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18.034 SOLUTIONS TO PROBLEM SET 9
Due date: Friday, April 30 in lecture. Late work will be accepted only with a medical note or for another Instituteapproved reason. You are strongly encouraged to work with others, but the final writeup should be entirely your own and based on your own understanding.
Each of the following problems is from the textbook. The point value of the problem is next to the problem.
(1)(5 points) p. 403, Problem 5 Solution: The trace is Trace(A) = 6, and the determinant is det(A) = 2· 4 −(−1)1 = 9. Therefore the characteristic polynomial is,
pA(λ) = λ^2 − Trace(A)λ + det(A) = λ^2 − 6 λ + 9 = (λ − 3)^2.
Therefore there is one eigenvalue λ = 3 with multiplicity 2. Because the matrix is not diagonal, the eigenspace is deficient. Therefore there is a generalized eigenvector. The matrix A − 3 I is,
A − 3 I = −^1 − 1 1
A generalized eigenvector is, (^) � � 1 v 2 = 0 ,
and the corresponding vector v 1 = (A − 3 I)v 2 is,
v^ −^1 1 =^ − 1.
The changeofbasis matrix is, (^) � �
U = [v 1 |v 2 ] =
Then AU = U D where D is the matrix in Jordan canonical form,
D = 3 1 0 3
Therefore, exp(tA)U = U exp(tD).
The solution space of the system, z�(t) = Dz(t),
has basis, (^) � � � �
z 1 (t) = 1 e^3 t, z 2 (t) = 1
t 0 e^3 t^.
Therefore the basic solution matrix is,
Z(t) = [z 1 (t)|z 2 (t)] = 1 t^ e^3 t^. 0 1
Of course Z(0) = I, therefore exp(tD) = Z(t). Therefore,
exp(tA) = U exp(tD)U −^1. 1
By Cramer’s rule, (^) � � � �
U −^1 = (1/det(U ))
Therefore, (^) � � � � � � − 1 1 1 t (^3) t 0 − 1 exp(tA) = (^) − 1 0 0 1 e 1 − 1.
Simplifying, this gives,
3 t (−t^ + 1)e^3 t^ te^3 t exp(tA) = =
−t + 1 t −t t + 1 e^ −te^3 t^ (t + 1)e^3 t.
To doublecheck, observe the derivative is,
d (^) (− 3 t + 2)e^3 t^ (3t + 1)e^3 t 3 t. dt
exp(tA) = (− 3 t − 1)e^3 t^ (3t + 4)e
This is the same as,
2 1 (−t + 1)e^3 t^ te^3 t − 1 4 −te^3 t^ (t + 1)e^3 t =^ A^ exp(tA).
And exp(0A) = I.
(2)(5 points) p. 403, Problem 11
Solution: This is one of the very few situations where the power series definition of the matrix exponential is useful. Observe that, ⎡ ⎤ ⎡ ⎤ ⎡ ⎤ 0 0 0 0 0 0 0 0 0 A^2 = ⎣^2 0 0 ⎦ ⎣^2 0 0 ⎦^ = ⎣^0 0 0 ⎦^ , 3 4 0 3 4 0 8 0 0
and, (^) ⎡ ⎤ ⎡ ⎤ 0 0 0 0 0 0 A^3 = ⎣^2 0 0 ⎦ ⎣^0 0 0 ⎦^ = 0. 3 4 0 8 0 0
Therefore An^ = 0 for n ≥ 3. So the power series,
� (^) tn exp(tA) = An^ ,
∞
n=0^ n!
reduces to, (^) ⎡ ⎤
1 1 0 0 exp(tA) = I + t + t^2 A^2 = ⎣^2 t 1 0 ⎦^. (^2 4) t (^2) + 3t 4 t 1
Therefore the solution of the IVP, (^) ⎧ ⎪⎪⎨ x�(t) =^ A⎡ x(t)⎤, 1 ⎪ ⎪⎩ x(0) =^ ⎣^2 ⎦ 3
has solution, ⎡ ⎤ ⎡ ⎤ ⎡ ⎤ 1 0 0 1 1 x(t) = exp(tA)x(0) = ⎣^2 t 1 0 ⎦ ⎣^2 ⎦^ = ⎣^2 t + 2 ⎦^. 4 t^2 + 3t 4 t 1 3 4 t^2 + 8t + 3 2
0
Therefore, (^) � � � � � � (^1 1) e−t^ + et = =. −e−t^ + 3e^ t e−t^ − et 1 exp(tA)x (^2) − 3 e−t (^) + 3e t (^3) e−t (^) − et 2 2 3 e−t (^) + et
Also, (^) � � � � s (^3) es exp(−sA)F(s) =
(^1) −es^ + 3e−s^ e − e−s s^. 2 −^3 es^ + 3e−s^3 e^ −^ e−s^ s
Simplifying, this gives, s exp(−sA)F(s) =
(^1) − 3 e^2 s^ + 9 + se − se−s s^. 2 −^9 e^2 s^ + 9 + 3se^ −^ se−s
So the final answer is, � (^) t � � (^) t t � � � (^) s � (^1) e−t^ + e 1 −e−t^ + 3e e−t^ − e t^1 − 3 e^2 s^ + 9 + se − se−s x(t) = (^2 3) e−t (^) + e t + (^2) − 3 e−t (^) + 3et (^3) e−t (^) − et (^2) − 9 e 2 s (^) + 9 + 3ses − se−s ds. 0
The following was NOT ASKED FOR IN THE EXERCISE. However, I can’t resist mentioning how much simpler the solution is if one does not use the matrix exponential. Since we have the transition matrix, the simplest solution is to change the basis,
x(t) = U z(t).
