RK 6 method giving large errors

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Sahil Kumar
Sahil Kumar 2021-8-8
编辑： darova 2021-8-11
The following code solves for a projectile motion without drag using RK6. I am finding huge variations between values derived from RK6 and that from the exact solution. Can anyone explain why this is happening?
clear all;
%Projectile motion without drag solution using RK6
%x'' = 0, y'' = -g Equations to be solved
g=9.807;
vel=100; th_deg=30; %input
x0=0; y0=0; %initial condition
t0=0; tf=100; %time span
vx=vel*cosd(th_deg); %velocity along x
vy=vel*sind(th_deg); %velocity along y
%transforming second order differential equation to first order
%x'=u represented by fx
%u'=0 represented by fu
%y'=v represented by fy
%v'=-g represented by fv
fx=@(t,x,u) u;
fu=@(t,x,u) 0;
fy=@(t,y,v) v;
fv=@(t,y,v) -g;
t(1)=0;
x(1)=0;y(1)=0;
x_exact(1)=0;y_exact(1)=0; %exact solution
u(1)=vx;v(1)=vy;
h=0.001;
N=ceil((tf-t(1))/h);
for j=1:N
t(j+1)=t(j)+h;
x_exact(j+1)=vx*t(j+1);
y_exact(j+1)=vy*t(j+1)-g*0.5*(t(j+1))^2;
k1x=fx(t(j),x(j),u(j));
k1u=fu(t(j),x(j),u(j));
k1y=fy(t(j),y(j),v(j));
k1v=fv(t(j),y(j),v(j));
k2x=fx(t(j)+h,x(j)+k1x,u(j)+k1u);
k2u=fu(t(j)+h,x(j)+k1x,u(j)+k1u);
k2y=fy(t(j)+h,y(j)+k1y,v(j)+k1v);
k2v=fv(t(j)+h,y(j)+k1y,v(j)+k1v);
k3x=fx(t(j)+h/2,x(j)+((3*k1x+k2x)/8),u(j)+((3*k1u+k2u)/8));
k3u=fu(t(j)+h/2,x(j)+((3*k1x+k2x)/8),u(j)+((3*k1u+k2u)/8));
k3y=fy(t(j)+h/2,y(j)+((3*k1y+k2y)/8),v(j)+((3*k1v+k2v)/8));
k3v=fv(t(j)+h/2,y(j)+((3*k1y+k2y)/8),v(j)+((3*k1v+k2v)/8));
k4x=fx(t(j)+2*h/3,x(j)+(8*k1x+2*k2x+8*k3x)/27,u(j)+(8*k1u+2*k2u+8*k3u)/27);
k4u=fu(t(j)+2*h/3,x(j)+(8*k1x+2*k2x+8*k3x)/27,u(j)+(8*k1u+2*k2u+8*k3u)/27);
k4y=fy(t(j)+2*h/3,y(j)+(8*k1y+2*k2y+8*k3y)/27,v(j)+(8*k1v+2*k2v+8*k3v)/27);
k4v=fv(t(j)+2*h/3,y(j)+(8*k1y+2*k2y+8*k3y)/27,v(j)+(8*k1v+2*k2v+8*k3v)/27);
k5x=fx(t(j)+(7-(21)^0.5)*h/14,x(j)+(3*(3*(21)^0.5-7)*k1x-8*(7-(21)^0.5)*k2x+48*(7-(21)^0.5)*k3x-3*(21-(21)^0.5)*k4x)/392,u(j)+3*((3*(21)^0.5-7)*k1u-8*(7-(21)^0.5)*k2u+48*(7-(21)^0.5)*k3u-3*(21-(21)^0.5)*k4u)/392);
k5u=fu(t(j)+(7-(21)^0.5)*h/14,x(j)+(3*(3*(21)^0.5-7)*k1x-8*(7-(21)^0.5)*k2x+48*(7-(21)^0.5)*k3x-3*(21-(21)^0.5)*k4x)/392,u(j)+3*((3*(21)^0.5-7)*k1u-8*(7-(21)^0.5)*k2u+48*(7-(21)^0.5)*k3u-3*(21-(21)^0.5)*k4u)/392);
