Math 21b
Linear Differential Operators
Linear Spaces. Recall a linear space V is any set that has a zero element and where elements can be added
and scalar multiplied.
1. (zero) 0 is in V .
2. (addition) if f and g are in V then so is f + g.
3. (scalar multiplication) if f is in V and c is a scalar then cf is in V .
Examples. Which of the following examples of function spaces are linear spaces? (Remember: The zero
function is the function z where z(t) = 0 for every input t.)
1. C ∞ the set of smooth functions f : R → R.
Solution. This is a linear space. Recall that a function is smooth if it is continuous and
every derivative of it is also continuous We know from calculus that
dn
dn f
dn g
(f
+
g)
=
+
dtn
dtn
dtn
so if both of the derives on the right are continuous, so is the one on the left. This shows
that a sum of two functions is again smooth. Similarly we can show a scalar multiple of a
smooth function is smooth. The zero function is smooth, because it is a constant function.
2. P the set of polynomials.
Solution. This is a linear space. Constant functions, and in particular the zero function,
are polynomials. The sum of two polynomials is a polynomial, and a scalar multiple of a
polynomial is a polynomial.
∞
the set of smooth functions f : R → R which are 2π-periodic (that is, f (t + 2π) = f (t)).
3. Cper
Solution. This is a linear space. We already showed the set of smooth functions is a linear
space. so we just need to deal with the periodicity requirement. The zero function is constant,
so it is 2π-periodic. The sum of periodic functions is periodic, as is a scalar multiple of a
periodic function.
4. V the set of smooth functions f : [0, 2π] → R with f (0) = f (2π).
Solution. This is a linear space. (In fact, it is equivalent to the space of 2π-periodic functions—we are just focusing on one period.)
Differential Operators. A differential operator is a linear transformation T : C ∞ → C ∞ of the form
T (f ) = f (n) + an−1 f (n−1) + · · · + a1 f 0 + a0 f
where a0 , a1 , . . . , an−1 are (possibly complex) constants. The order of T is n, the highest derivative in the
above expression.
Note: We require a coefficient of 1 in front of f (n) for convenience. In practice, we can always scale things
to make this happen.
Fact. A linear operator T is a linear transformation T : C ∞ → C ∞ . That is,
1. T (f + g) = T (f ) + T (g)
2. T (cf ) = cT (f )
Examples. For each of the following differential operators find the kernel. What is the kernel’s dimension?
1. T (f ) = f 0
Solution. A function f is in the kernel of T precisely when f 0 = 0. So, the kernel is the
space of constant functions. This is one dimensional.
2. T (f ) = f 0 − λf
Solution. A function f is in the kernel of T precisely when it satisfies the differential equation
f 0 = λf , which has solutions f (t) = beλt where b is an arbitrary constant. This is again one
dimensional.
3. T (f ) = f 00 + k 2 f
Solution. The kernel of T consists of solutions to the harmonic oscillator, so f (t) = a cos(kt)+
b sin(kt). It is two dimensional, since cos(kt) and sin(kt) form a basis for ker T .
Theorem. The kernel of an n-th order differential operator T is n-dimensional. It consists of solutions to
the differential equation T (f ) = 0. Differential equations of this form are called homogeneous.
Eigenfunctions. An eigenfunction for a differential operator T is a non-zero function f so that
T (f ) = λf
for some constant λ called the eigenvalue of f .
Example. Consider the differential operators T (f ) = f 000 − 6f 00 − 4f 0 + 24f and D(f ) = f 0 .
1. What are the eigenvalues and eigenfunctions for D?
Solution. A function f is an eigenfunction for D with eigenvalue λ whenever f 0 = λf . This
always has solutions f (t) = beλt . So, every real number λ is an eigenvalue for D, and the
λ-eigenfunctions are f (t) = beλt for any constant b.
