Math 411 Week 5
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Math 411 week 5
This is a summary of topics from the textbooks (Nonlinear Dynamics and chaos by steven strogatz and The Theory of Diferential Equations by kelley and peterson) discussed in class.
Kelley and peterson 4.2: The Matrix Exponential Function
Recalling from Math 25o, we know the solution to the 1-d linear system
.
① = a①
is
① (t) = ceat
where c is any constant, and applying an initial condition ① (o) = ①0 allows us to solve for c = ①0 . Further- more, from Math 141 we remember the power series representation of eat :
eat = 1 + at + (at)2 + (at)3 + . . . — 钝 < t < 钝.
we would now like to think of solving the n-dimensional linear system
x = Ax
in a similar framework. Namely, we are interested in a matrix valued function eAt such that the solution is given by
x(t) = eAt c,
( c1 )
' c2 '
c = ' . '
' . '
(' cn(.) )'
is any constant vector. (If you,re solving an initial value problem you would use the initial condition x(o)
ea(0) = e0 = 1
for the familiar case, we will have that
eA(0) = I,
where I is the n X n identity matrix, and that furthermore
eAt = I + tA + A2 + A3 + . . . , —钝 < t < 钝.
This is nice because it provides a generalization to exponentiation in the case of matrices. Notice that since we know how to raise square matrices to any positive integer power and add them together the power series expression at least makes sense formally — but it turns out all the components are absolutely convergent series so that the above formula makes sense from the standpoint of convergence as well. So indeed this “matrix exponential” is a full generalization of the familiar exponential function eat.
The power series expression is valid in theory but does not lead to a useful concrete formula in general. There is a way to obtain a nice formula following the linear algebra technique of diagonalization (which is delicate in the case of repeated eigenvalues — if the dimension of the eigenspace of a repeated eigenvalue is strictly less than the algebraic multiplicity of the eigenvalue in the characteristic polynomial then diagonalization is not possible). However, there is a more ODE-based algorithm for getting a useful formula for a matrix exponential called the putzeT algoTithm which we can use. Notably, it uses the technique of undetermined coe伍cients from Math 25O to take care of any repeated eigenvalue issues using ideas you already know.
The putzer algorithm
For any n X n constant matrix A, for now we deine the matrix exponential function eAt to be the (matrix- valued) solution to the matrix-valued IVp
X、= AX, X(O) = I,
where I is the n X n identity matrix.
Theorem 1. putzer Algorithm for eAt Let λ1 , λ2 , . . . , λn be the (not necessaTilg distinct) eigenvalues of the
matTi从 A. Then,
eAt = pk+1(t)Mk ,
wheTe MO := I,
k
Mk :=Ⅱ (A — λiI),
i=1
foT 1 三 k 三 n — 1 and the vectoT function p defned bg
l p1 (t) 」
' '
p(t) = ' '
' . . . '
[' pn (t) l'
foT —钝 < t < 钝 is the solution of the IⅤp
l λ 1 O O . . . O 」 l 1 」
' 1 λ2 O . . . O ' ' O '
p、= ' O 1 λ3 . . . O ' p, p(O) = ' O ' .
' . . ' ' . '
' . . . . . ' ' . '
[' O(.) . . . O 1 λ.n l' [' O(.) l'
Example 1. use the putzer algorithm to ind eAt when
A = l 1
[ — 1
1 」
3 l.
For the n = 2 case, the algorithm reads
eAt = p1 (t)Mo + p2 (t)M1 ,
where
Mo = I = l 1 O 」,
1
we ind and label the eigenvalues by solving det(A - λI) = O for λ :
det [(l) 1 -1λ 3 1 λ l(」) = O,
and so we have
(1 - λ)(3 - λ) - (-1) = O,
λ2 - 4λ + 4 = O,
(λ - 2)2 = O,
and therefore λ1 = λ2 = 2 is a repeated eigenvalue with (algebraic) multiplicity 2. (If we were using the eigenvector method to solve this system, we,d need to deal with the possibly (and in this case actually) deicient eigenspace for this repeated eigenvalue. The advantage to the putzer algorithm for matrix expo- nentiation is that ODE techniques from 25O/251 will take care of any possible issues automatically using
what you,ve already learned.) we then immediately have
M1 = A - λ 1 I = A - 2I = [(l) - 1(1)
and the system for p reads
1 」
1 l ,
p1(、) = λ 1p1 = 2p1 , p1 (O) = 1,
p2(、) = p1 + λ2p2 = p1 + 2p2 , p2 (O) = O.