k5y=fy(t(j)+(7-(21)^0.5)*h/14,y(j)+(3*(3*(21)^0.5-7)*k1y-8*(7-(21)^0.5)*k2y+48*(7-(21)^0.5)*k3y-3*(21-(21)^0.5)*k4y)/392,v(j)+3*((3*(21)^0.5-7)*k1v-8*(7-(21)^0.5)*k2v+48*(7-(21)^0.5)*k3v-3*(21-(21)^0.5)*k4v)/392);
k5v=fv(t(j)+(7-(21)^0.5)*h/14,y(j)+(3*(3*(21)^0.5-7)*k1y-8*(7-(21)^0.5)*k2y+48*(7-(21)^0.5)*k3y-3*(21-(21)^0.5)*k4y)/392,v(j)+3*((3*(21)^0.5-7)*k1v-8*(7-(21)^0.5)*k2v+48*(7-(21)^0.5)*k3v-3*(21-(21)^0.5)*k4v)/392);
k6x=fx(t(j)+(7+(21)^0.5)*h/14,x(j)+(-5*(231+51*(21)^0.5)*k1x-40*(7+(21)^0.5)*k2x-320*(21)^0.5*k3x+3*(21+121*(21)^0.5)*k4x+392*(6+(21)^0.5)*k5x)/1960,u(j)+(-5*(231+51*(21)^0.5)*k1u-40*(7+(21)^0.5)*k2u-320*(21)^0.5*k3u+3*(21+121*(21)^0.5)*k4u+392*(6+(21)^0.5)*k5u)/1960);
k6u=fu(t(j)+(7+(21)^0.5)*h/14,x(j)+(-5*(231+51*(21)^0.5)*k1x-40*(7+(21)^0.5)*k2x-320*(21)^0.5*k3x+3*(21+121*(21)^0.5)*k4x+392*(6+(21)^0.5)*k5x)/1960,u(j)+(-5*(231+51*(21)^0.5)*k1u-40*(7+(21)^0.5)*k2u-320*(21)^0.5*k3u+3*(21+121*(21)^0.5)*k4u+392*(6+(21)^0.5)*k5u)/1960);
k6y=fy(t(j)+(7+(21)^0.5)*h/14,y(j)+(-5*(231+51*(21)^0.5)*k1y-40*(7+(21)^0.5)*k2y-320*(21)^0.5*k3y+3*(21+121*(21)^0.5)*k4y+392*(6+(21)^0.5)*k5y)/1960,v(j)+(-5*(231+51*(21)^0.5)*k1v-40*(7+(21)^0.5)*k2v-320*(21)^0.5*k3v+3*(21+121*(21)^0.5)*k4v+392*(6+(21)^0.5)*k5v)/1960);
k6v=fv(t(j)+(7+(21)^0.5)*h/14,y(j)+(-5*(231+51*(21)^0.5)*k1y-40*(7+(21)^0.5)*k2y-320*(21)^0.5*k3y+3*(21+121*(21)^0.5)*k4y+392*(6+(21)^0.5)*k5y)/1960,v(j)+(-5*(231+51*(21)^0.5)*k1v-40*(7+(21)^0.5)*k2v-320*(21)^0.5*k3v+3*(21+121*(21)^0.5)*k4v+392*(6+(21)^0.5)*k5v)/1960);
k7x=fx(t(j)+h,x(j)+(15*(22+7*(21)^0.5)*k1x+120*k2x+40*(7*(21)^0.5-5)*k3x-63*(3*(21)^0.5-2)*k4x-14*(49+9*(21)^0.5)*k5x+70*(7-(21)^0.5)*k6x)/180,u(j)+(15*(22+7*(21)^0.5)*k1u+120*k2u+40*(7*(21)^0.5-5)*k3u-63*(3*(21)^0.5-2)*k4u-14*(49+9*(21)^0.5)*k5u+70*(7-(21)^0.5)*k6u)/180);
k7u=fu(t(j)+h,x(j)+(15*(22+7*(21)^0.5)*k1x+120*k2x+40*(7*(21)^0.5-5)*k3x-63*(3*(21)^0.5-2)*k4x-14*(49+9*(21)^0.5)*k5x+70*(7-(21)^0.5)*k6x)/180,u(j)+(15*(22+7*(21)^0.5)*k1u+120*k2u+40*(7*(21)^0.5-5)*k3u-63*(3*(21)^0.5-2)*k4u-14*(49+9*(21)^0.5)*k5u+70*(7-(21)^0.5)*k6u)/180);
