Download PART II: LINEAR EQUATIONS 13. Basic concepts 13.1. Linear

31
PART II: LINEAR EQUATIONS
13. Basic concepts
13.1. Linear equations. The standard form of a second order linear equation is
L[x] ≡ ẍ + p(t)ẋ + q(t)x = g(t).
The map x 7→ L[x] is a differential operation. If g(t) ≡ 0, then the equation
L[x] = 0
is called homogenous. The operation L has constant coefficients if p(t) and q(t) are
constant functions.
The standard form of an n-th order equation is L[x] = g(t) with
L[x] = x(n) + p1 (t)x(n−1) + · · · + pn (t)x.
13.2. Initial value problem. To solve the IVP(t0 , x0 , ẋ0 ) for a second order equation (not necessarily linear) is to find a solution x(t) such that
x(t0 ) = x0 ,
ẋ(t0 ) = ẋ0 .
In the case of n-th order equations, the initial conditions have the form
x(t0 ) = x0 ,
ẋ(t0 ) = ẋ0 ,
...,
(n−1)
x(n−1) (t0 ) = x0
.
Theorem. Consider a linear equation L[x] = g. Suppose that g and the coefficients
of L are continuous functions in some interval I. Then
(i) all solutions extend to the whole interval,
(ii) every IVP with t0 ∈ I has a unique solution defined on I.
13.3. Linear properties. If the coefficients of L are in C(I), then we have a linear
map
L : C (n) (I) → C(I).
Denote
ker L = {x : L[x] = 0}.
This is the set of all solutions of the homogeneous equation.
Lemma. ker L is a linear space of dimension n.
Proof: Fix some t0 ∈ I and consider the map
(n−1)
Rn → ker L,
x0 , ẋ0 , . . . , x0
7→ x(t),
where x(t) is the solution of the corresponding IVP. Applying the theorem we see
that this map is linear, 1-to-1, and ”onto”.
2
Recall that n elements (or vectors) x1 , . . . , xn of a linear space are (linearly) independent if
C 1 x1 + · · · + C n xn = 0
⇒
C1 = · · · = Cn = 0.
32
If the linear space has dimension n, then n independent vectors form a basis: every
vector x has a unique representation as a linear combination of xj ’s. Any n + 1
vectors of an n dimensional space are dependent.
Let x1 , . . . , xn be solutions of an n-th order homogeneous equation L[x] = 0. Then
the collection x1 , . . . , xn is called a fundamental system of solutions if the functions
xj are independent as elements of ker L. It follows that every solution of L[x] = 0
has a unique representation as a linear combination
x(t) = C1 x1 (t) + · · · + Cn xn (t).
Example. The functions 1, t, . . . , tn−1 form a fundamental system of solutions of
the equation x(n) = 0. The general solution (≡ the set of all solutions) is the space
of polynomials of degree < n.
Lemma. If x∗ is a solution of L[x] = g, then the general solution of L[x] = g is
x ∈ x∗ + ker L,
(or x = x∗ +C1 x1 +· · ·+Cn xn , ”particular integral” + ”complementary function”.)
13.4. Reduction of order. The following substitution is quite useful.
Theorem. If we know some solution x1 of a homogeneous equation L[x] = 0 of
order n, then the substitution x = x1 y, where y = y(t) is a new unknown function,
transforms the equation L[x] = g into a linear equation of order n − 1 for y 0 .
Proof: Consider the case n = 3,
L[x] ≡ x000 + p1 x00 + p2 x0 + p3 x = g,
(the coefficients are not necessarily constant). We have
00 0
0 00
000
L[x] = x000
1 y + 3x1 y + 3x1 y + x1 y
+ p1 x001 y + 2p1 x01 y 0 + p1 x1 y 00
+ p2 x01 y + p2 x1 y 0
+ p3 x1 y = g.
The sum of the first terms is yL[x1 ] = 0; the other terms do not involve y.
2
14. The Wronskian
14.1. Definition. The Wronskian of two functions y1 (x), y2 (x) ∈ C 1 (I) is the function
y (x) y2 (x)
∈ C(I).
W (x) = 10
y1 (x) y20 (x)
The Wronskian of n functions yj ∈ C (n−1) is defined similarly, e.g.
y1 y2 y3 0
W = y1 y20 y30 ∈ C(I).
y100 y200 y300 Let us consider a linear homogeneous equation with continuous coefficients,
L[y] ≡ y (n) + p1 (x)y (n−1) + · · · + pn (x)y = 0,
pj ∈ C(I).
33
Theorem. Let y1 , . . . , yn be solutions of L[y] = 0. FAE:
(i) {y1 , . . . , yn } is a fundamental system of solutions;
(ii) ∀x ∈ I, W (x) 6= 0;
(iii) ∃x0 ∈ I, W (x0 ) 6= 0.
Proof: As we know, the map
ker L → Rn :
y 7→ (y(x0 ), . . . , y (n−1) (x0 ))
is a linear isomorphism for every x0 ∈ I.
(i)⇒(ii) If the solutions yj are independent, then the vectors of initial conditions
are independent in Rn , and therefore W (x0 ) 6= 0 for all x0 .
(iii)⇒(i) If W (x0 ) 6= 0, then the initial vectors are independent, and so are the
solutions.
2
The most interesting part of this theorem is the equivalence of (ii) and (iii).
Examples. (a) The Wronskian of y1 (x) = x and y2 (x) = sin x is zero at x = 0 but
not identically. The Wronskian of y1 (x) = x3 and y2 (x) = x2 |x| is identically zero
but the functions are independent.
Conclusion: these pairs of functions can not be solutions of the same second order
linear equation with continuous coefficients.
(b) The Wronskian of exponential functions eλj x evaluated at 0 is the so called
Vandermonde determinant
1
1
...
1 λ
λ2
...
λn Y
=
(λj − λk ).
W (0) = 1
.
.
.
.
.
.
.
.
.
.
. . n−1
j<k
λ
λn−1
. . . λn−1
n
1
2
14.2. Abel’s formula. We don’t need to solve the differential equation to compute
the Wronskian of its fundamental system of solutions.
Theorem. Let W (x) be the Wronskian of some fundamental system of solutions
of L[y] = 0. Then
Z x
W (x) = W (x0 ) exp −
p1 (s)ds .
x0
Proof: We will verify the statement for second order equations
y 00 + p(x)y 0 + q(x)y = 0.
It is sufficient to show that the Wronskian satisfies the equation
W 0 = −p(x)W.
(14.1)
34
We have
0
y
W = 10
y1
0
y20 y1
+
y20 y100
y2 y200 y2 y1
+
−py20 −qy1
y
= 0 + 1 0
−py1
y1 y2 = −pW.
= −p 0
y1 y20 .
y2 −qy2 2
Corollary. Let y1 be a (known) solution of the 2nd order equation (14.1). Define
Z x
W (x) = exp −
p(s)ds .
x0
Then
Z
W
y12
is also a solution, and y1 , y2 are independent.
y2 = y1
(14.2)
Proof: According to the above theorem, there is an independent solution y2 such
that W is the Wronskian of y1 , y2 . To find y2 consider the formula
y1 y20 − y10 y2 = W
as a first order linear equation for y2 (x); the coefficients y1 , y10 , and W are known
functions. We have
0
y2
W
y1 y20 − y10 y2
= 2,
=
2
y1
y1
y1
which gives us (14.2)
2
Exercise: Strictly speaking, the formula (14.2) makes sense only on the intervals
where y1 6= 0. Interpret the formula in the case where y1 has zeros. E.g. y 00 +y = 0,
y1 (t) = sin t.
Example. Consider the equation x2 (1 + x)y 00 − 2y = 0. One can verify that
y1 (x) = 1 + x−1 is a solution. Find the general solution and determine the domain
of the maximal solution of the IVP(−2; 0, 1).
According to the general theory of linear equations, the IVP solution extends at
least to (−∞, −1) (but in principle it may exist on a larger interval.) Since the
Wronskian is constant, we find
Z
Z
1
x2 dx
1
1
y2 = y1
= y1 (x)
=1+x− − 1+
log(1 + x)2 .
y12
(1 + x)2
x
x
The general solution is y = C1 y1 + C2 y2 , and we have C2 6= 0 for the IVP solution.
The IVP solution is not differentiable at x = −1, so its domain is (−∞, −1).
35
15. Homogeneous equations with constant coefficients
L[x] ≡ a0 x(n) + · · · + an x = 0,
aj ∈ R.
15.1. Characteristic polynomial. Denote
P (λ) = a0 λn + a1 λn−1 + · · · + an ,
so (symbolically)
L = P (D),
D=
d
.
dt
The polynomial P has exactly n roots (”eigenvalues”)
λ1 , . . . , λn ,
P (λj ) = 0,
which may be complex and/or multiple, and we have
P (λ) = a0 (λ − λ1 ) . . . (λ − λn ).
Since the characteristic polynomial P has real coefficients, complex roots appear
in conjugate pairs, i.e. if λ = α + iω is an eigenvalue, then λ = α − iω is also an
eigenvalue.
Lemma. L[eλt ] = P (λ)eλt .
Proof: If x(t) = eλt , then x(k) (t) = λk eλt .
2
Corollary. A real number λ is a root of the characteristic polynomial if and only
if x(t) = eλt is a solution of the homogeneous equation.
15.2. The case of distinct real roots.
Theorem. If the roots of the characteristic polynomial are all real and distinct,
then the general solution of the homogeneous equation is
x(t) = C1 eλ1 t + · · · + Cn eλn t .
Proof: The exponential functions with distinct exponents are linearly independent.
Indeed, suppose
C1 eλ1 t + · · · + Cn eλn t ≡ 0
