#$&*
course Mth 279
5/24 @2:21 amGetting the rest submitted as soon as I type them up,
gonna make one last long night out of it." "Query 29 Differential Equations
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Question: Find the propagator matrix Phi(t) = e^(t A) for the system
y ' = [5, -4; 5, -4] y
and use to find y(3) given that y(1) == [ 1, 0].
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Your solution:
P(lambda) = det[ A - lambda*I] = 0
=[ (5-lambda), -4; 5, (-4-lambda)]
= (5-lambda)(-4-lambda) - (-4)(5) = 0
= lambda^2 -5*lambda + 4*lambda -20 + 20
= lambda^2 - lambda
= lambda (lambda - 1) = 0
Lambda_1 = 0
Lambda_2 = 1
for lambda_1 = 0, we solve (A)x_1 = 0
[ 5, -4; 5, -4][x_1; x_2] = [ 0; 0]
RREF
[ 1, (-4/5); 0, 0][ x_1; x_2] = [ 0; 0]
Which yeilds the equation:
x_1 - (4/5)x_2 = 0
A solution set would be:
x_1 = [ 4; 5]
for lambda_2 = 1, we solve (A - I)x_2 = 0
[ 4, -4; 5, -5][x_1; x_2] = [ 0; 0]
RREF
[ 1, -1; 0, 0][ x_1; x_2] = [ 0; 0]
Which yeilds the equation:
x_1 - x_2 = 0
A solution set would be:
x_2 = [ 1; 1]
T = [ 4, 1; 5, 1]
det[T] = -1
T^(-1) = -1*[ 1, -1; -5, 4] = [ -1, 1; 5, -4]
T^(-1)AT = D
= [ 0, 0; 0, 1]
e^(tA) = T L(t) T^(-1), where L(t) is capital Lambda of t
= [ 4, 1; 5, 1][ 1, 0; 0, e^t][ -1, 1; 5, -4]
= [4, e^t; 5, e^t][ -1, 1; 5, -4]
= [ 5e^t - 4, -4e^t + 4; 5e^t - 5, -4e^t + 5]
As a check:
de^(tA)/dt = Ae^(tA)
(e^(tA))' = [ 5e^t, -4e^t; 5e^t, -4e^t]
Ae^(tA)
[ 5, -4; 5, -4][ 5e^t - 4, -4e^t + 4; 5e^t - 5, -4e^t + 5] = [ 5e^t, -4e^t; 5e^t, -4e^t]
Using y(s) = y(1) = [ 1; 0] to find y(t) = y(3)
y(t) = phi(t - s)y(s)
[ 5e^t - 4, -4e^t + 4; 5e^t - 5, -4e^t + 5][ 1; 0] = y(t) = [ 5e^t - 4; 5e^t - 5]
(t - s) = 3 -1 = 2
final solution becomes:
[ 5e^2 - 4; 5e^2 - 5]
confidence rating #$&*:
^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution:
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Self-critique (if necessary):
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Self-critique rating:
@&
This is correct for t = 3.
Be sure you see the general method, which will propagate the y (1) solution to any t and will therefore provide a specific solution to the equation for all t, subject to the given initial condition.
The propagator matrix is
phi(t, s) = psi(t) * psi^-1(s)
= [ 4, e^t; 5, e^t ] * (-1 / e^s) * [e^s, -e^s; -5, 4]
= [ 5 e^(t-s) - 4, -4 e^(t-s) + 4; 5 e^(t-s) - 5, -4 e^(t-s) + 5 ].
To propagate y(1) = (1, 0) we propagate from s = 1 to t using the matrix
phi(t, s) = phi(t, 1) = [ 5 e^(t-1) - 4, -4 e^(t-1) + 4; 5 e^(t-1) - 5, -4 e^(t-1) + 5 ], obtaining
y(t) = phi(t, 1) * y(1)
= [ 5 e^(t-1) - 4, -4 e^(t-1) + 4; 5 e^(t-1) - 5, -4 e^(t-1) + 5 ] * [1; 0 ]
= [ 5 e^(t-1) - 4; 5 e^(t-1) - 5 ].
You can easily verify that for t = 3 and s = 1 this reduces to your given solution.
We easily verify that when t = 1, this solution is y(1) = [1; 0].
We also verify that this is a solution to the original equation
y ' = [5, -4; 5, -4] y
obtaining
[5 e^(t-1); 5e^(t-1) ] = [5, -4; 5, -4] * [ 5 e^(t-1) - 4; 5 e^(t-1) - 5 ]
= [ 25 e^(t-1) - 20 +(-20 e^(t-1) + 20); 25 e^(t-1) - 20 +(-20 e^(t-1) + 20) ]
= [5 e^(t-1); 5e^(t-1) ] .