we solve in order — since the p1 ODE is decoupled from the rest and we always take p1 (O) = 1, we,ll always have p1 (t) = eλ1 t and so in this case
p1 (t) = e2t.
Then we substitute this into the p2 ODE and rearrange to obtain
p2(、) - 2p2 = e2t, p2 (O) = O.
This is an inhomogeneous constant coe伍cient linear ODE, and we solve using 25O/251 techniques. The characteristic equation is
T - 2 = O
leading to the complementary solution
p2,c = ce2t.
The right hand side / inhomogeneity of the p2 ODE has matchup with the complementary solution, and so our particular solution guess using the technique of undetermined coe伍cients is
p2,p = ate2t
(remember the rule is the multiply by t “enough”times so that there is no matchup between your undeter- mined coe伍cient guess and the complementary solution). we substitute p2,p into the p2 ODE to solve for the undetermined coe伍cient a:
(ae2t + 2ate2t) - 2(ate2t) = e2t)
ae2t = e2t)
a = 1
to obtain
p2 = p2,c + p2,p = ce2t + te2t.
Now we can use the initial condition p2 (O) = O to solve for the unknown constant coming as part of the complimentary solution:
O = c + O)
and so c = O and we have
p2 (t) = te2t.
putting it all together,
eAt = p1 (t)Mo + p2 (t)M1 = e2t i(l) O(1) |
O 」 1 l |
+ te2t |
l -1 i -1 |
1 」 1 l |
= e2t |
l i |
1 - t -t |
t 1 + t |
」 l. |
Example 2. use the putzer algorithm to ind eAt when
A = l 1
i 5
For the n = 2 case we again have
-1 」
-1 l.
eAt = p1 (t)Mo + p2 (t)M1)
where
Mo = I = l 1 O 」,
1
we ind and label the eigenvalues by solving det(A - λI) = O for λ :
and so we have
(1 - λ)(-1 - λ) - (-5) = O,
λ2 - 1λ + 5 = O,
λ2 = -4,
and therefore λ1 = 2i, λ2 = -2i are a conjugate pair of complex eigenvalues. (which you pick as λ1 or λ2 doesn,t matter as long as you,re consistent.) we then immediately have
M1 = A - λ 1 I = A - 2iI = [(l) 15(-) 2i 1-1 2i l(」) ,
and the system for p reads
p1(、) = λ 1p1 = 2ip1 , p1 (O) = 1,
p2(、) = p1 + λ2p2 = p1 - 2ip2 , p2 (O) = O.
we solve in order — since the p1 ODE is decoupled from the rest and we always take p1 (O) = 1, we,ll always have p1 (t) = e)1 t and so in this case
p1 (t) = e2it.
(we can keep our intermediate steps complex but everything will work out to be real valued provided A has real entries.) Then we substitute this into the p2 ODE and rearrange to obtain
p2(、) + 2ip2 = e2it, p2 (O) = O.
This is an inhomogeneous constant coe伍cient linear ODE, and we solve using 25O/251 techniques (they work the same even with i). The characteristic equation is
T + 2i = O
leading to the complementary solution
p2,c = ce-2it.
The right hand side / inhomogeneity of the p2 ODE has no matchup with the complementary solution (+2it vs -2it), and so our particular solution guess using the technique of undetermined coe伍cients is
p2,p = ae2it
we substitute p2,p into the p2 ODE to solve for the undetermined coe伍cient a:
(2iae2it) + 2i(ae2it) = e2it ,
4iae2it = e2it ,
a =
to obtain
p2 = p2,c + p2,p = ce-2it + e2it.