k7y=fy(t(j)+h,y(j)+(15*(22+7*(21)^0.5)*k1y+120*k2y+40*(7*(21)^0.5-5)*k3y-63*(3*(21)^0.5-2)*k4y-14*(49+9*(21)^0.5)*k5y+70*(7-(21)^0.5)*k6y)/180,v(j)+(15*(22+7*(21)^0.5)*k1v+120*k2v+40*(7*(21)^0.5-5)*k3v-63*(3*(21)^0.5-2)*k4v-14*(49+9*(21)^0.5)*k5v+70*(7-(21)^0.5)*k6v)/180);
k7v=fv(t(j)+h,y(j)+(15*(22+7*(21)^0.5)*k1y+120*k2y+40*(7*(21)^0.5-5)*k3y-63*(3*(21)^0.5-2)*k4y-14*(49+9*(21)^0.5)*k5y+70*(7-(21)^0.5)*k6y)/180,v(j)+(15*(22+7*(21)^0.5)*k1v+120*k2v+40*(7*(21)^0.5-5)*k3v-63*(3*(21)^0.5-2)*k4v-14*(49+9*(21)^0.5)*k5v+70*(7-(21)^0.5)*k6v)/180);
x(j+1)=x(j)+h*(9*k1x + 64*k3x + 49*k5x + 49*k6x + 9*k7x)/180;
u(j+1)=u(j)+h*(9*k1u + 64*k3u + 49*k5u + 49*k6u + 9*k7u)/180;
y(j+1)=y(j)+h*(9*k1y + 64*k3y + 49*k5y + 49*k6y + 9*k7y)/180;
v(j+1)=v(j)+h*(9*k1v + 64*k3v + 49*k5v + 49*k6v + 9*k7v)/180;
if(y(j+1)<0)
break;
end
end
hmax = max(y);
rmax = max(x);
plot(x,y);
hold on;
plot(x_exact,y_exact);
xlabel('X');
ylabel('Y');

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回答（2 个）

darova
darova 2021-8-9
First of all you should rewrite your code to make more readable. The code is very difficult to to check. Make it as clear as possible
t(j+1)=t(j)+h;
x_exact(j+1)=vx*t(j+1);
y_exact(j+1)=vy*t(j+1)-g*0.5*(t(j+1))^2;
tj = t(j);
xj = x(j);
yj = y(j);
uj = u(j);
vj = v(j);
k1x=fx(tj,xj,uj);
% ... analogically
k2x = fx(tj+h,xj+k1x,uj+k1u);
% ...
C3x = (3*k1x+k2x)/8;
C3u = (3*k1u+k2u)/8;
% ... another C3...
k3x=fx(tj+h/2,xj+C3x,uj+C3u);
% ...
% ...
C71 = 15*(22+7*(21)^0.5);
C73 = 40*(7*(21)^0.5-5);
C74 = 63*(3*(21)^0.5-2);
C75 = 14*(49+9*(21)^0.5);
C76 = 70*(7-(21)^0.5);
dx7 = (C71*k1x+120*k2x+C73*k3x-C74*k4x-C75*k5x+C76*k6x)/180;
du7 = (C71*k1u+120*k2u+C73*k3u-C74*k4u-C75*k5u+C76*k6u)/180;
k7x = fx(tj+h, xj+dx7, uj+du7);
% ...
  2 个评论
Sahil Kumar
Sahil Kumar 2021-8-9
Here, I declared the coefficients separately. Did not do it for the initial few stages as they were just single digits. Also, note that the formula for k1, k2 ... are similar, it is just that they are calculated separately for all the four first order diff equations (fx, fy, fu,fv) and are named respectively (k1x, k1y, k1u,k1v), (k2x, k2y, k2u,k2v), and so on. You may refer to the previous comment for the formulae.