and λ1 < · · · < λn . If Cn 6= 0, then the LHS divided by Cn eλn t converges to 1 as
t → ∞, a contradiction. If Cn = 0, we repeat this argument with Cn−1 instead of
Cn , etc.
2
36
15.3. Second order equations. There are 3 possible cases:
(i) the roots λ1 , λ2 are real and distinct;
(ii) λ1 = λ2 ∈ R;
(iii) λ1,2 = α ± iω with ω 6= 0.
Theorem. The general solution in the above three cases is
(i)
x(t) = C1 eλ1 t + C2 eλ2 t ;
(ii)
x(t) = C1 eλ1 t + C2 teλ1 t ;
(iii)
x(t) = C1 eαt cos ωt + C2 eαt sin ωt.
Proof: (i) is clear.
(ii) is the limiting case of (i) as λ2 → λ1 :
eλ2 t − eλ1 t
d →
eλt = teλ1 t .
λ2 − λ1
dλ λ=λ1
Alternatively, we can apply the reduction of order method. The equation is x00 −
2λ1 x0 + λ21 x = 0, and x1 = eλ1 t is a solution. Substitution x = x1 y gives the
equation y 00 = 0, so we can take y(t) = t and get the second solution x2 = teλ1 t .
(iii) easy to verify; also see the next section.
2
15.4. Difference equations. Fibonacci numbers
0 1 1 2 3 5 8 ...
satisfy the second order linear difference (or recursive) equation
xn = xn−1 + xn−2 .
(Fibonacci’s interpretation: xn is the population of rabbits in the n-th generation.)
The linear equation is homogeneous. Let us look for solutions of the form
xn = an ,
(discrete exponentials). Then a has to satisfy the equation
an = an−1 + an−2 ,
a2 − a − 1 = 0.
We have two roots
√
1± 5
a1,2 =
,
2
(a1 is the ”golden ratio”, see Problem 22.3). The general solution to the difference
equation is
xn = C1 an1 + C2 an2 .
From the initial conditions
x0 = 0, x1 = 1,
we find the values
1
C1 = −C2 = √ .
5
This method works for all linear homogeneous recursive equations with constant
coefficients. See Examples 22.1 and 22.2 for the cases of repeated and complex
roots.
37
16. Complex eigenvalues
16.1. Complex exponentials. A complex-valued function of one real variable,
z(t) = x(t) + iy(t),
t ∈ R,
(the functions x(t) and y(t) are real-valued) is differentiable if both x(t) and y(t)
are differentiable, and in this case
z 0 = x0 + iy 0 .
Example. Let λ = α + iω. By definition,
eλt = eαt cos(ωt) + ieαt sin(ωt),
(16.1)
and it is easy to check that
d λt
e = λeλt .
dt
A more conceptual approach to the formula (16.1) is to introduce the complex
exponential function
X λn
eλ =
,
λ ∈ C,
(16.2)
n!
n≥0
If λ = α is real, then we get the usual (real) exponential eα , and if λ = iω is
imaginary, then
eiω = cos ω + i sin ω
because the real and imaginary parts of the series (16.2) are Taylor’s series of cosine
and sine functions respectively. The series (16.2) converges in the whole complex
plane, and one can verify (by multiplication of the series) the functional equation
eλ1 +λ2 = eλ1 eλ2 .
In particular, for λ = α + iω we have
eλt = eαt eiωt = eαt cos ωt + ieαt sin ωt,
which is the same as (16.1).
16.2. Complex solutions. Consider the equation
L[z] ≡ P (D)[z] = 0
for complex-valued functions z = z(t). Similarly to the real case, we have
∀λ ∈ C,
L[eλt ] = P (λ)eλt ,
which proves
Lemma. z(t) = eλt is a solution iff P (λ) = 0.
We also have the following statement concerning linear equations L[z] = 0 with real
coefficients.
Lemma. z(t) = z(t) + iy(t) is a (complex) solution iff both x(t) and y(t) are (real)
solutions.
38
Proof: Since the coefficients of the equation are real, we have L[z] = L[x] + iL[y],
so L[z] = 0 iff L[x] = 0 and L[y] = 0.
2
Combining the two lemmas, we derive
Theorem. Consider the equation L[x] = 0 with constant real coefficients, and
suppose that all roots of the characteristic polynomial are distinct. Let λ1 , . . . be
the real roots, and let α1 ± iβ1 , . . . be the complex roots. Then the collection of
functions
eλ1 t , . . . ;
eα1 t cos(β1 t), eα1 t sin(β1 t), . . .
(16.3)
is a fundamental system of (real-valued) solutions.
Note. In the case of distinct eigenvalues λ1 , . . . , λn (real or complex), the general
complex-valued solution is
n
X
z(t) =
C j eλ j t ,
j=1
where Cj ’s are arbitrary complex numbers.
16.3. General homogeneous equations with constant coefficients. Consider
the equation L[x] = 0 with real constant coefficients. If λj is a repeated real root
of multiplicity k,
P (λ) = (λ − λj )k P̃ (λ),
P̃ (λj ) 6= 0,
then we add the following functions to the list (16.3) of the last theorem:
teλj t , . . . , tk−1 eλj t .
If αj ± iβj is the conjugate pair of complex roots of multiplicity k, then we add the
functions
teαj t cos(βj t), teαj t sin(βj t), . . . , tk−1 eαj t cos(βj t), tk−1 eαj t sin(βj t).
Using reduction of order, it is not difficult to justify the following statement.
Theorem. The resulting collection of functions is a fundamental system of solutions of the equation L[x] = 0.
Example. Let P (λ) = λ3 (λ2 + 1)2 . The roots are
0, 0, 0, i, −i, i, −i.
We have the following fundamental system of solutions:
1, t, t2 , cos t, sin t, t cos t, t sin t.
The general solution of the equation x(7) + 2x(5) + x(3) = 0 is therefore
x(t) = C1 + C2 t + C3 t2 + C4 cos t + C5 sin t + C6 t cos t + C7 t sin t.
39
17. Inhomogeneous linear equations
We will discuss here how to solve equations
L[x] = g.
We have two general methods: reduction of order (Section 15.4) and variation
of constants. These methods work for all linear equations, not necessarily with
constant coefficients. The third method, the method of undetermined coefficients,
is often the most efficient but it applies only to equations with constant coefficients
and to functions g of special type (quasi polynomials). Also, see Section *** below
for the discussion of the important Laplace method.
17.1. Variation of constants. We will show that if we know the general solution
of the homogeneous equation L[x] = 0, then we can solve any inhomogeneous
equation L[x] = g. For simplicity, we will only discuss the case of 2d order equations:
L[x] = x00 + p(t)x0 + q(t)x.
Let x1 and x2 be two independent solutions of L[x] = 0. We look for a particular
solution of the inhomogeneous equation of the form
x∗ = C1 (t)x1 + C2 (t)x2 .
We have
x0∗ = (C1 x01 + C2 x02 ) + (C10 x1 + C20 x2 ),
and if the functions C10 and C20 are such that
C10 x1 + C20 x2 ≡ 0,
then
x00∗ = (C1 x001 + C2 x002 ) + (C10 x01 + C20 x02 ),
so
L[x∗ ] = C10 x01 + C20 x02 .
Thus x∗ is a particular solution if C10 and C20 satisfy the system of linear algebraic
equations
C10 x1 + C20 x2 = 0,
Solving this system:
0 x2 g x02 gx2
0
C1 =
=−
,
W
W
C10 x01 + C20 x02 = g(t).
x1
0
x1
0
C1 =
W
0
g
=
gx1
,
W
we derive
Z
x∗ = −x1
gx2
+ x2
W
Z
gx1
.
W
x
W := 10
x1
x2 ,
x02 40
17.2. Undetermined coefficients. We will now assume that L[·] has constant
coefficients and g is a quasi polynomial.
By definition, a quasi polynomial is a linear combination of the functions of the
following form:
tm eµt ,
tm eαt cos(ωt),
tm eαt sin(ωt).
(17.1)
Here m is a non-negative integer, and µ, α, ω are real. We will write µ = α + iω in
the last two cases in (17.1).
We will show that the equation L[x] = g has a quasi polynomial solution. Because
of the following (obvious) lemma, we can consider the functions (17.1) separately.
Lemma. If g = g1 + g2 , and if x1 and x2 are solutions of L[x] = g1 and L[x] = g2
respectively, then x = x1 + x2 is a solution of L[x] = g.
Let g be one of the functions (17.1), and let k denote the multiplicity of µ as a root
of the characteristic polynomial; e.g. k = 0 if µ is not an eigenvalue.
Theorem. (i) If µ is real (i.e. g = tm eµt ), then there is a solution of the form
x∗ (t) = tk (some polynomial of degree m) eµt .
(ii) If µ is complex, then there is a solution of the form
x∗ (t) = tk (some polynomial of degree m) eαt cos(ωt)
+ tk (some other polynomial of degree m) eαt sin(ωt).
Proof: We’ll only discuss the simpler case k = 0, i.e. P (µ) 6= 0. If µ is real, then
we denote
Eµ,m = {real polynomials of degree ≤ m} · eµt .
This is a linear space (over R) of dimension m + 1. It is enough to show that the
linear map L : x 7→ L[x] acts and is invertible in Eµ,m . Let us compute the matrix
of
d
: Eµ,m → Eµ,m
D≡
dt
in the basis
ej = tj eµt ,
(0 ≤ j ≤ m).
Since
Dej = jtj−1 eµt + µtj eµt = jej−1 + µej ,
we have


