Introduction to Numerical Methods for Differential Equations

Introduction to Numerical Methods for Diﬀerential
Equations
The Euler Method

Matthew Henderson

matthew.james.henderson@gmail.com

21 December, 2011

Overview

1 Initial Value Problems

2 Approximate Solutions of Initial Value Problems

3 The Euler Method

Matthew Henderson () Numerical Methods 21 December, 2011 2 / 15

An Initial Value Problem

Deﬁnition
An initial value problem (IVP) consists of an ordinary diﬀerential
equation along with an initial condition.



Deﬁnition

Example
y = 2(x + 3) − y, y(−1) = 3 (1)



Definition

Example
y = 2(x + 3) − y, y(−1) = 3 (1)

Definition
A solution is a function y(x) which satisfies both the ODE and the
initial condition.


The Solution

The IVP (1) has a unique solution:
y = 2(x + 2) + e−x−1 (2)

Function y satisﬁes the ODE:
y = 2 − e−x−1

2(x + 3) − y = 2(x + 3) − 2(x + 2) + e−x−1
= 2x + 6 − 2x − 4 − e−x−1
= 2 − e−x−1


The Solution

The IVP (1) has a unique solution:
y = 2(x + 2) + e−x−1 (2)

Function y also satisﬁes the initial condition:

y(−1) = 2(−1 + 2) + e−(−1)−1
= 2(1) + e1−1
= 2 + e0
= 3


A Plot of the Solution

y
10
y =2(x +2) + e−x−1
8

6

4

2

x
-3 -2 -1 1 2 3


A Plot of the Solution?

y
10

8

6

4

2

x
-3 -2 -1 1 2 3


The Euler Method

Deﬁnition
Given an IVP of the form:

y = f(x, y), y(a) = c


The Euler Method

Deﬁnition
Given an IVP of the form:

y = f(x, y), y(a) = c

To ﬁnd the approximate value of y(x + h) for some small value of h:

y(x + h) = y(x) + hf(x, y) (3)


The Euler Method

Example
Going back to our example from before:

y = 2(x + 3) − y, y(−1) = 3


The Euler Method

Example

y = 2(x + 3) − y, y(−1) = 3

In this case f(x, y) = 2(x + 3) − y.


The Euler Method

Example

y = 2(x + 3) − y, y(−1) = 3

Put h = 0.4, x = −1.0 and y = 3.0 into (3):


The Euler Method

Example

y = 2(x + 3) − y, y(−1) = 3

Put h = 0.4, x = −1.0 and y = 3.0 into (3):

y(−1.0 + 0.4) = y(−1.0) + 0.4f(−1.0, 3.0)


The Euler Method

Example

y = 2(x + 3) − y, y(−1) = 3

Put h = 0.4, x = −1.0 and y = 3.0 into (3):

y(−1.0 + 0.4) = y(−1.0) + 0.4f(−1.0, 3.0)
y(−0.6) = 3.0 + 0.4(2(−1.0 + 3.0) − 3.0)


The Euler Method

Example

y = 2(x + 3) − y, y(−1) = 3

Put h = 0.4, x = −1.0 and y = 3.0 into (3):

y(−1.0 + 0.4) = y(−1.0) + 0.4f(−1.0, 3.0)
y(−0.6) = 3.0 + 0.4(2(−1.0 + 3.0) − 3.0)
= 3.0 + 0.4(4 − 3.0)


The Euler Method

Example

y = 2(x + 3) − y, y(−1) = 3

Put h = 0.4, x = −1.0 and y = 3.0 into (3):

y(−1.0 + 0.4) = y(−1.0) + 0.4f(−1.0, 3.0)
y(−0.6) = 3.0 + 0.4(2(−1.0 + 3.0) − 3.0)
= 3.0 + 0.4(4 − 3.0)
= 3.4


The Euler Method

We can compute more values in the same way . . .
x = −0.6 and y = 3.4:

y(−0.6 + 0.4) = 3.4 + 0.4(2(−0.6 + 3) − 3.4)
y(−0.2) = 3.4 + 0.4(4.8 − 3.4)
= 3.96


The Euler Method

We can compute more values in the same way . . .
x = −0.6 and y = 3.4:

y(−0.6 + 0.4) = 3.4 + 0.4(2(−0.6 + 3) − 3.4)
y(−0.2) = 3.4 + 0.4(4.8 − 3.4)
= 3.96

x = −0.2 and y = 3.96:

y(−0.2 + 0.4) = 3.96 + 0.4(2(−0.2 + 3) − 3.96)
y(0.2) = bluey + 0.4(5.6 − 3.96)
= 4.616


The Euler Method

We continue to construct a table row-by-row:

x y hf(x, y)
-1.0 3.000000 0.400000
-0.6 0.560000
-0.2 3.960000 0.656000
0.2 4.616000 0.713600
0.6 5.329600 0.748160
1.0 6.077760 0.768896
1.4 6.846656 0.781338
1.8 7.627994 0.788803
2.2 8.416796 0.793281
2.6 9.210078 0.795969
3.0 10.00605 0.797581


The Euler Method

We continue to construct a table row-by-row:

x y hf(x, y)
-1.0 3.000000 0.400000
-0.6 3.400000 0.560000
-0.2 3.960000 0.656000
0.2 4.616000 0.713600
0.6 5.329600 0.748160
1.0 6.077760 0.768896
1.4 6.846656 0.781338
1.8 7.627994 0.788803
2.2 8.416796 0.793281
2.6 9.210078 0.795969
3.0 10.00605 0.797581


The Euler Method: Comparison
Now when we plot those values we get something that begins to look
like our exact solution, albeit only on the interval [−1, 3].
y
10
y =2(x +2) + e−x−1
8

6

4

2

x
-3 -2 -1 1 2 3


The Euler Method: Comparison

The numbers which we computed for y are not exactly correct but the
errors involved are quite small:

x y (approx) y (exact) Error
-1.0 3.000000 3.000000 0.000000
-0.6 3.400000 3.470320 0.070320
-0.2 3.960000 4.049329 0.089329
0.2 4.616000 4.701194 0.085194
0.6 5.329600 5.401896 0.072296
1.0 6.077760 6.135335 0.057575
1.4 6.846656 6.890718 0.044062
1.8 7.627994 7.660810 0.032816
2.2 8.416796 8.440762 0.023966
2.6 9.210078 9.227324 0.017246
3.0 10.00605 10.018316 0.012269


The Euler Method: Where does it come from?

Given an initial value problem of the form:

y = f(x, y), y(a) = c

we want to ﬁnd the approximate value of y(b) for some b > a. From
the deﬁnition of derivative:
y(x + h) − y(x)
y (x) ≈
h
for h > 0 given and small. So,

y(x + h) − y(x)
f(x, y) ≈
h
which gives
y(x + h) ≈ y(x) + hf(x, y)


The Euler Method: Why does it work?

y =2(x +2) + e−x−1 10

8

6

4

2

-3 -2 -1 1 2 3


The End

Merry Christmas!


Introduction to Numerical Methods for Differential Equations

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