#$&*
course MTH 279
I am resubmitting my queries 29 and 30 that I submitted on 4/21 because I haven't seen a reply to them on my access page. If you've been busy, I understand and it's no big deal. I just thought I'd resubmit them in case something happened and you hadn't received them. 4/26 9am
Query 29 Differential Equations
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Question: Find the propagator matrix Phi(t) = e^(t A) for the system
y ' = [5, -4; 5, -4] y
and use to find y(3) given that y(1) == [ 1, 0].
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Your solution:
Phi = [-4 +5e^t , 4 - 4e^t ; -5+5e^t , 5-4e^t]
y(3) = [e^2 + 1 ; e^2]
using the fact that e^(tA) = T*lam*T_inv
I found my evals to be 0 and 1 and my evecs to be [4 ; 5] and [1 ; 1]
so T = [4 , 1 ; 5 , 1] T_inv = [-1 , 1 ; 5 , -4]
I then computed [4 , 1 ; 5 , 1]*[1 , 0 ; 0 , e^t]*[-1 , 1 ; 5 , -4] = [-4 +5e^t , 4 - 4e^t ; -5+5e^t , 5-4e^t]
to find y(3), I set up [-4 +5e^(t-s) , 4 - 4e^(t-s) ; -5+5e^(t-s) , 5-4e^(t-s)] = [y_t1 - y_s1 ; y_t2 - y_s2]
or [-4 +5e^2 , 4 - 4e^2 ; -5+5e^2 , 5-4e^2] = [ y_t1 - 1 ; y_t2 + 0]
and solved for y.
confidence rating #$&*:
^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution:
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Self-critique (if necessary):
I think I found my phi okay, but I'm not so sure I solved for y(3) correctly.
@&
@&
The characteristic equation is
det( [5-lambda, -4; 5, -4-lambda] )
= (5 - lambda)(-4-lambda) - (5)(-4)
= lambda^2 - lambda
The eigenvalues are 0 and 1.
Thus for lambda = 0
5 y_1 - 4 y_2 = 0 and
y_2 = 5/4 y_1
so an eigenvector is
[1; 5/4] .
If we want integer coefficients we can use
[4;5].
The solution vector corresponding to eigenvalue 0 is therefore
e^(0 t) [4; 5] = [4, 5].
For lambda = 1 the eigenvector [1; 1] works, leading to solution
e^t [1; 1].
The corresponding phi(t) matrix is
phi(t) = [4, e^t ; 5, e^t].
Checking that this matrix solves the equation we have
phi ' (t) = [0, e^t; 0, e^t]
and
[5, -4; 5, -4] * phi(t) = [0, e^t; 0, e^t]
so that phi(t) is a solution to the system.
For the moment let's do the solution for y(0) = [1; 0].
The general solution y(t) is
y(t) = phi(t) * [c_1; c_2],
which according to the condition we are using here must be equal to [1, 0] when t = 0.
The inverse of phi(t) is
[-e^t, e^t; 5, -4] / e^t = [-1, 1; 5 e^(-t), -4 e^(-t) ]
When t = 0 this becomes just
[-1, 1; 5, -4]
so that the solution for the c matrix is
[c_1; c_2] = [-1, 1; 5 , -4 ] * [1; 0] = [-1, 5].
Thus our desired solution is
[4, e^t; 5, e^t] * [-1; 5] = [-4 + 5 e^t; -5 + 5 e^t ]
It is easy to check that this is a solution.
Evaluated at t = 0 it does indeed give us [1; 0].
Its derivative is
[5 e^t; 5 e^t]
and its product with the matrix [5, -4; 5, -4] is equal to this result.
*@
@&
This gives us a good solution, and we could as easily have found the solution for which y(1) = [1; 0].
The requested method, however, was the propagator matrix. Continuing for the moment with the initial condition y(0) = [1; 0]:
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We will use PHI (in caps) to designate the propagator matrix and distinguish it from the solution matrix phi(t).
The propagator matrix is
phi(t) phi^-1(s) =
[4, e^t; 5, e^t] * [-1, 1; 5 e^(-s), -4 e^(-s) ] =
PHI(t, s) = [-4 + 5 e^(t - s), 4 - 4 e^(t - s); -5 +5 e^(t-s), 5 - 4 e^(t-s) ].
To propagate the solution y(0) = [1; 0] we multiply [1; 0] by PHI(t, 0) to get
[-4 + 5 e^(t-0), 4 - 4 e^(t-0); -5 + 5 e^(t-0), 5 - 4 e^(t-0) ] * [1; 0] =
[ -4 + 5e^t; -5 + 5 e^t ]
in agreement with the above result.