µ 1 0 ...
 0 µ 2 . . .

D=
 0 0 µ . . . .
. . . ...
It follows that


P (µ)
∗
∗
...
 0
P (µ)
∗
. . .
.
L ≡ P (D) = 
 0
0
P (µ) . . .
.
.
.
...
41
Since P (µ) 6= 0, the eigenvalues of the last matrix are non-zero, and the operator
L : Eµ,m → Eµ,m is invertible. If µ is complex, we consider complex polynomials
in the definition of Eµ,m and use the same argument.
2
17.3. Examples.
(i) x00 + x = t2 .
We have µ = 0, which is not a root of the characteristic polynomial λ2 +1, so k = 0.
There is a solution of the form
x(t) = At2 + Bt + C.
Let’s find the (undetermined) coefficients. We need to satisfy
x00 + x = At2 + Bt + (2A + C) = t2 ,
so
A = 1,
B = 0,
C = −2A = −2.
Thus we get a particular solution x(t) = t2 − 2. The general solution is then
x(t) = C1 cos t + C2 sin t + t2 − 2.
(ii) x00 + x = e2t .
We have µ = 2, which is not a root of the characteristic polynomial, so k = 0.
There is a solution of the form x(t) = Ae2t . We find A = 1/5.
(iii) x00 + x = te−t .
There is a solution of the form x(t) = (At + B)e−t .
(iv)
x00 + x = t3 sin t.
Now α = 0, ω = 1, and µ = i, which is a root of multiplicity k = 1. There is a
solution of the form
x(t) = (At4 + Bt3 + Ct2 + Dt) · sin t + (Ãt4 + B̃t3 + C̃t2 + D̃t) · cos t.
(v) x(4) + 4x00 = sin 2t + tet + 4.
The characteristic polynomial is
P (λ) = λ4 + 4λ2 = λ2 (λ2 + 4),
λ1,2 = 0, λ3 = 2i, λ4 = −2i.
We have
g1 = sin 2t
⇒
x1 = t(A sin 2t + B cos 2t);
t
⇒
x2 = et (Ct + D);
g1 = 4
⇒
x3 = Et2 ,
g2 = te
and so there is a solution of the form
x(t) = At sin 2t + Bt cos 2t + Ctet + Det + Et2 .
42
17.4. Inhomogeneous linear difference equations. Example: the sequence
1, 1, 6, 12, 29, 59, . . .
is described by the IVP
xn = xn−1 + 2xn−2 + n,
x1 = x2 = 1.
The associated homogeneous equation is xn = xn−1 + 2xn−2 , and if xn = an is its
solution, then
a2 = a + 2;
a1 = −1, a2 = 2.
Since n = n1n and 1 is not a root of a2 = a+2, we will look for a particular solution
of the inhomogeneous equation of the form
xn = An + B.
We have
An + B = An − A + B + 2An − 4A + 2B + n,
and therefore
A = A + 2A + 1,
Thus A = −1/2,
B = −A + B − 4A + 2B.
B = −5/4. Conclusion: the general solution is
xn = −(n + 5)/2 + (−1)n C1 + 2n C2 ,
and it remains to determine C1 and C2 from the initial conditions.
18. Oscillations
In this section we discuss applications to mechanical and electrical vibrations. We
will consider 2nd order equations
ẍ + 2aẋ + ω02 x = g(t)
(18.1)
with constant coefficients and periodic functions g(t). It is remarkable that such
simple equations have many meaningful applications.
18.1. Interpretations.
(a) Harmonic oscillator: the motion of a mass m on a Hooke’s law spring with
spring constant k > 0. Newton’s equation for the elongation of the spring x = x(t)
is
mẍ + γ ẋ + kx = g(t),
where γ is the damping coefficient, and g(t) is the external force.
(b) Electrical circuits. An RLC circuit consists of a resistor with resistance R
(ohms), an inductor with inductance L (henrys), and a capacitor with capacitance
C (farads).
The resistor converts electrical energy into heat or light, e.g. a light bulb element.
The inductor has a special geometry such as coils which creates a magnetic field
that induces a voltage drop. A typical capacitor consists of two plates separated
by an insulator; charges of opposite signs will build up on the two plates.
43
The impressed voltage E(t) is a given function of time (seconds). We consider the
following characteristics of the circuit: the charge Q(t) (coulombs) on the capacitor,
and the current I(t) = Q̇(t) (amperes). By Kirchhoff,
E(t) = LI˙ + RI + C −1 Q.
(E(t) is equal to the sum of voltage drops across R, L, and C; the respective drops
are RI by Ohm’s law, C −1 Q by Coulomb’s law, and LI˙ by Faraday’s law.) Thus
we get equations similar to the mass-spring model:
E(t) = LQ̈ + RQ̇ + C −1 Q,
or
Ė(t) = LI¨ + RI˙ + C −1 I,
where R is responsible for damping.
18.2. Free vibrations: g = 0 in (18.1).
(a) Undamped case: a = 0. The general solution is
x(t) = C1 cos ω0 t + C2 sin ω0 t = M cos(ω0 t − φ).
We have a periodic motion with period equal to 2π/ω0 ; ω0 > is the natural frequency,
M > 0 is the amplitude of the motion (maximal displacement), and φ is the phase
angle.
(b) Damped case. The eigenvalues are
λ1,2 = −a ±
q
a2 − ω02 .
There are no oscillations in the over-damped case a2 ≥ ω02 . Otherwise (”underdamping”), we have complex eigenvalues
λ1,2 = −a ± iµ,
and the motion is a damped vibration:
x(t) = Re−at cos(µt − δ);
µ is called quasi frequency and 2π/µ quasi period.
18.3. Forced vibrations: undamped case. Suppose the exterior force g in
(18.1) is T -periodic,
g(t + T ) ≡ g(t),
and denote ω = 2π/T > 0, the forcing frequency. We will study the question
of existence and uniqueness of periodic solutions and also the question of their
stability. We only discuss the case
g(t) = A cos(ωt − δ),
and refer to Arnol’d for general theory. Assume first that there is no damping, so
the equation is
ẍ + ω02 x = A cos(ωt − δ).
44
Theorem. (i) If the ratio ω0 /ω is not a rational number, then there exists a unique
periodic solution. The period of the periodic solution is T .
