport Equation)
is solved byu(x, t) =f(x−t) for any differentiable functionf of 1-variable.
We need boundary/initial conditions to be able to solve the PDE uniquely: by adding the initial condition
u(x,0) =g(x), now there is only one solution: u(x, t) =g(x−t) whereg is given by the initial condition.
Remark. Linearity and homogeneity also applies to boundary and initial conditions:
• u(0, x) =f(x) is a linear, but nonhomogeneous initial condition;
• ∂u
∂x(L, t) =g(t) is a linear but nonhomogeneous boundary condition;
• ∂u
∂x(0, t) = 0 is a linear and homogeneous boundary condition;
• −∂u
∂x(L, t) =u(L, t)−g(t) is a linear but nonhomogeneous boundary condition;
• ∂u
September 14
th1
Heat Equation
1.2
Derivation of the conduction of heat in a one-dimensional rod
First, we consider a rod of constant cross-sectional areaAoriented in thexdirection fromx= 0 tox=L.
x
x= 0 x=L
We assume that all quantities are constant across a section, so that for all practical reason the rod is one-dimensional. The simplest way to do this is by insulating perfectly the lateral surface of the rod.
Heat energy. Now we introduce the thermal energy density:
e(x, t) = thermal energy density at pointxand timet.
The rod is not uniformly heated, so the thermal energy density varies from one cross section to another. Consider ∆xinfinitesimally small. Then we approximatee(x, t) as a constant throughout the volume in the figure below.
x
x= 0 x x+ ∆x x=L
Theheat energyis then
heat energy =e(x, t)A∆x,
because the volume of the shaded section isA∆x.
Conservation of heat energy. The heat energy betweenxandx+ ∆xchanges in time only due to the heat
energy flowing through the edges and to the heat energy generated inside:
rate of change heat energy flowing of heat energy = across boundaries +
in time per unit time
heat energy generated
inside per unit time (#)
Because inside the shaded section, we consider the heat energy constant inx, we know that
∂ ∂t
e(x, t)A∆x
= rate of change of heat
Heat flux. Since the rod is one-dimensional, the thermal energy flows to the right or left only. We now
introduce theheat flux:
φ(x, t) =
amount of thermal energy per unit time flowing to theright per unit surface area
Ifφ(x, t)<0, then the heat energy is flowing to the left. We now know that
φ(x, t)−φ(x+ ∆x, t)
A= heat energy flowing across
the boundaries per unit time. (#φ)
Heat sources. We also consider internal sources of heat energy:
Q(x, t) = heat energy generated per unit volume per unit time.
Q(x, t)A∆x= heat energy generated inside per unit time. (#Q)
Conservation of heat energy (thin slice). We put the (#), (#t), (#φ) and (#Q) together to deduce
∂ ∂t
e(x, t)A∆x
≈
φ(x, t)−φ(x+ ∆x, t)
A+Q(x, t)A∆x.
We don’t get equality because we are approximating these quantities by constants in the thin slice. Divide by
A∆xand take the limit as ∆x→0+ to obtain an exact equation: ∂e
∂t =∆xlim→0+
φ(x, t)−φ(x+ ∆x, t)
∆x +Q(x, t)
which can be simplified to
∂e ∂t =−
∂φ
∂x +Q. (?)
This equation now is exact because when taking the limit ∆x→ 0+ we are evaluating these quantities at a
pointx, so there is no approximation now.
Conservation of heat energy (exact). Another way to derive , which avoids the approximations above
involves using integrals. Consider a finite segment of the rod, betweenx=aandx=b. Then (#) becomes
Z b
a
∂e
∂t(x, t)dx= d dt
Z b
a
e(x, t)dx=φ(a, t)−φ(b, t) + Z b
a
Q(x, t)dx (Rb
a)
We can write every term as an integral in (a, b):
Z b
a
∂e
∂t(x, t)dx=
Z b
a
−∂φ∂x(x, t) +Q(x, t)
dx,
so we deduce the equation
Z b
a
∂e
∂t + ∂φ ∂x−Q
dx= 0.
Since this is valid for arbitraryaandb, then the integrand itself must be 0 and we recover
∂e ∂t =−
∂φ
Remark. The negative sign in front of ∂φ∂x can be explained with an example: if ∂φ∂x >0, then φis higher on the right than on the left, which means that more heat is flowing (to the right) atx=bthan atx=a, so more heat is escaping than coming in. This means that the heat energy should decrease, hence the negative sign.
x
x=a x=b
0 L
φ(a, t) φ(b, t)
Temperature and specific heat. Here we introduce thetemperature:
u(x, t) = temperature.
Temperature and thermal energy density are different concepts. This can be understood by the fact that for different materials, it takes different amounts of thermal energy to raise the temperature from one value to another. So we introduce the notion ofspecific heat:
c= heat energy necessary to raise the temperature of a unit mass by one unit.
Remark.
• The specific heat depends on the temperature of the rod itself: c = c(u). However, this assumption complicates the problems considerably. If we restrict the model to a narrow range of temperatures, then it is reasonable to assume that the specific heat is independent of the temperature.
September 15
thThermal energy. The thermal energy can also be defined as the energy it takes to raise the temperature of
a unit of volume from a reference temperature 0oC to its actual temperature u(x, t). We know that the heat
energy per unit mass isc(x)u(x, t), so we introduce the mass density:
ρ(x) = mass per unit volume.
Then, the thermal energy per unit volume is given by:
e(x, t) =ρ(x)c(x)u(x, t).
The conservation of heat energy (1.2) becomes
c(x)ρ(x)∂u
∂t =− ∂φ
∂x +Q. (?)
Fourier law. In the previous equation, we still have two unknowns: u(x, t) andφ(x, t). What is the heat flux
φ(x, t)? How does energy flow?
What we are looking for is the dependence of the heat fluxφ(x, t) on the temperature u(x, t). We know the following properties about heat flux:
1. Heat flows from hotter to colder;
2. The greater the temperature difference, the greater the flow;
3. The flow of heat energy varies depending on the material, even with the same temperature differences.
Fourier summarized these properties in the simplest possible way with the equation:
φ=−K0∂u
∂x (Fourier law of heat conduction)
The termK0is called thethermal conductivity, which depends on the material, thusK0=K0(x). Also, the conductivity changes with the temperature, so we should haveK0 =K0(x, u), but as before, this complicates the equations, and as long as we restrict the model to narrow band of temperatures, we can assume that
K0=K0(x) doesn’t depend on the temperatureu(x, t).
Heat Equation. Putting all this together, we end up with the equation
cρ∂u ∂t =
∂ ∂x
K0∂u ∂x
+Q, (?)
where the source of internal heat energyQis given, and c, ρ, K0 depend onx. In the special case of a uniform rod, these are constants and we have:
cρ∂u ∂t =K0
∂2u ∂x2 +Q.
Moreover, if there are no sourcesQ= 0, then the PDE becomes
∂u ∂t =k
∂2u
∂x2, (Diffusion Equation)
Initial/boundary conditions. To solve these equations, that is to make sure there is a unique solution, we
need extra conditions. As is expected, to predict the temperature distribution of a rod, we need to know what is the initial temperature distribution, so we must be given
u(x,0) =f(x) (InitialCondition)
Moreover, we also need to know what is happening at the edges of the rodx= 0 andx=L, how the temperature is flowing in/out through the edges.
Remark. If the rod is infinite, then the initial condition is sufficient.
1.3
Boundary conditions
There are several kinds of boundary conditions, depending on the mechanism that affects each edge. It may depend on the conditions immediately inside and outside the rod. We always assume that the environment outside the rod is known and not altered by the rod.
