Query_27

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course Mth 279

5/21 at 3am

Query 27 Differential Equations*********************************************

Question: Find the eigenvalues of the matrix [3, 1; -2, 1] and find the corresponding eigenvectors.

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Your solution:

P(lambda) = det[ A - lambda*I] = 0

=[ (3-lambda), 1; -2, (1-lambda)]

= (3-lambda)(1-lambda) - (-2)(1)

= lambda^2 - 4*lambda + 3 + 2

= lambda^2 - 4*lambda + 5

using the Quadratic formula we obtain:

2 +,- i

for lambda_1 = 2 + i, we solve (A - (2 + i)I)x_1 = 0

[ (1+i), 1; -2, (-1+i)][ x_1; x_2] = [ 0; 0]

Using elementary row operations:

-R2 -> R2

[ (1+i), 1; 2, (1-i)][ x_1; x_2] = [ 0; 0]

yeilds equations:

(1+i)x_1 + x_2 = 0

and

2x_1 + (1-i)x_2 = 0

Solving these equations yeilds:

x_1 = [ i; (1-i)]

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for lambda_2 = 2 - i, we solve (A - (2 - i)I)x_2 = 0

[ (1-i), 1; -2, (-1-i)][ x_1; x_2] = [ 0; 0]

Using elementary row operations:

-R2 -> R2

[ (1-i), 1; 2, (1+i)][ x_1; x_2] = [ 0; 0]

yeilds equations:

(1-i)x_1 + x_2 = 0

and

2x_1 + (1+i)x_2 = 0

Solving these equations yeilds:

x_2 = [ -i; (1+i)]

psi(t) = [ i, -i; (1-i), (1+i)]

det[psi(t)] = 2i

since the determinant is not zero, this is a fundamental set.

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Given Solution:

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Self-critique (if necessary):

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Question: Suppose that i + 1 is an eigenvalue of a matrix A and [-1 + i, i ] is a corresponding eigenvector. Find a fundamental set of real solutions to the equation y ' = A y.

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Your solution:

Based on the previous problem this solution is actually quite simple.

Since (i + 1) is one eigenvalue (i+1) = (1+i)

Then (-i+1) must be another. (-i+1) = (1-i)

Which comes from the quadratic equation, (1+ or - i)

And if solution one has Eigenvector [ (-1+i); i ]

Then solution 2 has Eigenvector [ (-1-i); -i ]

I had actually did this problem 2 previous times before I realized the mistake.

The sign in front of the i makes all the difference.

psi(t) = [ (-1+i), (-1-i); i, -i ]

det[psi(t)] = 2i

Since W(t) is not zero, this is a fundamental set.

I made the mistake the first time I solved this to make the sign in front of the number opposite rather than in front of i.

And in that case the determinant is zero.

Another solution to this would be the real solution.

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y1(t) = e^(1+i)t [ (-1+i); i]

e^((1+i)t) = (e^t)e^(it) = (e^t)(cost + i*sin(t)) = (e^t)cos(t) + (e^t)sin(t)

y(t) = (e^t)cos(t) + (e^t)sin(t) [ (-1+i); i]

= [ -(e^t)cos(t) - (e^t)sin(t); -(e^t)sin(t)] + i [ (e^t)cos(t) - (e^t)sin(t); (e^t)cos(t)]

2 real funtions:

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U(t) = [ -(e^t)cos(t) - (e^t)sin(t); -(e^t)sin(t)]

V(t) = [ (e^t)cos(t) - (e^t)sin(t); (e^t)cos(t)]

Psi(t) = [ -(e^t)cos(t) - (e^t)sin(t), (e^t)cos(t) - (e^t)sin(t); -(e^t)sin(t), (e^t)cos(t)]

W(t) = det[psi(t)] = -(e^t)(cos(t) + sin(t))(e^t)cos(t) - (e^t)(cos(t) - sin(t))(-e^t)sin(t)

