Help with this homework

Question

Consider the ODE system below y ′ = y² − z² ; z′ = y² + z²

Which of the following is true?

a. Solvable by trig substitution

b. Solvable using Euler method only if initial conditions are stated

c. RK 4 is not suitable for this system

d. Solvable using Euler method

Answer 1

I am supposing both y and z are functions of t.

Trig substitutions in calc applications are generally used to eliminate root operations (or just to change on calc function into another one) to allow for easier approaches to performing integrals. I don't think it will help here.

Euler method and RK4 both require some initial condition such as y(t₀) = y₀; z(t₀) = z₀ to be specified, meaning they cannot be used. But option b allows for this problem to be remedied in order to suggest the Euler method if the problem statement is modified a bit.

It might not be clear right away whether we can use Euler's method just by looking at the problem. But there is a way to find out: Let's just make up our own initial conditions and see if we can get anywhere by pretending that y(0) = 2 and z(0) = 1, understanding t₀ = 0 and arbitrarily choosing the sampling interval h = 0.1. Let's start with y:

First, find an equation for the tangent line at t₀. We just need a slope and a y-intercept. Our tangent line will be as simple as

y₀(t) = mt + b

m: The slope of the line tangent to y(t) at t₀

b: The y-intercept of the tangent line (since the tangent line is taken at t₀ = 0, this is just the value of y(t₀)

The slope of the tangent line of y(t) is given by the function y'(t), by definition. This is what a derivative is, no matter how complicated the representation of it is. We don't know what y'(t) is in terms of t alone (if we did, this wouldn't be a very interesting diff-eq. problem), but we have been given y'(t) in terms of y(t) and z(t):

y'(t) = y(t)² - z(t)²

We are specifically interested in what is the value of y'(t₀), which is just y'(0):

y'(0) = y(0)² - z(0)²

Initial conditions we made up are:

y(0) = 2

z(0) = 1

so

y'(0) = (2)² - (1)² = 3

That is our slope of y(t) at t₀. Now we just need to know the y-intercept of the tangent line. Since we are taking the tangent line at the y-axis (t = 0), we can just find the value of y(t₀):

y(t₀) = y(0) = 2

This is again a given value. Now assemble the slope and y-intercept into the tangent line equation:

y₀ = mt + b = 3t + 2

The idea is that if we follow the tangent line in t just a tiny bit, the output y₀(t) will be really close to the exact value of y(t), because y(t) is presumed continuous and differentiable.

We slide along the tangent line by our chosen interval h and see what we get:

y₀(t₀ + h) = 3(t₀ + h) + 2 = 3((0) + (0.1)) + 2 = 2.3

The key assumption is that this value is close to the actual y(t₀ + h), which we will call y(t₁). We can move forward by taking y(t₁) to be 2.3

Before we can make further approximations, we also need to get an estimate for z(t) at t₁. We use the same method:

z₀ = mt + b

z'(t) = y(t)² + z(t)²

m = z'(0) = y(0)² + z(0)² = 5

b = z(0) = 1

z₀ = 5t + 1

Find an approximation of z(t₁) by sampling the tangent line:

z₀(t₁) = z₀(0.1) = 5(0.1) + 1 = 1.5

We assume a value of 1.5 for z(t₁).

There, since we have values to use for y(t₁) and z(t₁) and y'(t) and z'(t) are given in terms of y and z, we can do the process again to get another estimate by finding the tangent lines to y and z at t₁ and advancing along them in t by the amount h. If we do so:

y'(t₁) = y(t₁)² - z(t₁)² = (2.3)² - (1.5)² = 3.04

z'(t₁) = y(t₁)² + z(t₁)² = (2.3)² + (1.5)² = 7.54

y₁ = 2.3 + 3.04(t - 0.5)

z₁ = 1.5 + 7.54(t - 0.5)

y(t₂) ≈ y₁(t₂) = y₁(1) = 2.3 + 3.04(1 - 0.5) = 3.82

z(t₂) ≈ z₁(t₂) = z₁(1) = 1.5 + 7.54(1 - 0.5) = 5.27

It looks to me like Euler's method is working as long as initial conditions are provided, so the best answer is b.