how can i get an improved Euler's method code for this function?
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Ibrahem abdelghany ghorab
am 15 Dez. 2018
Kommentiert: Santiago Cerón
am 12 Nov. 2020
dy = @(x,y).2*x*y;
f = @(x).2*exp(x^2/2);
x0=1;
xn=1.5;
y=1;
h=0.1;
fprintf ('x \t \t y (euler)\t y(analytical) \n') % data table header
fprintf ('%f \t %f\t %f\n' ,x0,y,f(x0));
for x = x0 : h: xn-h
y = y + dy(x,y)*h;
x = x + h ;
fprintf (
'%f \t %f\t %f\n' ,x,y,f(x));
end
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FastCar
am 16 Dez. 2018
Euler has its limit to solve differential equations. You can change the integration step going towards the optimum step that is given by the minimum of the sum of the truncation error and step error, but you cannot improve further. What do you mean by improve?
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Are Mjaavatten
am 17 Dez. 2018
There are two problems with your code:
- The analytical solution is incorrect
- You increment x inside the for loop. Don't. The for loop does this automatically.
Here is a corrected version:
a = 0.2;
y0 = 1;
x0 = 1;
xn = 1.5;
h = 0.1;
dy = @(x,y)a*x*y; % dy/dx
f = @(x) y0*exp(a/2*(x.^2-1)); % Correct analytic solution
y = y0;
fprintf ('x \t \t y (euler)\t y(analytical) \n') % data table header
fprintf ('%f \t %f\t %f\n' ,x0,y,f(x0));
for x = x0+h : h: xn
y = y + dy(x,y)*h;
fprintf ('%f \t %f\t %f\n' ,x,y,f(x));
end
Choose a smaller step length h to for better accuracy. Alternatively try a higher order method like Runge-Kutta.
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James Tursa
am 17 Dez. 2018
Bearbeitet: James Tursa
am 17 Dez. 2018
The "Modified" Euler's Method is usually referring to the 2nd order scheme where you average the current and next step derivative in order to predict the next point. E.g.,
dy1 = dy(x,y); % derivative at this time point
dy2 = dy(x+h,y+h*dy1); % derivative at next time point from the normal Euler prediction
y = y + h * (dy1 + dy2) / 2; % average the two derivatives for the Modified Euler step
See this link:
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