Archive Digitization

Mathematics Course Notes

Processed February 20, 2026 at 09:41 AM
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2.5 Improved Euler's Method

Forward Euler has some serious issues

\( y' = -10y \), \( y(0) = 1 \)     (solution: \( y = e^{-10t} \))

A coordinate graph with t and y axes. A green curve representing the true solution y = e^{-10t} starts at (t0, y0) and decays exponentially towards the t-axis. A dashed black line starts at (t0, y0) and represents a Forward Euler step. This line has a steep negative slope and crosses the t-axis, ending at a point (t1, y1*) where y1* is negative, illustrating numerical instability when the step size is too large.
Figure: A coordinate graph with t and y axes. A green curve representing the true solution y = e^{-10t} starts at (t0, y0) and decays exponentially towards the t-axis. A dashed black line starts at (t0, y0) and represents a Forward Euler step. This line has a steep negative slope and crosses the t-axis, ending at a point (t1, y1*) where y1* is negative, illustrating numerical instability when the step size is too large.

true solution \( y = e^{-10t} > 0 \)

step too large: \( y_1^* < 0 \) (wrong)

look further:

Note: \( y' = -10y = f \)

\[ y_{n+1} = y_n + f(t_n, y_n)h \]\[ = y_n - 10y_n h \]\[ = y_n(1 - 10h) \]

if \( h < \frac{1}{10} \), \( y_n > 0 \) (OK)

if \( h = \frac{1}{5} \), \( y_{n+1} = y_n(-2) \)

each new \( y \) changes sign, doubles in magnitude

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A coordinate graph with t and y axes. A dashed blue line starts near the origin and oscillates with increasing amplitude as t increases. Black dots are placed at various peaks and troughs.
Figure: A coordinate graph with t and y axes. A dashed blue line starts near the origin and oscillates with increasing amplitude as t increases. Black dots are placed at various peaks and troughs.

blows up, clearly not exponential decay

Backward Euler (Adams-Moulton order 1)

use slope at end of step instead of start of step

A coordinate graph with t and y axes illustrating the Backward Euler method. A green curve labeled 'true y' represents the exact solution. Points t_0 and t_1 are marked on the t-axis with a distance h between them. On the y-axis, y_0, y_1, and y_1* are marked. A dashed black line connects (t_0, y_0) to (t_1, y_1). A red dashed line shows the 'slope at (t_1, y_1)'. A blue note says 'use this at (t_0, y_0) to move forward'.
Figure: A coordinate graph with t and y axes illustrating the Backward Euler method. A green curve labeled 'true y' represents the exact solution. Points t_0 and t_1 are marked on the t-axis with a distance h between them. On the y-axis, y_0, y_1, and y_1* are marked. A dashed black line connects (t_0, y_0) to (t_1, y_1). A red dashed line shows the 'slope at (t_1, y_1)'. A blue note says 'use this at (t_0, y_0) to move forward'.

this is implicit method (Euler is explicit)

no stability issue like w/ Euler

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\[ y_{n+1} = y_n + f(t_{n+1}, y_{n+1}) h \]

need \( y_{n+1} \) to estimate \( y_{n+1} \) ? ! ?

for some equations, it's ok.

for example, \( y' = -2y \)

\[ y_{n+1} = y_n + f(t_{n+1}, y_{n+1}) h \]

\[ = y_n - 2y_{n+1} h \]

\[ (1 + 2h) y_{n+1} = y_n \rightarrow y_{n+1} = \frac{y_n}{1 + 2h} \quad \text{no issue!} \]

but generally problematic

\( y' = \sin(y) \)

\[ y_{n+1} = y_n + \sin(y_{n+1}) h \]

needs root-finding method such as Newton's method

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Improved Euler (Heun's method) is a combination of forward and backward Euler

basic idea: use forward Euler to estimate \( y_{n+1} \)

then use that to estimate \( f(t_{n+1}, y_{n+1}) \): slope at the end of the step

then take the average of the two slopes to move forward

A coordinate graph illustrating Heun's method. The horizontal axis is labeled t and the vertical axis is labeled y. A solid green curve represents the 'true y' solution. Starting from a point (t0, y0), a black dashed line shows the forward Euler step with slope f(tn, yn) reaching a point (t1, y-tilde-1). A red dashed line shows the slope f(tn+1, y-tilde-n+1) at that estimated next point. A blue dotted line represents the average of these two slopes, labeled 1/2 [f(tn, yn) + f(tn+1, y-tilde-n+1)], starting from (t0, y0) and ending at a point (t1, y1-star) which is closer to the true solution than the forward Euler estimate. The y-axis has tick marks for y0, y-tilde-1, and y1-star.
Figure: A coordinate graph illustrating Heun's method. The horizontal axis is labeled t and the vertical axis is labeled y. A solid green curve represents the 'true y' solution. Starting from a point (t0, y0), a black dashed line shows the forward Euler step with slope f(tn, yn) reaching a point (t1, y-tilde-1). A red dashed line shows the slope f(tn+1, y-tilde-n+1) at that estimated next point. A blue dotted line represents the average of these two slopes, labeled 1/2 [f(tn, yn) + f(tn+1, y-tilde-n+1)], starting from (t0, y0) and ending at a point (t1, y1-star) which is closer to the true solution than the forward Euler estimate. The y-axis has tick marks for y0, y-tilde-1, and y1-star.
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Algorithm

\( t_0, y_0 \) given

  • decide \( h \)
  • \( t_1 = t_0 + h \)
  • \( \tilde{y}_1 = y_0 + f(t_0, y_0)h \) “predictor” step
  • find slope there: \( f(t_1, \tilde{y}_1) \)
  • \( y_1^* = y_0 + \frac{1}{2} [f(t_0, y_0) + f(t_1, \tilde{y}_1)]h \) “corrector” step
  • repeat until \( t = \text{target } t \)

each step has two sub-steps

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Example

\( y' = 2y - 3t \), \( y(0) = 1 \)

\( h = 0.05 \), estimate \( y(0.1) \)

\( t_0 = 0 \)

\( y_0 = 1 \)

\( t_1 = 0 + 0.05 = 0.05 \)

predictor: \( \tilde{y}_1 = y_0 + f(t_0, y_0)h = 1 + [2(1) - 3(0)](0.05) = 1.1 \)

corrector: \( y_1^* = y_0 + \frac{1}{2} [f(t_0, y_0) + f(t_1, \tilde{y}_1)]h \)

\( = 1 + \frac{1}{2} [(2(1) - 3(0)) + (2(1.1) - 3(0.05))](0.05) = 1.10125 \)

one more major step

\[ \vdots \]

\[ y_2^* = 1.205256 \]

Comparison (using same \( h \)):

  • Forward Euler: 1.2025 (2 steps)
  • Backward Euler: 1.1867 (2 steps)
  • Imp. Euler: 1.205256 (2 (x2) steps)
  • true : 1.20535

after exam: Runge-Kutta order 4 (RK4)

(Imp. Euler is RK2)