(ii) If ω0 /ω is a rational number but ω 6= ω0 , then all solutions are periodic.
(iii) If ω = ω0 , then every solution is unbounded and non-periodic
The case (iii) is called (pure) resonance: ”bounded input, unbounded output”
Proof: For simplicity assume δ = 0. If ω 6= ω0 , then the general solution is
A
cos ωt,
x(t) = C1 cos ω0 t + C1 sin ω0 t + 2
ω0 − ω 2
(use the method of undetermined coefficients). The choice C1 = C2 = 0 gives a
periodic solution. All other solutions are not periodic unless the frequencies are
commensurable. In the resonance case, we have
A
x(t) = C1 cos ω0 t + C1 sin ω0 t +
t sin ω0 t,
2ω0
(again, use undetermined coefficients).
2
Comments.
(a) The results in the case of a general periodic exterior force are similar. In
particular, if the frequency of the exterior force is equal to the natural frequency,
then all solutions are unbounded. Soldiers used to break step when crossing a
bridge to eliminate the periodic force of their marching that could resonate with a
natural frequency of the bridge.
(b) Beats. In the non-resonance case, every solution is the sum of two periodic
functions,
x(t) = A cos(ω0 t + α) − B cos(ωt + β).
The graphs of such functions could be quite intriguing. For example, consider the
case where the amplitudes are equal, A = B, (or almost equal), and the frequencies
are not very different in the sense that
|ω − ω0 | |ω + ω0 |.
(Think of two flutes playing slightly out of tune with each other.) We have (for
α = β = 0, otherwise up to a phase shift)
ω − ω0
ω + ω0
x(t) = 2A sin
t sin
t.
2
2
The first term oscillates slowly compared to the second one. We can think of the
function
2A sin ω − ω0 t
2
as a slowly changing amplitude of the vibrations with frequency |ω + ω0 |/2. We say
that the solution exhibits the phenomenon of beats.
(c) Example of case (ii):
ẍ + 36x = 3 sin 4t.
Every solution is the sum of a 2π/6-periodic function and a 2π/4-periodic function,
so every solution is π-periodic.
45
Relation to Lissajous’ curves: consider the orbit of the motion {x(t), ẋ(t)} in the
”phase space” R2 = {(x, ẋ)}. For example, draw the curve
1
3
sin 6t +
sin 4t,
10
20
which is the orbit of the IVP(0;0,0).
3
3
ẋ(t) = − cos 6t + cos 4t,
5
5
x(t) = −
18.4. Forced vibrations: damped case.
ẍ + 2aẋ + ω02 x = A cos(ωt − δ).
Theorem. If a > 0, then there is a unique periodic solution x∗ (t). This solution
has period 2π/ω and is a steady state.
The last sentence means that x∗ (t) is globally asymptotically stable: for any solution
x(t), we have
|x(t) − x∗ (t)| → 0 as t → ∞
(convergence is exponentially fast). In other words, no matter what initial conditions are, we asymptotically get the same function x∗ (t). Note that there are
no steady states in the undamped case – the effect of the initial conditions would
persist at all times.
p
Proof: The eigenvalues are λ1,2 = −a ± a2 − ω02 . Clearly, <λ1 < 0 and <λ2 < 0,
in particular iω is not an eigenvalue. By the method of undetermined coefficients,
there is a particular solution x∗ (t) of the form B cos(ωt − φ). On the other hand,
every ”complementary function” (solution to the associated homogeneous equation)
is exponentially small as t → +∞, so the general solution is
x(t) = B cos(ωt − φ) + o(1),
t → +∞.
2
This theorem generalizes to arbitrary linear systems with constant coefficients such
that all eigenvalues have negative real parts. If the non-homogeneous term is periodic, then there is a unique periodic solution, which is a steady state.
Terminology: the periodic force g(t) is the input, and the steady state is the output
or forced response of the system. The o(1) part of the solution is the transient
solution. The ratio of the amplitudes G = B/A is the gain.
Exercise: show
G= p
1
(ω 2 − ω02 )2 + 4a2 ω 2
.
(18.2)
Proof: If x = B̃eiωt with |B̃| = B, then L[x] = Ãeiωt with |Ã| = A and p(iω)B̃ = Ã.
We have
|B̃|
1
1
G=
=
=
2 − ω 2 ) + 2aωi| .
|p(iω)|
|(ω
|Ã|
0
2
The gain can be large if the frequencies are almost equal and the damping is small,
the case of ”practical resonance”.
46
Applications. (a) For a given circuit or spring (ω0 and a are fixed), the gain is a
function of ω, G = G(ω). Sometimes we want to find the ”resonant frequency” ωr
for which the gain is maximal. Assuming a < ω02 (small damping), we have
1
,
ωr2 = ω02 − 2a2 ,
Gmax = p 2
2a ω0 − a2
(differentiate the expression under the square root in (18.2) with respect to ω 2 ).
(b) Tuning. Consider the equation.
LI¨ + RI˙ + C −1 I = A1 cos ω1 t + A2 cos ω2 t,
where L and R are fixed but C is a tuning parameter. Think of two radio station
broadcasting on different frequencies. We want to tune the radio to one of the
stations by reducing the amplitude of the other frequency. The output (up to
phase shifts) is
x∗ (t) = G1 A1 cos ω1 t + G2 A2 cos ω2 t,
where the gains G1 and G2 are given by (18.2). By tuning C −1 ≈ Lω1 we can often
make G2 G1 .
19. Linear systems with constant coefficients
ẋ = Ax + g(t),
x(t) ∈ Rn .
Here A is a constant n × n matrix, g(t) is a given vector-valued function, and x(t)
is the unknown vector-valued function,


x1 (t)
 ... 

