Chap 44 Exercises

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Exercise 1 We have seen that the solution \(x(t)\) to the linear dynamical system in two state variables \[\begin{array}{c}\partial_t x = ax + b y \\ \partial_t y = c x + dy\end{array} = \left[\begin{array}{cc}a & b\\c & d\end{array}\right] \left[\begin{array}{c}x\\y\end{array}\right]\] can be written as a linear combination of two exponentials:

\(x(t) = A e^{\lambda_1 t} + B e^{\lambda_2 t}\).

Let’s call the two components of this linear combination the “A-part” and the “B-part.”

In each of the following, you are given two specific numerical values for \(\lambda_1\) and \(\lambda_2\). Your task is to determine whether the A-part or the B-part (or both or neither) is transient.

For $\lambda_1 = -1$ and $\lambda_2 = -0.5$, which parts are transient?

```{mcq}
#| label: fms02-1
#| show_hints: true
1. A-part 
2. B-part 
3. both parts [ correct hint: right-o  ]
4. neither part 
```
For $\lambda_1 = -0.01$ and $\lambda_2 = 0.01$, which parts are transient?

```{mcq}
#| label: fms02-2
#| show_hints: true
1. A-part [ correct hint: Nice.  ]
2. B-part 
3. both parts 
4. neither part 
```
For $\lambda_1 = 0.01$ and $\lambda_2 = -0.3$, which parts are transient?

```{mcq}
#| label: fms02-3
#| show_hints: true
1. A-part 
2. B-part [ correct hint: Yes  ]
3. both parts 
4. neither part 
```

In answering the next two questions, keep in mind that for large \(t\), \[A e^{\lambda_1 t} + B e^{\lambda_2 t} \approx A e^{\lambda_1 t}\ \ \text{when}\ \ \lambda_2 < \lambda_1\ .\]

For $\lambda_1 = 0.01$ and $\lambda_2 = 0.001$, which parts are transient?

```{mcq}
#| label: fms02-4
#| show_hints: true
1. A-part 
2. B-part [ correct hint: You're right  ]
3. both parts 
4. neither part 
```
For $\lambda_1 = -0.1 $ and $\lambda_2 = -0.001$, which parts are transient?

```{mcq}
#| label: fms02-5
#| show_hints: true
1. A-part 
2. B-part 
3. both parts [ correct hint: Correct Both parts are exponential decay. ]
4. neither part 
```

Exercise 2 The linear dynamical system in two variables is

\[\begin{array}{c}\partial_t x = ax + b y \\ \partial_t y = c x + dy\end{array} = \left[\begin{array}{cc}a & b\\c & d\end{array}\right] \left[\begin{array}{c}x\\y\end{array}\right]\ .\] The matrix abcd can be turned into two values, \(\lambda_1\) and \(\lambda_2\) such that $x(t) = A e^{_1 t} + B e^{_2 t}. Use the formula for \(\lambda\) in the text to find \(\lambda_1\) and \(\lambda_2\) for each of the following:

  1. \(\left[\begin{array}{rr}1 & 1\\0 & 2 \end{array}\right]\)

  2. \(\left[\begin{array}{rr}1 & 1\\1 & 2 \end{array}\right]\)

  3. \(\left[\begin{array}{rr}-1 & 1\\0 & -2 \end{array}\right]\)

  4. \(\left[\begin{array}{rr}-1 & 1\\1 & -2 \end{array}\right]\)

  5. \(\left[\begin{array}{rr}-1 & 1\\0 & 2 \end{array}\right]\)

  6. \(\left[\begin{array}{rr}0 & 1\\-1 & -1 \end{array}\right]\)

Exercise 3 For the two-state variable linear dynamical system \[\begin{array}{c}\partial_t x = ax + b y \\ \partial_t y = c x + dy\end{array} = \left[\begin{array}{cc}a & b\\c & d\end{array}\right] \left[\begin{array}{c}x\\y\end{array}\right]\] the solution can be written as a linear combination of exponentials \[x(t) = A e^{\lambda_1 t} + B e^{\lambda_2 t}\ .\] For each of the following systems and initial conditions, find the coefficients \(A\) and \(B\).

  1. \(\left[\begin{array}{rr}1 & 1\\0 & 2 \end{array}\right]\) with \(x(0) = 3\) and \(\partial_t x(0) = 1\).

  2. \(\left[\begin{array}{rr}1 & 1\\0 & 2 \end{array}\right]\) with \(x(0) = 3\) and \(\partial_t x(0) = -1\).

  3. \(\left[\begin{array}{rr}1 & 1\\0 & 2 \end{array}\right]\) with \(x(0) = -3\) and \(\partial_t x(0) = -1\).

  4. \(\left[\begin{array}{rr}-1 & 1\\0 & -2 \end{array}\right]\) with \(x(0) = 0\) and \(\partial_t x(0) = 5\).

  5. \(\left[\begin{array}{rr}-1 & 1\\1 & -2 \end{array}\right]\) with \(x(0) = 5\) and \(\partial_t x(0) = 0\).

Exercise 4 For the linear dynamical system in two state variables \[\begin{array}{c}\partial_t x = ax + b y \\ \partial_t y = c x + dy\end{array} = \left[\begin{array}{cc}a & b\\c & d\end{array}\right] \left[\begin{array}{c}x\\y\end{array}\right]\] the two values of \(\lambda\) are \[\lambda_1 = \frac{1}{2}\left(a + d\right) + \frac{1}{2}\sqrt{\left(a - d\right)^2 - 4 b c}\\\lambda_1 = \frac{1}{2}\left(a + d\right) - \frac{1}{2}\sqrt{\left(a - d\right)^2 - 4 b c}\]

  1. Show that the sum \(\lambda_1 + \lambda_2 = a + d\).

  2. Show that the square of the difference \(\left(\lambda_1 - \lambda_2\right)^2 = (a - d)^2 - 4bc\).

These two facts provide the path to finding a set of values \(a, b, c, d\) that will generate any given set \(\lambda_1\) and \(\lambda_2\). First, pick any \(a\) and \(d\) to match the sum, then use these values in the formula for the square of the difference to choose an appropriate \(b\) and \(c\).

  1. Find an appropriate set \(a, b, c, d\) to give \(\lambda_1 = 4\) and \(\lambda_2 = -2\).

  2. Find an appropriate set \(a, b, c, d\) to give \(\lambda_1 = -1\) and \(\lambda_2 = 1\).

  3. Find an appropriate set \(a, b, c, d\) to give \(\lambda_1 = -2\) and \(\lambda_2 = 2\).

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