= -e^(2t)(cos^2(t) + sin(t)cos(t)) + e^(2t)(-sin^2(t) + sin(t)cos(t))

= -e^(2t)cos^2(t) - e^(2t)sin(t)cos(t) - e^(2t)sin^2(t) + e^(2t)sin(t)cos(t)

= -e^(2t)cos^2(t) - e^(2t)sin^2(t)

= -e^(2t)(cos^2(t) + sin^2(t))

= -e^(2t)(1)

= -e^(2t)

Since W(t) is not zero, this is a fundamental set.

confidence rating #$&*:

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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: Solve the equation

y ' = [0, -9; 1, 0] y

with initial condition

y(0) = [6, 2].

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Your solution:

P(lambda) = det[A - lambda*I] = 0

[ -lambda, -9; 1, -lambda]

= (-lambda)(-lambda) + 9

= lambda^2 + 9

Using the quadratic equation we get:

lambda = + or - 3i

lambda_1 = 3i

lambda_2 = -3i

for lambda_1 = 3i we can solve (A-3i*I)x_1 = 0

[ -3i, -9; 1, -3i][ x_1; x_2] = [ 0; 0]

This yeilds equations:

-3i(x_1) - 9(x_2) = 0

and

x_1 - 3i(x_2) = 0

Solved by inspection of these equations:

x_1 = [ i, (1/3)]

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for lambda_2 = -3i we can solve (A+3i*I)x_1 = 0

[ 3i, -9; 1, 3i][ x_1; x_2] = [ 0; 0]

This yeilds equations:

3i(x_1) - 9(x_2) = 0

and

x_1 + 3i(x_2) = 0

Solved by inspection of these equations:

x_1 = [ -i, (1/3)]

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Solutions:

(lambda_1, x_1) = (3i, x_1)

y_1(t) = e^(3i*t) [ i, (1/3)] = [ ie^(3i*t); (1/3)e^(3i*t)]

and

(lambda_2, x_1) = (-3i, x_1)

y_2(t) = e^(-3i*t) [ -i, (1/3)] = [ -ie^(-3i*t); (1/3)e^(-3i*t)]

y(t) = (c_1)y_1(t) + (c_2)y_2(t)

= [ ie^(3i*t), -ie^(-3i*t); (1/3)e^(3i*t), (1/3)e^(-3i*t)][ c_1; c_2]

W(0) = [ i, -i; (1/3), (1/3)] = (1/3)i - (-i)(1/3) = (1/3)i + (1/3)i = (2/3)i

Since not zero, this is a fundamental set.

y(0) = [ i, -i; (1/3), (1/3)][ c_1; c_2] = [ 6; 2]

Proving difficult to solve for the constants, may need to go back and do real solutions only.

@&

Simply invert the matrix and solve the matrix equation.

The determinant of the matrix is 2/3 i, so the inverse is

1 / (2/3 i) [ 1/3, i; -1/3, i]

= 3/2 i [ -1/3, -i; 1/3, -i]

and

[c_1; c_2] = 3/2 i [ -1/3, -i; 1/3, -i] [ 6; 2].

Note also that

e^(-3 i t) = cos(3 t) - i sin(3 t)

and

e^(3 i t) = cos(3 t) + i sin(3 t).

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Real Solutions:

y_1 = e^(3i*t) [ i, (1/3)]

e^(3i*t) = (cos(3t) + i*sin(3t)) [ i; (1/3)]

= [ i*cos(3t) - sin(3t); (1/3)cos(3t) + (1/3)i*sin(3t)]

U(t) + i*V(t)

[ -sin(3t); (1/3)cos(3t)] + i*[ cos(3t); (1/3)sin(3t)]

Psi(t) = [ -sin(3t), cos(3t); (1/3)cos(3t), (1/3)sin(3t)]

W(t) = det[ psi(t)] = -sin(3t)*(1/3)sin(3t) - cos(3t)*(1/3)cos(3t)

= (-1/3)(sin^2(3t) + cos^2(3t))

= (-1/3)(1)

= -1/3

Since its not zero, this is also a fundamental set.