x(t) = 
 ... .
xn (t)
19.1. Converting the system into a single n-th order linear equation. This
is usually the most practical method of finding an explicit solution, especially in
the inhomogeneous case. Let us consider the 2D system,
(
ẋ = ax + by + f (t)
,
ẏ = cx + dy + h(t)
i.e.
a b
A=
,
c d
To solve the system, do the following:
f
g=
.
h
(1) use the first equation to express y in terms of ẋ, x, t, namely
y=
ẋ ax f (t)
−
−
;
b
b
b
(19.1)
(2) find ẏ in terms of ẍ, ẋ, t by differentiating (19.1);
(3) use the second equation of the system and the expressions for y and ẏ to derive
a 2nd order equation for x(t);
(4) solve this 2nd order equation and find x(t);
47
(5) use (19.1) to find y(t).
[See Examples 26.1, 26.2 in the text.]
19.2. Vector exponentials. This method applies to homogeneous systems
ẋ = Ax,
x(t) ∈ Rn .
We look for solutions of the form
x(t) = eλt w,
(19.2)
where λ is a real number and w is a constant vector in Rn . Since
d λt e w = λeλt w,
A eλt w = eλt Aw,
dt
we see that (19.2) is a solution if and only if
λw = Aw,
i.e. if and only if λ is an eigenvalue of A and w is a corresponding eigenvector.
Recall that the eigenvalues of A are the roots of the characteristic polynomial
P (λ) = det(λI − A),
where I is the identity matrix. We have n eigenvalues, real or complex, possibly
multiple. For simple eigenvalues, the eigenvectors (in Cn ) are uniquely determined
up to a scalar multiplicative constant. Since the matrix A is real, we can choose
w ∈ Rn if the eigenvalue is real. Eigenvectors corresponding to distinct eigenvalues
are linearly independent.
Theorem. Suppose the eigenvalues λ1 , . . . , λn of A are real and distinct, and let
w1 , . . . wn ∈ Rn be corresponding eigenvectors. Then
x(t) = C1 eλ1 t w1 + · · · + Cn eλn t wn ,
Cj ∈ R,
is the general solution of the system ẋ = Ax.
Proof: The set of solutions is a linear space of dimension n (same argument as in
the case of higher order linear differential equations, see Section 15 of the notes).
The n solutions eλj t wj are linearly independent, and therefore form a basis in the
space of solutions.
2
Complex eigenvalues. Let
w = u + iv,
u ≡ <w ∈ Rn ,
u ≡ =w ∈ Rn ,
be an eigenvector corresponding to λ = α + iω. Then
z(t) = eλt w = eαt (cos ωt + i sin ωt) (u + iv)
is a solution of the complexified equation
ż = Az,
z(t) ∈ Cn .
Since A is real, we get two real solutions:
<z(t) = eαt [cos ωt u − sin ωt v],
=z(t) = eαt [sin ωt u + cos ωt v].
48
If n = 2 and if the eigenvalues are complex, then the general solution of ẋ = Ax is
x(t) = C1 <z(t) + C2 =z(t).
Note. The characteristic polynomial has real coefficients, so complex eigenvalues
appear in conjugate pairs and w̄ := u − iv is an eigenvector corresponding to
λ̄ = α − iω. The complex solution eλ̄t w̄ produces the same real solutions as eλt w.
20. The exponential of a matrix
20.1. Definition. Let A be a constant n by n matrix. Consider the following IVP
for the unknown matrix-valued function X(t):
Ẋ = AX,
X(0) = I
If xjk (t) are the matrix elements of X(t),

ẋ11 (t)
Ẋ(t) =  . . .
ẋn1 (t)
(identity matrix).
(20.1)
then
...
...
...

ẋ1n (t)
... ,
ẋnn (t)
2
so (20.1) is a linear system in Rn . By definition,
etA = X(t),
Theorem.
etA =
in particular
∞
X
Ak
k=0
k!
tk = lim
k→∞
eA = X(1).
I+
tA
k
k
.
Remarks.
(a) Relate to Picard’s and Euler’s approximations respectively, see Section 6 of the
notes.
(b) In the above statement, the limit is understood in terms of the limits of all
matrix elements; same for the infinite sum.
(c) It is not difficult to justify the following formal computations:
!
!
∞
∞
X d
X
d X Ak k
2t 2 3t2 3
Ak k
t
=
=A+ A +
A + ··· = A
t ,
dt
k!
dt
2!
3!
k!
k=0
k=0
and
"
"
k #
k−1
k #
d
tA
d
tA
A
tA
lim I +
= lim
= lim k I +
= A lim I +
.
dt
k
dt
k
k
k
(d) It is not difficult to prove that if AB = BA (which is not always the case!),
then
eA+B = eA eB ,
in particular,
−1
eA
= e−A .
49
20.2. Solving homogeneous and inhomogeneous systems of equations.
Theorem. For every x0 ∈ Rn , the vector-valued function
x(t) = eAt x0
is the solution of the IVP
ẋ = Ax,
x(0) = x0 .
Proof:
ẋ(t) =
d At d At
e x0 =
e
x0 = AeAt x0 = Ax(t).
dt
dt
2
Variation of constants. Consider now the inhomogeneous system
ẋ = Ax + g(t),
x(t) ∈ Rn .
Since etA C with C ∈ Rn is the general solution of the associated homogeneous
system of equations, we look for a particular solution [of the inhomogeneous system]
of the form
x(t) = etA C(t),
where C(t) is the unknown vector-valued function. We have
ẋ = AetA C + etA Ċ,
and this has to be equal to
Ax + g = AetA C + g,
so C(t) satisfies
i.e. Ċ = e−tA g.
etA Ċ = g,
20.3. Computation of etA .
(a) Diagonal matrices:

λ1
0 ...
 0
λ
...
2
A=
. . . . . . . . .
0
0 ...

0
0

. . .
λn
⇒
etA
 λt
e 1
 0
=
...
0
...
...
...
...

0
0
.
. . .
λkn
0
eλ2 t
...
0
...
...
...
...
Proof:
λk1

0
Ak = 
. . .
0

(b) Nilpotent matrices:

0 1
0 0
N =
0 0
0 0
0
1
0
0

0
0

1
0
0
λk2
...
0
⇒
etN

1 t
0 1
=
0 0
0 0
t2
2!
t
1
0
t3
3!
t2 
2!  .

t
1

0
0 
.
... 
eλn t
50
Proof:

0

0
N2 = 
0
0
0
0
0
0
1
0
0
0
(c) Jordan cells:

λ0
0
A=
0
0
1
λ0
0
0

0
1
,
0
0

0

0
N3 = 
0
0

0
0

1
λ0
0
1
λ0
0
0
0
0
0

1
0
,
0
0
0
0
0
0
N 4 = N 5 = · · · = 0.

