Download Topic 19 Notes 19 Variation of parameters; exponential inputs; Euler’s method Jeremy Orloff

Topic 19 Notes
Jeremy Orloff
19 Variation of parameters; exponential inputs;
Euler’s method
19.1 Variation of parameters
As with ordinary first order linear equations we have the method of variation of parameters for solving inhomegeneous equations. Since this involves integration, matrix
inverses and matrix multiplication it is our last choice. Nonetheless, sometimes it’s
the only method available. In addition, the derivation of the formula is really very
pretty.
Suppose we have a fundamental matrix Φ(t) for the homogeneous linear equation
x0 = A(t)x
(H)
Remember this means that Φ has columns which are solutions to (H).
19.2 Variation of parameters formula
Now suppose we want to solve
x0 = A(t)x + F(t).
(I)
The equation (I) has solution given by the variation of parameters formula
Z
−1
x(t) = Φ(t) ·
Φ(t) · F(t) dt + C .
Proof. (Remember this.) We will use a form of the method of optimism to derive
this formula.
We know the general homogeneous solution is x = Φ · c for a constant vector c. The
vector c is called a parameter. Variation of parameters is an old-fashioned way of
saying let’s optimistically make it a (dependent) variable u(t). So, we try a solution
of the form x(t) = Φ(t) · u(t). The function u(t) is unknown, we substitute our guess
into (I) and see what u(t) needs to be:
Φ0 · u + Φ · u0 = AΦ · u + F
So, (don’t forget Φ0 = AΦ.)
AΦ · u + Φ · u0 = AΦ · u + F ⇒ Φ · u0 = F.
1
19 Variation of parameters; exponential inputs; Euler’s method
This last equation is easy to solve
0
−1
u =Φ
Z
· F ⇒ u(t) =
Φ−1 (t) · F(t) dt + C.
We take this formula for u(t) and use it in our trial solution
Z
−1
x(t) = Φ(t) · u(t) = Φ(t) ·
Φ(t) · F(t) dt + C .
QED.
0
Example 19.1. Solve x =
6 5
1 2
x+
et
e5t
.
1
answer: Let’s introduce some notation to save typing: A =
,F=
.
t
et 5e7t
.
We know the fundamental matrix from an earlier example: Φ(t) =
−et e7t
7t
−8t
e −5e7t
So, Φ−1 (t) = e 6
.
et
et
6 5
1 2
Calculating with the variation of parameters we get
Z
x = Φ(t) Φ−1 (t) · F(t) dt
t
Z −8t 7t
e
e −5e7t
e
= Φ(t)
· 5t dt
t
t
e
e
e
6
Z 1
1 − 5e4t
dt
= Φ(t)
6 e−6t + e−2t
1
t − 45 e4t + c1
= Φ(t)
− 1 e−6t − 1 e−2t + c2
6
t 56 5t 5 2t 5 5t
1 te − 4 e − 6 e − 2 e + c1 et + 5c2 e7t
=
6 −tet + 54 e5t − 16 et − 12 e5t − c1 et + c2 e7t
1
1
1
t
5t −15/4
t −5/6
t
7t 5
=
te
+e
+e
+ c1 e
+ c2 e
.
−1
3/4
−1/6
−1
1
6
Notice the homogeneous solution appearing with the constants of integration.
19.2.1 Definite integral version of variation of parameters
The equation (I) with initial condition xt0 = b has definite integral solution
Z t
−1
x(t) = Φ(t)
Φ (u) · F(u) du + C where C = Φ−1 (t0 ) · b.
t0
2
19 Variation of parameters; exponential inputs; Euler’s method
19.3 Exponential input theorem
The above example of the variation of parameters formula should be enough to convince you that we’d like better techniques for solving inhomogeneous systems. Fortunately, for many of the systems we care about in 18.03 this is the case. The first
formula we’ll look at is for exponential input.
Exponential input theorem. For a constant matrix A and a constant vector k the DE
x0 = Ax + eat k
has a particular solution:
xp = −eat (A − aI)−1 k
This formaula is valid as long as A − aI is invertible, i.e. as long as a is not an
eigenvalue of A.
Proof. Not surprisingly we discover this formula by the method of optimism. We
try a solution of the form xp (t) = eat v, where v is a constant vector.
Plug the guess into the DE and solve for v:
xp 0 = aeat v = eat Av + eat k ⇒ −(A − aI)v = k ⇒ v = −(A − aI)−1 k.
