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 $$\begin{array}{c}{\partial }_{t}x=ax+by\\ {\partial }_{t}y=cx+dy\end{array}=\left[\begin{array}{cc}a& b\\ c& d\end{array}\right]\left[\begin{array}{c}x\\ y\end{array}\right]$$ 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.

1. For ${\lambda }_1=-1$ and ${\lambda }_2=-0.5$, which parts are transient?

A-part       B-part       both parts       neither part      

question id: fms02-1

2. For ${\lambda }_1=-0.01$ and ${\lambda }_2=0.01$, which parts are transient?

A-part       B-part       both parts       neither part      

question id: fms02-2

3. For ${\lambda }_1=0.01$ and ${\lambda }_2=-0.3$, which parts are transient?

A-part       B-part       both parts       neither part      

question id: fms02-3

In answering the next two questions, keep in mind that for large $t$, $$A{e}^{{\lambda }_{1}t}+B{e}^{{\lambda }_{2}t}\approx A{e}^{{\lambda }_{1}t}\text{when}{\lambda }_2<{\lambda }_1.$$

4. For ${\lambda }_1=0.01$ and ${\lambda }_2=0.001$, which parts are transient?

A-part       B-part       both parts       neither part      

question id: fms02-4

5. For $_1 = -0.1 $ and ${\lambda }_2=-0.001$, which parts are transient?

A-part       B-part       both parts       neither part      

question id: fms02-5

Activities

Exercise 2 The linear dynamical system in two variables is

$$\begin{array}{c}{\partial }_{t}x=ax+by\\ {\partial }_{t}y=cx+dy\end{array}=\left[\begin{array}{cc}a& b\\ c& d\end{array}\right]\left[\begin{array}{c}x\\ y\end{array}\right].$$ 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 $$\begin{array}{c}{\partial }_{t}x=ax+by\\ {\partial }_{t}y=cx+dy\end{array}=\left[\begin{array}{cc}a& b\\ c& d\end{array}\right]\left[\begin{array}{c}x\\ y\end{array}\right]$$ 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 $$\begin{array}{c}{\partial }_{t}x=ax+by\\ {\partial }_{t}y=cx+dy\end{array}=\left[\begin{array}{cc}a& b\\ c& d\end{array}\right]\left[\begin{array}{c}x\\ y\end{array}\right]$$ the two values of $\lambda$ are $$\begin{gathered} {\lambda }_1=\frac{1}{2}(a+d)+\frac{1}{2}\sqrt{{(a-d)}^2-4bc} \\ {\lambda }_1=\frac{1}{2}(a+d)-\frac{1}{2}\sqrt{{(a-d)}^2-4bc} \end{gathered}$$

1. Show that the sum ${\lambda }_1+{\lambda }_2=a+d$.

2. Show that the square of the difference ${({\lambda }_1-{\lambda }_2)}^2=(a-d{)}^2-4bc$.

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 $c$.

3. Find an appropriate set $a,b,c,d$ to give ${\lambda }_1=4$ and ${\lambda }_2=-2$.

4. Find an appropriate set $a,b,c,d$ to give ${\lambda }_1=-1$ and ${\lambda }_2=1$.

5. Find an appropriate set $a,b,c,d$ to give ${\lambda }_1=-2$ and ${\lambda }_2=2$.

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