Prescribed temperature. Sometimes the temperature at the end of the rod is fixed:
u(0, t) =uB(t), (BCD)
whereuB(t) is given. This kind of condition is also known as a Dirichlet Boundary Condition.
Insulated boundary. We may also prescribe the heat flux across the edge:
−K0(0)∂u
∂x(0, t) =φ(t), (BCN)
whereφ(t) is given. This kind of condition is also known as a Neumann Boundary Condition. The simplest of this kind of boundary condition is when the rod is perfectly insulated:
∂u
∂x(0, t) = 0.
Newton’s Law of cooling. When the rod is in contact at the boundary with a moving fluid (e.g., air), then
the movement of the fluid will carry the heat into/away from the rod. This process is calledconvection. And the boundary condition is
−K0(0)∂u
∂x(0, t) =−H
u(0, t)−uB(t)
, (BCR)
whereH (the heat transfer coefficient) anduB(t) are given. This kind of condition is known asNewton’s law
of coolingor a Robin Boundary Condition.
Remark. The minus sign on the right-hand side is explained through a simple example. On one hand, the
left-hand side is the heat flux (flowing to the right when positive and to the left when negative). If the temperature of the rod is hotter than the fluid outside,u(0, t)> uB(t), then heat should flow to the outside, which is to the
left. In fact, because of the minus sign, the heat flux−K0(0)∂u
∂x(0, t)<0, which means that heat is flowing to
the left.
Analogously, for the right edge we have
−K0(L)∂u
∂x(L, t) =H
u(L, t)−uB(t)
2
Method of Separation of Variables
2.3
Heat Equation with frozen ends
2.3.1 Introduction
First, we assume that all the coefficients in (1.2) are constant, i.e., that the rod is made of a uniform material. The heat equation becomes:
∂u ∂t =k
∂2u ∂x2 +
Q(x, t)
cρ .
This equation is linear ifQ(x, t) = 0, so we will assume this too:
∂u ∂t =k
∂2u
∂x2. (PDE)
The rod has frozen ends, so the temperature at the ends is prescribed as 0, so we have the boundary conditions:
u(0, t) =u(L, T) = 0, (BC)
and we have an initial condition
u(x,0) =f(x), (IC)
September 17
th2.3.2 Separation of Variables
In this method, we attempt to find solutions where the dependence in timetand spacexare separable:
u(x, t) =φ(x)G(t),
whereφ(x) is a function only ofxandG(t) is only a function oft.
Becauseusatisfies (PDE) with the conditions (BC) and (IC). At first, we will ignore the initial condition (IC). Later we will see how to satisfy the initial conditions.
With this assumption onu(x, t), we compute its partial derivatives:
∂u
∂t =φ(x)G
0(t),
∂2u ∂x2 =φ
00(x)G(t),
so (PDE) becomes
φ(x)G0(t) =kφ00(x)G(t).
Notice that we can separate the variablest andxby dividing both sides by kφ(X)G(t):
G0(t)
kG(t) =
φ00(x)
φ(x) .
Now the left-hand side depends only ontand the right-hand side depends only onx. How can a function of time equal a function of space for all time and space?The only possible way is if both sides are equal and constant:
G0(t)
kG(t) =
φ00(x)
φ(x) =−λ,
for some constant λ (the minus sign is just for convenience). So this equation now splits into two separate equations of one-variable:
φ00(x) =−λφ(x), (ODEx)
G0(t) =−λkG(t). (ODEt)
The product solution must also satisfy the boundary conditions:
u(0, t) =φ(0)G(t) = 0,
u(L, t) =φ(L)G(t) = 0.
IfG(t)≡0, then the solution isu(x, t)≡0 which is called thetrivial solution that we knew existed because the problem is homogeneous. To obtain more interesting solutions, we assume instead that
2.3.3 Time-dependent function
Here we solve the differential equation (ODEt). This is a first-order linear homogeneous differential equation
with constant coefficients, so we can easily solve it with an exponential solution:
G(t) =Ce−kλt,
for some constantC.
Remark.
1. Ifλ >0, then the solution will decay exponentially;
2. Ifλ= 0, then the solution is constant;
3. Ifλ <0, then the solution will grow exponentially. This case means that the temperature of the rod will increase (decrease) exponentially, which is not an expected behavior for the heat distribution of a rod;
4. As we will find out, only some values ofλare allowed, so that in factλ>0;
5. This is the reason why we introduced the minus sign withλ.
2.3.4 Boundary-value problem
We now solve (ODEx) with the boundary conditions (BCx). We will see, that for most values ofλ, only trivial
solutions exist, so the values of λ for which we can find nontrivial solutions are called eigenvalues of the boundary-value problem and the nontrivial solutions are calledeigenfunctions.
This is a second-order linear homogeneous differential equation, so we look for exponential solutionsφ(x) =erx
and we deduce thatris a root of the characteristic polynomial:
r2=−λ.
We now have 3 cases:
Case 1: λ >0
In this case,r=±i√λ, which is complex, so the solution takes the form
φ(x) =C1cos √λx
+C2sin √λx
.
Applying the boundary conditions, we have:
0 =φ(0) =C1,
0 =φ(L) =C2sin √λL
.
So eitherC2= 0 or sin √λL
= 0. IfC2= 0, then we obtain a trivial solution, so the nontrivial solution must satisfy:
which implies that√λL=nπ for some positive integer n. We then deduce that we have
λ=nπ
L
2
, (Eigenvalues)
φ(x) = sinnπx
L
, (Eigenfunctions)
forn= 1,2,3, . . .. We putC2= 1, because we are only looking for one eigenfunction – since it is a solution of a linear homogeneous differential equation, it can always be multiplied by a constant.
Case 2: λ= 0
In this case,r= 0 is a double root, so the solution is
φ(x) =C1+C2x,
which using the boundary conditions becomes
φ(x)≡0.
September 21
stCase 3: λ <0
In this case, let µ=−λ >0, and we haver=±√−λ=±√µ, so the solution is
φ(x) =C1e−√µx+C2e√µx
Using the boundary conditions we obtain
0 =φ(0) =C1+C2
0 =φ(L) =C1e−√µL+C2e√µL →
C1=−C2
−C2e−√µL =−C2e√µL →
C1=−C2
C2= 0 ore−√µL=e√µL
Becauseµ, L >0, we know thate−√µL
6
=eõL, so we conclude thatC1=C2= 0 and we only obtain the
trivial solution. So again,λ <0 is not an eigenvalue.
Summary. The nontrivial solutions of (ODEx) and (BCx):
φ00(x) =−λφ(x)
φ(0) = 0
φ(L) = 0
are found for the eigenvalues and eigenfunction:
λn=
nπ
L
2
, (Eigenvalues)
φn(x) = sin
nπx
L
, (Eigenfunctions)
forn= 1,2,3, . . ..
2.3.5 Principle of superposition
From 2.3.3 and 2.3.4, we deduced an infinite number of product solutions of the heat equation (PDE) satisfying the boundary conditions (BC) corresponding to specific values ofλ >0. The solutions are:
un(x, t) =Bnsin
nπx L e
−k(nπ L)
2
t
forn= 1,2,3, . . .
Using the superposition principle, we obtain we may add all these solutions to obtain a general solution of the heat equation satisfying the boundary conditions:
u(x, t) = ∞
X
n=1 Bnsin
nπx L e
−k(nπ L)
2
t.
