July 2013
Chapter 25
Runge-Kutta Methods
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This chapter is devoted to solving ordinary differential equations of the
form
𝑑𝑦
𝑑𝑥
= 𝑓(𝑥, 𝑦)
We can use a numerical method to solve such an equation for the velocity
of the falling parachutist. The method is of the general form
New value = old value + slope × step size
or, in mathematical terms,
𝑦𝑖+1 = 𝑦𝑖 + ∅ℎ
According to this equation, the slope estimate of ∅ is used to extrapolate
from an old value 𝑦𝑖 to a new value 𝑦𝑖+1 over a distance h (next figure).
All one-step methods can be expressed in this general form, with the only
difference being the manner in which the slope is estimated. As in the
falling parachutist problem, the simplest approach is to use the differential
equation to estimate the slope in the form of the first derivative at xi.
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25.1 Euler’s Method
The first derivative provides a direct estimate of the slope at 𝑥𝑖
(next figure)
∅ = 𝑓(𝑥𝑖 , 𝑦𝑖 )
where 𝑓(𝑥𝑖 , 𝑦𝑖 ) is the differential equation evaluated at 𝑥𝑖 and 𝑦𝑖 .
This estimate can be substituted into
𝑦𝑖+1 = 𝑦𝑖 + 𝑓(𝑥𝑖 , 𝑦𝑖 )ℎ
(25.2)
This formula is referred to as Euler’s Method. A new value of y is predicted
using the slope (equal to the first derivative at the original value of x)
to extrapolate linearly over the step size h. (previous figure)
‫احجز خمسه وأربعون دقيقه فى جدول‬
. ‫أعمالك كل أسبوع لالسترخاء العميق‬
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EXAMPLE 25.1 Euler’s Method
Problem Statement:
Use Euler’s method to numerically integrate
𝑑𝑦
𝑑𝑥
= −2𝑥 3 + 12𝑥 2 − 20𝑥 + 8.5
from x = 0 to x = 4 with a step size of 0.5. The initial condition at x = 0 is
y = 1. Recall that the exact solution is given by
𝑦 = −0.5𝑥 4 + 4𝑥 3 + 10𝑥 2 − 8.5𝑥 + 1
Solution:
Equation (25.2) can be used to implement Euler’s Method:
𝑦(0.5) = 𝑦(0) + 𝑓(0, 1) 0.5
where y(0) = 1 and the slope estimate at x = 0 is
𝑓(0, 1) = −2(0)3 + 12(0)2 − 20(0) + 8.5 = 8.5
𝑦(0.5) = 1.0 + 8.5(0.5) = 5.25
The true solution at x = 0.5 is
𝑦 = −0.5(0.5)4 + 4(0.5)3 − 10(0.5)2 − 8.5(0.5) + 1 = 3.21875
𝐸𝑡 = 𝑡𝑟𝑢𝑒 − 𝑎𝑝𝑝𝑟𝑜𝑥𝑖𝑚𝑎𝑡𝑒 = 3.21875 − 5.25 = −2.03125
Or, expressed as percent relative error, εt = -63.1%. For the second step,
𝑦(1) = 𝑦(0.5) + 𝑓(0.5, 5.25)
= 5.25 + [−2(0.5)3 + 12(0.5)2 − 20(0.5) + 8.5]0.5
= 5.875
The true solution at x = 1.0 is 3.0, and therefore, the percent relative error
is -95.8%. The computation is repeated, and the results are compiled in
the next table and the next figure. Note that the error can be reduced by
using a smaller step size
‫مصااااابيف الفلوريسااااون الحدي ااااة تااااوفر‬
‫ مما تستهلكه المصابيف التقليدية‬%80
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25.2 Improvements of Euler’s Method
A fundamental source of error in Euler’s method is that the derivative
at the beginning of the interval is assumed to apply across the entire
interval. Two simple modifications are available to help overcome this
shortcoming. Both modifications actually belong to a larger class of
solution techniques called Runge-Kutta methods.
‫إذا كان هم الربان المحافظة‬
‫ فسفينته‬،‫على سالمه السفينة‬
.‫لن تبرح الميناء‬
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However,
because
they
have
a
very
straightforward
graphical
interpretation, we will present them prior to their formal derivation as
Runge-Kutta methods.
25.2.1 Heun’s Method
One method to improve the estimate of the slope involves the
determination of two derivatives for the interval – one at the initial point
and another at the end point. The two derivatives are then averaged to
obtain an improved estimate of the slope for the entire interval.