*@
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Question: Use a propagator matrix to find y(1), given y ( 0 ) = [1; 1; 0], for the system
y ' = [ 1, 1, 1; 0, 2, 1; 0, 0, -1 ] y.
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Your solution:
P(lambda) = det[A - lambda*I] = 0
= [(1 - lambda), 1, 1; 0, (2 - lambda), 1; 0, 0, (-1-lambda)]
= (1-lambda)[(2-lambda), 1; 0, (-1-lambda)] - 1[ 0, 1; 0, (-1-lambda)] + 1[ 0, (2-lambda); 0, 0]
= (1-lambda)(2-lambda)(-1-lambda)
= (-lambda + 1)(-lambda + 2)(-lambda - 1)
= - (lambda - 1)(lambda - 2)(lambda + 1)
Lambda_1 = 1
Lambda_2 = 2
Lambda_3 = -1
for Lambda_1 = 1, we solve (A - I)x_1 = 0
[0, 1, 1; 0, 1, 1; 0, 0, -2][ x_1; x_2; x_3] = [ 0; 0; 0]
RREF
[ 0, 1, 0; 0, 0, 1; 0, 0, 0][ x_1; x_2; x_3] = [ 0; 0; 0]
Which yeilds equations:
x_2 = 0
x_3 = 0
A solution set could be:
x_1 = [ 1; 0; 0]
Since x_1 can be any number.
for Lambda_2 = 2, we solve (A - 2I)x_2 = 0
[-1, 1, 1; 0, 0, 1; 0, 0, -3][ x_1; x_2; x_3] = [ 0; 0; 0]
RREF
[ 1, -1, 0; 0, 0, 1; 0, 0, 0][ x_1; x_2; x_3] = [ 0; 0; 0]
Which yeilds equations:
x_1 - x_2 = 0
x_3 = 0
A solution set could be:
x_2 = [ 1; 1; 0]
for Lambda_3 = -1, we solve (A + I)x_3 = 0
[2, 1, 1; 0, 3, 1; 0, 0, 0][ x_1; x_2; x_3] = [ 0; 0; 0]
RREF
[ 1, 0, (1/3); 0, 1, (1/3); 0, 0, 0][ x_1; x_2; x_3] = [ 0; 0; 0]
Which yeilds equations:
x_1 + (1/3)x_3 = 0
x_2 + (1/3)x_3 = 0
A solution set could be:
x_3 = [ 1; 1; -3]
T = [ 1, 1, 1; 0, 1, 1; 0, 0, -3]
det[T] = [ 1, 1; 0, -3] - [ 0, 1; 0, -3] + [ 0, 1; 0, 0] = -3
T^(-1) = 1/det [adj(M)]
M = [ 1, 1, 1; 0, 1, 1; 0, 0, -3]
M^T = [ 1, 0, 0; 1, 1, 0; 1, 1, -3]
mini matricies determinants of M^T
M_11 = [ 1, 0; 1, -3] = -3
M_12 = [ 1, 0; 1, -3] = -3
M_13 = [ 1, 1; 1, 1] = 0
M_21 = [ 0, 0; 1, -3] = 0
M_22 = [ 1, 0; 1, -3] = -3
M_23 = [ 1, 0; 1, 1] = 1
M_31 = [ 0, 0; 1, 0] = 0
M_32 = [ 1, 0; 1, 0] = 0
M_33 = [ 1, 0; 1, 1] = 1
adj(M) = [ -3, -3, 0; 0, -3, 1; 0, 0, 1][+, -, +; -, +, -; +, -, +] = [ -3, 3, 0; 0, -3, -1; 0, 0, 1]
T^(-1) = 1/det [adj(M)]
= -1/3[ -3, 3, 0; 0, -3, -1; 0, 0, 1] = [ 1, -1, 0; 0, 1, (1/3); 0, 0, (-1/3)]
T^(-1)AT = D
[ 1, -1, 0; 0, 1, (1/3); 0, 0, (-1/3)][ 1, 1, 1; 0, 2, 1; 0, 0, -1 ][ 1, 1, 1; 0, 1, 1; 0, 0, -3] = [ 1, 0, 0; 0, 2, 0; 0, 0, -1]
Phi(t) = e^(tA) = TL(t)T^(-1) where L(t) is used to represent capital lambda.