We now see that this solution satisfies the initial condition:
u(x,0) =f(x) = ∞
X
n=1 Bnsin
Example. We can already solve some simple problems. Solve the problem:
∂u ∂t =k
∂2u
∂x2, (PDE)
u(0, t) =u(L, t) = 0, (BC)
u(x,0) = 2 sin3πx
L −sin
5πx
L . (IC)
Solution. Using the formula above, we know that the solution can be written as
u(x, t) = ∞
X
n=1 Bnsin
nπx L e
−k(nπ L)
2
t,
and we know that the initial condition is
2 sin3πx
L −sin
5πx L =
∞
X
n=1
Bnsinnπx
L
withBn = 0 forn6= 3,5 andB3= 2, B5=−1.
We conclude that the solution is:
u(x, t) = 2 sin3πx
L e
−k(3π L)
2
t
−sin5πx
L e
−k(5π L)
2
t.
In fact, we can find solution of the heat equation for any initial condition that can be written as:
f(x) = ∞
X
n=1 Bnsin
nπx L .
This expansion of the functionf(x) is called aFourier series expansion.
Then, the initial condition determines the values of the constants Bn which we use to find the corresponding
solution:
u(x, t) = ∞
X
n=1
Bnsinnπx
L e
−k(nπ L)
2
t.
2.3.6 Orthogonality of Sines
Given an initial condition
u(x,0) =f(x).
How do we find the coefficientsBn such that
f(x) = ∞
X
n=1 Bnsin
nπx
L for 06x6L?
Recall that
1. We can define an inner product
(f, g) = Z L
0
2. And a norm:
kfk=p
(f, f) = Z L
0
f2(x)dx
!12 ;
3. Two functionsf, g are orthogonal if (f, g) = 0, i.e.
Z L
0
f(x)g(x)dx= 0;
4. A set of functions{φn(x)}n=1,2,... is orthogonal if
(φn, φm) = 0 for alln6=m;
5. An orthogonal set of functions{φn(x)}n=1,2,...is complete if for anyf piecewise smooth,
(f, φn) = 0 for alln= 1,2, . . . ⇒ f ≡0;
6. If{φn(x)}n=1,2,... is complete, then any piecewise smooth functionf(x) can be written as
f(x) = ∞
X
n=1
bnφn(x) with bn=
(f, φn)
kφnk2
;
Now we apply this theory to our case. First note that
{sinnπx
L }n=1,2,... is complete in [0, L]
andksinnπx L k=
q
L
2.
So, from 6, we know that any piecewise smooth functionf(x) can be written as
f(x) = ∞
X
n=1
Bnsinnπx
L with Bn=
2
L
Z L
0
f(x) sinnπx
September 22
nd2.3.7 Example
In this example, immerse a rod perfectly insulated on the lateral surface in a bath of boiling water (100oC) and
let it sit for a long time. We expect the temperature on the whole rod to be at 100oC. Then suddenly immerse
the rod in a bath of ice water (0oC). Follow the temperature distribution after the second immersion.
The mathematical problem is
∂u ∂t =k
∂2u
∂x2 t >0,0< x < L (PDE)
u(0, t) =u(L, t) = 0 t >0 (BC)
u(x,0) = 100 0< x < L (IC)
According to what was done before, the solution is
u(x, t) = ∞
X
n=1
Bnsinnπx
L e
−k(nπ L)
2
t,
where
Bn=
2
L
Z L
0
f(x) sinnπx
L dx,
andf(x) = 100.
We now compute the coefficientsBn:
Bn =
200
L
Z L
0
sinnπx
L dx=
200
L
−nπL cosnπx
L
L
0
=200
nπ (1−cosnπ) =
200
nπ (1−(−1)
n) =
0 neven
400
nπ nodd
Notice thatf(x) = 100 for 0< x < Landf(0) =f(L) = 0. This is due to the boundary conditions and to the fact that the series
∞
X
n=1 Bnsin
nπx L
is actually equal to
100
L 2L
−L
−2L
The “wiggly” graph is an approximation of the series up to n= 15. That’s why the whole series can be 100 inside the interval (0, L) and 0 at the endpoints.
The graph of the solution is:
2.4
Worked Examples with the Heat Equation
2.4.1 Heat conduction in a rod with insulated ends
We will now work out the following mathematical problem:
∂u ∂t =k
∂2u
∂x2 (PDE)
∂u
∂x(0, t) = 0 (BC1)
∂u
∂x(L, t) = 0 (BC2)
u(x,0) =f(x) (IC)
fort>0 and 06x6L.
Apply the method of separation of variables: u(x, t) =φ(x)G(t), which implies the ODEs:
G0(t) =−kλG(t) (ODEt)
φ00(x) =−λφ(x) (ODEx)
whereλ= separation constant. Then,
G(t) =e−λkt
and the boundary conditions imply
September 24
thWe now solve the problem inx. The characteristic polynomial is the same: r2=
−λ, so once again we have 3 cases:
λ >0: The general solution is:
φ(x) =C1cos√λx+C2sin√λx
so
φ0(x) =−C1√λsin√λx+C2√λcos√λx
and the boundary conditions imply
0 =φ0(0) =C2√λ
0 =φ0(L) =−C1√λsin√λL
So the eigenvalues are the same:
λn =
nπ
L
2
(Eigenvalues)
φn(x) = cos
nπx
L (Eigenfunctions)
forn= 1,2,3, . . .
λ= 0: The general solution is:
φ(x) =C1+C2x
so
φ0(x) =C2
and the boundary conditions implyC2= 0. So we have another eigenvalue are the same:
λ0= 0 (Eigenvalue)
φ0(x) = 1 (Eigenfunction)
λ <0: The general solution is
φ(x) =C1e−√µx+C2e√µx
where µ=−λ >0, so
φ0(x) =−C1√µe−√µx+C2√µe√µx
and the boundary conditions imply
0 =φ0(0) =
−C1√µ+C2√µ
0 =φ0(L) =
−C1√µe−√µL+C2√µe√µL →
C1=C2
C2= 0 ore−√µL=e√µL
Using the superposition principle, we conclude that the solution is:
u(x, t) =A0+ ∞
X
n=1
Ancosnπx
L e
−k(nπ L)
2
t.
The initial condition is now:
f(x) =u(x,0) =A0+ ∞
X
n=1 Ancos
nπx L .
As we did in 2.3.6, but applying to cosines, we note that
{1,cosnπxL }n=1,2,... is complete in [0, L]
andkcosnπxL k=qL2 ifn= 1,2, . . .andk1k=√L.
So, from 6 in 2.3.6, we know that any piecewise smooth functionf(x) can be written as
f(x) =A0 2 +
∞
X
n=1 Ancos
nπx
L with An=
2
L
Z L
0
f(x) cosnπx
L dx.
Note. We divideA0 by 2 so that it follows the same formula.
2.4.2 Heat conduction in a thin circular ring
Now we have a circular ring of length 2L as in the figure on the right. The trick in this example is to formulate the boundary conditions correctly. First, since the circumference of a circle is 2πr= 2L, the radius of the ring isr=L
π.
