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2.5 Improved Euler

Forward Euler can have stability issues

\[ y' = -10y \quad y(0) = 1 \quad \text{(solution } y = e^{-10t} \text{)} \]

Figure: A coordinate graph illustrating the stability issue of the Forward Euler method. The vertical axis is y and the horizontal axis is t. A green curve labeled 'true y' starts at (t0, y0) on the y-axis and decays exponentially towards the t-axis, remaining positive. A dashed line also starts at (t0, y0) but has a steeper downward slope, crossing the t-axis at time t1 to reach a point labeled y1 below the axis. This illustrates that if the step size is too large, the numerical solution y1 becomes negative, which is incorrect since the true solution is always positive.

true \( y > 0 \)

notice if step is too big, \( y_1 < 0 \) (clearly wrong)

for this eq., we can find a limit on step size

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

if \( h < \frac{1}{10} \), \( y_n, y_{n+1}, y_{n+2}, \dots \) all have same sign and decreasing in magnitude (ok)

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

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Figure: A graph on a horizontal \(t\)-axis showing a solid green curve decaying smoothly. A dashed blue curve oscillates with increasing amplitude, with red dots marking the extreme points.

stability issue!

Backward Euler (Adams-Moulton order 1)

use the slope at the end of the step instead of at the start

Figure: A graph illustrating the Backward Euler method. The horizontal axis is labeled \(t\) and has points \(t_0\) and \(t_1\) with a bracketed interval labeled \(h\). The vertical axis has points \(y_0\), \(y_1\), and \(y_1^*\). A solid green curve represents the true solution, labeled 'true y'. A red dotted line represents the slope \(f(t_1, y_1)\) at the end of the step, showing the backward-looking approximation.

this "looking back" eliminates the stability issue noted above

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

need \( y_{n+1} \) to find \( f(t_{n+1}, y_{n+1}) \) which is used to estimate \( y_{n+1} \)

Surprisingly, not always a problem

\[ y' = -2y \]

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

\[ (1 + 2h) y_{n+1} = y_n \rightarrow y_{n+1} = \frac{y_n}{1 + 2h} \]

but not always doable

\[ y' = \sin(y) \]

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

need, for example, a root-finding technique such as Newton's method

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Improved Euler ( Heun's method) is a combination of these
two

basic idea: use forward Euler to find \(\tilde{y}_{n+1}\)

use that to find slope at the end : \(f(t_{n+1}, \tilde{y}_{n+1})\)

so back to \((t_n, y_n)\) , use average of the two
slopes to move to \((t_{n+1}, y_{n+1}^*)\)

Figure: A coordinate graph illustrating Heun's method (Improved Euler). The horizontal axis is labeled 't' and the vertical axis is labeled 'y'. A solid green curve represents the 'true y' solution. On the t-axis, two points t0 and t1 are marked, with the interval between them labeled 'h'. On the y-axis, points y0, y_tilde_1, and y1_star are marked. From the starting point (t0, y0), a black dashed line with slope f(t0, y0) (labeled 'Forward') extends to (t1, y_tilde_1). A red dotted line with slope f(t1, y_tilde_1) (labeled 'Backward') is shown starting from (t0, y0). A blue dashed line, representing the average of these two slopes (labeled '1/2 [f(t0, y0) + f(t1, y_tilde_1)]'), starts at (t0, y0) and ends at the improved estimate (t1, 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 \( y(0.1) \)

\( t_0 = 0 \)
\( y_0 = 1 \)
\( t_1 = t_0 + h = 0.05 \)

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

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

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

\( t_2 = t_1 + h = 0.1 \) (target)
\( \vdots \)
repeat the predictor, corrector sub-steps
\( \vdots \)
\( y_2^* = 1.205256 \)

Comparison (Same \( h = 0.05 \))