= [ 1, 1, 1; 0, 1, 1; 0, 0, -3][ e^t, 0, 0; 0, e^(2t), 0; 0, 0, e^(-t)][ 1, -1, 0; 0, 1, (1/3); 0, 0, (-1/3)]
= [e^t, e^(2t), e^(-t); 0, e^(2t), e^(-t); 0, 0, -3e^(-t)][ 1, -1, 0; 0, 1, (1/3); 0, 0, (-1/3)]
= [e^t, -e^t + e^(2t), (1/3)e^(2t) - (1/3)e^(-t); 0, e^(2t), (1/3)e^(2t) - (1/3)e^(-t); 0, 0, e^(-t)]
y(t) = Phi(t-s)y(s)
y(s) = y ( 0 ) = [1; 1; 0]
y(t) = y(1) = ?
[e^t, -e^t + e^(2t), (1/3)e^(2t) - (1/3)e^(-t); 0, e^(2t), (1/3)e^(2t) - (1/3)e^(-t); 0, 0, e^(-t)][1; 1; 0]
= [e^t - e^t + e^(2t); e^(2t); 0] = [ e^(2t); e^(2t); 0]
(t - s) = 1 - 0 = 1
= [e^2; e^2; 0]
confidence rating #$&*:
^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
3
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Given Solution:
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Self-critique (if necessary):
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Self-critique rating:
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Question: Given the solution matrix
psi(t) = [t, t^2; 1, 2 t]
find the propagator matrix phi(t, s), t > 0, s > 0.
Is the propagator matrix a function of t - s?
Find y(3) given that y(1) = [1, -1].
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Your solution:
psi(t) = [t, t^2; 1, 2 t]
det[psi(t)] = t(2t) - t^2 = 2t^2 - t^2 = t^2
psi^(-1)(t) = 1/t^2 [ 2t, -t^2; -1, t] = [ 2/t, -1; -1/t^2, 1/t]
psi^(-1)(s) = [ 2/s, -1; -1/s^2, 1/s]
Phi(t,s) = Phi(t - s) = psi(t)psi^(-1)(s)
= [t, t^2; 1, 2 t][ 2/s, -1; -1/s^2, 1/s] = [2ts^(-1) - t^2s^(-2), -t + t^2s^(-1); 2s^(-1) - 2ts^(-2), -1 + 2ts^(-1)]
@&
Good.
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y(t) = Phi(t-s)y(s)
[2ts^(-1) - t^2s^(-2), -t + t^2s^(-1); 2s^(-1) - 2ts^(-2), -1 + 2ts^(-1)] [ 1; -1]
= [ 2ts^(-1) - t^2s^(-2) + t - t^2s^(-1); 2s^(-1) - 2ts^(-2) + 1 - 2ts^(-1)] = y(t)
Plug in t, s
t = 3
s = 1
[ 2(3)(1)^(-1) - (3)^2(1)^(-2) + (3) - (3)^2(1)^(-1); 2(1)^(-1) - 2(3)(1)^(-2) + 1 - 2(3)(1)^(-1)]
= [ -9; -9]
confidence rating #$&*:
^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
3
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Given Solution:
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Self-critique (if necessary):
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Self-critique rating:
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Question: Suppose that T^-1 A T = [lambda_1, 0; 0, lambda_2].
Let p(A) be the matrix polynomial 2 A^3 - A + 3 I.
Find the matrix B such that p(A) = T B T^-1.
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Your solution:
confidence rating #$&*:
^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution:
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Self-critique (if necessary):
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Self-critique rating:
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(T^-1 A T)^3 = [lambda_1^3, 0; 0, lambda_2^3].
Also, (T^-1 A T)^3 = (T^-1 A T) * (T^-1 A T) * (T^-1 A T) = T^-1 A T T^-1 A T T^-1 A T = T^-1 A I A I A T = T^-1 * A * A * A * T = T^-1 A^3 T.
By the above, invoking also the established linearity of matrix products, we have
T^-1 P(A) T
= T^-1( 2 A^3 - A + 3 I ] T
= 2 * T^-1( A^3) T- T^-1 A T + 3 * T^-1 I T
= [ 2 labmda_1^3 - lambda_1 + 3, 0; 0, 2 labmda_2^3 - lambda_2 + 3 ].
If
P(A) = T B T^-1
then
T^-1 P(A) T = B
so B is our matrix
[ 2 labmda_1^3 - lambda_1 + 3, 0; 0, 2 labmda_2^3 - lambda_2 + 3 ].
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Question: Let A be invertible and diagonalizable, and let D = T^-1 A T be the matrix as diagonalized by a similarity transform.
Show that D is invertible.
Show that A^-1 is diagonalizable by the same similarity transform that diagonalizes A (i.e., show that D^-1 = T^-1 A^-1 T).