The edges of the ringx=−L andx=Lshould be touching, so the tempera-tures should match:
u(−L, t) =u(L, t). (BC1)
Moreover, heat flows freely betweenx=−Landx=L, so the heat flux should also match:
∂u
∂x(−L, t) = ∂u
∂x(L, t). (BC2)
x= 0
x=L
x=−L
The rest of the problem is known:
∂u ∂t =k
∂2u
∂x2 (PDE)
September 28
thApply the method of separation of variables: u(x, t) =φ(x)G(t), which implies the ODEs:
G0(t) =−kλG(t) (ODEt)
φ00(x) =−λφ(x) (ODEx)
whereλ= separation constant. Then,
G(t) =e−λkt
and the boundary conditions imply
φ(0) =φ(L) and φ0(0) =φ0(L). (BCx)
We now solve the problem inx. The characteristic polynomial is the same: r2=−λ, so once again we have 3
cases:
λ >0: The general solution is:
φ(x) =C1cos√λx+C2sin√λx
so
φ0(x) =
−C1√λsin√λx+C2√λcos√λx
and the boundary conditions imply
C1cos√λ(−L) +C2sin√λ(−L) =φ(−L) =φ(L) =C1cos√λL+C2sin√λL
−C1√λsin√λ(−L) +C2√λcos√λ(−L) =φ0(−L) =φ0(L) =−C1√λsin√λL+C2√λcos√λL
which is equivalent to
C2sin√λL= 0
C1√λsin√λL= 0
And we have the eigenvalues:
λn=
nπ
L
2
, n= 1,2, . . . (Eigenvalues)
For each eigenvalueλn, we have two (linearly independent) eigenfunctions:
φn(x) = cos
nπx L , sin
nπx
L (Eigenfunctions)
λ= 0: The general solution is:
φ(x) =C1+C2x
so the boundary conditions imply thatC1=C1+C2L, which means that C2= 0 and we have one more eigenvalue:
λ0= 0 (Eigenvalue)
with the eigenfunction:
φ0(x) = 1 (Eigenfunction)
Using the superposition principle, we conclude that the solution is:
u(x, t) =a0+ ∞
X
n=1
ancosnπx
L e
−k(nπ L) 2 t+ ∞ X n=1
bnsinnπx
L e
−k(nπ L)
2
t.
The initial condition is now:
f(x) =u(x,0) =a0+ ∞
X
n=1 ancos
nπx L +
∞
X
n=1 bnsin
nπx L .
As we did in 2.3.6, but applying to sines and cosines in the interval [−L, L], we note that
{1,sinnπx L ,cos
nπx
L }n=1,2,... is complete.
So, from 6 in 2.3.6, we know that any piecewise smooth functionf(x) can be written as
f(x) = a0 2 +
∞
X
n=1 ancos
nπx L +
∞
X
n=1 bnsin
nπx L .
with
an=
1
L
Z L
−L
f(x) cosnπx
L dx,
bn=
1
L
Z L
−L
f(x) sinnπx
L dx.
2.5
Laplace’s Equation
We didn’t see it, but the heat equation in two dimensions is:
∂u ∂t =k
∂2u ∂x2 +
∂2u ∂y2
=k∇2u=k∆u.
So if we are looking for the equilibrium distribution of heat, that’s when the temperatures don’t change with time, thus ∂u∂t = 0 and we have theLaplace equation:
∂2u ∂x2 +
∂2u
September 29
th2.5.1 Laplace’s equation in a rectangle
In a rectangle 06x6L, 06y6H, we have the following boundary conditions:
u(0, y) =g0(y) (BC1) u(L, y) =gL(y) (BC2) u(x,0) =f0(x) (BC3) u(x, H) =fH(x) (BC4)
wheref0(x), fH(x), g0(y), gL(y) are given functions.
Note. Although the PDE is linear and homogeneous, the boundary conditions are nonhomogeneous, so we
can’t apply the method of separation of variables in the usual form.
So we break the problem into 4 easier problems that can be solved using the method of separation of variables. We break the solution in the following form:
u(x, y) =u1(x, y) +u2(x, y) +u3(x, y) +u4(x, y)
where
∆u1= 0
u1(0, y) =g0(y)
u1(L, y) = 0
u1(x,0) = 0
u1(x, H) = 0
∆u2= 0
u2(0, y) = 0
u2(L, y) =gL(y)
u2(x,0) = 0
u2(x, H) = 0
∆u3= 0
u3(0, y) = 0
u3(L, y) = 0
u3(x,0) =f0(x)
u3(x, H) = 0
∆u4= 0
u4(0, y) = 0
u4(L, y) = 0
u4(x,0) = 0
u4(x, H) =fH(x)
We will work out the problem foru3 using the method of separation of variables:
u3(x, y) =φ(x)g(y).
As before, we first ignore the nonhomogeneous boundary condition (before we ignored the initial condition) and the other three homogeneous conditions yield:
φ(0) =φ(L) = 0 g(H) = 0.
We substitute the new form ofu3 into the PDE:
φ00(x)g(y) +φ(x)g00(y) = 0,
so we separate the variables by dividing everything byφ(x)g(y):
−φ 00(x)
φ(x) =
g00(y)
Now the right-hand side depends only ony while the left-hand side depends only onx. Then both must equal a separation constant
−φ 00(x)
φ(x) =
g00(y)
g(y) =λ.
We now use λ because the term with both boundary conditions is going to give us the eigenvalues, and like before, the exponentials won’t work, since you can’t have an exponential equal 0 at both endpoints, so we want a separation constantλ <0 to generate the exponentials. If we chooseλinstead of−λwe can do that, although we don’t assume thatλ>0.
We now have two problems to solve:
φ00(x) =
−λφ(x)
φ(0) =φ(L) = 0 and
g00(y) =λg(y)
g(H) = 0
The problem in xis a boundary problem, so we can use it to determine the eigenvaluesλ. The characteristic polynomial is: r2=
−λ, so once again we have 3 cases:
λ >0: The general solution is:
φ(x) =C1cos√λx+C2sin√λx
so
φ0(x) =−C1√λsin√λx+C2√λcos√λx
and the boundary conditions imply
0 =φ(0) =C1
0 =φ(L) =C1cos√λL+C2sin√λL
which is equivalent to
C1= 0
C2sin√λL= 0
And we have the eigenvalues:
λn=
nπ
L
2
, n= 1,2, . . . (Eigenvalues)
For each eigenvalueλn, we have the eigenfunction:
φn(x) = sin
nπx
L (Eigenfunctions)
As before, the casesλ= 0 and λ <0 don’t admit nontrivial solutions.
Now we solve the problem in y for each eigenvalue λn. The characteristic polynomial is r2 =λn >0, so we
obtain the solution:
g(y) =Ae−√λny+Be √
Using the conditiong(H) = 0 we obtain:
0 =g(H) =Ae−
√
λnH+Be √
λnH
→ Ae−
√
λnH= −Be
√
λnH= C 2
so the solution is
gn(y) =Ce
−√λn(y−H)−e √
λn(y−H)
2 =Csinh(
p
λn(y−H))
By the superposition principle, we conclude that the solution is:
u3(x, y) = ∞
X
n=1 ansin
nπx L sinh
nπ
L(y−H).
Now we introduce the “initial” condition intou3:
f0(x) =u3(x,0) =− ∞
X
n=1 ansin
nπx L sinh
nπH L =
∞
X
n=1
−ansinh
nπH L
sinnπx
L .
We know the formula for the coefficients:
−ansinh
nπH L =
2
L
Z L
0
f0(x) sinnπx
L dx,
so the coefficients are:
an=− 2
LsinhnπH L
Z L
0
f0(x) sinnπx
L dx.
October 1
st2.5.2 Laplace equation for a circular disk
Now we have a circular disk of radiusa as in the figure on the right. Assume that thermal properties are constant and there are no heat sources. The trick in this example is to formulate the problem in polar coordinates.
