CHAPTER 5 - موقع المهندس حماده شعبان CH25.docx · Web viewThis chapter is...

July 2013

Chapter 25

فيرجى العام للنفع مجانية عن المساهمةالنوتات خطأ باإلبالغ ضرورية مالحظات أوأي نصية تراها برسالةاإللكتروني أو 9 4444 260 بالبريد

Physics I/II, English 123, Statics, Dynamics, Strength, Structure I/II, C++, Java, Data, Algorithms, Numerical, Economyشعبان. حمادة محلولة info@ eng-hs.com 260 4444 9م ومسائل , eng-hs. com بالموقعين مجاناشرح

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Runge-Kutta Methods

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This chapter is devoted to solving ordinary differential equations of the

formdydx

=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,y i+1= y i+∅ h

According to this equation, the slope estimate of ∅ is used to extrapolate

from an old value y i to a new value y i+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.فيرجى العام للنفع مجانية عن المساهمةالنوتات خطأ باإلبالغ ضرورية مالحظات أوأي نصية تراها برسالة

اإللكتروني أو 9 4444 260 بالبريد Physics I/II, English 123, Statics, Dynamics, Strength, Structure I/II, C++, Java, Data, Algorithms, Numerical, Economy

شعبان. حمادة محلولة info@ eng-hs.com 260 4444 9م ومسائل , eng-hs. com بالموقعين مجاناشرحeng-hs. net


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25.1 Euler’s Method

The first derivative provides a direct estimate of the slope at x i

(next figure)

∅=f (xi , y i )

where f (x i , y i ) is the differential equation evaluated at x i and y i.

This estimate can be substituted into

y i+1= y i+ f (x i , y i )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)



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EXAMPLE 25.1 Euler’s Method

Problem Statement:

Use Euler’s method to numerically integrate dydx

=−2 x3+12 x2−20 x+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+4 x3+10 x2−8.5 x+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



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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.



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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.



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Recall that in Euler’s methods, the slope at the beginning of an interval

y i'=f (x i , y i)

is used to extrapolate linearly to y i+1 :

y i+10 = y i

❑+ f (x i , y i )h

For the standard Euler method we would stop at this point.

However, in Heun’s method the y i+10 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:

y i+1' =f (x i+ 1 , y i+10 )

Thus, the two slopes can be combined to obtain an average slope for the

interval:

y '=y i'+ y i+1

'

2 =f (x i , y i )+ f ( x i+1 , y i+10 )

2

This average slope is then used to extrapolate linearly from y i to y i+1using

Euler’s method:

y i+1= y i+f ( xi , y i )+f (x i+1 , y i+10 )

2 h



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which is called a corrector equation.

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

Predictor (previous figure a): y i+10 = y i+ f (x i , y i)h

Corrector (figure b): y i+1= y i+f ( xi , y i )+f (x i+1 , y i+10 )

2 h (25.16)

Note that because the previous equation has y i+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 y i+1. The process is shown in the next figure.



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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

|εa|=| y i+1j − y i+1j−1

y i+1j |

where y i+1j−1 and y i+1j are the result from the prior and the present iteration of

the corrector, respectively.



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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= 413

(e0.8x−e−0.5 x )+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'=4e0−0.5 (2 )=3



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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.

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 y i+1, we use the value y10 to predict

the slope at the end of the interval

y1'= f ( x1 , y10 )=4 e0.8 (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+6.4021642

=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



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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.

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+4e0.8(1)−0.5 (6.701082 )]

21=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+4e0.8(1)−0.5 (6.275811)]

21=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 فيرجى العام للنفع مجانية عن المساهمةالنوتات خطأ باإلبالغ ضرورية مالحظات أوأي نصية تراها برسالة



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. حداد لنقفدقيقة


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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.

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

y i+1= y i+f ( xi )+ f (xi+1)

2h (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 equationdxdy

=f (x)

This equation can be solved for y by integration:

∫yi

y i+1

dy=∫x i

x i+1

f ( x ) dx

which yields

y i+1− y i=∫x i

xi+1

f ( x )dx



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or

y i+1= y i+∫x i

x i+1

f ( x )dx (25.21)

Now, recall that the trapezoidal rule is defined as

∫x i

xi+1

f ( x )dx≅f ( xi )+ f (x i+1)

2h

where h = x i+1−x i. Substituting equations yields

y i+1= y i+f ( xi )+ f (xi+1)

2h

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 x i and x i+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.



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The next figure, which shows the result of using Heun’s method to

solve the polynomial from example 25.1 demonstrates this behavior.

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):

y i+1 /2= y i+ f (x i, y i )h2

Then, this predicted value is used to calculate a slope at the midpoint:

y i+1 /2' =f (x i+1 /2 , y i+1/2 )



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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

x i to x i+1 (figure b):

y i+1= y i+ f (x i+1 /2 , yi+1/2 )h (25.27)

Observe that because y i+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



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∫a

b

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

∫x i

xi+1

f ( x )dx≅ h f (x i+1 /2)

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

y i+ λ= y i+ϕ (xi , y i ,h )h



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where ϕ (x i , y i , 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 asϕ=a1 k1+a2 k2+…+an kn

where the a’s are constants and the k’s are

k1=f (x i , y i )

k 2=f (x i+p1h , y i+q11k1h )

k3=f (x i+ p2h, y i+q21k1h+q22 k2h )

.

.

.

k n=f (x i+ pn−1h , y i+qn−1,1k1h+qn−1,2 k2h+…+qn−1 , n−1k n−1h )

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.



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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.