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Your solution:
Perhaps since A = TDT^(-1)
A^(-1) = (TDT^(-1)) = 1/(TDT^(-1)) = T^(-1)D^(-1)T
and since D = T^(-1)AT
D^(-1) = (T^(-1)AT)^(-1) = 1/(T^(-1)AT) = TA^(-1)T^(-1)
@&
Right idea, but 1 / (T^-1 A T) is not defined. You can't divide the number 1 by a matrix.
You would simply note that (A B)^-1 = B^-1 A^-1, which follows simply from the fact that
AB * (B^-1 A^-1) = A * B * B^-1 * A^-1 = A * I * A^-1 = A * A^-1 = I.
Here we have used the associative property of matrix multiplication without explicitly stating that we have done so.
Having established this it is a straightforward step to invert D:
D^-1 = (T^-1 A T)^-1 = T^-1 A^-1 (T^-1)^-1 = T^-1 A^-1 T.
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confidence rating #$&*:
^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution:
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Self-critique (if necessary):
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Self-critique rating:
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Question: Let A be a diagonalizable 2 x 2 matrix with
lambda_1 = 1/4, x_1 = [2, 5]
and
lambda_2 = 1/2, x_2 = [1, 3].
Find cos(pi A) and sin(pi A).
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Your solution:
T = [ 2, 1; 5, 3]
det = 2(3) - 1(5) = 6 -5 = 1
T^(-1) = [ 3, -1; -5, 2]
Cos(tA) = T [cos(lambda_1 * t), 0; 0, cos(lambda_2 * t)] T^(-1)
cos(pi*A) = [ 2, 1; 5, 3][cos(pi/4), 0; 0, cos(pi/2)][ 3, -1; -5, 2]
= [ 2cos(pi/4) - 5cos(pi/2), -2cos(pi/4) + 2cos(pi/2); 15cos(pi/4) - 15cos(pi/2), -5cos(pi/4) + 6cos(pi/2)]
= [ 4.24, -1.41; 10.6, -3.54] approximatly.
Sin(tA) = T [sin(lambda_1 * t), 0; 0, sin(lambda_2 * t)] T^(-1)
sin(pi*A) = [ 2, 1; 5, 3][sin(pi/4), 0; 0, sin(pi/2)][ 3, -1; -5, 2]
= [ 6sin(pi/4) - 5sin(pi/2), -2sin(pi/4) + 2sin(pi/2); 15sin(pi/4) - 15sin(pi/2), -5sin(pi/4) + 6sin(pi/2)]
= [ -0.76, 0.58; -4.39, 2.46] approximatly.
@&
Your cosine calculates correctly.
Your sine is set up right but doesn't calculate correctly.
[ 2, 1; 5, 3][sin(pi/4), 0; 0, sin(pi/2)][ 3, -1; -5, 2]
is not equal to
[ 6sin(pi/4) - 5sin(pi/2), -2sin(pi/4) + 2sin(pi/2); 15sin(pi/4) - 15sin(pi/2), -5sin(pi/4) + 6sin(pi/2)]
I believe the correct matrix product is
[3·sqrt(2), - sqrt(2); 15·sqrt(2)/2, - 5·sqrt(2)/2]
which approximates to
[4.24, -1.41; 10.6, -3.54]
but check this out.
*@
confidence rating #$&*:
^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution:
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Self-critique (if necessary):
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Self-critique rating:
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Question: Let A be a diagonalizable 2 x 2 matrix with
lambda_1 = 1/4, x_1 = [2, 5]
and
lambda_2 = 1/2, x_2 = [1, 3].
Solve the equation
y '' + y ' + A y = 0.
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Your solution:
T = [ 2, 1; 5, 3]
D = [(1/4), 0; 0, (1/2)]
Psi(t) = [ 2e^(1/4), e^(1/2); 5e^(1/4), 3e^(1/2)]
x(t) = Tw(t)
Tw'' + ATw = 0
w'' + Dw = 0
In this case:
y(t) = Tw(t)
Tw'' + Tw' + ATw = 0
w'' + w' + Dw = 0
w_1'' + w_1' + (1/4)w = 0
w_2'' + w_1' + (1/2)w = 0
solutions
w_1 = Ate^(-t/2) + Be^(-t/2)
w_2 = e^(-t/2)(Asin(t/2) + Bcos(t/2))
confidence rating #$&*:
^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
2
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Given Solution:
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Self-critique (if necessary):
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Self-critique rating:"
Self-critique (if necessary):
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Self-critique rating:
Self-critique (if necessary):
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Self-critique rating:
#*&!
@&
You're doing well but check my notes for some details and a little broader view of the propagator matrix.
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The fact that integration by parts doesn't work doesn't mean that you can't evaluate the integral.
However the fact that the integral diverges, as can be see from the fact that e^(-s t) * e^(t sqrt(t)) is after a certain point an increasing positive function, does.
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