This approach, called Heun’s method, is shown graphically in the next
figure.
‫اقبااال الحاااق ولاااو ممااان تكرهاااه‬
‫وابغض الباطل ولو من صديق‬
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Recall that in Euler’s methods, the slope at the beginning of an interval
𝑦𝑖′ = 𝑓(𝑥𝑖 , 𝑦𝑖 )
is used to extrapolate linearly to 𝑦𝑖+1 :
0
𝑦𝑖+1
= 𝑦𝑖 + 𝑓(𝑥𝑖 , 𝑦𝑖 )ℎ
For the standard Euler method we would stop at this point.
0
However, in Heun’s method the 𝑦𝑖+1
calculated in the previous equation is
not the final answer, but an intermediate prediction.
This is why we have distinguished it with a superscript 0.
The previous equation is called a predictor equation. It provided an
estimate of yi+1 that allows the calculation of an estimated slope at the end
of the interval:
′
0
𝑦𝑖+1
= 𝑓(𝑥𝑖+1 , 𝑦𝑖+1
)
Thus, the two slopes can be combined to obtain an average slope for the
interval:
′
𝑦̅ =
′
𝑦𝑖′ + 𝑦𝑖+1
2
=
0
𝑓(𝑥𝑖 , 𝑦𝑖 )+ 𝑓(𝑥𝑖+1 , 𝑦𝑖+1
)
2
This average slope is then used to extrapolate linearly from 𝑦𝑖 to 𝑦𝑖+1 using
Euler’s method:
𝑦𝑖+1 = 𝑦𝑖 +
0
𝑓(𝑥𝑖 , 𝑦𝑖 )+ 𝑓(𝑥𝑖+1 , 𝑦𝑖+1
)
2
which is called a corrector equation.
ℎ
،‫إن أردت أن تعرف نفسك جيدا‬
‫بمجرد استيقاظك تذكر ما كنت‬
‫تحلم به فهذا هو أنت‬
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The Heun method is a predictor-corrector approach.
The Heun method is the only one-step predictor-corrector method
described in this book. As derived above, it can be expressed concisely as
0
Predictor (previous figure a): 𝑦𝑖+1
= 𝑦𝑖 + 𝑓(𝑥𝑖 , 𝑦𝑖 )ℎ
Corrector (figure b): 𝑦𝑖+1 = 𝑦𝑖 +
0
𝑓(𝑥𝑖 ,𝑦𝑖 )+ 𝑓(𝑥𝑖+1 ,𝑦𝑖+1
)
2
ℎ
(25.16)
Note that because the previous equation has 𝑦𝑖+1 on both sides of
the equal sign, it can be applied in an iterative fashion.
That is, an old estimate can be used repeatedly to provide an improved
estimate of 𝑦𝑖+1 . The process is shown in the next figure.
‫اجعل لنفسك فترات راحة‬
،‫قصيرة خالل وقت دراستك‬
‫ال تنسى كلمة قصيرة‬
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It should be understood that this iterative process does not
necessarily converge on the true answer but will converge on an estimate
with a finite truncation error, as will be demonstrated in the following
example.
As with the similar iterative methods, a termination criterion for
convergence of the corrector is provided by
𝑗
|𝜀𝑎 | = |
𝑗−1
𝑗−1
𝑦𝑖+1 − 𝑦𝑖+1
𝑗
𝑦𝑖+1
|
𝑗
where 𝑦𝑖+1 and 𝑦𝑖+1 are the result from the prior and the present
iteration of the corrector, respectively.
‫ تماما كما‬،‫دل غيرك على الخير‬
‫تحب أن يدلك اآلخرون عليه‬
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EXAMPLE 25.5 Heun’s Method
Problem Statement:
Use Heun’s method to integrate y’ = 4e0.8x – 0.5y from x = 0 to x = 4 with a
step size of 1. The initial condition at x = 0 is y = 2.
Solution:
Before solving the problem numerically, we can use calculus to determine
the following analytical solution:
𝑦=
4
13
(𝑒 0.8𝑥 − 𝑒 −0.5𝑥 ) + 2𝑒 −0.5𝑥
This formula can be sued to generate the true solution values in the next
table
First, the slope at at (𝑥0 , 𝑦0 ) is calculated as
𝑦0′ = 4𝑒 0 − 0.5(2) = 3
This result is quite different from the actual average slope for the interval
from 0 to 1.0, which is equal to 4.1946, as
calculated from the differential equation.
