Assignment 5

Due Thursday, October 8

1Solving from a basis we find

Consider the equation

𝑦 ″ - 3 𝑦' + 2 𝑦 = 0 .

(a)

Before looking for any solutions, explain why the solution space $S$ of this equation is two-dimensional. Then look for solutions by guessing and checking: try an exponential $𝑦 = 𝑒 𝑟 𝑡$ . Find two linearly independent solutions.

(b)

Use your two solutions from the previous part as an ordered basis $B$ of the solution space. Write down the evaluation map

𝐸 0 : S ⟶ ℝ 2

onto the space of initial conditions. Using the basis $B$ for $S$ and the standard basis $(1, 0), (0, 1)$ for $ℝ 2$ , express $𝐸 0$ as a matrix.

(c)

The inverse of the evaluation map is the solution map $𝐸 - 1 0 : ℝ 2 \to S$ . It takes an initial condition $(𝑦 0, 𝑣 0)$ to the solution starting from it. In these same bases, write down the matrix of the solution map, and use it to write the solution from $(𝑦 0, 𝑣 0)$ .

(d)

Use your work to solve $𝑦 ″ - 3 𝑦' + 2 𝑦 = 0$ with each of the following initial conditions:

$𝑦 (0) = 3$ , $𝑦' (0) = 4$ ;
$𝑦 (0) = 1$ , $𝑦' (0) = - 1$ ;
$𝑦 (0) = 0$ , $𝑦' (0) = 2$ .

2Discovering the Hyperbolic Functions

Consider the equation

𝑦 ″ = 𝑦 .

In class, we found the basis $B = (𝑒 𝑡, 𝑒 - 𝑡)$ for its solution space $S$ , and the evaluation map taking each solution to its initial data:

𝐸 0 : 𝑎 𝑒 𝑡 + 𝑏 𝑒 - 𝑡 ⟼ (𝑎 + 𝑏, 𝑎 - 𝑏) .

Form its matrix $𝑀$ , using the basis $B$ for $S$ and the standard basis for $ℝ 2$ . To solve an initial-value problem in this basis, we have to invert $𝑀$ . In this problem we look for a better basis.

(a)

Find a basis $(𝑅 1, 𝑅 2)$ of $S$ in which the matrix of $𝐸 0$ is the identity. What must $𝐸 0 (𝑅 1)$ and $𝐸 0 (𝑅 2)$ be? Write $𝑅 1$ and $𝑅 2$ as combinations of $𝑒 𝑡$ and $𝑒 - 𝑡$ .

(b)

Check that in the basis $(𝑅 1, 𝑅 2)$ , the matrix of evaluation is the identity. Explain why the matrix of the solution map is then the identity too, and write down the solution with initial condition $(𝑦 0, 𝑣 0)$ . Use it to solve $𝑦 ″ = 𝑦$ with $𝑦 (0) = 2$ , $𝑦' (0) = - 1$ .

(c)

The functions $𝑅 1$ and $𝑅 2$ are called the hyperbolic cosine and sine, written $c o s h 𝑡$ and $s i n h 𝑡$ . They get this name because they parametrize a hyperbola. Just as the point $(c o s 𝑡, s i n 𝑡)$ moves along the circle $𝑥 2 + 𝑦 2 = 1$ , the point $(c o s h 𝑡, s i n h 𝑡)$ moves along the hyperbola $𝑥 2 - 𝑦 2 = 1$ .

To see this, we need to show that the hyperbolic functions satisfy the identity

c o s h 2 𝑡 - s i n h 2 𝑡 = 1 .

To show that a function is constant, we can show that its derivative is zero. Then we can find which constant it is by evaluating it at a single point. Carry out this strategy for $c o s h 2 𝑡 - s i n h 2 𝑡$ :

Show that $c o s h' = s i n h$ and $s i n h' = c o s h$ .
Use these to show that the derivative of $c o s h 2 𝑡 - s i n h 2 𝑡$ is zero.
Evaluate $c o s h 2 𝑡 - s i n h 2 𝑡$ at $𝑡 = 0$ to find the constant it equals.

3A missing solution

Consider the third-order equation

𝑥 ‴ - 𝑥 ″ = 0 .

(a)

Before looking for any solutions, explain why the solution space $S$ is three-dimensional. Then try $𝑥 = 𝑒 𝑟 𝑡$ and find every value of $𝑟$ that works. How many linearly independent solutions does this give? Why does the dimension of $S$ tell you that these cannot be all of the solutions?

(b)

Look for a third solution by trying powers $𝑥 = 𝑡 𝑛$ . Which powers work?

Then check that your three solutions are linearly independent. You do not need to work with the functions directly. The evaluation map $𝐸 0 (𝑥) = (𝑥 (0), 𝑥' (0), 𝑥 ″ (0))$ is a linear isomorphism, and a linear isomorphism preserves linear independence in both directions. So solutions are linearly independent exactly when their initial data are linearly independent vectors in $ℝ 3$ . Compute the initial data of your three solutions, and show that these vectors are independent.

