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12/06/11 - Homogeneous Autonomous Systems of First-Order ODEs with Constant Coefficients
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 | Linear Algebra Review -- 5-minute version of Math 270
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| Convert to matrix equation Ax = b:
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| Solve by multiplying by inverse of matrix x = A-1b
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| Matrix as linear operator
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| A function can be pictured as a machine with an input and output:
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 | The argument of the function f is x, the input value. The output of this function is the square of its input value.
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| Similarly, in the equation Ax = b, we can speak of A as being a linear operator that takes an input vector x and and turns it into the output vector b.
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| Example: The matrix A as a rotation and length-scaling operator:
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| In the figure below, the input vector x is a unit vector (shown in red). Operating on the input vector x with the matrix A produced the output vector b (shown in blue). We can see that operating on the input vector x with the matrix A has the effect of rotating the input vector and extending its length.
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| As we change the input vector, we see the rotation and length scaling varies.
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| Eigenvectors and Eigenvalues
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| Eigenvectors are a special case of the example above. When the output vector is parallel (or anti-parallel) to the input vector, the input vector is called an eigenvector of the matrix. The ratio of the length of the output vector to the input vector is called the eigenvalue of the matrix corresponding to that specific eigenvector.
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| We can write this special parallel relationship between input and output vectors as:
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which says that the result of operating on the input vector x with the matrix A is to produce an output vector which is parallel to x and extended in length by the factor λ. If λ is positive and greater than 1, the output vector is longer than the input vector. If λ is positive and less than 1, the output vector is shorter than the input vector. If λ is negative, the output vector is in the opposite direction of the input vector.
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| We can find the eigenvalues in the following manner:
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Note that wherever 0 appears below it represents the zero vector:
Introducing I (the identify matrix) allows us to factor out the vector x:
If (A-λI) has an inverse, then
x = 0 (the zero vector) is called the "trivial" solution, which implies that it's valid but not particularly useful. The only way to avoid the trivial solution in this situation is to require that (A-λI) does not have an inverse, and that means that the determinant of (A-λI) = |(A-λI)| = 0.
Finding the eigenvalues of the matrix above:
The equation we found by setting |(A-λI)| = 0 is called the "characteristic" equation, and its solutions are the eigenvalues of the matrix. In this case, there are two distinct eigenvalues: λ = 5 and λ = -1.
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| We can find the eigenvector corresponding to each eigenvalue in the following manner:
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| Find the eigenvector corresponding to the eigenvalue λ = 5:
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which says that x - (1/2)y = 0 or x = (1/2)y. For simplicity, pick y = 2, x = 1.
Check that this result really is the eigenvector corresponding to λ = 5:
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| Find the eigenvector corresponding to the eigenvalue λ = -1:
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which says that x = -y. For simplicity, pick y = 1, x = -1.
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| Shortcut for finding eigenvalues of 2 x 2 matrices
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Define the Trace of the matrix A as the sum of the diagonal elements T = a + d.
Define the determinant D = ad - bc.
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| Applying these equations for the eigenvalues to the matrix above:
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T = 1 + 3 = 4, D = (1)(3)-(4)(2) = 3 - 8 = -5
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| System of homogeneous autonomous first-order ODEs with constant coefficients
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| The solution to a system of first-order differential equations:
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has the general form:
where λ1 and λ2 are eigenvalues of the matrix
and
and
are the eigenvectors of the matrix corresponding to the eigenvalues λ1 and λ2, respectively.
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| Convert the second-order differential equation for the spring-mass system with damping to a system of two first-order differential equations:
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| Find eigenvalues of matrix:
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| Find eigenvectors corresponding to each eigenvalue:
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| Write the solution to the system of equations:
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Assume initial conditions:
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| Note that in this phase plot, the horizontal axis is the displacement of the mass from equilibrium, and the vertical axis is the velocity of the mass. This phase plot shows that the velocity decreases rapidly due to the heavy damping while the displacement barely changes, and then both the displacement and the velocity go to zero (at the origin).
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| Displacement and velocity vs. time
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| Let b = 0, k = 4, m = 1
For complex eigenvalues of the form λ = a + ib and corresponding eigenvector w = v1 + iv2, the solution is y = Aeat(cos(bt) v1 - sin(bt) v2) + B eat(sin(bt) v1 + cos(bt) v2). In our case, λ = a + ib = 0 + i2 (so a = 0 and b = 2) and the corresponding eigenvector is [0 2]T + i[-1 0]T (so v1 = [0 2]T and v2 = [-1 0]T). Hence:
Assume initial conditions:
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| Again, note that the horizontal axis is displacement, and the vertical axis is velocity. The phase plot shows the oscillation continues indefinitely (since there is no damping) with the displacement varying from -1.414 to 1.414 and the velocity varying from -2.828 to 2.828.
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| Displacement and Velocity vs. Time
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| Let b = 1, k = 6.5, m = 1
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| This time the phase plot shows the mass swirling around the origin of the x-x' (displacement-velocity) axes as it loses kinetic energy with both the displacement and velocity decaying exponentially to zero.
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| Displacement and Velocity vs. Time
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| Case 4 - Critical damping:
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| Let b = 1, k = 0.25, m = 1
In this case, with repeated eigenvalues but only one eigenvector, we need to generate another independent vector. Theorem 2.40 (p. 388) tells us how to do that. λ is the repeated eigenvalue and v1 is the corresponding eigenvector. The second vector is not a multiple of v1, but it is related to v1. Specifically, (A-λI)v2 = v1 (see pp. 386-389 for the explanation). In our example, we proceed as follows:
Our text suggests (p. 389) that we pick a simple vector for v2 that has zeroes (but not all zeroes), such as [0 1]T and when this is multiplied by (A-λI), we should get a multiple of v1. We then divide our original estimate for v2 by that factor, and we're almost done. Watch closely. We want:
We guess:
We test it:
Sure enough, it's a multiple of v1. We now divide our first estimate of v2 by (-1/2) to get our revised estimate of v2:
We now test the revised v2:
And it meets our requirement:
Theorem 2.40 then says the solution has the form:
Assume the initial condition:
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| This phase plot shows that the mass slows down as the displacement increases and then both the displacement and velocity quickly decay to zero.
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| Displacement and Velocity vs. Time
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| Read Sections 9.1 and 9.2 of the text, pp. 372-376 and pp. 378-389.
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| Do exercises p. 376 (1,14) and p. 389 (5,11,21,27,33,39,42,55).
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