‫غالبا تكون الواجبات التي عليك أك ر من‬
‫ أبدأ بالواجبات‬،‫الحقوق التي تستحقها‬
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Continue:
The numerical solution is obtained by using the predictor to obtain an
estimate of y at 1.0:
𝑦10 = 2 + 3(1) = 5
Note that this is the result that would be obtained by the standard Euler
method. The true value in the previous table shows that it corresponds to
a percent relative error of 19.3 percent.
Now, to improve the estimate for 𝑦𝑖+1 , we use the value 𝑦10 to predict
the slope at the end of the interval
𝑦1′ = 𝑓(𝑥1 , 𝑦10 ) = 4𝑒 0.8(1) − 0.5(5) = 6.402164
Which can be combined with the initial slope to yield an average slope
over the interval from x = 0 to 1
𝑦′ =
3+ 6.402164
2
= 4.701082
which is closer to the true average slope of 4.1946.
This result can then be substituted into the corrector to give the prediction
at x = 1
𝑦1 = 2 + 4.701082(1) = 6.701082
which represents a percent relative error of -8.18 percent.
Thus, the Heun method without iteration of the corrector reduces the
absolute value of the error by a factor of 2.4
as compared with Euler’s method.
‫ ماذا‬،‫ اليوم يوافق تاريخ زواجنا‬:‫الزوجة‬
.‫ لنقف دقيقة حداد‬:‫ الزوج‬،‫سنفعل‬
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Continue:
Now this estimate can be used to refine or correct the prediction of
𝑦1 by substituting the new result back into the right-hand side of equation
(25.16):
𝑦1 = 2 +
[ 3+ 4𝑒 0.8(1) − 0.5(6.701082)]
2
1 = 6.275811
which represents an absolute percent relative error of 1.31 percent.
This result, in turn, can be substituted back into equation (25.16) to
further correct:
𝑦1 = 2 +
[ 3+ 4𝑒 0.8(1) − 0.5(6.275811)]
2
1 = 6.382129
which represents an |𝜀𝑡 | of 3.03%.
Notice how the errors sometimes grow as the iterations proceed.
Such increases can occur, especially for large step sizes, and they prevent
us from drawing the general conclusion that an additional iteration will
always improve the result.
However, for a sufficiently small step size, the iterations should
eventually converge on a single value. For our case, 6.360865, which
represents a relative error of 2.68 percent, is attained after 15 iterations.
The previous table shows results for the remainder of the computation
using the method with 1 and 15 iterations per step.
‫الشخص الناجف هو في األساس‬
‫شخص يستطيع أن يتخيل‬
‫ويحول خياله إلى واقع‬
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In the previous example, the derivative is a function of both the
dependent variable y and the independent variable x.
For cases such as polynomials, where the ODE is solely a function of the
independent variable, the predictor step is not required and the corrector
is applied only once for each iteration. For such cases, the technique is
expressed concisely as
𝑦𝑖+1 = 𝑦𝑖 +
𝑓(𝑥𝑖 )+ 𝑓(𝑥𝑖+1 )
2
ℎ
(25.18)
Notice the similarity between the right-hand side of the previous
equation and the trapezoidal rule. The connection between the two
methods can be formally demonstrated by starting with the ordinary
differential equation
𝑑𝑥
𝑑𝑦
= 𝑓(𝑥)
This equation can be solved for y by integration:
𝑦𝑖+1
∫𝑦
𝑖
𝑥
𝑑𝑦 = ∫𝑥 𝑖+1 𝑓(𝑥 )𝑑𝑥
𝑖
which yields
𝑥
𝑦𝑖+1 − 𝑦𝑖 = ∫𝑥 𝑖+1 𝑓(𝑥 )𝑑𝑥
𝑖
or
𝑥
𝑦𝑖+1 = 𝑦𝑖 + ∫𝑥 𝑖+1 𝑓(𝑥 )𝑑𝑥
𝑖
(25.21)
‫طلبه بيكتبوا محاضرة مع الدكتور كل‬
‫ما يمسف السبورة يقطعوا الورقة‬
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Now, recall that the trapezoidal rule is defined as
𝑥𝑖+1
∫𝑥
𝑖
𝑓(𝑥 )𝑑𝑥 ≅
𝑓(𝑥𝑖 )+ 𝑓(𝑥𝑖+1 )
2
ℎ
where h = 𝑥𝑖+1 − 𝑥𝑖 . Substituting equations yields
𝑦𝑖+1 = 𝑦𝑖 +
𝑓(𝑥𝑖 )+ 𝑓(𝑥𝑖+1 )
2
ℎ
which is equivalent to equation 25.18
Because the previous equation is a direct expression of the
trapezoidal rule, the local truncation error is given by
𝐸𝑡 = −
𝑓 ′′ (𝜉)
12
ℎ3
where 𝜉 is between 𝑥𝑖 and 𝑥𝑖+1 .
