LINEAR DIFFERENTIAL EQUATIONS

MINSEON SHIN

1. Existence and Uniqueness

Theorem 1 (Existence and Uniqueness). [1, NSS, Section 6.1, Theorem 1]1 Sup-pose p1(x), . . . , pn(x) and g(x) are continuous real-valued functions on an inter-val (a, b) that contains the point x0. Then, for any choice of (initial values)γ0, . . . , γn−1, there exists a unique solution y(x) on the whole interval (a, b) to thenonhomogeneous differential equation

y(n)(x) + p1(x)y(n−1)(x) + · · ·+ pn(x)y(x) = g(x) (1)

for all x ∈ (a, b) and y(i)(x0) = γi for i = 0, . . . , n− 1.

Corollary 2. Suppose p1(x), . . . , pn(x) are continuous real-valued functions on aninterval (a, b) containing the point x0. Let V be the set of solutions2 to the homo-geneous differential equation

y(n)(x) + p1(x)y(n−1)(x) + · · ·+ pn(x)y(x) = 0 . (2)

Then V is a vector space, and the function T : V → Rn defined by

y(x) 7→

y(x0)y′(x0)

...y(n−1)(x0)

is a linear isomorphism. In particular, V is an n-dimensional vector space.

Proof. Check that V is closed under addition, scalar multiplication, and containsthe zero vector (zero function). Check that T is linear. Surjectivity of T followsfrom the existence part of Theorem 1. Injectivity of T follows from the uniquenesspart of Theorem 1. �

As was the case for systems of linear equations (see [1, Lay, Section 1.5]), the set ofsolutions to a homogeneous system (1) is a subspace of the vector space of contin-uous real-valued functions on (a, b), and the set of solutions to a nonhomogeneoussystem (2) is a translation of this subspace by a particular solution yp.

Corollary 3. Suppose p1(x), . . . , pn(x) are differentiable real-valued functions (forexample, constants) on an interval (a, b). Then any solution y to (2) is smooth(i.e. y is infinitely differentiable). Furthermore, the values of y(k)(x0) for k ≥ nare determined by y(x0), y′(x0), . . . , y(n−1)(x0).

1Notation: y(n) = dnydxn .

2It is implicit that any solution y(x) to (2) is n-times differentiable (i.e. y′, y′′, . . . , y(n) all

exist).

1

2 MINSEON SHIN

Proof. Rearrange (2) to get

y(n)(x) = −(p1(x)y(n−1)(x) + · · ·+ pn(x)y(x))

whose RHS is differentiable (since it is obtained as the sum of products of dif-ferentiable functions). Thus y(n)(x) is differentiable. Differentiate inductively toget

y(k)(x) = −(p1(x)y(k−1)(x) + · · ·+ pn(x)y(k−n)(x))

for any k ≥ n. �

Note that [1, NSS, Section 4.2, Theorem 1] is a special case of [1, NSS, Section6.1, Theorem 1] when n = 2, the function g(x) is identically zero, and (a, b) =(−∞,∞) = R.

Proposition 4 (Extension of solutions to a larger interval). Suppose p1(x), . . . , pn(x)and g(x) are continuous real-valued functions on R. Suppose −∞ ≤ a2 ≤ a1 < b1 ≤b2 ≤ ∞, and suppose y : (a1, b1)→ R is a solution to (1). Then there exists a uniquefunction y : (a2, b2) → R satisfying (1) for all x ∈ (a2, b2) and y(x) = y(x) for allx ∈ (a1, b1).

Proof. Choose t0 ∈ (a1, b1). By the existence part of Theorem 1 (applied to thecase a = a2, b = b2), there exists a function y : (a2, b2)→ R which satisfies (1) andhas the same initial conditions as y, i.e. y(k)(t0) = y(k)(t0) for all k = 0, . . . , n− 1.Since y and y have the same initial conditions at t0 and satisfy the same differentialequation (1), the uniqueness part of Theorem 1 (applied to the case a = a1, b = b1)shows that y(t) = y(t) for all t ∈ (a1, b1). Thus such y having the desired propertiesexists.

To show that such y is unique, suppose that y : (a2, b2) → R also satisfies (1) andhas the same initial conditions as y. The uniqueness part of Theorem 1 (applied toa = a2, b = b2) shows that y(t) = y(t) for all t ∈ (a2, b2), so that y = y. �

2. The Wronskian

Definition 5. Let y1, . . . , yn be (n−1)-times differentiable functions. The Wronskianof y1, . . . , yn is the function

W [y1, . . . , yn](x) := detMy1,...,yn(x)

where

My1,...,yn(x) :=

y1(x) y2(x) · · · yn(x)y′1(x) y′2(x) · · · y′n(x)

......

. . ....

y(n−1)1 (x) y

(n−1)2 (x) · · · y

(n−1)n (x)

=[T (y1(x)) T (y2(x)) · · · T (yn(x))

]where T is the linear isomorphism in Corollary 2.

Proposition 6. Let y1, . . . , yn be (n − 1)-times differentiable functions on (a, b).Suppose that {y1, . . . , yn} is linearly dependent. Then W [y1, . . . , yn](x) = 0 for allx ∈ (a, b).3

3Note that Definition 5 and Proposition 6 make no reference to the equation (1).

LINEAR DIFFERENTIAL EQUATIONS 3

Proof. Let c1, . . . , cn be scalars, not all zero, such that

c1y1(x) + · · ·+ cnyn(x) = 0 (3)

for all x ∈ (a, b). Differentiating (3) k times gives

c1y(k)1 (x) + · · ·+ cny

(k)n (x) = 0 .