In polar coordinates (r, θ), we have
x=rcosθ, y=rsinθ,
r=px2+y2, θ= arctany x,
so
∂u ∂x =
∂u ∂r
∂r ∂x +
∂u ∂θ
∂θ ∂x =
∂u
∂rcos(θ)− ∂u ∂θ
sinθ r ∂u
∂y = ∂u ∂r
∂r ∂y +
∂u ∂θ
∂θ ∂y =
∂u
∂rsin(θ) + ∂u ∂θ
cosθ r ,
∆
u
= 0
a
u
(
a, θ
) =
f
(
θ
)
θ=±π
so continuing, we can deduce that
∆u= 1
r ∂ ∂r
r∂u ∂r
+ 1
r2 ∂2u
∂θ2 = 0 (PDE)
with boundary condition
u(a, θ) =f(θ). (BCr=a)
The domain of the problem is
06r6a, −π6θ6π,
so we need boundary conditions on all edges of the domain, and we only have conditions on one edge: r=a. We find the missing boundary conditions starting with r= 0. This “edge” reduces to the center point of the domain, where we have no condition, so we just assume that the solutionu(r, θ) is bounded there:
|u(0, θ)|<∞ (BCr=0)
The “edge” θ=±πis also in the interior of the domain, but in this case, we observe that both edges are the same line, so the solution and its flux across the lineθ=±πmust be the same:
u(r,−π) =u(r, π)
∂u
∂θ(r,−π) = ∂u ∂θ(r, π).
(BCθ)
These kind of conditions are called periodicity conditions.
Now we have 4 boundary conditions, and only one of these conditions is nonhomogeneous, so we don’t even have to split our solution in different parts as before. Using the method of separation of variables, we assume
u(r, θ) =φ(θ)G(r),
From (BCθ), we deduce that
φ(−π) =φ(π) and φ0(−π) =φ0(π)
The (PDE) becomes:
1
r
rG0(r)0
φ(θ) + 1
r2G(r)φ
00(θ) = 0.
We want to separate the terms inr from the terms inθ, so we divide by G(rr)2φ(θ) and we obtain r
G(r)
rG0(r)0
=− 1
φ(θ)φ
00(θ) =λ.
We obtain two problems:
φ00=−λφ
φ(−π) =φ(π)
φ0(−π) =φ0(π).
(BVPθ)
and
r
rG0(r)0
=λG(r)
|G(0)|<∞. (IVPr)
The first problem inθ is the same problem solved in 2.4.2 withL=π, and the eigenvalues are
λn=
nπ
L
2
=n2, n= 0,1,2, . . . (Eigenvalues)
with the corresponding eigenfunctions
φ0(θ) = 1
φn(θ) = cosnθ and sinnθ.
(Eigenfunctions)
For the problem inr, we expand the derivative on the LHS:
r2G00+rG0−n2G= 0 (DEr)
This is a Cauchy-Euler equation. Forn= 1,2, . . ., the characteristic equation is
p(p−1) +p−n2= 0
p2=n2,
so the solutions are
Gn(r) =C1r−n+C2rn
The initial condition, implies thatC1= 0, so we obtain
Gn(r) =rn.
For the eigenvalueλ0= 0, we have the solution
G0(r) =C1+C2lnr
and the initial condition implies thatC2= 0, so we have
Using the superposition principle, we obtain the solution
u(r, θ) =A0 2 +
∞
X
n=1
Anrncosnθ+
∞
X
n=1
Bnrnsinnθ.
Now we are ready to introduce the condition (BCr=a):
f(θ) =u(a, θ) =A0 2 +
∞
X
n=1
Anancosnθ+
∞
X
n=1
Bnansinnθ,
and we already know the form of the coefficients:
Anan=
1
π
Z π
−π
f(θ) cosnθ dθ
Bnan=
1
π
Z π
−π
October 5
th2.5.4 Qualitative properties of the Laplace’s equation
We now study some properties of the solutions.
Mean Value Theorem. In the previous example, we deduced that
u(0, θ) = A0 2 =
1 2π
Z π
−π
f(θ)dθ= 1 2π
Z π
−π
u(a, θ)dθ,
i.e., the temperature at the center of the circle is the average of the temperatures at the perimeter of the circle.
Now, assume a domain where we want to solve the Laplace equation. Take any pointP inside that domain, as in the figure. Take a radiusa >0 such that the circle centered atP with radiusais still inside the domain.
Then we can solve the same problem in 2.5.2 in polar coordinates centered at
P withf(θ) =u(a, θ). We deduce that
u(P) = 1 2π
Z π
−π
u(a, θ)dθ,
and so the same property holds for anyP andain the conditions above.
a
u(a, θ) =f(θ)
P
Maximum Principle. The previous property allows us to easily prove a maximum principle:
In steady state, the temperature cannot attain its maximum (and minimum) in the interior.
Proof. To prove this property, assume by contradiction that the maximum temperature is attained at a point P in the interior. Then, the mean value theorem states that this temperature is the average of the temperatures on any circle still in the inside of the domain. Since the temperature on the circle can’t be higher than the temperature at P, the only way that the average can be the maximum is if the temperature on the whole circle is constant and equal to the maximum. Then we can use the points in the circle to get more circles until we cover the whole domain.
This proves that if the maximum is attained in the interior, then the temper-ature is constant throughout the domain.
P
Wellposedness and Uniqueness. Using the maximum principle, we can prove that the Laplace equation is
wellposed, i.e.,
small changes in data → small changes in the solution.
Proof. Suppose that we have the Laplace equation with boundary dataf(x, y)
∆u= 0
Assume also that we have the same problem with boundary data g(x, y) where g is very close to f on the boundary:
∆v= 0
v=g(x, y) on the boundary.
We want to prove thatuandv are very similar.
Takew=u−v, so thatwsolves
∆w= 0
w=f(x, y)−g(x, y) on the boundary.
The maximum (and minimum) principles state that the maximum (and minimum) ofwoccur on the boundary:
min(f−g)6w6max(f−g).
This proves that the result.
We can also use the maximum principle to prove uniqueness of solution.
Proof. Letube solve the Laplace equation with boundary dataf(x, y):
∆u= 0
u=f(x, y) on the boundary,
and letvbe another solution of the same problem:
∆v= 0
v=f(x, y) on the boundary.
Then as before, letw=u−v, which satisfies
∆w= 0
w= 0 on the boundary,
so by the maximum principle:
0 = min on boundary6w6max on boundary = 0,
October 6
thSolvability condition. When we prescribe the heat flow on the boundary instead of the temperature, the
equation might have no solutions. The reason is that the prescribed heat flow must satisfy asolvability con-dition.
To find this condition, integrate ∆u= 0 in the domain:
0 = Z Z
∆u dx dy= Z Z
∇ ·(∇u)dx dy.
Apply the divergence theorem to deduce
0 = I
∇u·n ds. (SolvabilityCondition)
This last term∇u·nis proportional to the heat flow, so it implies that thenet heat flow must be000.
Note. nis the vector normal to the boundary pointing outwards, called the outwards unit normal.
3
Fourier Series
3.1
Introduction
Definition. A functionf(x) is called piecewise smooth if it is possible to divide the domain in a finite number
of pieces where bothf(x) and its derivativef0(x) are continuous.
All functions in this chapter are assumed to be piecewise smooth unless noted otherwise.
Definition. Given a functionf(x), its Fourier series expansion in the interval [−L, L] is
FS(f)(x) = a0 2 +
∞
X
n=1
ancosnπx
L +
∞
X
n=1
bnsinnπx
L
where the Fourier coefficients are
an= 1
L
Z L
−L
f(x) cosnπx
L dx,
bn=
1
L
Z L
−L
f(x) sinnπx
L dx,
and we write
f(x)∼ a0 2 +
∞
X
n=1 ancos
nπx L +
∞
X
n=1 bnsin
nπx L ,
3.2
Convergence
Theorem 3.1 (Fourier’s Theorem). If f(x) is piecewise smooth in [−L, L], then the Fourier series of f(x)
converges
1. to the periodic extension off(x)where it is periodic; 2. to the average of the jump:
FS(f)(x) =f(x
−) +f(x+)
2
where the periodic extension is discontinuous.
This means that
f(x) = a0 2 +
∞
X
n=1 ancos
nπx L +
∞
X
n=1 bnsin
nπx L ,
for all−L < x < Lwhere the functionf(x) is continuous.