⇒
1 t

λ0 t 0 1
=e 
0 0
0 0
etA
t2
2!
t
1
0
t3
3!
t2 
2!  .

t
1
Proof: A = λ0 I + N . Since IN = N I, we have
etA = etλ0 I etN = eλ0 t etN .
Note. If P (λ) is the characteristic polynomial of A, then P (A) = 0 (CayleyHamilton theorem). In particular, if P (λ) = (λ − λ0 )n , then (A − λ0 I)n = 0 and
the computation of
etA = eλ0 t et(A−λ0 I)
is very simple (similar to (b)).
(d) Complex eigenvalues:
0 −1
J=
1 0
⇒
eβJ =
cos β
sin β
− sin β
.
cos β
Proof: Since J 2 = −I, J 3 = −J, J 4 = I, . . . , we have
eβJ = I + βJ −
β2
β3
β4
I−
J+
I + · · · = (cos β)I + (sin β)J.
2!
3!
4!
Corollary:
A=
α
ω
−ω
α
⇒
etA = eαt
cos ωt − sin ωt
.
sin ωt cos ωt
Pf: A = αI + ωJ and IJ = JI.
(e) Linear change of variables.
If B = SAS −1 , where S is an invertible n × n matrix, then
etB = SetA S −1 .
Pf: B 2 = SA2 S −1 , B 3 = SA3 S −1 , . . .
Remark. The last observation plus Examples (a)–(d) allow us to compute etA for
all matrices: use S that puts the matrix A into its Jordan canonical form.
51
21. Linear equations with analytic coefficients
21.1. Example: binomial expansion.
y 0 + xy 0 = νy,
(ν ∈ R).
y(0) = 1,
Suppose that the solution y = yν has a power series representation
∞
X
y(x) =
an xn .
n=0
Let us find the coefficients an . We have
∞
X
y0 =
(n + 1)an+1 xn ,
xy 0 =
n=0
∞
X
nan xn ,
n=0
and therefore
ν−n
an
n+1
for all n ≥ 0. From the initial condition y(0) = 1 we find a0 = 1, and so
(n + 1)an+1 + nan = νan ,
an+1 =
ν(ν − 1) 2 ν(ν − 1)(ν − 2) 3
x +
x + ...
2!
3!
If ν ∈ N then yν is a polynomial, and if ν = −1 then yν is the sum of a geometric
progression with radius of convergence equal to 1.
yν (x) = 1 + νx +
Note: yν (x) = (1 + x)ν , of course.
Exercise.
(a) Show that if ν 6∈ N, then the radius of convergence is 1.
(b) Solve the following IVPs by the power series method:
y 0 = y,
y(0) = 1;
y 00 + y = 0,
y(0) = 0, y 0 (0) = 1.
We’ll be considering equations
y 00 + p(x)y 0 + q(x)y = 0
(21.1)
such that both coefficients p and q are analytic in some neighborhood of x0 .
21.2. Power series solutions. Recall that analyticity of p(x) at x0 means that
there is a number R > 0 such that p(x) has a power series representation
X
p(x) =
cn (x − x0 )n
in |x − x0 | < R.
n
We must have R ≤ ρ, where ρ is the radius of convergence of the series,
ρ = lim inf |cn |−1/n .
Note. Sometimes the easiest way to find the radius of convergence is to use the
fact that ρ is the radius of the maximal disc in C in which the function represented
by the series is analytic. In other words, ρ is the distance from x0 to the nearest
singularity. For example, ρ = 1 for p(x) = (1 + x2 )−1 ; note that there are no
singularities on R.
52
Theorem. Suppose that the coefficients of (21.1) have power series representations
in the interval |x − x0 | < R. Then every solution of the equation has a power series
representation in this interval.
Corollary. If p and q are polynomials, then all solution of (21.1) are entire functions, (i.e. they have a power series representation on the whole line).
Idea of proof:
• Write y as a power series with undetermined coefficients.
• From the equation find a recurrence relation between the coefficients.
• This relation plus the initial conditions determine the coefficients uniquely.
• Check that the radius of convergence is a least R.
• The sum of the power series automatically satisfies the equation.
21.3. Example: Airy’s equation.
y 00 − xy = 0.
If
y=
∞
X
an xn ,
n=0
then
xy =
∞
X
y 00 =
an−1 xn ,
∞
X
(n + 2)(n + 1) an+2 xn .
n=0
n=1
Equating the coefficients of xn , we get
a2 = 0,
(n + 2)(n + 1) an+2 = an−1
(n ≥ 1).
It follows that
a0
a3 =
,
3·2
a1
a4 =
,
4·3
a2 = a5 = · · · = 0;
a3
a0
a6 =
=
,
6·5
6·5·3·2
a4
a1
a7 =
=
,
7·6
7·6·4·3
...
...
Clearly,
C
n!
so the radius of convergence is infinite and all solutions are entire functions. The
coefficients are uniquely determined by the initial conditions
|a3n |, |a3n+1 | <
a0 = y0 ,
a1 = y00 .
Let A1 (x) be the solution of IVP(0; 1, 0),
A1 (x) = 1 +
x3
x6
+
+ ...
3·2 6·5·3·2
53
and let A2 (x) be the solution of IVP(0; 0, 1),
x4
x7
+
+ ...
4·3 7·6·4·3
Then A1 and A2 are independent, and so the general solution of the Airy equation
is
y(x) = C1 A1 (x) + C2 A2 (x).
A2 (x) = x +
Note. The functions A1 (x), A2 (x), or rather their certain linear combinations
Ai(x) and Bi(x) known as the Airy functions, are examples of special functions.
Airy functions can not be expressed in terms of elementary functions, but since
they appear in many important applications, their properties have been extensively
studied.
See Fig. 20.3 in the textbook for the graph of A1 (x). Note the difference in the
behavior for negative and positive x. The function oscillates on R− but has a
monotone, super-exponential growth on R+ . This can be explained (informally) as
follows: think of the Airy equation y 00 − xy = 0 on the negative axis as a harmonic
oscillator y 00 + ω(x)2 y = 0 whose frequency depends on x, ω 2 = −x.
21.4. Aside: Riccati equation. Second order linear equations are closely related
to the (first order, non-linear) Riccati equation
u0 = a(x)u2 + b(x)u + c(x).
(Riccati’s equation is linear if a(x) = 0 and Bernoulli if c(x) = 0.)
Lemma. If y is a solutions of y 00 + py 0 + qy = 0, then the function
u=
y0
y
satisfies the first order equation
u0 + pu + q + u2 = 0.
Proof:
u0 =
y 00 y − y 0 y 0
−py 0 y − qy 2 − y 0 y 0
=
.
2
y
y2
2
00
Example. The Airy equation y − xy = 0 corresponds to Riccati’s equation
u0 = x − u2 .
(21.2)
The general solution of (21.2) is therefore
u=
C1 A01 + C2 A02
;
C1 A1 + C2 A2
it depends on just one parameter C2 /C1 . Earlier in the course we mentioned (21.2)
as an example of a ”non-solvable” (in elementary functions) equation.
54
22. Power series method for equations with meromorphic
coefficients
We will now consider equations
P (x)y 00 + Q(x)y 0 + R(x)y = 0
such that the coefficients P, Q, R are analytic at x0 . We can rewrite the equations
in the normal form,
y 00 + p(x)y 0 + q(x)y = 0,
but the coefficients p, q will be meromorphic functions (the ratios of two analytic
functions).
If P (x0 ) 6= 0, then the functions p and q are analytic at x0 and we say that x0 is an