Thus, the method is second order because the second derivative of
the ODE is zero when the true solution is quadratic.
In addition, the local and global errors are 𝑂(ℎ3 ) and 𝑂(ℎ2 ), respectively.
Therefore, decreasing the step size decreases the error at a faster rate
than for Euler’s method.
The next figure, which shows the result of using Heun’s method to
solve the polynomial from example 25.1 demonstrates this behavior.
،‫ لكااان ماااا تكلماااه‬،‫شاااخص تحباااه‬
‫وإذا كلمته يقولوا عليك مجنون؟‬
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25.2.2 The Midpoint (or Improved Polygon) Method
The next figure illustrates another simple modification of Euler’s
method. Called the mid-point method, this technique uses Euler’s method
to predict a value of y at the midpoint of the interval (figure a):
𝑦𝑖+1/2 = 𝑦𝑖 + 𝑓 (𝑥𝑖 , 𝑦𝑖 )
ℎ
2
Then, this predicted value is used to calculate a slope at the midpoint:
′
𝑦𝑖+1/2
= 𝑓(𝑥𝑖+1/2 , 𝑦𝑖+1/2 )
which is assumed to represent a valid approximation of the average slope
for the entire interval. This slope is then used to extrapolate linearly from
𝑥𝑖 to 𝑥𝑖+1 (figure b):
𝑦𝑖+1 = 𝑦𝑖 + 𝑓(𝑥𝑖+1/2 , 𝑦𝑖+1/2 ) ℎ
(25.27)
‫ حرف غير‬24 ‫جملة مكونة من‬
‫منقوط البد أن نكررها يوميا‬
‫أك ر من عشر مرات؟‬
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Observe that because 𝑦𝑖+1 is not on both sides, the corrector cannot be
applied iteratively to improve the solution. This approach can also be
linked to Newton-Cotes integration formulas.
Recall that the simplest Newton-Cotes open integration formula,
which is called the midpoint method, can be represented as
𝑏
∫𝑎 𝑓(𝑥 )𝑑𝑥 = ≅ (𝑏 − 𝑎)𝑓(𝑥1 )
where 𝑥1 is the midpoint of the interval (a, b). Using the nomenclature for
the present case, it can be expressed as
𝑥𝑖+1
∫𝑥
𝑖
𝑓(𝑥 )𝑑𝑥 ≅ ℎ 𝑓(𝑥𝑖+1/2 )
،‫ األول رعد‬:‫أم أحمد عندها أبناء‬
‫ فماذا‬،‫ مطر‬:‫ ال الث‬،‫ برق‬:‫ال اني‬
‫تتوقع أن يكون اسم الرابع؟‬
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Substitution of this formula into equation (25.21) yields the equation
(25.27). Thus, just as the Heun method can be called the trapezoidal rule,
the midpoint method gets its name from the underlying integration
formula upon which it is based.
The midpoint method is superior to Euler’s method because it utilizes
a slope estimate at the midpoint of the prediction interval.
25.3 Runge-Kutta Methods
Runge-Kutta (RK) methods achieve the accuracy of a Taylor series
approach without requiring the calculation of higher derivatives.
Many variations exist but all can be cast in the generalized form of
𝑦𝑖+𝜆 = 𝑦𝑖 + 𝜙(𝑥𝑖 , 𝑦𝑖 , ℎ)ℎ
where 𝜙(𝑥𝑖 , 𝑦𝑖 , ℎ) is called an increment function, which can be
interpreted as a representative slope over the interval.
The increment function can be written in general form as
𝜙 = 𝑎1 𝑘1 + 𝑎2 𝑘2 + … + 𝑎𝑛 𝑘𝑛
‫شيء أنت تمكله ومع ذلك يستخدمه‬
‫اآلخرون دون أن يستئذنوك؟‬
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where the a’s are constants and the k’s are
𝑘1 = 𝑓(𝑥𝑖 , 𝑦𝑖 )
𝑘2 = 𝑓(𝑥𝑖 + 𝑝1 ℎ, 𝑦𝑖 + 𝑞11 𝑘1 ℎ)
𝑘3 = 𝑓(𝑥𝑖 + 𝑝2 ℎ, 𝑦𝑖 + 𝑞21 𝑘1 ℎ +𝑞22 𝑘2 ℎ)
.