To sketch the Fourier series of a function, follow the steps:
1. Sketchf(x) for−L < x < Lwheref(x) is continuous;
2. Sketch its 2L-periodic extension;
3. Mark the average of all the points where there are
jumps. -10 -7.5 -5 -2.5 0 2.5 5 7.5 10
-2.5 2.5 5 7.5
Continuous Fourier Series. The Fourier series is continuous and converges to f(x) if and only if f(x) is
October 8
th3.3
Cosine and Sine Series
3.3.1 Fourier Sine Series
Odd functions. An odd function is a function that satisfies the property: f(−x) =−f(x) for allx.
The Fourier series of an odd function in [−L, L] is
FS(f)(x) = a0 2 +
∞
X
n=1 ancos
nπx L +
∞
X
n=1 bnsin
nπx L
with
an=
1
L
Z L
−L
f(x) cosnπx
L dx= 0,
bn=
1
L
Z L
−L
f(x) sinnπx
L dx=
2
L
Z L
0
f(x) sinnπx
L dx,
so we can simplify to:
FS(f)(x) = ∞
X
n=1 bnsin
nπx
L with bn =
2
L
Z L
0
f(x) sinnπx
L dx.
Note. For odd functions, we only need to know the function in [0, L].
Fourier sine series. If the domain is only [0, L], like in the case of the heat conduction in a one-dimensional
rod with frozen ends, it might be convenient to obtain a sine series expansion. In that same example, the method of separation of variables forced us to use a sine expansion for the initial condition f(x), which was likely not odd. Because we don’t care about the value of the initial condition outside the rod, we can extend
f(x) to−L6x6Las an odd function ˜f(x) and use Fourier’s Theorem to write
FS( ˜f)(x) = ∞
X
n=1 bnsin
nπx L
But for 06x6L, the functions ˜f =f(x), so
FS(f)(x) = ∞
X
n=1 bnsin
nπx
L for 06x6L. (Fourier sine series)
To sketch the Fourier sine series of a function, follow the steps:
1. Sketchf(x) for 0< x < Lwheref(x) is continuous;
2. Sketch the odd extension ˜f(x) for−L < x <0;
3. Sketch its 2L-periodic extension;
4. Mark the average of all the points where there are jumps.
p 2p
3.3.2 Fourier Cosine Series
Even functions. An even function is a function that satisfies the property: f(−x) =f(x) for all x.
The Fourier series of an even function in [−L, L] is
FS(f)(x) = a0 2 +
∞
X
n=1
ancosnπx
L +
∞
X
n=1
bnsinnπx
L
with
an = 1
L
Z L
−L
f(x) cosnπx
L dx=
2
L
Z L
0
f(x) cosnπx
L dx,
bn =
1
L
Z L
−L
f(x) sinnπx
L dx= 0,
so we can simplify to:
FS(f)(x) = a0 2 +
∞
X
n=1 ancos
nπx
L with an=
2
L
Z L
0
f(x) cosnπx
L dx.
Note. For even functions, we only need to know the function in [0, L].
Fourier cosine series. If the domain is only [0, L], like in the case of the heat conduction in a one-dimensional
rod with insulated ends, it might be convenient to obtain a cosine series expansion. In that same example, the method of separation of variables forced us to use a cosine expansion for the initial condition f(x), which was likely not even. Because we don’t care about the value of the initial condition outside the rod, we can extend
f(x) to−L6x6Las an even function ˜f(x) and use Fourier’s Theorem to write
FS( ˜f)(x) =a0 2 +
∞
X
n=1 ancos
nπx L
But for 06x6L, the functions ˜f =f(x), so
FS(f)(x) =a0 2 +
∞
X
n=1 ancos
nπx
L for 06x6L. (Fourier cosine series)
To sketch the Fourier cosine series of a function, follow the steps:
1. Sketchf(x) for 0< x < Lwheref(x) is continuous;
2. Sketch the even extension ˜f(x) for−L < x <0;
3. Sketch its 2L-periodic extension;
4. Mark the average of all the points where there are jumps.
p 2p
−p
October 13
thReview
October 15
thOctober 19
thComputations using Fourier series
Example. Find the Fourier series for the function f(x) =xin [0, π] and use it to compute ∞
X
n=1
nodd
1
n2.
Solution. Because we only want the Fourier series in the interval [0, π], we can choose to extend the function as an even function or an odd function. If we choose the even extension, then we obtain:
FS(f)(x) =a0 2 +
∞
X
n=1
ancosnx (L=π)
with
an=
2
π
Z π
0
xcosnx dx.
We compute the coefficients:
a0= 2
π
Z π
0
x dx=π
an=
2
π
Z π
0
xcosnx dx (integrate by parts withu=x, dv= cosnx)
= 2
π
x
nsinnx
π 0 − 1 n Z π 0
sinnx dx
= 2
πn2cosnx
π 0 = 2
πn2(cosnπ−1) =
2
πn2((−1)
n
−1)
0 ifnis even
− 4
πn2 ifnis odd
So
FS(f)(x) =π 2 − ∞ X n=1 nodd 4
πn2cosnx= π
2 − 4
πcosx−
4
9πcos 3x−
4
25πcos 5x− · · ·
Observe that, if we findx0 such that cosnx0= 1 for allnodd, then
FS(f)(x0) = π 2 − ∞ X n=1 nodd 4 πn2
and we can compute the series. In fact, x0 = 0 satisfies the condition, and the even extension ˜f(x) of f(x) is continuous atx0= 0 and satisfies ˜f(0) = 0, so
0 = ˜f(0) =FS(f)(0) = π 2 − ∞ X n=1 nodd 4 πn2
and we conclude that
∞
X
n=1
nodd
1
n2 = π2
3.4
Term-by-term Differentiation of Fourier Series
When we use the method of separation of variables to solve a PDE we obtain a solution in the form of an infinite series. This solution is obtained by making sure the each individual term solves the PDE, and when we join all the terms together they solve the initial condition as well.
The problem with infinite series expansions is that sometimes it is not possible to differentiate term-by-term. So we question, does the solution obtained actually solve the problem?
Example. Take the example of the heat conduction problem with frozen ends:
∂u ∂t =k
∂2u ∂x2, u(0, t) =u(L, t) = 0,
u(x,0) =f(x).
The solution by the method of separation of variables is
u(x, t) = ∞
X
n=1 Bnsin
nπx L e
−(nπ L)
2
kt.
We can verify that this solution actually solves the problem. If the infinite Fourier sine series can be differentiated term-by-term, then we have
∂u ∂t =−
∞
X
n=1 knπ
L
2
Bnsin
nπx L e
−(nπ L)
2
kt,
∂2u ∂x2 =−
∞ X n=1 nπ L 2
Bnsin
nπx L e
−(nπ L)
2
kt,
so ∂u
∂t =k ∂2u
∂x2 is satisfied.
Fourier Series. A continuous Fourier series can be differentiated term-by-term iff0(x) is piecewise smooth.
Which implies that:
If f0(x) is piecewise smooth, then the Fourier series of a continuous function f(x) can be differentiated term-by-term iff(−L) =f(L).
So
f0(x)∼ − ∞
X
n=1 Annπ
L sin nπx L + ∞ X n=1 Bnnπ
L cos nπx
L .
Fourier cosine series. Iff0(x) is piecewise smooth, then the Fourier series of a continuous functionf(x) can
be differentiated term-by-term.