*@
@&
Now we use the propagator matrix to solve the given problem:
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The given problem asked that we propagate y(1) = [1; 0], which would give us
PHI(t, 1) = [-4 + 5 e^(t-1), 4 - 4 e^(t-1); -5 + 5 e^(t-1), 5 - 4 e^(t-1) ] * [1; 0] =
[-4 + 5 e^(t-1); -5 + 5 e^(t - 1)]
or
[-4 + 5/e * e^t; -5 + 5 / e * e^t ]
*@
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Self-critique rating: OK
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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:
y(1)= [ 1/3e^2 +1 ; 1/3e^2 - 1/3e^-1 +1 ; e^-1]
I think I ended up doing this the hardest way possible, but I found evals of 1, 2, -1 and corresponding evecs of [1 ; 0 ; 0] , [ 1 ; 1 ; 0] and [-1 ; -1 ; 3]
So psi(t) = [e^t , e^2t , -e^-t ; - e^2t , -e^-t ; 0 , 0 , 3e^-t]
psi^-1(s) = [ e^-s , -e^-s , 0 ; 0 , 0 , 1/3e^-2s ; 0 , 0 , 1/3e^s]
I then used the relationship [e, -e , 1/3e^2 ; 0 , 0 , 1/3e^2-1/3e^-1 ; 0 , 0 , e^-1 ] = [ y_1 -1 ; y_2 - 1 ; y_3 + 0]
to find my y(3) values.
confidence rating #$&*:
^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: 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].
phi = [2ts^-1-t^2s^-2 , -t+t^2s^-1 ; 2s^-1-2ts^-2 , -1+2ts^-1]
I do not think the propagator matrix is a function of t-s. I can't really see a way to factor the compnents out in a way that could reflect this relationship.
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Self-critique (if necessary):
I'm confident I made the correct propagator matrix, (since it's just psi(t)*psi_inv(s)), but I can't make out how to find y(3)
@&
Eigenvectors are
v_1=[1;0;0] corresponding to eigenvalue lambda = 1, yielding solution e^t [ 1; 0; 0]
v_2=[-1;-1; 3] corresponding to eigenvalue lambda = -1, yielding solution e^(-t) * [-1; -1; 3]
v_3=[1;1;0] corresonding to eigenvalue lambda = 2, yielding solution e^(2 t) [ 1; 1; 0].
This yields solution matrix
psi(t) = [ê^t, - ê^(-t), ê^(2·t); 0, - ê^(-t), ê^(2·t); 0, 3·ê^(-t), 0]
psi^-1(t) = [ê^(-t), - ê^(-t), 0; 0, 0, ê^t/3; 0, ê^(- 2·t), ê^(- 2·t)/3]
The propagator matrix is
PSI(t, s) = psi(t) psi^-1(s) =[ê^(t - s), ê^(2·(t - s)) - ê^(t - s), ê^(2·(t - s))/3 - ê^(s - t)/3; 0, ê^(2·(t - s)), ê^(2·(t - s))/3 - ê^(s - t)/3; 0, 0, ê^(s - t)]
Multiplying PSI(t, 0) by [1; 1; 0] we get the solution vector function
[ê^t, ê^(2·t) - ê^t, ê^(2·t)/3 - ê^(-t)/3; 0, ê^(2·t), ê^(2·t)/3 - ê^(-t)/3; 0, 0, ê^(-t)] * [1; 1; 0] =
[ê^(2·t); ê^(2·t); 0]
The derivative of this solution is
2 [ê^(2·t); ê^(2·t); 0]
This should equal the original matrix times the solution vector.
You can easily verify that
[1, 1, 1; 0, 2, 1; 0, 0, -1]·[ê^t; ê^(-t); 0] [ê^(2·t); ê^(2·t); 0] = 2 [ê^(2·t); ê^(2·t); 0]
confirming this solution.
*@
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Self-critique rating: 1
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Question:
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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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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:
I'm not sure how to start out on this one. I think I need to find a matrix B that makes p(A) diagonlizable.
???? Should I try to find the matrix A?, then calculate p(A)? Then what could I do to find B? premultiply p(A) by T^-1 and postmultiply by T? Are the T for A and B the same?