2. Suppose f is an eigenfunction for D with eigenvalue λ. What is T (f )?
Solution. Suppose f (t) = beλt is a λ-eigenfunction for D. Then
T (beλt ) = (beλt )000 − 6(beλt )00 − 4(beλt )0 + 24beλt
= λ3 beλt − 6λ2 beλt − 4λbeλt + 24beλt
= (λ3 − 6λ2 − 4λ + 24)beλt .
In other words, f (t) = beλt is also an eigenfunction for T but with eigenvalue λ3 −6λ2 −4λ+24.
3. Use your answer to part (2) to find all functions of the form f (t) = eλt which solve the differential
equation f 000 − 6f 00 − 4f 0 + 24f = 0
Solution. The functions f (t) = eλt are eigenfunctions of T with eigenvalue λ3 −6λ2 −4λ+24.
Therefore
f 000 − 6f 00 − 4f 0 + 24f = 0
(λ3 − 6λ2 − 4λ + 24)eλt = 0
λ3 − 6λ2 − 4λ + 24 = 0
λ2 (λ − 6) − 4(λ − 6) = 0
(λ2 − 4)(λ − 6) = 0
So for f (t) = eλt to be a solution, either λ = ±2 or λ = 6. Therefore we get three solutions
of the prescribed form, namely e2t , e−2t and e6t .
4. Find the generic solution to f 000 − 6f 00 − 4f 0 + 24f = 0.
Solution. The solutions to this differential equation are the same as ker T , which is 3dimensional, as T has order 3. But in part (3) we found three linearly independent solutions
e2t , e−2t and e6t (they are linearly independent because eigenfunctions with different eigenvalues are linearly independent). These solutions then span ker T and so the generic solution
to the differential equation is
f (t) = c1 e2t + c2 e−2t + c3 e6t
for constants c1 , c2 , c3 .
Characteristic Polynomial. The characteristic polynomial of a differential operator T (f ) = f (n) +
an−1 f (n−1) + · · · + a1 f 0 + a0 f is
pT (λ) = λn + an−1 λn−1 + · · · + a1 λ + a0 .
Theorem. Suppose T is an order n differential operator and pT (λ) its characteristic polynomial.
1. The function eλt is an eigenfunction for T with eigenvalue pT (λ), i.e., T (eλt ) = pT (λ)eλt .
2. If pT (λ) has n distinct roots then the functions eλ1 t , . . . , eλn t is a basis for ker T .
Example. Consider the differential equation f 00 + 4f 0 + 4f = 0.
1. Find the solutions of the form f (t) = eλt .
Solution. The characteristic polynomial is pT (λ) = λ2 + 4λ + 4 = (λ + 2)2 which has a single
root λ = −2 with multiplicity 2. So the only solution of the form eλt is e−2t .
2. Show that teλt is also a solution.
Solution. In the first part we found that λ = −2. To verify the two above functions are also
solutions, we need to compute a bunch of derivatives:
(te−2t )0 = e−2t − 2te−2t
(tet )00 = −2e−2t + −2(e−2t − 2te−2t ) = −4e−2t + 4te−2t
(tet )000 = 12e−2t − 8te−2t
Then
f 00 + 4f 0 + 4f = (12e−2t − 8te−2t ) + 4(−4e−2t + 4te−2t ) + 4(e−2t − 2te−2t ) = 0
so te−2t is also a solution.
3. What is the general solution to the differential equation?
Solution. The differential operator D2 + 4D + 4 has order 2, so its kernel is 2-dimensional.
The functions e−2t and te−2t are linearly independent and in the kernel, so are a basis for
it. Therefore the general solution is
f (t) = c1 e−2t + c2 te−2t .
Recap. Suppose T is an order n differential operator and pT (λ) its characteristic polynomial.
1. If pT (λ) has n distinct roots λ1 , . . . , λn then the functions eλ1 t , . . . , eλn t are linearly independent
solutions to the differential equation T (f ) = 0.
2. If λ is a repeated root of pT (λ) with multiplicity m, then the functions eλt , teλt , t2 eλt , . . . , tm−1 eλt are
linearly independent solutions to the differential equation T (f ) = 0.