(c)

Use your three solutions as the basis $B$ of $S$ , and the standard basis for $ℝ 3$ . Express the evaluation map $𝐸 0$ as a matrix. Then invert it to find the matrix of the solution map, which takes an initial condition $(𝑥 0, 𝑣 0, 𝑤 0)$ to its solution.

(d)

Use your work to solve $𝑥 ‴ - 𝑥 ″ = 0$ with each of the following initial conditions:

$𝑥 (0) = 1$ , $𝑥' (0) = 0$ , $𝑥 ″ (0) = 2$ ;
$𝑥 (0) = 2$ , $𝑥' (0) = 1$ , $𝑥 ″ (0) = 1$ ;
$𝑥 (0) = 0$ , $𝑥' (0) = 3$ , $𝑥 ″ (0) = - 1$ .

4The Hyperbolic Functions from a Matrix Exponential

We return to the equation

𝑦 ″ = 𝑦,

this time solving it with the matrix exponential. Along the way, we will find the power series of the hyperbolic functions.

(a)

Set $𝑣 = 𝑦'$ and write the equation as a first-order system $𝑋' = 𝐴 𝑋$ with state $𝑋 = (𝑦, 𝑣)$ . Compute $𝐴 2$ , $𝐴 3$ , and $𝐴 4$ . What pattern do the powers of $𝐴$ follow?

(b)

Substitute your pattern into the series

𝑒 𝑡 𝐴 = \infty \sum 𝑘 = 0 𝑡 𝑘 𝐴 𝑘 𝑘! .

Collect the even and odd powers to write

𝑒 𝑡 𝐴 = 𝑐 (𝑡) 𝐼 + 𝑠 (𝑡) 𝐴,

where $𝑐 (𝑡)$ and $𝑠 (𝑡)$ are scalar power series. Find these two series, and write $𝑒 𝑡 𝐴$ as a $2 \times 2$ matrix with entries $𝑐 (𝑡)$ and $𝑠 (𝑡)$ .

(c)

The matrix $𝑒 𝑡 𝐴$ is the flow matrix of the system, so its first column is the state $(𝑦 (𝑡), 𝑣 (𝑡))$ of the solution starting from $(1, 0)$ . Use your work in Discovering the Hyperbolic Functions to identify that solution and its state. What functions are $𝑐 (𝑡)$ and $𝑠 (𝑡)$ ? What power series have you found?

5The Exponential of a Sum

For numbers, $𝑒 𝑎 + 𝑏 = 𝑒 𝑎 𝑒 𝑏$ . In this problem we test whether the same law holds for matrices, using

𝐵 = (0100), 𝐶 = (0010) .

(a)

Compute $𝐵 2$ and $𝐶 2$ . Use your answers to find $𝑒 𝑡 𝐵$ and $𝑒 𝑡 𝐶$ from the exponential series.

(b)

Compute $𝐵 + 𝐶$ . Where have you seen this matrix before, and what is $𝑒 𝑡 (𝐵 + 𝐶)$ ? Now compute the products $𝑒 𝑡 𝐵 𝑒 𝑡 𝐶$ and $𝑒 𝑡 𝐶 𝑒 𝑡 𝐵$ . Does $𝑒 𝑡 (𝐵 + 𝐶) = 𝑒 𝑡 𝐵 𝑒 𝑡 𝐶$ ? Does the order of the product matter?

(c)

Why does the law fail? Let $𝐵$ and $𝐶$ now be any two square matrices of the same size. Write out the terms of $𝑒 𝑡 (𝐵 + 𝐶)$ and of $𝑒 𝑡 𝐵 𝑒 𝑡 𝐶$ up through $𝑡 2$ . Show that the $𝑡 2$ terms agree exactly when $𝐵 𝐶 = 𝐶 𝐵$ . Compute $𝐵 𝐶$ and $𝐶 𝐵$ for our matrices. Why does this issue never arise for numbers?

6A Series That Stops

Consider the third-order equation

𝑥 ‴ = 0 .

(a)

Solve this equation with initial data $𝑥 (0) = 𝑥 0$ , $𝑥' (0) = 𝑣 0$ , $𝑥 ″ (0) = 𝑤 0$ by integrating three times.

(b)

Set $𝑣 = 𝑥'$ and $𝑤 = 𝑥 ″$ , and write the equation as a first-order system $𝑋' = 𝑁 𝑋$ with state $𝑋 = (𝑥, 𝑣, 𝑤)$ . Compute $𝑁 2$ and $𝑁 3$ .

(c)

Compute $𝑒 𝑡 𝑁$ from the exponential series. Use it to solve the system from the initial state $𝑋 0 = (𝑥 0, 𝑣 0, 𝑤 0)$ , and check that you recover your answer from the first part.