.
.
𝑘𝑛 = 𝑓(𝑥𝑖 + 𝑝𝑛−1 ℎ, 𝑦𝑖 + 𝑞𝑛−1,1 𝑘1 ℎ +𝑞𝑛−1,2 𝑘2 ℎ + ⋯ +𝑞𝑛−1,𝑛−1 𝑘𝑛−1 ℎ )
where the p’s and q’s are constants.
Notice that the k’s are recurrence relationships. That is, k1 appears in
the equation for k2, which appears in the equation for k3, and so forth.
Because each k is a functional evaluation, this recurrence makes RK
methods efficient for computer calculations.
Various types of Runge-Kutta methods can be devised by employing
different numbers of terms in the increment function as specified by n.
Note that the first-order RK method with n = 1 is, in fact, Euler’s method.
Once n is chosen, values for the a’s, p’s and q’s are evaluated by setting
equation (25.28) equal to terms in a Taylor series expansion.
Thus, at least for the lower-order versions,
the number of terms, n, usually represents the
order of the approach.
‫ فهناك دائما‬،‫األمر ال ينتهي أبدا‬
‫ والهدف التالي‬،‫الغاية التالية‬
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For example, second-order RK methods use an increment function with
two terms (n=2). These second-order methods will be exact if the solution
to the differential equation is quadratic, the local truncation error is O(h 3)
and the global error is O(h2). The third- and forth- order RK methods
( n=3 and 4, respectively) are developed. For these cases, the global
truncation errors are O(h3) and O(h4) respectively.
25.3.1 Second-Order Runge-Kutta Methods
The second-order version of equation (25.28) is
𝑦𝑖+1 = 𝑦𝑖 + 𝑓(𝑎1 𝑘1 + 𝑎2 𝑘2 )ℎ
where
𝑘1 = 𝑓(𝑥𝑖 , 𝑦𝑖 )
𝑘2 = 𝑓(𝑥𝑖 + 𝑝1 ℎ, 𝑦𝑖 + 𝑞11 𝑘1 ℎ)
Values for a1, a2, p1, and q11 are evaluated by setting the previous equation
equal to a Taylor series expansion to the second-order term.
By doing this, we derive three equations to evaluate the four
unknown constants. The three equations are
𝑎1 + 𝑎2 = 1
𝑎2 𝑝1 =
𝑎2 𝑞11 =
1
2
1
2
،‫إذا لم يكن ما ستفعله صحيحا‬
‫ وإذا لم يكن ما‬،‫فال تفعله‬
‫ فال تقله‬،‫ستقوله صادقا‬
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Because we have three equations with four unknowns, we must
assume a value of one of the unknowns to determine the other three.
Suppose that we specify a value for a2. Then the previous three equations
can be solved simultaneously for
𝑎1 = 1 − 𝑎2
𝑝1 = 𝑞11 =
1
2𝑎2
Because we can choose an infinite number of values for 𝑎2 , there are
an infinite number of second-order RK methods. Every version would yield
exactly the same results if the solution to the ODE were quadratic, linear
or
a
constant.
However,
they
yield
different
results
when
(as is typically the case) the solution is more complicated. We present
three of the most commonly used and preferred versions:
Heun Method with a Single Corrector (a2 = ½)
Thus, a1 = ½ and p1 = q11 = 1. These parameters when substituted, yield
1
𝑦𝑖+1 = 𝑦𝑖 + ( 𝑘1 +
2
1
𝑘 )ℎ
2 2
where
𝑘1 = 𝑓(𝑥𝑖 , 𝑦𝑖 )
𝑘2 = 𝑓(𝑥𝑖 + ℎ, 𝑦𝑖 + 𝑘1 ℎ)
،‫العقل يستطيع اإلجابة على األسئلة‬
‫ولكن يجب على الخيال توجيهها أوال‬
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Note that 𝑘1 is the slope at the beginning of the interval and 𝑘2 is the
slope at the end of the interval. Consequently, this second-order
Runge-Kutta is actually Heun’s technique without iteration.
The Midpoint Method (a2 = 1).
