Euler修正法
For a given differential equation 
with initial condition 
find the approximate solution using Predictor-Corrector method.
with initial condition find the approximate solution using Predictor-Corrector method.
Predictor-Corrector Method :
The predictor-corrector method is also known as Modified-Euler method.
In the Euler method, the tangent is drawn at a point and slope is calculated for a given step size. Thus this method works best with linear functions, but for other cases, there remains a truncation error. To solve this problem the Modified Euler method is introduced. In this method instead of a point, the arithmetic average of the slope over an interval
is used.
The predictor-corrector method is also known as Modified-Euler method.
In the Euler method, the tangent is drawn at a point and slope is calculated for a given step size. Thus this method works best with linear functions, but for other cases, there remains a truncation error. To solve this problem the Modified Euler method is introduced. In this method instead of a point, the arithmetic average of the slope over an interval
is used.
Thus in the Predictor-Corrector method for each step the predicted value of 
is calculated first using Euler’s method and then the slopes at the points 
and 
is calculated and the arithmetic average of these slopes are added to 
to calculate the corrected value of .
is calculated first using Euler’s method and then the slopes at the points and is calculated and the arithmetic average of these slopes are added to to calculate the corrected value of .
So,
,here h is step size for each increment
then go to step – 1.
As, in this method, the average slope is used, so the error is reduced significantly. Also, we can repeat the process of correction for convergence. Thus at every step, we are reducing the error thus by improving the value of y.
Input : eq = 
, y(0) = 0.5, step size(h) = 0.2
To find: y(1)
Output: y(1) = 2.18147
, y(0) = 0.5, step size(h) = 0.2To find: y(1)
Output: y(1) = 2.18147
// C++ code for solving the differential equation
// using Predictor-Corrector or Modified-Euler method
// with the given conditions, y(0) = 0.5, step size(h) = 0.2
// to find y(1)
#include <bits/stdc++.h>
using namespace std;
// consider the differential equation
// for a given x and y, return v
double f(double x, double y)
{
double v = y - 2 * x * x + 1;
return v;
}
// predicts the next value for a given (x, y)
// and step size h using Euler method
double predict(double x, double y, double h)
{
// value of next y(predicted) is returned
double y1p = y + h * f(x, y);
return y1p;
}
// corrects the predicted value
// using Modified Euler method
double correct(double x, double y,
double x1, double y1,
double h)
{
// (x, y) are of previous step
// and x1 is the increased x for next step
// and y1 is predicted y for next step
double e = 0.00001;
double y1c = y1;
do {
y1 = y1c;
y1c = y + 0.5 * h * (f(x, y) + f(x1, y1));
} while (fabs(y1c - y1) > e);
// every iteration is correcting the value
// of y using average slope
return y1c;
}
void printFinalValues(double x, double xn,
double y, double h)
{
while (x < xn) {
double x1 = x + h;
double y1p = predict(x, y, h);
double y1c = correct(x, y, x1, y1p, h);
x = x1;
y = y1c;
}
// at every iteration first the value
// of for next step is first predicted
// and then corrected.
cout << "The final value of y at x = "
<< x << " is : " << y << endl;
}
int main()
{
// here x and y are the initial
// given condition, so x=0 and y=0.5
double x = 0, y = 0.5;
// final value of x for which y is needed
double xn = 1;
// step size
double h = 0.2;
printFinalValues(x, xn, y, h);
return 0;
}
輸出畫面
The final value of y at x = 1 is : 2.18147 ...Program finished with exit code 0
Press ENTER to exit console.
C語言
// C code for solving the differential equation
// using Predictor-Corrector or Modified-Euler method
// with the given conditions, y(0) = 0.5, step size(h) = 0.2
// to find y(1)
#include <stdio.h>
#include <math.h>
// consider the differential equation
// for a given x and y, return v
double f(double x, double y)
{
double v = y - 2 * x * x + 1;
return v;
}
// predicts the next value for a given (x, y)
// and step size h using Euler method
double predict(double x, double y, double h)
{
// value of next y(predicted) is returned
double y1p = y + h * f(x, y);
return y1p;
}
// corrects the predicted value
// using Modified Euler method
double correct(double x, double y, double x1, double y1,double h)
{
// (x, y) are of previous step
// and x1 is the increased x for next step
// and y1 is predicted y for next step
double e = 0.00001;
double y1c = y1;
do {
y1 = y1c;
y1c = y + 0.5 * h * (f(x, y) + f(x1, y1));
} while (fabs(y1c - y1) > e);
// every iteration is correcting the value
// of y using average slope
return y1c;
}
void printFinalValues(double x, double xn, double y, double h)
{
while (x < xn) {
double x1 = x + h;
double y1p = predict(x, y, h);
double y1c = correct(x, y, x1, y1p, h);
x = x1;
y = y1c;
}
// at every iteration first the value
// of for next step is first predicted
// and then corrected.
printf("The final value of y , at x = %2.4lf is :%2.4lf " ,x ,y);
}
int main()
{
// here x and y are the initial
// given condition, so x=0 and y=0.5
double x = 0, y = 0.5;
// final value of x for which y is needed
double xn = 1;
// step size
double h = 0.2;
printFinalValues(x, xn, y, h);
return 0;
}
沒有留言:
張貼留言