The problem I am facing is that the exact and the numerical solution are different, which should not happen. Is the problem in the code or the logic?
clear all;
%Projectile motion without drag solution using RK6
%x'' = 0, y'' = -g Equations to be solved
g=9.807;
vel=100; th_deg=30; %input
x0=0; y0=0; %initial condition
t0=0; tf=100; %time span
vx=vel*cosd(th_deg); %velocity along x
vy=vel*sind(th_deg); %velocity along y
%transforming second order differential equation to first order
%x'=u represented by fx
%u'=0 represented by fu
%y'=v represented by fy
%v'=-g represented by fv
fx=@(t,x,u)u;
fu=@(t,x,u)0;
fy=@(t,y,v)v;
fv=@(t,y,v)-g;
%initial values
t(1)=0;
x(1)=0;y(1)=0;
x_exact(1)=0;y_exact(1)=0; %exact solution
u(1)=vx;v(1)=vy;
h=0.01; % time step
N=ceil((tf-t(1))/h); % number of iterations
%loop for RK6 calculation
for j=1:N
t(j+1)=t(j)+h;
x_exact(j+1)=vx*t(j+1); % for exact solution x= vx*t
y_exact(j+1)=vy*t(j+1)-g*0.5*(t(j+1))^2; %for exact solution y=vy*t + 0.5*g*t^2
k1x=fx(t(j),x(j),u(j));
k1u=fu(t(j),x(j),u(j));
k1y=fy(t(j),y(j),v(j));
k1v=fv(t(j),y(j),v(j));
k2x=fx(t(j)+h,x(j)+k1x,u(j)+k1u);
k2u=fu(t(j)+h,x(j)+k1x,u(j)+k1u);
k2y=fy(t(j)+h,y(j)+k1y,v(j)+k1v);
k2v=fv(t(j)+h,y(j)+k1y,v(j)+k1v);
k3x=fx(t(j)+h/2,x(j)+((3*k1x+k2x)/8),u(j)+((3*k1u+k2u)/8));
k3u=fu(t(j)+h/2,x(j)+((3*k1x+k2x)/8),u(j)+((3*k1u+k2u)/8));
k3y=fy(t(j)+h/2,y(j)+((3*k1y+k2y)/8),v(j)+((3*k1v+k2v)/8));
k3v=fv(t(j)+h/2,y(j)+((3*k1y+k2y)/8),v(j)+((3*k1v+k2v)/8));
k4x=fx(t(j)+2*h/3,x(j)+(8*k1x+2*k2x+8*k3x)/27,u(j)+(8*k1u+2*k2u+8*k3u)/27);
k4u=fu(t(j)+2*h/3,x(j)+(8*k1x+2*k2x+8*k3x)/27,u(j)+(8*k1u+2*k2u+8*k3u)/27);
k4y=fy(t(j)+2*h/3,y(j)+(8*k1y+2*k2y+8*k3y)/27,v(j)+(8*k1v+2*k2v+8*k3v)/27);
k4v=fv(t(j)+2*h/3,y(j)+(8*k1y+2*k2y+8*k3y)/27,v(j)+(8*k1v+2*k2v+8*k3v)/27);
C5 = (7-(21)^0.5)/14; % coefficient for independent var
%coeff for dependent var
C51 = 3*(3*(21)^0.5-7);
C52 = -8*(7-(21)^0.5);
C53 = 48*(7-(21)^0.5);
C54 = -3*(21-(21)^0.5);
k5x=fx(t(j)+C5*h,x(j)+(C51*k1x+C52*k2x+C53*k3x+C54*k4x)/392,u(j)+(C51*k1u+C52*k2u+C53*k3u+C54*k4u)/392);
k5u=fu(t(j)+C5*h,x(j)+(C51*k1x+C52*k2x+C53*k3x+C54*k4x)/392,u(j)+(C51*k1u+C52*k2u+C53*k3u+C54*k4u)/392);
k5y=fy(t(j)+C5*h,y(j)+(C51*k1y+C52*k2y+C53*k3y+C54*k4y)/392,v(j)+(C51*k1v+C52*k2v+C53*k3v+C54*k4v)/392);
k5v=fv(t(j)+C5*h,y(j)+(C51*k1y+C52*k2y+C53*k3y+C54*k4y)/392,v(j)+(C51*k1v+C52*k2v+C53*k3v+C54*k4v)/392);
C6 = (7+(21)^0.5)/14; % coefficient for independent var
%coeff for dependent var
C61 = -5*(231+51*(21)^0.5);
C62 = -40*(7+(21)^0.5);