Then, (^) � �
0 1 1 z = U −^1 x^0 = , 2 1 and, (^) � � � � (^1 ) G(t) = U −^1 F(t) =
et^ + 2 −^1
t. 2 9
The original inhomogeneous IVP is equivalent to the IVP,
z�(t) = Dz(t) + G(t) z(0) = z^0
Now this is a set of two uncoupled inhomogeneous 1st^ order linear IVPs, ⎧ (^3) 2
1 2
t (^) + z 1 (0)^1 ⎨ z 1 z 1 = − e t, = 2
⎩ (^9 2 1) z 1 2 �^2 (0)
By the method of undetermined coefficients, particular solutions of the inhomogeneous ODEs (but not of the initial conditions!) are, ⎧
z − z 2 = 2 e − 2 t, = 2
3 4 e^
1 2
1 ⎨ z 1 (t) = − (^2) t (^) + t −
⎩ (^) z 2 (t)^ =^92 tet^ +^12 t^ +^12
The general solution of the homogeneous equation is z 1 (t) = C 1 e−t^ , z 2 (t) = C 2 et^. Therefore the solution of the system of IVPs is, ⎧ (^3) 4 e
t (^) + 1 2 t^ −^
1 2 +^
7 4 e ⎨ z 1 (t)^ =^ − −t ⎩ (^) z 2 (t)^ =^92 tet^ +^12 t^ +^12
In vector form, (^) � �
z(t) =
(^1) − 3 et^ + 2t − 2 + 7e−t 4 18 tet^ + 2t^ + 2^
4
Therefore, (^) � � � �
x(t) = U z(t) =
(^1 1 1) − 3 et^ + 2t − 2 + 7e−t 4 3 1 18 tet^ + 2t^ + 2^
Simplifying, this gives,
x(t) =
(^1 18) tet^ − 3 et^ + 4t + 7e−t 4 18 tet^ −^9 et^ + 8t^ −^ 4 + 21e−t^
(4)(10 points) p. 403, Problem 25
Solution: First of all, (^) � � 1 t exp(tA) = I + tA = 0 1 ,
and, (^) � � et 0 exp(tB) = 0 1
Thus, (^) � � � � � � 1 t et 0 et t exp(tA) exp(tB) = 0 1 0 1 = 0 1.
Also, (^) � � � � � � et 0 1 t et tet exp(tB) exp(tA) = 0 1 0 1
Therefore exp(tA) exp(tB) does not equal exp(tB) exp(tA).
Also, (^) � � 1 1 A + B = 0 0.
Because this is an upper triangular matrix, pA+B (λ) = (λ − 1)(λ − 0). So the eigenvalues are λ 1 = 0 and λ 2 = 1. The eigenvector for λ 1 = 0 is,
1 v 1 = (^) − 1.
The eigenvector for λ 2 = 1 is,
v^1 1 =^0.
So the changeofbasis matrix is,
1 1 U = [v 1 |v 2 ] = (^) − 1 0.
And (A + B)U = U D where D is the diagonal matrix,
0 0 D = 0 1.
Of course, (^) � � 1 0 exp(tD) = (^0) et.
By Cramer’s rule,
U −^1 = 0 −^1. 1 1
Therefore, (^) � � � � � � 1 1 1 0 0 − 1 exp(t(A + B)) = U exp(tD)U −^1 = (^) − 1 0 0 et 1 1. 5
The response of the system is unbounded iff ω = −2.
(6)(10 points) p. 420, Problem 11
Solution: The goal is to prove Theorem 6.8.5 (which, incidentally, is essentially equivalent to our Green’s kernel solution). Let c(t) be a continuously differentiable vectorvalued function. Consider the continuously differentiable vectorvalue function,
x(t) = Φ(t, t 0 )c(t).
This is a solution of the IVP iff Φ�(t, t 0 )c(t) + Φ(t, t 0 )c�(t) = A(t)Φ(t, t 0 )c(t) + F(t), Φ(t 0 , t 0 )c(t 0 ) = 0
By hypothesis, Φ�(t, t 0 ) = A(t)Φ(t, t 0 ) and Φ(t 0 , t 0 ) = In. Therefore x(t) is a solution of the IVP iff,
Φ(t, t 0 )c�(t) = F(t), c(t 0 ) = 0
By the last part of Theorem 6.8.4, Φ(t 0 , t)Φ(t, t 0 ) = In. Therefore x(t) is a solution of the IVP iff,
c�(t) = Φ(t 0 , t)F(t), c(t 0 ) = 0
By the Fundamental Theorem of Calculus, there is a unique solution of this IVP, and it is given by, t c(t) = Φ(t 0 , s)F(s)ds. t 0 Therefore, the unique solution of the IVP is given by, t x(t) = Φ(t, t 0 )c(t) = Φ(t, t 0 ) Φ(t 0 , s)F(s)ds. t 0
The second part of Theorem 6.8.5 follows easily from the first part.
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