Fourier sine series. If f0(x) is piecewise smooth, then the Fourier series of a continuous function f(x) can
When we try to prove this result (see pages 120-121 in the textbook), we obtain a formula for the derivative of the Fourier sine series of a continuous functionf(x) when the conditionf(0) =f(L) = 0 is not satisfied. Then
f0(x)∼ f(L)−f(0)
L +
∞
X
n=1
"
nπ L Bn+
2 (−1)nf(L)
−f(0)
L
#
cosnπx
October 20
th3.5
Term-by-term Integration of Fourier Series
A Fourier series of a piecewise smoothf(x) can always be integrated term-by-term. Moreover, the result is a
convergent infinite series that always converges to the integral off(x): Z x
−L
f(t)dt for all−L6x6L.
If
f(x)∼a0 2 +
∞
X
n=1
ancos
nπx
L +bnsin nπx
L
then
Z x
−L
f(t)dt= a0
2 (x+L) + ∞
X
n=1
an
Z x
−L
cosnπt
L dt+bn
Z x
−L
sinnπt
L dt
.
8
Nonhomogeneous Problems
8.2
Heat Flow with Sources and Nonhomogeneous Boundary Conditions
Example (Time Independent Boundary Conditions). Consider the heat equation:
∂u ∂t =k
∂2u ∂x2
and with the boundary conditions
u(0, t) =A
u(L, t) =B
and initial condition
u(x,0) =f(x)
Find the solution.
Solution. We divide the solution in steps.
Step 1. Equilibrium temperature.
First we obtain an equilibrium temperatureuE(x) satisfying the boundary conditions:
d2u
E
dx2 = 0 uE(0) =A
uE(L) =B
This can be easily found:
uE(x) =A+
Step 2. Displacement from equilibrium.
Now we introduce the difference between the actual temperature and the equilibrium temperature:
v(x, t) =u(x, t)−uE(x).
This functionvsatisfies:
∂v ∂t =
∂u ∂t ∂2v ∂x2 =
∂2u ∂x2 −
∂2u
E
∂x2 = ∂2u ∂x2
Sov(x, t) satisfies the problem:
∂v ∂t =k
d2u
E
dx2 v(0, t) =v(L, t) = 0
v(x,0) =f(x)−uE(x)
This is theheat equation with frozen ends, which we have solved:
v(x, t) = ∞
X
n=1 bnsin
nπx L e
−(nπ L)
2
kt
with
bn :=
2
L
Z L
0
f(x)−uE(x)sin
nπx L dx.
We conclude that the solution is:
u(x, t) =uE(x) +
∞
X
n=1 bnsin
nπx L e
−(nπ L)
2
kt.
October 22
nd8.3
Method of Eigenfunction Expansion with Homogeneous Boundary Conditions
The results on the differentiation term-by-term, allow us to solve the heat conduction with heat sources.
Example (Exercise 3.4.11 or 8.3.2). Consider the heat equation with a steady heat source:
∂u ∂t =k
∂2u
∂x2 +Q(x)
and with the boundary conditions
u(0, t) =u(L, t) = 0
and initial condition
u(x,0) =f(x)
Find the solution, assuming that a continuous solution with continuous derivatives exist.
Solution. We divide the solution in steps:
Step 1. Ignoring the heat sourceQ(x), find the eigenvalues and eigenfunctions.
Using the method of separation of variables, we have:
u(x, t) =G(t)φ(x),
so ignoring the heat sources, we obtain the eigenvalue problem:
φ00(x) +λφ(x) = 0
φ(0) =φ(L) = 0
We solve the problem and obtain:
λn=
nπ
L
2
, n= 1,2, . . . (Eigenvalues)
φn(x) = sinnπx
L (Eigenfunctions)
Step 2. Use superposition principle.
We obtain
u(x, t) = ∞
X
n=1
Gn(t) sin
Step 3. Find differential equation forGn(t).
Now we plug this expression for u(x, t) back into the original PDE (with the heat source Q(x)): First we calculate ∂u ∂t = ∞ X n=1 G0
n(t) sin
nπx L
∂2u ∂x2 =−
∞ X n=1 nπ L 2
Gn(t) sin
nπx L
and we expand the heat sourceQ(x) using the Fourier sine series expansion:
Q(x)∼ ∞
X
n=1 qnsin
nπx L ,
which means that
qn=
2
L
Z L
0
Q(x) sinnπx
L dx.
We now rewrite the PDE:
∞
X
n=1
G0n(t) sin
nπx L =−
∞
X
n=1 knπ
L
2
Gn(t) sin
nπx L +
∞
X
n=1 qnsin
nπx L
which simplifies to
∞
X
n=1
G0n(t) +
nπ
L
2
kGn(t)−qn
sinnπx
L = 0
So we conclude thatGn(t) satisfies:
G0n(t) +
nπ
L
2
kGn(t)−qn= 0
Step 4. FindGn(t).
The DE forGn(t) is nonhomogeneous, so we first find the homogeneous solution:
Ghn(t) =e−(
nπ L)
2
kt
and we find one particular solution:
Gpn(t) =
qnL2
(nπ)2k.
We obtain:
Gn(t) = qnL
2
(nπ)2k +Dne
−(nπ L)
2
kt
Step 5. Write solution.
The solution is
u(x, t) = ∞
X
n=1
qnL2
(nπ)2k+Dne
Step 6. Match the initial condition.
The initial condition is
f(x) =u(x,0) = ∞
X
n=1
C
nL2
(nπ)2k+Dn
| {z }
Bn
sinnπx
L
so we know the expression forBn:
Bn=
2
L
Z L
0
f(x) sinnπx
L dx,
thus
Dn=Bn−
CnL2
(nπ)2k Dn=
2
L
Z L
0
f(x) sinnπx
L dx− CnL2
(nπ)2k Dn= 2
L
Z L
0
f(x) sinnπx
L dx−
2L
(nπ)2k
Z L
0
g(x) sinnπx
L dx
The solution can be written as:
u(x, t) = ∞
X
n=1
CnL2
(nπ)2k+Dne
−(nπ L) 2 kt sinnπx L
with the constants calculated above.
Note. In the case of the general heat equation with prescribed temperature at the ends:
∂u ∂t =k
∂2u
∂x2 +Q(x, t)
and with the boundary conditions
u(0, t) =A(t)
u(L, t) =B(t)
and initial condition
u(x,0) =f(x).
We proceed in steps:
Step 1. Reference temperaturer(x, t).
∂2r ∂x2 = 0 r(0, t) =A(t)
r(L, t) =B(t)
So one possible choice is
r(x, t) =A(t) +B(t)−A(t)
October 26
thStep 2. Displacement from reference. As before, we let
v(x, t) =u(x, t)−r(x, t)
and compute its derivatives:
∂v ∂t =
∂u ∂t −
∂r ∂t ∂2v
∂x2 = ∂2u ∂x2 −
∂2r ∂x2 =
∂2u ∂x2
Sov(x, t) satisfies the problem:
∂v ∂t =k
d2v
dx2 +Q(x, t) v(0, t) =v(L, t) = 0
v(x,0) =g(x)
where
Q(x, t) =Q(x, t)−∂r∂t(x, t)
g(x) =f(x)−r(x,0) =f(x)−A(0)−B(0)−A(0)
L x
Step 3. Findv(x, t) using the method of eigenfunction expansion.
The difference with the example we solved is thatQnow depends onxandt, so the differential equation used to calculateGn(t) is a little more complicated.
4
Wave Equation: Vibrating Strings and Membranes
4.2
Derivation of a vertically vibrating string
Consider a tightly horizontally stretched string with the ends tied down in some way (as you can imagine, the way which we tie the ends will be part of the boundary conditions). In principle, a point Aof the string may move in any direction, so
Assumption 1: The perturbation from horizontal is very
small. This way, we can neglect the horizontal displacement.
The displacement is entirely vertical:
y=u(x, t). x
u(x, t)
Newton’s Law. We consider an infinitesimally thin segment of string betweenxand ∆xin the unperturbed
horizontal position.