C63 = -320*(21)^0.5;
C64 = 3*(21+121*(21)^0.5);
C65 = 392*(6+(21)^0.5);
k6x=fx(t(j)+C6*h,x(j)+(C61*k1x+C62*k2x+C63*k3x+C64*k4x+C65*k5x)/1960,u(j)+(C61*k1u+C62*k2u+C63*k3u+C64*k4u+C65*k5u)/1960);
k6u=fu(t(j)+C6*h,x(j)+(C61*k1x+C62*k2x+C63*k3x+C64*k4x+C65*k5x)/1960,u(j)+(C61*k1u+C62*k2u+C63*k3u+C64*k4u+C65*k5u)/1960);
k6y=fy(t(j)+C6*h,y(j)+(C61*k1y+C62*k2y+C63*k3y+C64*k4y+C65*k5y)/1960,v(j)+(C61*k1v+C62*k2v+C63*k3v+C64*k4v+C65*k5v)/1960);
k6v=fv(t(j)+C6*h,y(j)+(C61*k1y+C62*k2y+C63*k3y+C64*k4y+C65*k5y)/1960,v(j)+(C61*k1v+C62*k2v+C63*k3v+C64*k4v+C65*k5v)/1960);
%coeff for dependent var
C71 = 15*(22+7*(21)^0.5);
C72 = 120;
C73 = 40*(7*(21)^0.5-5);
C74 = -63*(3*(21)^0.5-2);
C75 = -14*(49+9*(21)^0.5);
C76 = 70*(7-(21)^0.5);
k7x=fx(t(j)+h,x(j)+(C71*k1x+C72*k2x+C73*k3x+C74*k4x+C75*k5x+C76*k6x)/180,u(j)+(C71*k1u+C72*k2u+C73*k3u+C74*k4u+C75*k5u+C76*k6u)/180);
k7u=fu(t(j)+h,x(j)+(C71*k1x+C72*k2x+C73*k3x+C74*k4x+C75*k5x+C76*k6x)/180,u(j)+(C71*k1u+C72*k2u+C73*k3u+C74*k4u+C75*k5u+C76*k6u)/180);
k7y=fy(t(j)+h,y(j)+(C71*k1y+C72*k2y+C73*k3y+C74*k4y+C75*k5y+C76*k6y)/180,v(j)+(C71*k1v+C72*k2v+C73*k3v+C74*k4v+C75*k5v+C76*k6v)/180);
k7v=fv(t(j)+h,y(j)+(C71*k1y+C72*k2y+C73*k3y+C74*k4y+C75*k5y+C76*k6y)/180,v(j)+(C71*k1v+C72*k2v+C73*k3v+C74*k4v+C75*k5v+C76*k6v)/180);
x(j+1)=x(j)+h*(9*k1x + 64*k3x + 49*k5x + 49*k6x + 9*k7x)/180;
u(j+1)=u(j)+h*(9*k1u + 64*k3u + 49*k5u + 49*k6u + 9*k7u)/180;
y(j+1)=y(j)+h*(9*k1y + 64*k3y + 49*k5y + 49*k6y + 9*k7y)/180;
v(j+1)=v(j)+h*(9*k1v + 64*k3v + 49*k5v + 49*k6v + 9*k7v)/180;
if(y(j+1)<0)
break;
end
end
hmax = max(y); % max altitude
rmax = max(x); %range
%plotting results
plot(x,y);
hold on;
plot(x_exact,y_exact);
xlabel('X');
ylabel('Y');
darova
darova 2021-8-11
编辑：darova 2021-8-11
Ran in:
I couldn't fidn a mistake. Here is comparison of ode45 and exact solution. Looks like there is a mistake in the algorithm RK6
g=9.807;
vel=100; th_deg=30; %input
x0=0; y0=0; %initial condition
t0=0; tf=100; %time span
vx=vel*cosd(th_deg); %velocity along x
vy=vel*sind(th_deg); %velocity along y
t = linspace(0,9);
x_exact = vx*t; % for exact solution x= vx*t
y_exact = vy*t-g*0.5*t.^2; %for exact solution y=vy*t + 0.5*g*t^2
f = @(t,f) [f(3);f(4);0;-g];
[t1,f1] = ode45(f,t,[0 0 vx vy]);
plot(x_exact,y_exact,'.r')
line(f1(:,1),f1(:,2))
legend('exact solution','ode45')

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Wan Ji
Wan Ji 2021-8-9
Did you compare it with solution from ode45 or ode23s in the matlab?

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