Thus, a1 = 0, p1 = q11 = ½, and that gives
𝑦𝑖+1 = 𝑦𝑖 + 𝑘2 ℎ
Where
𝑘1 = 𝑓(𝑥𝑖 , 𝑦𝑖 )
1
1
2
2
𝑘2 = 𝑓 (𝑥𝑖 + ℎ, 𝑦𝑖 + 𝑘1 ℎ)
This is the midpoint method.
Ralston’s Method (a2 = 2/3)
Choosing such value provides a minimal bound on the truncation error for
the second-order RK algorithms. For this version, a1 = 1/3, and p1 = 3/2 and
yields
1
𝑦𝑖+1 = 𝑦𝑖 + ( 𝑘1 +
3
2
𝑘 )ℎ
3 2
Where
𝑘1 = 𝑓(𝑥𝑖 , 𝑦𝑖 )
3
3
4
4
𝑘2 = 𝑓 (𝑥𝑖 + ℎ, 𝑦𝑖 + 𝑘1 ℎ)
‫عادة ما تتخفى الفرص في شكل عمل‬
‫ لذلك ال يتعرف عليها معظم الناس‬،‫شاق‬
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EXAMPLE 25.6 Comparison of Various Second-Order RK Schemes
Problem Statement:
Use the midpoint method and Ralston’s method to numerically integrate
𝑓(𝑥, 𝑦) = −2𝑥 3 + 12𝑥 2 − 20𝑥 + 8.5
from x = 0 to x = 4 using a step size of 0.5. The initial condition at x = 0 is
y = 1. Compare the results with the values obtained using another
second-order RK algorithm, that is, the Heun method without corrector
iteration (next table).
Solution:
The first step in the midpoint method is to compute
𝑘1 = −2(0)3 + 12(0)2 − 20(0) + 8.5 = 8.5
However, because the ODE is a function of x only, this result has no
bearing on the second step, and compute
𝑘2 = −2(0.25)3 + 12(0.25)2 − 20(0.25) + 8.5 = 4.21875
‫ حتااى عناادما‬،‫إننااي ال أستساالم أباادا‬
‫يقول لي الناس إنني لن أنجف أبدا‬
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Notice that this estimate of the slope is much closer to the average value
for the interval (4.4375) than the slope at the beginning of the interval
(8.5) that would have been used for Euler’s method.
The slope at the midpoint can then be substituted to predict
𝑦(0.5) = 1 + 4.21875(0.5) = 3.109375
𝜀𝑡 = 3.4%
The computation is repeated, and the results are summarized in the next
figure and the previous table
For Ralston’s method, k1 for the first interval also equals 8.5 and
𝑘2 = −2(0.375)3 + 12(0.375)2 − 20(0.375) + 8.5 = 2.58203125
The average slope is computed by
𝜙=
1
3
(8.5) +
2
3
(2.58203125)
𝑦(0.5) = 1 + 4.5546875(0.5) = 3.27734375
𝜀𝑡 = −1.82%
The computation is repeated, and the results are summarized in the
previous figure and table.
‫هناك طريقة واحدة يمكنك الفشل‬
‫ وتلك هي االستسالم‬،‫بها‬
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Notice how all the second-order RK methods are superior to Euler’s
method.
25.3.2 Third-Order Runge-Kutta Methods
For n = 3, a derivation similar to the one for the second-order method
can be performed. The result of this derivation is six equations with eight
unknowns. Therefore, values for two of the unknowns must be specified a
priori in order to determine the remaining parameters.
One common version that results is
𝑦𝑖+1 = 𝑦𝑖 +
1
6
(𝑘1 + 4𝑘2 + 𝑘3 )ℎ
where
𝑘1 = 𝑓(𝑥𝑖 , 𝑦𝑖 )
1
1
2
2
𝑘2 = 𝑓 (𝑥𝑖 + ℎ, 𝑦𝑖 + 𝑘1 ℎ)
𝑘3 = 𝑓(𝑥𝑖 + ℎ, 𝑦𝑖 − 𝑘1 ℎ + 2𝑘2 ℎ)
Note that if the derivative is a function of x only, this third-order
method reduces to Simpson’s 1/3 rule. The third-order RK methods have
local and global errors of O(h4) and O(h3), respectively, and yield exact
results when the solution is cubic. When dealing with polynomials, the
previous equations will be exact when the differential equation is cubic
and the solution is quadratic. This is because Simpson’s 1/3 rule provides
exact integral estimates for cubics.