Thus, we have found a particular solution xp (t) = eat v = −eat (A − aI)−1 k. QED
2t x
6 5
x
e
Example 19.2. Find a solution to
=
+
3e2t
y
1 2
y
0
2t 1
. The exponential input
answer: We rewrite the equation as x = Ax + e
3
−1 1
4 5
1
2t
−1
2t
theorem gives us a solution xp = −e (A − 2I)
= −e
3
1 0
3
2t
0
5
1
15
= − 51 e2t
= − e5
1 −4
3
−11
2t x
6 5
x
3e
Example 19.3. Solve
=
+
.
y
1 2
y
5e3t
0
3
2t
3t
answer: Write the input as e
+e
. Now you can solve the equation
0
5
for each input term and then use superposition.
There are more examples in the next section.
19.4 Exponential input theorem examples
Example 19.4. Find the general solution to
3
x0 = 3x − y + e2t
y 0 = 4x − y − e2t
19 Variation of parameters; exponential inputs; Euler’s method
1
3 −1
2t
. The expoanswer: In matrix form the equation is x =
x+e
−1
4 −1
nential input theorem tells us a particular solution is
−3 1
1
1
2t 4
2t
2t
−1
.
=e
= −e
xp = −e (A − 2I)
5
−4 1
−1
−1
0
We’ll let you verify the calculation of the inverse. Likewise we’ll let you find the
homogeneous solution needed for the general solution.
Example 19.5. Find a particular solution to
x0 = 3x − y + 3
y 0 = 4x − y + 2
answer: Note that we could use the exponential input theorem, where the exponent
is a = 0. Instead we’ll just try a constant solution and solve for its exact value.
3
0
Try x = v. Substitution into the DE gives x = 0 = Av +
.
2
−1 3 −1
3
1
1
−1 3
So, v = −A
=−
=
. That is xp (t) =
.
2
4 −1
2
6
6
Again, we’ll let you verify the calculation of the inverse.
0 1 2
x
cos(t)
x
=
+
.
Example 19.6. Find a particular solution to
y0
2 1
y
0
answer: To use the exponential input theorem we first need to use complex replacement. The complexified equation is
1 2
0
it 1
z =
z+e
, where x = Re(z).
2 1
0
Now we set up the exponential input theorem
−1
1
1−i
2
1 − i −2
−1
=
(A − iI) =
2
1−i
−2i − 4 −2 1 − i
−2
1
1
1
it
−1 1
it 1 − i
it 1 − i
= 2i+4 e
.
So, zp = −e (A − iI)
= − −2i−4 e
0
−2 1 − i
0
−2
To find the real part of zp we work in polar coordinates. First we write the various
complex numbers in polar form:
√
2i + 4 = 2 5eiφ1 , where φ1 = Arg(2i + 4) = tan.5 in the first quadrant.
√
Likewise 1 − i = 2eiφ2 ; , where φ2 = −π/4 .
√ iφ √ i(t+φ −φ ) 2
1
−eit
1
2e
2e 2
So zp = √
=− √
.
−2
−2ei(t−φ1 )
2 5eiφ1
2 5
Taking the real part:
√
xp = Re(zp ) =
2 cos(t + φ2 − φ1 )
−2 cos(t − φ1 )
4
19 Variation of parameters; exponential inputs; Euler’s method
Here is the same calculation in rectangular coordinates. I think the arithmetic is
more error prone and the answer is harder to interpret.
4 − 2i 1 − i
1
1
1−i
1 − 3i
=
=
.
−2
2i + 4 −2
20
10 −4 + 2i
So,
1
1
1 − 3i
cos(t) + 3 sin(t) + i(sin(t) − 3 cos(t))
=
zp = (cos(t)+i sin(t))
−4 + 2i
10
10 −4 cos(t) − 2 sin(t) + i(−4 sin(t) + 2 cos(t))
Thus,
1
xp (t) = Re(zp (t)) =
10
cos(t) + 3 sin(t)
.
−4 cos(t) − 2 sin(t)
19.5 Euler’s method
Euler’s method works without change for systems.
Consider a first order system: x = F(x, t), x(t0 ) = x0 .
Using stepsize h we have the same algorithm as for ordinary DE’s.
m = F(xn , tn ) ⇒ xn+1 = xn + hm, tn+1 = tn + h.
Again, just like for ordinary DE’s, there are other, better, algorithms for choosing m
or varying h.
0
y
1
x
=t
, x(1) =
.
Example 19.7. Consider
x
0
y
x
Let x =
and use h = .5 to estimate x(2).
y
n
0
1
2
tn
xn
1
0
1
.5
1.375
1.25
1.0
1.5
2.0
m = F(xn , tn )
0
1
.75
1.5
5