Numerical solutions and error

Chapter 10 - Follow the local rule in finite steps

Many useful differential equations have no elementary closed-form solution. Numerical methods turn the rate rule into a sequence of approximations.

Euler's method uses the current slope for one short step:

$$\boxed{y_{n+1}=y_n+hf(t_n,y_n)},t_{n+1}=t_n+h.$$

It is the tangent-line approximation repeated.

Lab: shrink the step

For each equation, compare (h=1), (0.5), (0.25), and (0.1). Smaller steps usually reduce error, but they require more work. "Numerical" does not mean "unreliable"; it means error must be estimated and controlled.

Worked example: show every step

A table is part of the reasoning

The table prevents a common error: evaluating the new slope at a mixture of old and new coordinates.

Local and global error

Euler replaces a curved segment by a tangent step. If the exact solution has bounded second derivative, one step's local truncation error is proportional to (h^2). Over about (1/h) steps across a fixed time interval, these errors accumulate into global error proportional to (h).

Practical consequence: halving (h) should roughly halve Euler's global error once the steps are small enough for the asymptotic pattern to appear.

Better slopes: midpoint and Runge-Kutta

Euler uses the slope at the start of a step. A midpoint method samples a slope near the middle. Classical fourth-order Runge-Kutta blends four slope samples. The central design idea is the same: spend more rate evaluations to approximate the curve across each step more accurately.

You do not need to memorize RK4 here. You should be able to ask:

• where does the method sample slopes?
• what order of global error does it claim?
• does halving the step produce the expected convergence?
• does the numerical curve respect known equilibria, signs, and bounds?

Numerical stability is not physical stability

For (y'=-10y), the true solution decays. Euler gives

$$y_{n+1}=(1-10h)y_n.$$

If (h=0.3), the multiplier is (-2): the approximation alternates and grows even though the true system decays. The step is too large for Euler's stability region.

Fade the computation

Use Euler with (h=0.25) for (y'=y(1-y)), (y(0)=0.2).

1. (y_1=0.2+0.25[0.2(0.8)]=\underline{\hspace{4em}}).
2. Evaluate the next slope at ((t_1,y_1)=(0.25,\underline{\hspace{3em}})).
3. Compute (y_2) and confirm that the approximation stays between 0 and 1.

Practice

1. Practice - Euler table

   10.1 Use Euler with \(h=0.2\) to approximate \(y(0.6)\) for \(y'=y-t\), \(y(0)=1\). Show a slope table.

   Detailed solution

2. Practice - convergence

   10.2 For \(y'=y\), \(y(0)=1\), derive Euler's approximation \(y_n=(1+h)^n\). Compare \(h=1,1/2,1/4\) at \(t=1\) with \(e\).

   Detailed solution

3. Integrated - qualitative check

   10.3 Design a test that would catch a numerical solution of \(P'=P(1-P/10)\), \(P(0)=2\), crossing above 10.

   Detailed solution

4. Challenge - step stability

   10.4 For \(y'=-ay\) with \(a>0\), find all \(h>0\) for which Euler approximations decay in magnitude. Which subset decays without alternating sign?

   Detailed solution

Retrieve the whole loop

Choose one equation from any earlier chapter and state:

1. its structural fingerprint;
2. a qualitative prediction;
3. an analytic method, if available;
4. a numerical fallback; and
5. two independent checks.

I compared at least four step sizes and checked the numerical results against qualitative behavior.

Next: Enter the mixed practice studio, where the method is never named for you.