Now we can use the initial condition p2 (O) = O to solve for the unknown constant coming as part of the
complimentary solution:
O = c + 1
and so c = - and we have
p2 (t) = (e2it - e-2it).
The fact that eit = cost + isin t, cosine is even, and sine is odd mean that we have the useful identities
cost = , sint = ,
and so we can more compactly write
1
p2 (t) =
putting it all together,
eAt = p1 (t)Mo + p2 (t)M1
= e2it [(l) O(1) 1(O) l(」) + 2(1) sin(2t) [(l) 15(-) 2i -1--1 2i l(」) = = ( cos(2t) + isin(2t)) [(l) O 1(1 O) l(」) + 2(1) sin(2t) [(l) 15(-) 2i
= [ (Notice the imaginary parts canceled!) |
1 - 2 cos(2t) |
sin(2t) 1 2 |
」 l. |
Example 3. Find all solutions to the diferential equation
= Ax,
l 2 O
'
A = ' 1 2
[' 1 O
-1
-1 - 2i
」
l
For n = 3, there is one more step to the algorithm. we have
eAt = p1 (t)Mo + p2 (t)M1 + p3 (t)M2
where
Mo = I,
1
M1 = Ⅱ (A - λKI) = (A - λ 1 I),
K=1
2
M2 = Ⅱ (A - λKI) = (A - λ 1 I)(A - λ2 I),
K=1
p1(、) = λ 1p1 , p1 (O) = 1,
p2(、) = p1 + λ2p2 , p2 (O) = O,
p3(、) = p2 + λ3p3 , p3 (O) = O.
First we ind and label the eigenvalues, the roots to det(A - λI) = O:
l 2 - λ
det ' 1 2 [' 1
(2 - λ) det |
l 2 - λ [ O |
3 O λ l(」) - Odet [(l) 1 3(1) O λ l(」) |
+ Odet |
O 」
' O '
3 - λ l' 2 - λ 」 O l
= O,
= O,
(2 - λ)2 (3 - λ) = O,
and so we can take λ1 = λ2 = 2, λ3 = 3. Then,
l
M1 = A - λ 1 I = A - 2I = '
[
M2 = (A - λ 1 I)(A - λ2 I) =
['
Then
p1(、) = λ 1p1
」 ' ' , l(') O 」 l O ' ' O ' ' 1 1 l' [' 1 |
O O O |
O O 1 |
p1 (O) = 1
」 l O
' '
' = ' O
l' [' 1
O 」
'
O ' .
1 l'
means
p1 (t) = e2t.
The ODE for p2 becomes
p2(、) = p1 + λ2p2 = e2t + 2p2 , p2 (O) = O,
which using the exact same undetermined coe伍cients calculation as the previous example yields
p2 (t) = te2t.
The ODE for p3 becomes
p3(、) = p2 + λ3p3 = te2t + 3p3 , p3 (O) = O,
and so we rearrange to obtain
p3(、) - 3p3 = te2t.
The characteristic equation is T - 3 = O and so the complementary solution isp3,c = ce3t. For a particular solution we use undetermined coe伍cients and guess
p3,p = ate2t + be2t
(remember your guess has to include all linearly independent terms arising from diferentiation — since the right hand side inhomogeneity is te2t and because of the product rule, our undetermined coe伍cients guess has to include terms te2t and e2t). Then, plugging in p3,p into the p3 ODE to solve for a,b yields
(ae2t + 2ate2t + 2be2t) - 3ate2t - 3be2t = te2t ,
and so we have
(-a)te2t + (a - b)e2t = te2t + Oe2t ,
and so a = -1, b = -1 upon matching coe伍cients. Therefore,
p3 = p3,c + p3,p = ce3t - te2t - e2t ,
and p3 (O) = O means c = 1 and therefore
p3 (t) = e3t - te2t - e2t.
putting it all together,
eAt = p1 (t)Mo + p2 (t)M1 + p3 (t)M2
l 1 = e2t ' O [' O |
O 1 O |
O 」 l O O ' + te2t ' 1 1 l' [' 1 |
|
2023-07-26