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#1 Re: Help Me ! » fourth order RungeKutta in C# - two differential equations » 2011-07-21 02:12:26
Should I paste them here?
So no errors calculating?
#2 Re: Help Me ! » fourth order RungeKutta in C# - two differential equations » 2011-07-21 02:02:04
I think I just solved the problem:
[CODE]
public void runge(double x, double y, double t)
{
double K1x = dx(x, y, t);
double K1y = dy(x, y);
double K2x = dx((x + 0.5 * H), (y + 0.5 * K1x * H), t + 0.5 * H);
double K2y = dy((x + 0.5 * H), (y + 0.5 * K1y*H));
double K3x = dx((x + 0.5 * H), (y + 0.5 * K2x * H), t + 0.5 * H);
double K3y = dy((x + 0.5 * H), (y + 0.5 * K2y * H));
double K4x = dx((x + H), (y + K3x * H), t + 0.5 * H);
double K4y = dy((x + H), (y + K3y * H));
double restX = (K1x + 2 * K2x + 2 * K3x + K4x) * H / 6;
double restY = (K1y + 2 * K2y + 2 * K3y + K4y) * H / 6;
this.x = x + restX;
this.y = y + restY;
}
[CODE]
However, now I must calculate some errors, because graphs are too little. Could you give me some piece of advice what errors and how to calculate? Thank you 
#3 Re: Help Me ! » fourth order RungeKutta in C# - two differential equations » 2011-07-21 01:39:26
Initial conditions:
public double x = 0.81;
public double y = 0.055;
#4 Re: Help Me ! » fourth order RungeKutta in C# - two differential equations » 2011-07-21 01:15:27
0 to 5, with step of 0.01
#5 Re: Help Me ! » fourth order RungeKutta in C# - two differential equations » 2011-07-21 00:51:50
Yes.
#6 Re: Help Me ! » fourth order RungeKutta in C# - two differential equations » 2011-07-21 00:17:00
Consts are as here: (FI = PHI)
private const int ALFA = 7600;
private const int MU = 7200;
private int BETA = 1430;
private const int LAMBDA = 2470;
private int V = 139000;
private const double FI = 0.51; //PHI
private const double TETA = 2.5;
private const int QL = 8400;
private const double H = 0.01;and U(t) is simple function:
public double doplyw(double t)
{
if (t >= 0.5 && t < 0.75)
return 100000;
else return 0;
}#7 Re: Help Me ! » fourth order RungeKutta in C# - two differential equations » 2011-07-20 23:58:09
I think the correct answer is what Simulink produces, as an author of the model prepared a simulation in Simulink. I can visually compare my C# graphs and those in Simulink and they are different. Could you take a look at my rungeDX method and validate?
#8 Re: Help Me ! » fourth order RungeKutta in C# - two differential equations » 2011-07-20 23:48:32
I changed solution a little, however still not what expected...
public Solution Start()
{
xList = new List<Double>();
yList = new List<Double>();
double t = 0;
Solution solution = new Solution();
for (double krok = 0; krok <= 500; krok++)
{
t = krok / 100;
solution.xList.Add(x*100);
solution.yList.Add(y*1000);
solution.time.Add(t);
rungeDX(x, t);
}
return solution;
}
public class Solution
{
public Solution()
{
time = new List<Double>();
xList = new List<Double>();
yList = new List<Double>();
}
public List<Double> time;
public List<Double> yList;
public List<Double> xList;
}
private double dx(double x, double y, double t)
{
double result;
if (x > TETA)
result = (doplyw(t) + QL - LAMBDA * x - V * x * y - MU * (x - TETA)) / 15000;
else
result = (doplyw(t) + QL - LAMBDA * x - V * x * y) / 15000;
return result;
}
private double dy(double x, double y)
{
double result;
if (x > FI)
result = (-ALFA * y + BETA * (x - FI)) / 15000;
else
result = (-ALFA * y) / 15000;
return result;
}
public void rungeDX(double x, double t)
{
double K1x = dx(x, y, t);
double K1y = dy(x, y);
double K2x = dx((x + 0.5 * K1x), (y + 0.5 * K1y), t + 0.5 * H);
double K2y = dy((x + 0.5 * K1x), (y + 0.5 * K1y));
double K3x = dx((x + 0.5 * K2x), (y + 0.5 * K2y), t + 0.5 * H);
double K3y = dy((x + 0.5 * K2x), (y + 0.5 * K2y));
double K4x = dx((x + K3x), (y + K3y), t + 0.5 * H);
double K4y = dy((x + K3x), (y + K3y));
double restX = (K1x + 2 * K2x + 2 * K3x + K4x) * H;
double restY = (K1y + 2 * K2y + 2 * K3y + K4y) * H;
this.x = x + restX;
this.y = y + restY;
}#9 Re: Help Me ! » fourth order RungeKutta in C# - two differential equations » 2011-07-20 23:43:44
"for"
and Cg = 15000 and Ci = 15000 (divisions in dx and dy).
