ODEs
Second Linear order linear equations
Many traditional problems involving ordinary equations arise as second order linear equations
au + bu + cu = f ,
more briey as Lu = f .
The problem is, given f , nd u ; often we will want to nd u that satises some auxiliary initial or
boundary conditions.
Here we have used the notation
Lu := a(x)u + b(x)u + c(x)u,
(1)
so L takes a function u and gives a new function Lu . This operator L is a linear map because it has
the two properties
L(u) = Lu and L(u + v) = Lu + Lv,
where is any constant and u and v are functions. One consequence is that if Lu = 0 and Lv = 0,
then L(Au + Bv) = 0 for any constants A and B . The solutiond of Lu = 0 are often called the
nullspace or kernel of L . These properties show that the nullspace of a linear map is a linear space.
For instance, in the special case where Lu = u + u we know that L cos x = 0 and L sin x = 0. Thus
L(A cos x + B sin x) = 0 for any constants A and B .
E XAMPLE Well show that any solution of Lu := u + u = 0 has the form
u(x) = A cos x + B sin x.
This will show that the nullspace of Lu := u + u = 0 has dimension two.
First we must pick the constants A and B . Letting x = 0 we see that (if this is to work) A = u(0) .
Similarly, taking the derivative we get B = u (0) . Let v(x) := u(0) cos x + u (0) sin x . Our task is to
show that u(x) = v(x) . Equivalently, if we let w(x) := u(x) v(x) , we must show that w(x) 0. A
key observation motivating us is that by linearity, w + w = 0, and w(0) = 0, w (0) = 0.
Introduce the function
1
E (x) = [w 2 + w2 ].
2
Then since w = w ,
E (x) = w w + ww = w (w) + ww = 0,
so E (x) = constant (in many physical examples, this is conservation of energy). But from w(0) = 0
and w (0) = 0 we nd E (0) = 0. Since E (x) is a sum of squares, the only possibility is w(x) 0,
as claimed.
This Example generalizes. Assuming a(x) = 0, the nullspace of (1) always has dimension 2. Let
(x) and (x) be the solutions of the homogeneous equation Lu = 0 with (0) = 1, (0) = 0,
and (0) = 0, (0) = 1, then every solution of the homogeneous equation Lu = 0 has the form
u(x) = A(x) + B(x) for some constants A and B . The proof, which we dont give, has two parts.
The rst is the existence of the solutions and , the second is their uniqueness. While these
proofs are not obvious, they are not killers.
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For (1) and other linear ordinary and partial differential equations, it is surprising that if one
knows the general solution of the homogeneous equation Lu = 0 one can nd an explicit formula
for a particular solution of the inhomogeneous equation Lu = f .
Based on our experience with the rst order linear inhomogeneous equation u + au = f it is plausable to seek u in the form u(x) = p(x)v(x) where p(x) is chosen cleverly to make the equation for
v simple to solve. We do this for Lu := u + u = f (the general case of (1) is then routine). Clearly
u = pv + p v and u = pv + 2 p v + p v
so
Lu = pv + 2 p v + ( p + p)v.
This clearly simplies if we pick p as a solution of the homogeneous equation p + p = 0. But we
know two solutions of this, cos x and sin x . Which should we use? After some experimentation,
Lagrange decided he should use both and instead sought u in the more general form
u = pv + qw,
(2)
where for our example p(x) = cos x and q(x) = sin x . Now he had one equation, f = Lu = pv +
2 p v + qw + 2q w for the two unknowns, v(x) and w(x) so he could impose another condition.
After some experimenting he imposed the condition
pv + qw = 0,
(3)
which resulted in the two linear equations:
f = Lu = p v + q w
0 = pv + qw ,
and
that is,
f = ( sin x)v + (cos x)w
0 = (cos x)v + (sin x)w .
and
He solved these for v and w :
v (x) = sin x f (x) and w (x) = cos x f (x).
Thus integrating and using (2), we obtain the simple formula for a particular solution, upart , of the
inhomogeneous equation Lu = f :
Zx
upart (x) = cos x
sin s f (s) ds + sin x
Zx
0
Zx
cos s f (s) ds =
0
sin(x s) f (s) ds
(4)
0
To get the general solution of the inhomogeneous equation we simply add the general solution of
the homogeneous equation:
Zx
u(x) = A cos x + B sin x +
sin(x s) f (s) ds.
0
In honor of George Green we often write (4) in the symbolic form
Zx
u(x) =
G(x, s) f (s) ds
0
2
(5)
and call G(x, s) := sin(x s) Greens function for the equation Lu = f . The point is that (5) can
be thought of as writing u = L1 f so we have a conceptually satisfying formula for the inverse
operator L1 .
Lagranges procedure for nding the formula (4) for a particular solution of the inhomogeneous
equation is called variation of parameters. The key step is to seek u in the form (2) with Lp = 0
and Lq = 0.
We now carry out the details for the general case of the general second order equation
Lu := u + b(x)u + c(x)u = f (x).
(6)
Note that here the coefcient of u is 1 (if not, then divide by it).
Using equation (2) and the condition (3) he found that
u = p v + q w,
and u = p v + q w + pv + qw.
Substitute these into the equation (6) for L . After a computation short that uses Lp = Lq = 0, we
get the simple formula
Lu = p v + q w .
(7)
To solve Lu = f we thus need to nd v and w that satisfy this and (3):
pv + qw = 0
p v +q w = f.