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(h3)

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

y i+1= y i+ f (a1k1+a2k 2)h

where

k1=f (x i , y i )

k 2=f (x i+p1h , y i+q11k1h )



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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 area1+a2=1

a2 p1=12

a2q11=12

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 fora1=1−a2

p1=q11=12a2

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:



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Heun Method with a Single Corrector (a2 = ½)

Thus, a1 = ½ and p1 = q11 = 1. These parameters when substituted, yield

y i+1= y i+( 12 k1+ 12 k 2)hwhere

k1=f (x i , y i )

k 2=f (x i+h , y i+k1h )

Note that k1 is the slope at the beginning of the interval and k 2 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 givesy i+1= y i+k2h

Where

k1=f (x i , y i )

k 2=f ( xi+ 12 h , y i+12 k1h)This is the midpoint method.

Ralston’s Method (a2 = 2/3)



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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

y i+1= y i+( 13k¿¿1+ 2

3k¿¿2)h¿¿

Where

k1=f (x i , y i )

k 2=f ( xi+ 34 h , y i+ 34 k1h)

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 )=−2 x3+12 x2−20 x+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).



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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

k 2=−2 (0.25 )3+12 (0.25 )2−20 (0.25 )+8.5=4.21875

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



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For Ralston’s method, k1 for the first interval also equals 8.5 and

k 2=−2 (0.375 )3+12 (0.375 )2−20 (0.375 )+8.5=2.58203125

The average slope is computed by

ϕ=13

(8.5 )+ 23(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.

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

y i+1= y i+16(k¿¿1+4 k¿¿2+k3)h¿¿

where

k1=f (x i , y i )

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k3=f (x i+h , y i−k1h+2k2h )

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.

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:

y i+1= y i+16(k¿¿1+2k¿¿2+2 k3+k4)h¿¿

where

k1=f (x i , y i )

k 2=f ( xi+ 12 h , y i+12 k1h)k3=f ( xi+ 12 h , y i+12 k2h)k3=f (x i+h , y i+k3h )



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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.

EXAMPLE 25.7 Classical Fourth-Order RK Method

Problem Statement:

(a) Use the classical fourth-order RK method to integrate

f ( x , y )=−2 x3+12 x2−20 x+8.5



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using a step size of = 0.5 and an initial condition of y = 1 at x = 0.

(b) Similarly, integrate

f ( x , y )=4 e0.8x−0.5 y

using h = 0.5 with y(0) = 2 from x = 0 to 0.5

Solution:

(a)Compute k1=8.5, k 2=4.21875, k3=4.21875 and k 4=1.25

y (0.5 )=1+{16 [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.

(b) The slope at the beginning of the interval is computed as

k1=f (0 ,2 )=4e0.8 (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

k 2=f (0.25 ,2.75 )=4e0.8(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 )=4e0.8 (0.25)−0.5 (2.877653 )=3.44678



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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

k 4=f (0.5 ,3.723392 )=4 e0.8(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=16 [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.

Problem 25.7

Use the (a) Euler and (b) Heun (without iteration) methods to solved2 yd t2

=0.5 t+ y=0

where y(0) = 2 and y’(0) = 0. Solve from x = 0 to 4 using h = 0.1.

Solution

The second-order ODE is transformed into a pair of first-order ODEs as in dydt =z y (0 )=2

dzdt

=0.5 x− y z (0 )=0

(a) The first few steps of Euler’s method areفيرجى العام للنفع مجانية عن المساهمةالنوتات خطأ باإلبالغ ضرورية مالحظات أوأي نصية تراها برسالة




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x y z dy/dx dz/dx0 2 0 0 -20.1 2 -0.2 -0.2 -1.950.2 1.98 -0.395 -0.395 -1.880.3 1.9405 -0.583 -0.583 -1.79050.4 1.8822 -0.76205 -0.76205 -1.68220.5 1.805995 -0.93027 -0.93027 -1.556

(b) For Heun (without iterating the corrector) the first few steps are

x y z dy/dx dz/dx yend zend dy/dx dz/dx ave slope0 2 0 0 -2 2 -0.2 -0.2 -1.95 -0.10.1 1.99 -0.1975 -0.1975 -1.94 1.97025 -0.3915 -0.3915 -1.87025 -0.29450.2 1.96055 -0.38801 -0.38801 -1.86055 1.921749 -0.57407 -0.57407 -1.77175 -0.481040.3 1.912446 -0.56963 -0.56963 -1.76245 1.855483 -0.74587 -0.74587 -1.65548 -0.657750.4 1.846671 -0.74052 -0.74052 -1.64667 1.772619 -0.90519 -0.90519 -1.52262 -0.822860.5 1.764385 -0.89899 -0.89899 -1.51439 1.674486 -1.05043 -1.05043 -1.37449 -0.97471

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:dydt

= y sin3(t) y (0 )=1

Solution

(a) The Heun method without iteration can be implemented as in the

following table:فيرجى العام للنفع مجانية عن المساهمةالنوتات خطأ باإلبالغ ضرورية مالحظات أوأي نصية تراها برسالة




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t y k1 yend k2

0 1 0 1 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

Continue

(b) The Ralston 2nd order RK method can be implemented as in the

following table

t y k1 yint k2

0 1 0 1 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 1.00395 0.049332 0.041508

0.4 1.006162 0.059418 1.010618 0.096672 0.084254



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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 3.787946 0.010646 3.788744 0.001116 0.004293

Problem 25.10

Solve the following problem numerically from t = 0 to 3:dydt

=− 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 y k1 k2 k3



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0 1 -1 -0.6875 -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 4.938848 4.061152 4.608364 4.733364 4.537995



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