ordinary point. If P (x0 ) = 0, then x0 is a singular point. In this case, we should
consider the equation on the intervals {x > x0 } and {x < x0 } separately.
22.1. Cauchy-Euler equation. This is a model example:
x2 y 00 + αxy 0 + βy = 0,
or
β
α 0
y + 2 y = 0.
x
x
Note that in the last equation, the coefficient of y 0 has a pole of order (at most) 1,
and the coefficient of y has a pole of order (at most) 2.
y 00 +
Consider the interval R+ = {x > 0}, and make the following change of the independent variable:
x = et ,
log x = t,
R+ ↔ R,
0 ↔ −∞.
We have
dy
ẏ
ẏ
= = ,
dx
ẋ
x
(ẏ/x)˙
ÿx − ẏ ẋ
ÿ − ẏ
y 00 =
=
=
.
2
ẋ
x ẋ
x2
The new equation is linear with constant coefficients:
y0 =
(ÿ − ẏ) + αẏ + βy = 0,
or
ÿ + (α − 1)ẏ + βy = 0.
The eigenvalue equation is
λ2 + (α − 1)λ + β = 0.
Traditional notation and terminology: λi = rj are called the exponents at the
singularity, and the equation
F (r) ≡ r(r − 1) + αr + β = 0
is called the indicial equation.
(Shortcut: to derive the indicial equation, substitute y = xr in the C.-E. equation;
note ert = xr .)
55
The exponents at the singularity are
p
(1 − α)2 − 4β
.
2
Theorem. (i) If the exponents are real and distinct, then the general solution is
r1,2 =
1−α±
y(t) = C1 xr1 + C2 xr2 ,
(ii) If they are equal, them
y(x) = C1 xr + C2 xr log x.
(iii) If the exponents are complex a ± iω, then
y(x) = xa C1 cos(ω log x) + xa C2 sin(ω log x).
We can describe the behavior of solutions near the singular point. For example, in
the first case, if both exponents are positive, then all solutions tend to 0 at 0, if
both are negative, then all solutions are unbounded, and if r1 > 0, r2 < 0, then one
solution tends to zero, and the others are unbounded.
The case {x < 0} is similar: just write |x| instead of x.
22.2. Frobenius theory. Consider the equation
y 00 + p(x)y 0 + q(x)y = 0
with meromorphic coefficients, and let x0 be a singular point. The singular point
is called regular if the singularities of p and q at x0 are not worse than those in the
CE equation, i.e. p has a pole of order at most 1, and q has a pole of order at most
2:
α
const
β
p(x) =
+
+ ...,
q(x) =
+ ...,
x − x0
(x − x0 )2
x − x0
where the dots denote analytic functions. Let rj be the roots (”exponents”) of the
indicial equation
F (r) := r(r − 1) + αr + β = 0.
We are interested in the behavior of the solutions near the singular point, and
it is natural to expect that it is similar to the behavior of the solutions of the
corresponding CE equation. This turns out to be true except for the resonance
case
r1 − r2 ∈ Z.
Here is a precise statement. We will only describe the case of real exponents (the
complex case is similar and even simpler since there’s no resonance). Without loss
of generality, we will assume x0 = 0.
Theorem. Suppose that the functions p̃(x) := xp(x) and q̃(x) := x2 q(x) have a
power series representation on the interval (−ρ, ρ), and suppose that the exponents
rj are real. Then the following is true in each of the two intervals (−ρ, 0) and (0, ρ).
(i) If r1 − r2 6∈ Z, then there are two (independent) solutions of the form
yj = |x|rj (1 + . . . ),
where the dots stand for a power series which converges on (−ρ, ρ) and is zero at
zero.
(ii) If 0 ≤ r1 − r2 ∈ Z, then there is a solution y1 of the form as above.
56
22.3. Remarks. (a) The proof (and the use of the theorem) is similar to the analytic case: we introduce undetermined coefficients and find a recurrence relation.
In the equation
x2 y 00 + p̃(x)xy 0 + q̃(x)y = 0,
(x > 0),
we set
X
X
X
an xr+n ,
p̃(x) =
pn xn ,
q̃(x) =
qn xn .
y(x) =
n≥0
Since
xy 0 =
X
(r + n)an xr+n
n≥0
2 00
x y =
X
(r + n)(r + n − 1)an xr+n
n≥0
!
0
p̃(x)xy =
X
X
n≥0
k+l=n
X
X
n≥0
k+l=n
xr+n
(r + l)al pk
!
q̃(x)y =
al qk
xr+n
we get the equation
(r + n)(r + n − 1)an +
X
[(r + l)pk + qk ]al = 0,
k+l=n
i.e.
[(r + n)(r + n − 1) + (r + n)p0 + q0 ]an = −
n−1
X
[(r + l)pn−l + qn−l ]al .
l=0
Note that
(r + n)(r + n − 1) + (r + n)p0 + q0 = F (r + n).
In particular, for n = 0 the recurrence relation gives
F (r)a0 = 0.
We see that the condition F (r) = 0 is necessary for the existence of a non-trivial
solution of the form y = xr · (analytic function). This condition is also sufficient –
we only need to note that if r is the largest exponent, then F (r + n) 6= 0 for n ≥ 1,
and the same is true for the smallest exponent in the non-resonance case. (We also
need to check the convergence of the series.)
(b) In the resonance case, we can find the second (independent) solution using the
Wronskian formula:
Z
R
W
− p
y2 = y1
,
W
=
e
.
y12
One can show that if r1 = r2 , then y2 has a logarithmic term (as in the CE
equation), but if 0 < r1 − r2 ∈ Z, then we may or may not have a logarithmic term.
In fact, it is known that in the resonance case, there is a second solution of the
form
y2 (x) = const y1 (x) log |x| + |x|r2 (1 + . . . ),
where the constant can be zero (or non-zero) if r1 6= r2 .
57
23. Hypergeometric functions
23.1. Infinity as a singular point. In the case of rational coefficients, we can
think of infinity as a singular point. The change of variables t = x−1 transforms
x0 = ∞ to t0 = 0. We have
ẏ
y 0 = = −t2 ẏ,
ẋ
and
2t
ÿ
y 00 = − ẏ − t2 = 2t3 ẏ + t4 ÿ,
ẋ
ẋ
so the new equation is
ÿ + P (t)ẏ + Q(t)y = 0
with
2
1
1
1
1
P (t) = − 2 p
,
Q(t) = 4 q
.
t
t
t
t
t
In particular, ∞ is a regular singular point iff
1
1
, q(x) = O
at ∞.
p(x) = O
x
x2
[Exercise: compute the exponents at infinity in the case of CE equation.]
23.2. Hypergeometric equation. Consider equations with rational coefficients
such that all singular points (in Ĉ, including ∞) are regular. There are no such
equations if the number of singular points is 0 or 1. If the number of singular
points is 2, and the points are 0 and ∞, then the only such equation is CE. The
first non-trivial case is when we have three regular singular points. Without loss of
generality, the points are 0, 1, ∞ (use linear-fractional transformations).
Theorem. In this case, the equation is
x(1 − x)y 00 + [γ − (1 + α + β)x]y 0 − αβy = 0
for some values of the parameters α, β, γ.
The singularities are:
x = 0,
r1 = 0,
r2 = 1 − γ,
x = 1,
r1 = 0,
r2 = γ − α − β,
x = ∞,