Total mass =ρ0(x)∆x
For a point mass, we have:
~
F =m~a (Newton’s Law)
Which forces are acting on the segment of string?
There are forces acting vertically on the body of the string, e.g. gravity, and forces acting on the ends of the segment of string.
Assumption 2: The string is perfectly flexible – it offers no resistance to bending.
This means that the force acting on the ends has the direction tangential to the string. The magnitude of this force is calledtension:
T(x, t) = force exerted at pointxand timetin the direction tangent to the string (Tension)
x x+ ∆x
T(x, t) T(x+ ∆x, t)
θ(x, t)
θ(x+ ∆x, t)
To express the tension, we need to find an expression for the angle of the stringθ(x, t):
tanθ(x, t) = dy
dx = ∂u ∂x.
From Newton’s Law, we obtain two equations:
Horizontal component of Force = very small = negligible
Vertical component
of Force =
Vertical component of tensile forces +
Vertical component of body forces
Now we look at each term in detail for the segment of string:
Vertical component
of Force =F2=ma2=ρ0(x)∆x
∂2u
∂t2 (by Newton’s Law)
Vertical component
of tensile forces =T(x+ ∆x, t) sinθ(x+ ∆x, t)−T(x, t) sinθ(x, t)
Vertical component
of body forces =ρ0(x)∆xQ(x, t)
We have:
ρ0(x)∆x∂ 2u
∂t2 =T(x+ ∆x, t) sinθ(x+ ∆x, t)−T(x, t) sinθ(x, t) +ρ0(x)∆xQ(x, t)
Divide everything by ∆x:
ρ0(x)∂
2u ∂t2 =
T(x+ ∆x, t) sinθ(x+ ∆x, t)−T(x, t) sinθ(x, t)
∆x +ρ0(x)Q(x, t)
and take the limit as ∆x→0+:
ρ0(x)∂
2u ∂t2 =
∂
∂x[T(x, t) sinθ(x, t)] +ρ0(x)Q(x, t)
Assumption 3: The anglesθare small, i.e. θ≈0, so
∂u
∂x = tanθ=
sinθ
cosθ ≈sinθ
The equation becomes:
ρ0(x)∂
2u ∂t2 =
∂ ∂x
T∂u ∂x
October 27
thPerfectly Elastic Strings. Most strings are perfectly elastic, which means that the tensile force T(x, t)
depends only on how stretched the string is near the pointx.
We assumed thatθis very small, so the displacement is very small too, thus the string is stretched as much as in the unperturbed setting. This implies that
T(x, t)≈T0constant.
The small vertical vibrations of a highly stretched string satisfies:
ρ0(x)∂
2u ∂t2 =T0
∂2u
∂x2 +ρ0(x)Q(x, t) (Highly stretched string)
One-dimensional Wave Equation. Often, gravity is the only body force, and as often the tensile force is
much stronger than the gravitational one: ρ0g T0
∂2u
∂x2
, so we can neglect the influence of the body forces:
Q(x, t) = 0.
ρ0(x)∂
2u ∂t2 =T0
∂2u ∂x2
or
∂2u ∂t2 =c
2∂2u
∂x2 (Wave Equation)
wherec2= T0
ρ0(x).
4.3
Boundary Conditions
The PDE obtained is of second-order, so we expect two boundary conditions: one at each end.
Fixed end. If the string is fixed at one end, as in a musical instrument, then the vertical displacementu= 0
at that end:
u(0, t) = 0.
However, we may prescribe the displacement at the end which changes with time:
u(0, t) =f(t).
Free end. If the string can move freely in the vertical direction, as in a frictionless track, then the vertical
component of the tensile force is 0 at the end:
T0∂u
String attached to a Spring. If the string is attached to a spring-mass system, as in the figure.
m
y(t)
ys(t)
Then the displacement on the end of the string is the same as the displacement y(t) of the mass attached to the spring.
u(0, t) =y(t)
On the other hand,y(t) satisfies its own dynamical system (ODE):
md 2y
dt2 =−k(y(t)−ys(t)−`) + other forces applied to the mass (Hooke’s Law)
where`is the unstretched length of the spring and ys(t) is the position of the support of the spring.
The other force applied to the spring is the tensile force applied by the stringT(0, t) sinθ(0, t)≈T0∂u
∂x(0, t) and
a forceg(t) representing other external forces:
m∂ 2u
∂t2(0, t) =−k(u(0, t)−ys(t)−`) +T0 ∂u
∂x(0, t) +g(t) (Hooke’s Law)
after substitutingy(t) =u(0, t).
String and Massless-Spring. If the mass is very small,m= 0 and there are no external forces g(t) = 0,
then
T0∂u
∂x(0, t) =k(u(0, t)−uE(t))
whereuE(t) =ys(t)−`is the equilibrium position of the spring-mass system.
Moreover, if the equilibrium position matches the unperturbed position of the string: uE(t) = 0 and
T0∂u
∂x(0, t) =k u(0, t)
which is homogeneous.
This condition changes sign when applied to the other end:
T0∂u
October 29
th4.4
Vibrating String with Fixed Ends
We will now work out the following problem:
∂2u ∂t2 =c
2∂2u
∂x2 (PDE)
u(0, t) = 0 (BC1)
u(L, t) = 0 (BC2)
u(x,0) =f(x) (IC1) ∂u
∂t(x,0) =g(x) (IC2)
fort>0 and 06x6L.
We need two initial conditions because the problem is of second-order in time. Apply the method of separation of variables: u(x, t) =φ(x)h(t), which implies
φ(x)h00(t) =c2φ00(x)h(t)
Divide both sides byc2φ(x)h(t) to obtain:
h00(t)
c2h(t) = φ00(x)
φ(x) =−λ
and we obtain two ODEs:
h00(t) =
−c2λh(t) (ODE
t)
φ00(x) =−λφ(x) (ODEx)
whereλ= separation constant. The boundary conditions yield:
φ(0) =φ(L) = 0,
so we solve the eigenvalue problem inφ(x):
λn=
nπ
L
2
, n= 1,2, . . .
φn(x) = sin
nπx L
For the problem int, we obtain:
hn(t) =C1cosncπt
L +C2sin ncπt
L
Using the superposition principle, we conclude that the solution is:
u(x, t) = ∞
X
n=1
Ancos
ncπt
L +Bnsin ncπt
L
sinnπx
We now introduce the initial conditions:
f(x) =u(x,0) = ∞
X
n=1
Ansinnπx
L
g(x) =∂u
∂t(x,0) =
∞
X
n=1 Bn
ncπ L sin
nπx L
This means that we expand the functionsf(x) andg(x) using a Fourier sine series:
An=
2
L
Z L
0
f(x) sinnπx
L dx
Bn
ncπ L =
2
L
Z L
0
g(x) sinnπx
L dx
Note. If this is about a musical instrument, then what information can we tell about the sound?
1. Each term of the series is called a normal mode of vibration
2. To find the amplitude and frequency, we need to write
Ancos
ncπt
L +Bnsin ncπt
L =Asin(ωt+θ)
which implies that
A=pA2
n+Bn2 = amplitude
ω= ncπ
2πL = frequency
θ= arctanAn
Bn
3. The first normal mode is called the first harmonic or fundamental, and it decides the frequency of the sound, so the frequency is ω=2cL.
4. Recall thatc=qT0
ρ0, whereT0is the tension on the string and ρ0 the mass density.
Since in a musical instrument, the mass density is fixed and so is the length of the string, the best way to tune it is by adjusting the tension T0.
When the player clamps down a string with a finger, he/she is shortening the lengthL, thus producing a higher pitched sound.