#10 Re: Help Me ! » fourth order RungeKutta in C# - two differential equations » 2011-07-20 23:27:17
Thank you bobbym for your answer.
I uploaded the equations (sorry, I was working late at night...).
Do you have any idea how to solve them properly?
Thank you in advance!
#11 Help Me ! » fourth order RungeKutta in C# - two differential equations » 2011-07-20 10:57:12
- Itosu
- Replies: 20
Hello!
I am trying to solve two differential, non-linear equations in C# using fourth-order Runge-Kutha method.
I implemented them in C#:
private const int ALFA = 7600;
private const int MU = 7200;
private int BETA = 1430;//int.Parse(HttpContext.Current.Session["BETA"].ToString());
private const int LAMBDA = 2470;
private int V = 139000;//int.Parse(HttpContext.Current.Session["V"].ToString());
private const double FI = 0.51;
private const double TETA = 2.5;
private const int QL = 8400;
private const double H = 0.01;
public double x = 0.81;
public double y = 0.055;
// ;
public ArrayList yList;
public ArrayList xList;
public void Start()
{
xList = new ArrayList();
yList = new ArrayList();
double t = 0;
// xList.
// y = Y0;
for (double krok = 0; krok <= 500; krok++)
{
t = krok / 100;
xList.Add(x);
yList.Add(y);
x = rungeDX(x, t);
y = rungeDY(y, t);
//x = X0 + (i * H);
//y = runge(x, y);
}
}
public double doplyw(double t)
{
if (t >= 0.5 && t < 0.75)
return 100000;
else return 0;
}
private double dx(double x, double y, double t)
{
double result;
if (x > TETA)
result = (doplyw(t) + QL - LAMBDA * x - V * x * y - MU * (x - TETA)) / 15000;
else
result = (doplyw(t) + QL - LAMBDA * x - V * x * y) / 15000;
return result;
}
private double dy(double x, double y)
{
double result;
if (x > FI)
result = (-ALFA * y + BETA * (x - FI)) / 15000;
else
result = (-ALFA * y) / 15000;
return result;
}
double rungeDX(double x, double t)
{
double K1 = dx(this.x, y, t);
double K2 = dx((this.x + 0.5 * H), (y + 0.5 * K1), t);
double K3 = dx((this.x + 0.5 * H), (y + 0.5 * K2), t);
double K4 = dx((this.x + H), (y + K3), t);
double rest = (K1 + 2 * K2 + 2 * K3 + K4) * H;
double runge = this.x + rest;
return runge;
}
double rungeDY(double x, double t)
{
double K1 = dy(x, y);
double K2 = dy((x + 1 / 2 * H), (y + 1 / 2 * K1));
double K3 = dy((x + 1 / 2 * H), (y + 1 / 2 * K2));
double K4 = dy((x + H), (y + K3));
double rest = (K1 + 2 * K2 + 2 * K3 + K4) * H;
double runge = y + rest;
return runge;
}
}However the results are not that that should be (in comparison to Simulink model).
Do you have any ideas?
Thank you in advance!
I am sorry, the image didn't upload. These are the equations:
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