This means

My1,...,yn(x)

c1...cn

=

0...0

(4)

for all x ∈ (a, b). ThusMy1,...,yn(x) is not invertible for all x. ThusW [y1, . . . , yn](x) =

detMy1,...,yn(x) = 0 for all x. �

Corollary 7. Let y1, . . . , yn be (n−1)-times differentiable functions on (a, b). Sup-pose W [y1, . . . , yn](x0) 6= 0 for some x0 ∈ (a, b). Then {y1, . . . , yn} is linearlyindependent on (a, b).

Proof. This is the contrapositive of Proposition 6. �

Example 8. The converse to Proposition 6 is false. Consider y1(x) = x2 and y2(x) =x · |x|. Then y′1(x) = 2x and y′2(x) = 2|x|. Thus

W [y1, y2](x) = det

[x2 x · |x|2x 2|x|

]= 0

for all x ∈ R, but x2 and x · |x| are linearly independent on R. (They are linearlydependent on (−∞, 0) and (0,∞), separately, though.)4

Theorem 9 (Wronskian Criterion). [1, NSS, Section 6.1, Theorem 2] Supposep1(x), . . . , pn(x) are continuous real-valued functions on an interval (a, b). Supposey1, . . . , yn are n arbitrary solutions to (2). Then the following are equivalent:

(i) {y1, . . . , yn} is linearly independent;5

(ii) the Wronskian W [y1, . . . , yn](x) 6= 0 for all x ∈ (a, b).

In other words, the following are equivalent:

(i’) {y1, . . . , yn} is linearly dependent;

(ii’) the Wronskian W [y1, . . . , yn](x0) = 0 for some x0 ∈ (a, b).

Proof. (i’) =⇒ (ii’): Follows directly from Proposition 6.

(ii’) =⇒ (i’): Since W [y1, . . . , yn](x0) = 0, we can find scalars c1, . . . , cn, notall zero, satisfying (4). This means c1T (y1(x)) + · · · cnT (yn(x)) = 0 (where Tis the linear isomorphism defined in Corollary 2). Since T is injective, we havec1y1(x) + · · ·+ cnyn(x) = 0. Thus {y1(x), . . . , yn(x)} is linearly dependent. �

4Why couldn’t we do this with x and |x|? This is because we need a second power of x to“smooth out” |x| at the origin (i.e. make it differentiable).

5It is important to note that linear independence is conditional on the domain of the functions.For example, {t, |t|} is linearly dependent when the domain is (0,∞) but linearly independent whenthe domain is (−∞,∞).

4 MINSEON SHIN

Corollary 10. Suppose p1(x), . . . , pn(x) are continuous real-valued functions onan interval (a, b). Suppose y1, . . . , yn are n arbitrary solutions to (2). Then theWronskian W [y1, . . . , yn](x) is either always zero or always nonzero.

Example 11. Theorem 9 requires that you take the Wronskian of n solutions, wheren is the order of the differential equation. The criterion is no longer true if youcompute the Wronskian of less than n solutions. For example, consider y2(x) = xand y3(x) = ex, which are both solutions to the 3rd order differential equationy′′′ − y′′ = 0. We have

W [y2, y3](x) = det

[x ex

1 ex

]= (x− 1)ex

which has a root at x = 1 even though y2, y3 are linearly independent on anyinterval (which, at first glance, seems to contradict Theorem 9). Only when youthrow in the third basis element (of the space of solutions to y′′′ − y′′ = 0), forexample y1(x) = 1, do you get

W [y1, y2, y3](x) = det

1 x ex

0 1 ex

0 0 ex

= ex

which has no roots (i.e. always nonzero).

3. Linear second-order constant-coefficient differential equations

(This is from [1, NSS, Section 4.2, 4.3].) We consider the vector space V of solutionsy(x) to the differential equation

ay′′ + by′ + cy = 0 (5)

where a 6= 0. By Corollary 2, we have that V is a 2-dimensional vector space. Theassociated characteristic equation is

ar2 + br + c = 0 (6)

which has discriminant b2 − 4ac.

(i) If b2 − 4ac > 0, then (6) has two distinct real roots, say r1 and r2. Weobserve that y1(x) = er1x and y2(x) = er2x are two elements of V (i.e.they are solutions to (5)), and {y1, y2} is linearly independent (which canbe shown by computing the Wronskian). Since dimV = 2, we have V =Span{y1, y2}.

(ii) If b2−4ac = 0, then (6) has one real root of multiplicity 2, say r. We observethat y1(x) = erx and y2(x) = xerx are two elements of V (i.e. they aresolutions to (5)), and {y1, y2} is linearly independent (which can be shownby computing the Wronskian). Since dimV = 2, we have V = Span{y1, y2}.

(iii) If b2 − 4ac < 0, then (6) has two complex conjugate roots, say r + ciand r − ci. We observe that y1(x) = e(r+ci)x = erx(cos cx + i sin cx) andy2(x) = e(r−ci)x = erx(cos cx− i sin cx) are two elements of V (i.e. they aresolutions to (5)), and {y1, y2} is linearly independent (which can be shownby computing the Wronskian). Since dimV = 2, we have V = Span{y1, y2}.

LINEAR DIFFERENTIAL EQUATIONS 5

References

[1] Lay, D. C., Nagle, R. K., Saff, E. B., Snider, A. D. Linear Algebra & Differential Equations,Second Custom Edition for University of California, Berkeley. Pearson, 2012.