These are two linear equations for v and w . Their solution is
v=
q f
W
and w =
pf
,
W
where W (x) := pq p q (called the Wronskian of p and q ). Integrating we nd v and w and
thus from (2), a particular solution upart
upart = p(x)
Zx
q(s) f (s)
0
W (s)
ds + q(x)
where
G(x, s) :=
Zx
p(s) f (s)
0
W (s)
Zx
ds =
G(x, s) f (s) ds,
(8)
0
q(x) p(s) p(x)q(s)
W (s)
is Greens function for the problem. In the special case of u + u = f done earlier, p(x) = cos x and
q(x) = sin x so W (x) = 1 and g(x, s) = sin(x s) , just as in (4).
First order linear systems
Next consider the rst order system of equations
LU := U (x) + A(x)U (x) = F (x),
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(9)
where U and F are vectors with n components and A(x) is an n n matrix. We assume that both
A and F depend continuously on x .
A typical problem is to seek a solution of (9) that satises some initial condition U (0) = C , where
C Rn is a given vector. If A(x) and F (x) are both periodic with period P , another typical problem
is to seek a periodic solution U (x) [the simplest scalar example u = 1 has no periodic solutions
with any period and shows that answering this question may involve some work].
The homogeneous equation
A general theorem, which well not prove (it is not a killer) is
Theorem 1 . Given any constant C Rn , the homogeneous equation LU = 0 has a unique solution
satisfying U (0) = C .
Note that has a unique solution means the same as has one and only one solution.
Let e1 := (1, 0, . . . , 0) , e2 := (0, 1, 0, . . . , 0) , . . . , en := (0, 0, . . . , 0, 1) be the standard basis vectors in
Rn . It is useful to use the special solutions 1 (x) ,. . . , n (x) that satisfy the homogeneous equation
L j (x) = 0 with j (0) = e j and use them to construct the n n matrix (x) whose columns are
the vectors 1 (x) ,. . . , n (x) . Then satises
(x) + A(x)(x) = 0,
and the initial condition (0) = I .
This matrix (x) is sometimes called the fundamental solution matrix.
E XAMPLE 1 Let
U (x) =
u1 (x)
,
u2 (x)
A :=
0 1
,
1
0
and
F (x) =
f1 (x)
,
f2 (x)
so the system of equations LU := U + AU = F is
u1 u2 = f1
(10)
u2 + u1 = f2 .
The vectors
1 (x) :=
cos x
,
sin x
2 (x) :=
sin x
cos x
both satisfy the homogeneous equation L j = 0 with initial conditions 1 (0) = e1 , 2 (0) = e2 ,
so the fundamental solution matrix is
(x) :=
cos x sin x
sin x cos x
(11)
The inhomogeneous equation
Next we show that if you know a fundamental matrix solution (x) for the homogeneous equation,
then you can nd a formula for a particular solution of the inhomogeneous equation LU = F , that
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is, U + AU = F . As before, seek U in the special form U (x) = S(x)V (x) , where S(x) is an n n
matrix. The goal is to choose a clever S so the resulting differential equation for V (x) is simple.
Clearly
LU = SV + (S + AS)V.
This evidently simplies dramatically if S + AS = 0, so we let S(x) = (x) be the fundamental
matrix solution of the homogeneous equation L = 0. Because (0) = I , we know that S(x) is
invertible, at lease for x near 0 (In fact, it is invertible for all x . We leave that for you).
The equation LU = F is thus SV = F so V (x) = S1 (x)F (x) . Integrating this we can obtain the
desired particular solution, Upart of LU = F . Since we just want a particular solution, we can let
Upart (0) = 0, which implies V (0) = 0. Thus the desired formula is:
Zx
Upart (x) =S(x)V (x) = S(x) V (0) +
Zx
=
S(x)S1 (s)F (s) ds =
0
S1 (s)F (s) ds
(12)
G(x, t )F (s) ds,
(13)
0
Zx
0
where G(x, t ) := S(x)S1 (s) is Greens function for this problem.
E XAMPLE 1 ( CONTINUED ) We are now in a position to write a formula for a particular solution
of LU = F for Example 1 above. Then (11) is the fundamental matrix solution for the homogeneous equation, S(x) = (x) . Since this happens to be an orthogonal matrix, its inverse is just the
transpose. Consequently
G(x, s) = (x)1 (s) =
=
cos x sin x
sin x cos x
cos s sin s
sin s
cos s
cos(x s) sin(x s)
.
sin(x s) cos(x s)
(14)
(15)
Consequently
Zx
Upart (x) =
0
cos(x s) sin(x s)
sin(x s) cos(x s)
f1 (s)
d s,
f2 (s)
(16)
E XAMPLE 2 We can write any second order equation u + bu + cu = f as a rst order system by
letting u1 (x) = u(x) and u2 (x) = u (x) . Then, using the differential equation,
u1 = u2
and u2 = u = bu2 cu1 + f .
that is,
u1 (x)
u2 (x)
+
0 1
c
b
u1 (x)
=
u2 (x)
0
f (x)
In the special case of u + u = f we have b = 0 and c = 1 so the previous equation becomes
u1 (x)
u2 (x)
+
0 1
1
0
u1 (x)
=
u2 (x)
5
0
,
f (x)
which is exactly (10) with f1 = 0 and f2 = f . Now (16) gives a formula for a particular solution of
this inhomogeneous equation. It is
Zx
Upart (x) =
0
Zx
=
0
cos(x s) sin(x s)
sin(x s) cos(x s)
0
ds
f (s)
sin(x s)
f (s) ds
cos(x s)
Since u1 (x) = u(x) , this formula is exactly the same as (4) found earlier.
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