‫إن كل فكرة من أفكارك هي لبنة تضيفها‬
‫إلى قصر أحالمك أو إلى قبر مشاريعك‬
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25.3.3 Fourth-Order Runge-Kutta Methods.
The most popular RK methods are fourth order.
As with the second-order approaches, there are an infinite number of
versions. The following is the most commonly used form, and we
therefore call it the classical fourth-order RK method:
𝑦𝑖+1 = 𝑦𝑖 +
1
6
(𝑘1 + 2𝑘2 + 2𝑘3 + 𝑘4 )ℎ
where
𝑘1 = 𝑓(𝑥𝑖 , 𝑦𝑖 )
1
1
2
2
1
1
2
2
𝑘2 = 𝑓 (𝑥𝑖 + ℎ, 𝑦𝑖 + 𝑘1 ℎ)
𝑘3 = 𝑓 (𝑥𝑖 + ℎ, 𝑦𝑖 + 𝑘2 ℎ)
𝑘3 = 𝑓(𝑥𝑖 + ℎ, 𝑦𝑖 + 𝑘3 ℎ)
Notice that for ODEs that are a function of x alone, the classical
fourth-order RK method is similar to Simpson’s 1/3 rule.
In addition, the fourth-order RK method is similar to the Heun approach in
that multiple estimates of the slope are developed in order to come up
with an improved average slope for the interval.
As shown in the next figure, each of the k’s represents a slope.
‫ ولكن ثمن‬,‫ثمن الكرامة والحرية فادح‬
. ‫السكوت عن الذل واالستعباد أفدح‬
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EXAMPLE 25.7 Classical Fourth-Order RK Method
Problem Statement:
(a) Use the classical fourth-order RK method to integrate
𝑓(𝑥, 𝑦) = −2𝑥 3 + 12𝑥 2 − 20𝑥 + 8.5
using a step size of = 0.5 and an initial condition of y = 1 at x = 0.
(b) Similarly, integrate
𝑓(𝑥, 𝑦) = 4𝑒 0.8𝑥 − 0.5𝑦
using h = 0.5 with y(0) = 2 from x = 0 to 0.5
Solution:
(a) Compute 𝑘1 = 8.5, 𝑘2 = 4.21875, 𝑘3 = 4.21875 and 𝑘4 = 1.25
𝑦(0.5) = 1 + {
1
6
[8.5 + 2(4.21875) + 2(4.21875) + 1.25] } 0.5
= 3.21875
which is exact. Thus, because the true solution is a quartic, the
fourth-order method gives an exact result.
‫ال تساااتطيع أن تطيااال حياتاااك ولكااان‬
‫يمكن أن تجعلها أك ر عرضا وعمقا‬
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(b) The slope at the beginning of the interval is computed as
𝑘1 = 𝑓(0, 2) = 4𝑒 0.8(0) − 0.5(2) = 3
This value is used to compute a value of u and a slope at the
midpoint,
𝑦(0.25) = 2 + 3(0.25) = 2.75
𝑘2 = 𝑓(0.25, 2.75) = 4𝑒 0.8(0.25) − 0.5(2.75) = 3.510611
This slope in turn is used to compute another value of y and another
slope at the mid-point,
𝑦(0.25) = 2 + 3.510611 (0.25) = 2.877653
𝑘3 = 𝑓(0.25,2.877653) = 4𝑒 0.8(0.25) − 0.5(2.877653) = 3.44678
Next, this slope is used to compute a value of y and a slope at the
end of the interval,
𝑦(0.5) = 2 + 3.071785(0.5) = 3.723392
𝑘4 = 𝑓(0.5, 3.723392) = 4𝑒 0.8(0.5) − 0.5(3.723392) = 4.105603
Finally, the four slope estimates are combined to yield an average
slope. This average slope is then used to make the final prediction at
the end of the interval.
𝜙=
1
6
[2 + 2(3.510611) + 2(3.446785) + 4.106603] = 3.503399
𝑦(0.5) = 2 + 3.503399(0.5) = 3.751699
which compares favorably with the true solution of 3.751521.
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July 2013
Problem 25.7
Use the (a) Euler and (b) Heun (without iteration) methods to solve
𝑑2 𝑦
= 0.5 𝑡 + 𝑦 = 0
where y(0) = 2 and y’(0) = 0. Solve from x = 0 to 4 using h = 0.1.