@&
Observe that
p(T^-1 A T) = 2 (T^-1 A T)^3 - (T^-1 A T) + 3 I =
2 [ lambda_1, 0; 0, lambda_2 ] ^3 - [ lambda_1, 0; 0, lambda_2 ] + [3, 0; 0, 3] =
[2 lambda_1^3 - lambda_1 + 1, 0; 0, 2 lambda_2^3 - lambda_2 + 1]
*@
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 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:
Let D be any diagonal nxn matrix that is formed from T^-1 A T by a similarity transform.
By definition, D is invertible if it has a rank equal to its size (n). Since all rows/columns that make up D are nonzero, its rank will always be equal to its size.
Hence, D is invertible.
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Nice argument. Much more elegant than the following:
(T^-1 A T) ^-1 = T^-1 A^-1 T, since
(T^-1 A T) * (T^-1 A^-1 T) = T^-1 A T T^-1 A^-1 T
= T^-1 A I A^-1
= T^-1 A A^-1 T
= T^-1 I T
= T^-1 T
= I.
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If you multiply a diagonal matrix by a nondiagonal matrix, the product is not diagonal.
So T^-1 A^-1 T is diagonal, showing that A^-1 has been diagonalized by the same similarity transform as A.
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The eigenvalues of the matrix D^-1 (its diagonal entries) are also the eigenvalues of A^-1. We can rewrite D^-1 = T^-1 A^-1 T as TD^-1 = A^-1T
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If D has diagonal elements lambda_1, lambda_2, ..., then D^-1 has diagonal elements 1 / lambda_1, 1 / lambda_2, ... .
These are the reciprocals of the eigenvalues of D, not equal to the eigenvalues of D (unless all the eigenvalues of D have magnitude 1).
*@
Since these two matrices are equal, we can see that expanded out each jth element of T: t_j corresponds as t_j*d_j = A*t_j, which can be rearranged as d_j*t_j=A*t_j, which shows that A is equal to d_j for every jth element, which means that A^-1 is in fact diagonalizable.
confidence rating #$&*:
^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution:
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Self-critique (if necessary):
I'm not very good at proofs, but I think I got the highlights in there.
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Self-critique rating: 3
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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:
cos(piA) = [3sqrt(2) , -5sqrt(2) ; 3sqrt(2)/2 , -5sqrt(2)/2]
sin(piA)= [3sqrt(2) -5 , -5sqrt(2) +10 ; 3sqrt(2)/2 -3 , -5sqrt(2)/2 +6]
T = [2 , 5; 1 , 3] and T^-1 = [3 , -5 ; -1 , 2]
Set up: [2 , 5; 1 , 3]*[cos/sin(1/4t) , 0 ; 0 , cos/sin(1/2t)]
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We have the equation
D = T^-1 A T
where D = [1/4, 0; 0, 1/2].
A is the solution to this equation:
A = T^-1 D T = [ -1, 1/2; -15/4, 7/4 ]
cos(pi A) = T^-1 cos(pi D) T
= T^-1 [ cos(-pi), cos(-pi/2); cos(-15/4 pi), cos(7/4 pi) ]
= T^-1 [ -1, 0; sqrt(2)/2, sqrt(2) / 2 ] * T.
This is easily calculated, since we know T and T^-1.
A similar analysis shows that
sin(pi A) = T^-1 [ 0, 1; sqrt(2)/2, -sqrt(2) / 2] T,
again easily calculated.
*@
confidence rating #$&*:
^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution:
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Self-critique (if necessary): OK
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Self-critique rating: OK
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Question: Let A be a diagonalizable 2 x 2 matrix with
lambda_1 = 1/4, x_1 = [2, 5]
and
lambda_2 = 1/2, x_2 = [1, 3].
Solve the equation
y '' + y ' + A y = 0.
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Your solution:
y = [2e^(-t/2) +2te^(-t/2) + 5e^(-t/2)cos(1/2*t) + 5e^(-t/2)sin(1/2*t) ; e^(-t/2)+te^(-t/2) + 3e^-t/2)cos(1/2*t+3e^(-t/2)sin(1/2*t)
I made the initial substitution y=Tz to rearrange the equation into z'' +z' +Dz = 0
That gave me two separate equations to solve: z_1''+z_1' +1/4z_1 = 0 and z_2'' +z_2' + 1/2z=0
which had solutions of z_1 = Ae^(-1/2t) + Bte^(-1/2t) (a case of repeated roots) and z_2 = e^(-1/2t)*(Ccos(1/2t) + Dsin(1/2t))
confidence rating #$&*:
^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution:
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Self-critique (if necessary): OK
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Self-critique rating: OK"
&#Good responses. See my notes and let me know if you have questions. &#