r1 = α,
r2 = β.
Consider the singular point x0 = 0. If γ 6= 0, −1, −2, . . . , then there is an analytic
solution (Frobenius series corresponding to r1 = 0)
X
y=
an xn
n≥0
The recurrence relation is
an+1 =
(α + n)(β + n)
an .
(n + 1)(n + γ)
58
Choosing a0 = 1, we get the solution
F (α, β, γ; x) := 1 +
α·β
α · (α + 1) · β · (β + 1) 2
x+
x + ...
1·γ
1 · 2 · γ · (γ + 1)
This is called hypergeometric series. The series converges in (−1, 1). If we denote
(a)n := a(a + 1) . . . (a + n − 1) =
Γ(a + n)
,
Γ(a)
then
F (α, β, γ; x) =
X (α)n (β)n
xn .
(γ)n n!
n≥0
23.3. Special cases. (i) If β = −N ∈ −N, then aN +1 = 0 and the function is a
polynomial of degree N . These polynomials are related to the so called Legendre
polynomials PN , which are defined by the equation
(1 − x2 )y 00 − 2xy 0 + α(α + 1)y = 0.
Here the singular points are ±1, ∞. If we change the variable x 7→ (1 − x)/2, then
we find
PN (x) = const · F (N + 1, −N, 1, (1 − x)/2).
The case α = −N is similar. Exercise: relate to Chebychev’s polynomials defined
by the equation
(1 − x2 )y 00 − xy 0 + α2 = 0.
(ii) There are other elementary hypergeometric functions. For example,
F (1, 1, 2; x) = 1 +
x x2
1
1
+
+ · · · = log
,
2
3
x
1−x
which follows from the relation
an+1 =
n+1
an .
n+2
Another example is
F (α, β, α; x) = (1 − x)−β ,
an+1 =
β+n
.
n+1
We also have
arcsin x
,
x
arctan x
F (1/2, 1/2, 3/2; x2 ) =
.
x
(Hint: differentiate the arc-functions.)
F (1/2, 1/2, 3/2; x2 ) =
(iii) The above examples are exceptional. A typical hypergeometric function is not
elementary. An important example is the elliptic integral
Z π/2
dφ
p
= F (1/2, 1/2, 1; k).
1 − k(sin φ)2
0
59
24. Bessel’s functions
The Bessel equation with parameter ν ≥ 0 is
x2 y 00 + xy 0 + (x2 − ν 2 )y = 0,
(x > 0).
The point x0 = 0 is regular singular with exponents ±ν:
r(r − 1) + r − ν 2 = r2 − ν 2 = 0.
24.1. First solution. We always have the Frobenius solution with r = ν,
∞
X
y1 (x) = xν
an xn .
n=0
The recurrence relation
an = −
an−2
,
n(n + 2ν)
(n ≥ 2);
a1 = 0,
shows that
aodd = 0,
and
a2m =
(−1)m a0
,
22m m!(1 + ν)m
where
Γ(1 + ν + m)
.
Γ(1 + ν)
Note that formula for an ’s also makes sense for non-integer negative ν’s, and the
corresponding series satisfies the Bessel equation.
(1 + ν)m := (1 + ν) . . . (m + ν) =
24.2. Definition. For ν 6= −1, −2, . . . , the Bessel function of order ν is
∞
x 2m
X
(−1)m
.
Jν (x) = xν
m! Γ(1 + ν + m) 2
m=0
This function is the solution with
a0 =
1
.
2ν Γ(1 + ν)
The power series in the definition of Jν has infinite radius of convergence (why?),
so Jν extends to an analytic function in C \ R̄− . If ν ∈ Z+ , then Jν is an entire
function:
∞
z 2m+ν
X
(−1)m
Jν (z) =
.
m! (m + ν)! 2
m=0
E.g., [D. Bernoulli 1732, Bessel 1824]
∞
X
(−1)m z 2m
J0 (z) =
.
[m!]2 2
m=0
If ν 6∈ Z, then the functions J±ν are independent solutions, so we have the following
statement.
Theorem. If ν 6∈ Z, then the general solution of the Bessel equation of order ν is
given by the formula
y = C1 Jν + C2 J−ν .
60
(Note that the resonance case is 2ν 6∈ Z. The second solution has a logarithmic
term iff ν ∈ Z.)
EX: If ν ≥ 0, then all solutions excdept for the multiples of Jν are unbounded at
0. Hint:
Z
dx
y2 = y1
.
xy12 (x)
Several usefull identities:
(xJ1 )0 = xJ0 ,
J2 =
2
J1 − J0 .
x
More generally,
Jν+1 =
2ν
Jν − Jν−1 ,
x
(ν ≥ 1).
Jν+1 = −2Jν0 + Jν−1 ,
(xν Jν )0 = xν Jν−1 ,
(x
−ν
0
Jν ) = −x
−ν
(ν ≥ 1).
(ν ≥ 1).
Jν+1 ,
(ν ≥ 0).
24.3. Elementary Bessel’s functions.
Theorem. If ν = 1/2 + n and n ∈ Z, then Jν is an elementary function. E.g.,
r
r
2
2
J1/2 (x) =
sin x,
J−1/2 (x) =
cos x,
(x > 0).
πx
πx
Proof: If ν = 1/2, then
an = −
an−2
,
n(n + 1)
a2m =
(−1)n a0
,
(2m + 1)!
and so
(−1)m 2m
sin x
x = const √ .
(2m + 1)!
x
m≥0
√
To get the right constant use the value Γ(1/2) = π.
√ X
J1/2 (x) = const x
To get the expressions for J±3/2 , J±5/2 etc use the recurrence formula.
Fact (asymptotics at infinity):
r
h
2
πi
Jν (x) =
cos x − (2ν + 1)
+ O(x−3/2 )
πx
4
as
2
x → ∞.
E.g.,
r
J0 (x) ≈
2
π
cos x −
,
πx
4
r
J1/2 (x) ≈
2
sin x,
πx
r
J1 (x) ≈
2
π
sin x −
.
πx
4
An informal explanation of oscillations: almost constant frequency and negligible
damping if |x| 1.
61
24.4. Bessel’s functions on the imaginary axis. Modified Bessel’s functions Iν
are defined as follows:
x 2m
X
1
Jν (ix) = iν Iν (x),
Iν (x) := xν
,
(x > 0).
m! Γ(m + ν + 1) 2
Note that all coefficients are positive, so Iν (x) → ∞ as x → ∞, no oscillations. For
example,
r
r
2
2
I1/2 =
sinh x,
I1/2 =
cosh x.
πx
πx
Theorem. Iν satisfies the ”modified” Bessel’s equation
x2 y 00 + xy 0 − (x2 + ν 2 )y = 0.
Proof: The expressions x2 (d2 y/dx2 ) and x(dy/dx) don’t change if x 7→ ix but x2 y
becomes −x2 y.
2
If ν 6∈ Z, then the general solution of the modified equation is y = C1 Iν + C2 I−ν .
24.5. Change of variables. Many equations can be solved in terms of Bessel’s
functions. Just two examples:
Lemma. Suppose w(z) is a solution of the Bessel equation with parameter ν. Then
the function
y(x) = xα w(kxβ )
satisfies the equation
x2 y 00 + Axy 0 + (B + Cx2β )y = 0
with
A = 1 − 2α,
B = α2 − β 2 ν 2 ,
C = β 2 k2 .
Example. Airy’s equation is y 00 = xy, or
x2 y 00 − x3 y = 0.
We have
A = 0,
so
α=
B = 0,
C = −1,
1
,
2
1
,
3
ν=
k=
β=
3
,
2
2
i,
3
and the general solution is
√
√
2 3/2
2 3/2
y = C1 xI1/3
x
+ C2 xI−1/3
x
.
3
3
Example. (”Aging spring”)
ÿ + ω 2 e−εt y = 0,
where 0 < ε 1. Change: x = aebt , the constants to be chosen later. We have
ẏ = y 0 ẋ = y 0 bx,
ÿ = y 00 (bx)2 + y 0 b2 x.
Since
e−εt = (x/a)−ε/b ,
62
it follows that if we choose
b = −ε/2,
then the new equation is:
x2 y 00 + xy 0 +
ω2 2
x y = 0.
a2 b2
Finally, we set
a = −ω/b = (2ω)/ε,
so the equation is Bessel of order 0. Its general solution is
2ω −(εt)/2
y(t) = C1 J0 (x) + C2 Y0 (x), x =
e
,
ε
where Y0 is the second independent solution.
Exercise (Riccati equations). We already mentioned that the substitution u = y 0 /y
transforms y 00 + py 0 + qy = 0 into
u0 + pu + q + u2 = 0.
Similarly, the substitution v = −y 0 /y transforms y 00 + py 0 + qy = 0 into
v 0 = pv + q + v 2 .
Solve the following equations in terms of Bessel’s functions:
u0 = x2 − u2 ,
u0 = x − u2
v 0 = v 2 ± x2 ,
v 0 = v 2 ± x.
and