𝑑𝑡 2
Solution
The second-order ODE is transformed into a pair of first-order ODEs as in
𝑑𝑦
=𝑧
𝑦(0) = 2
𝑑𝑡
𝑑𝑧
𝑑𝑡
𝑧(0) = 0
= 0.5 𝑥 − 𝑦
(a) The first few steps of Euler’s method are
x
0
0.1
0.2
0.3
0.4
0.5
y
z
2
2
1.98
1.9405
1.8822
1.805995
0
-0.2
-0.395
-0.583
-0.76205
-0.93027
dy/dx
0
-0.2
-0.395
-0.583
-0.76205
-0.93027
dz/dx
-2
-1.95
-1.88
-1.7905
-1.6822
-1.556
(b) For Heun (without iterating the corrector) the first few steps are
x
0
0.1
0.2
0.3
0.4
0.5
y
z
dy/dx
2
1.99
1.96055
1.912446
1.846671
1.764385
0
-0.1975
-0.38801
-0.56963
-0.74052
-0.89899
0
-0.1975
-0.38801
-0.56963
-0.74052
-0.89899
dz/dx
-2
-1.94
-1.86055
-1.76245
-1.64667
-1.51439
yend
2
1.97025
1.921749
1.855483
1.772619
1.674486
zend
-0.2
-0.3915
-0.57407
-0.74587
-0.90519
-1.05043
dy/dx
-0.2
-0.3915
-0.57407
-0.74587
-0.90519
-1.05043
dz/dx
-1.95
-1.87025
-1.77175
-1.65548
-1.52262
-1.37449
ave slope
-0.1
-0.2945
-0.48104
-0.65775
-0.82286
-0.97471
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July 2013
Problem 25.9
Solve from t = 0 to 3 with h = 0.1 using
(a) Heun (without corrector) and
(b) Ralston’s 2nd-order RK method:
𝑑𝑦
𝑑𝑡
= 𝑦 sin3 (𝑡)
𝑦(0) = 1
Solution
(a) The Heun method without iteration can be implemented as in the
following table:
t
0
y
1
k1
0
yend
1

k2
0.000995 0.000498
0.1 1.00005 0.000995 1.000149 0.007843 0.004419
0.2 1.000492 0.007845 1.001276 0.025841 0.016843
0.3 1.002176 0.025865 1.004762 0.059335 0.0426
0.4 1.006436 0.059434 1.012379 0.11156 0.085497
0.5 1.014986 0.111847 1.02617
0.184731 0.148289






2.9 3.784421 0.051826 3.789604 0.01065 0.031238
3
3.787545 0.010644 3.78861
0.000272 0.005458
‫لو كان على الكالم ضرائب لتخلى‬
‫ال رثارون عن ثالثة أرباع كالمهم‬
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July 2013
Continue
(b) The Ralston 2nd order RK method can be implemented as in the
following table
t
0
y
1
k1
0
yint
1

k2
0.000421 0.00028
0.1 1.000028 0.000995 1.000103
0.005278 0.003851
0.2 1.000413 0.007845 1.001001
0.020043 0.015977
0.3 1.002011 0.02586
0.049332 0.041508
1.00395
0.4 1.006162 0.059418 1.010618
0.096672 0.084254
0.5 1.014587 0.111803 1.022972
0.164537 0.146959






2.9 3.785066 0.051835 3.788954
0.017276 0.028796
3
0.001116 0.004293
3.787946 0.010646 3.788744
‫نعرف األمور الصحيحة وننادي‬
‫ لكن الصعب أن نلتزم بها‬،‫بها‬
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July 2013
Problem 25.10
Solve the following problem numerically from t = 0 to 3:
𝑑𝑦
𝑑𝑡
= −𝑦 + 𝑡 2
𝑦(0) = 1
Use the third-order RK method with a step size of 0.5
Solution
The solution results are as in the following table:
t
0
y
1
k1
-1
k2
-0.6875

k3
-0.5625
-0.71875
0.5 0.640625
-0.39063 0.019531 0.144531 -0.02799
1
0.626628
0.373372 0.842529 0.967529 0.78517
1.5 1.019213
1.230787 1.735591 1.860591 1.67229
2
1.855358
2.144642 2.670982 2.795982 2.604092
2.5 3.157404
3.092596 3.631947 3.756947 3.562889
3
4.061152 4.608364 4.733364 4.537995
4.938848
‫ العبيد أذالء‬:‫معظم الناس أذالء‬
‫ والسادة أذالء ألهوائهم‬،‫لسادتهم‬
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