Query 29

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course MTH 279

Query 29 Differential Equations*********************************************

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:

First, we find the eigenpairs of A.

λ = 0,1.

For λ = 0, v_1 = [4;5].

For λ = 1, v_2 = [1;1].

y(t) = sum(t^m/m!*y^(m)(0)), m=0 to ∞ = (I+t^k/k!A^k)y_0.

S_k = (I+…+t^k/k!*A^k).

lim(S_k) k->∞ = e^(At) = sum(t^m/m!*A^m) from m=0 to ∞.

d/dt(e^(At)) = Ae^(At).

Φ’ = AΦ, Φ(0) = I.

Φ(t) = e^(At).

A = TDT^(-1)

e^(At) = Te^(Dt)T^(-1)

Φ(t) = Te^(Dt)T^(-1) = [4,1; 5,1]*e^([0,0;0,1]t)*[-1,1; 5,-4] = [4,1; 5,1]*[1,0; 0, e^(t)]*[-1,1; 5,-4] = [5e^(t)-4, 4-4e^(t); 5e^(t) -5, 5-4e^(t)].

We plug in t = 3, to get: Φ(3) = [5e^(3)-4, 4-4e^(3); 5e^(t)-5, 5-4e^(t)].

@&

You've used diagonalization, which is also a good approach.

However the propagator matrix is

Phi(t, s) = psi(t) psi^-1(s)

where psi(t) is a fundamental matrix.

Phi(t, s) * y(s) gives you y(t), so

y(t) = Phi ( t, 1 ) * y(1) = Phi(t, 1) * [1; 0]

*@

confidence rating #$&*:

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Given Solution:

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Self-critique (if necessary):

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Self-critique rating: 3

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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:

A = [ 1, 1, 1; 0, 2, 1; 0, 0, -1 ].

We find the eigenpairs of A. det(A-λI) =0: λ=-1,1,2.

For λ=-1: A-(-1)I=0. The eigenvector for this eigenvalue is v_1=[-1;-1;3].

For λ=1: A-I=0. The eigenvector for this eigenvalue is v_2=[1;0;0].

For λ=2: A-2I=0. The eigenvector for this eigenvalue is v_3=[1;1;0].

T is the ordered eigenvector matrix: T= [-1,1,1; -1,0,1; 3,0,0].

T^(-1) = [0,0,1/3; 1,-1,0; 0,1,1/3].

D is the diagonal eigenvalue matrix: D=[-1,0,0; 0,1,0; 0,0,2].

We assume our solution will come in a form of Φ=e^(tA). Therefore, Φ’=Ae^(tA) = AΦ.

The similarity transform process says that A=TDT^(-1). Therefore, we have e^(tA) = T*e^(tD)*T^(-1) = Φ.

e^(tA) = T*[e^(-t),0,0; 0,3e^(t),0; 0,0,e^(2t)]*T^(-1) = [e^(t), e^(2t)-e^(t), e^(2t)/3-e^(-t)/3; 0, e^(2t), e^(2t)/3-e^(-t)/3; 0, 0, e^(-t)] = Φ = y.

We multiply this by a constant matrix [c_1;c_2;c_3].

We now have: y = [e^(t), e^(2t)-e^(t), e^(2t)/3-e^(-t)/3; 0, e^(2t), e^(2t)/3-e^(-t)/3; 0, 0, e^(-t)]*[c_1;c_2;c_3].

We impose the initial condition y(0) = [1;1;0]. y(0) = [1;1;0] = I*[c_1;c_2;c_3]. c_1=1, c_2=1, c_3=0.

Our solution is y(t) = [e^(t), e^(2t)-e^(t), e^(2t)/3-e^(-t)/3; 0, e^(2t), e^(2t)/3-e^(-t)/3; 0, 0, e^(-t)]*[1;1;0].

Finding y(1), we get y(1) = [e^(2); e^(2); 0].

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: 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:

Φ(t,s) = Ψ(t)*Ψ^(-1)(s) = [t,t^2; 1, 2t]*[2/s, -1; -1/s^2, 1/s] = [2t/s-t^2/s^2, -t+t^2/; 2/s-2t/s^2, -1+2t/s].

The propagator matrix is a function of t-s.

y = Ψ(t)*[c_1; c_2]. y(1) = [1;-1] = Ψ(t)*[c_1; c_2]. c_1 = 3, c_2 = -2.

y(3) = [3, 3^2; 1, 2(3)]*[3;-2] = [-9; -9].

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: 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:

B = [λ_1, 0; 0, λ_2] = T^(-1)AT. For any two similar matrices, A and B, the characteristic polynomials are the same. If they have the same polynomial, they have the same eigenvalues.

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 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:

Using the similarity transform algorithm, we know that D = T^(-1)AT. Just looking at the formula, we know that T has an inverse, and is obviously invertible by that fact. The invertibility of D is dependent on the invertibility of A. Since we know that A is invertible and diagonalizable, we can have D^(-1) = TA^(-1)T^(-1).

We can show using an example:

A = [1,2;2,1]. T=[1,-1; 1,1]. T^(-1)=[1/2,1/2; -1/2, ½]. A^(-1) = [-1/3, 2/3; 2/3, -1/3].

D=T^(-1)AT = [3,0; 0, -1]. We can invert this to show that it is in fact invertible.

Now by our transform algorithm, D^(-1) = T^(-1)A^(-1)T = [1/3, 0; 0, -1]. D^(-1) is the diagonal of A^(-1).

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].

Find cos(pi A) and sin(pi A).

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Your solution:

T = [2,1; 5,3]. T^(-1) = [3, -1; -5, 2]. D=[1/4, 0; 0, ½]. A=TDT^(-1) = [-1, ½; -15/4, 7/4].

cos(πA) = I + (-1)^k*(πA)^(2k)/(2k)! . In our A, k is 2.

cos(πA) = I + (-1)^1*(πA)^(2)/2! + (-1)^2*(πA)^4/4! ≈ [4.24, -1.41; 10.61, -3.54].

sin(πA) = A + (-1)^k/(2k+1)!*A^(2k+1).

sin(πA) ≈ [-0.76, 0.59; -4.4, 2.5].

confidence rating #$&*:

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Given Solution:

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Self-critique (if necessary): I’m not entirely possible about the cos(πA) and sin(πA) parts. I’ve never seen this series expansion before. I believe that k = the number of eigenvalues for the matrix A?? Is that correct?? From what I read, this series expansion is supposed to be more meaningful than just doing the cos() and sin() of the individual elements.

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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].

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] T^(-1)=[3,-1;-5,2].

A=TDT^(-1)=[-1, ½; -15/4, 7/4].

λ^2 + λ + A = 0;

λ = -1/2±[sqrt(5)/2, i/2; 2, isqrt(6)/2].

y = Ae^(λ_1t) + Be^(λ_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"

&#Your work looks good. See my notes. Let me know if you have any questions. &#