Although only half...

But after looking back at example 1, in 6.6

we can see that the last note states the final solution

equals a matrix of lambda_1 values, and lambda_2 values

times another matrix, so both solutions may be here after all.

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: Find all values of mu such that any fundamental set [ y_1, y_2 ] of the system

y ' = [1, 3; mu, -2] y

have the property that the limit of the expression (y_1(t))^2 + (y_2(t))^2, as t -< infinity, is zero.

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Your solution:

@&

The eigenvalues of the system are the solutions to the equation

(lambda - 1) ( lambda + 2) + 3 mu = 0

or

lambda^2 + lambda + 2 + 3 mu = 0.

The solutions are

lambda = (-1 +- sqrt( 1 - 4 - 12 mu) ) / 2

= -1/2 +- sqrt(3)/2 * sqrt( 1 - 4 mu)

If mu is small enough the discriminant could be positive, giving us real solutions for lambda. There is a possibility that some of the real solutions will be positive, leading to solutions of the form

[y1; y2] = e^(lambda t) [x1; x2] for positive lambda. Such solutions would get larger as t approaches infinity.

For example if mu = 0 we have

lambda = -1/2 +- sqrt(3) / 2,

and since sqrt(3) > 1 it is certainly possible for one of the eigenvalues lambda to be positive.

More specifically,

lambda > 0 if

-1/2 + sqrt(3) / 2 * sqrt( 1 - 4 mu) > 0.

Solving we get

sqrt(1 - 4 mu) > 1 / sqrt(3) = sqrt(3) / 3

1 - 4 mu > 1/3

mu < 1/6.

Specifically if mu < 1/6, one of the eigenvalues lambda is positive, leading to an eigenvector of the form

[y1; y2] = e^(lambda t) [ x1; x2 ].

As long as [x1; x2 ] is not the zero vector, its magnitude is positive, so that

|| [y1; y2 ] || ^ 2 = e^(2 lambda t) || x1; x2 || = e^(2 lambda t) * (x1^2 + x2^2)

approaches infinity as e^(2 lambda t) approaches infinity.

*@

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: A particle moves in an unspecified force field in such a way that its position vector r(t) = x(t) i + y(t) j and the corresponding velocity vector v(t) = r ' (t) satisfy the equation

v ' = 2 k X v

Write this condition as a system

v ' = A v,

with v = [v_x; v_y].

If the particle starts at position r(0) = 2 i + j, v(0) = i + 2 j, find its position at t = 3 pi / 2.

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Your solution:

@&

`r(t) = x(t) `i + y(t) `i is represented as

`r = [x; y],

where we understand x and y as representing the functions x(t) and y(t).

`v = `r ' = x ' `i + y ' `j is represented as

`v = [ x ' ; y ']

`k stands for the unit vector in the z direciton, `i and `j for the unit vectors in the x and y directions.

v ' = 2 `k X `v = 2 x ' (t) * `k X `i + 2 y ' (t) * `k X `j

= 2 x ' (t) `j - 2 y ' (t) `i.

If this vector is represented as a column vector we have

`v ' = [-2 y ' (t); 2 x ' (t) ].

Thus the system `v ' = A `v is

[-2 y ' ; 2 x ' ] = A [ x' ; y' ].

So the matrix A is

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

Using v_x and v_y for x ' and y ' our system is therefore

[vx '; vy '] = [0, -2; 2, 0 ] [ vx, vy ]

The eigenvalue equation is easily solved, yielding eigenvalues -2 i and +2 i with respective eigenvectors

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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:"

Self-critique (if necessary):

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Self-critique rating:

Self-critique (if necessary):

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Self-critique rating:

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&#Good responses. See my notes and let me know if you have questions. &#