When I run my code, I keep getting this error: "Undefined function or variable 'y'."

1 view (last 30 days)
Nauryzbay
Nauryzbay on 26 May 2023
Commented: Nauryzbay on 30 May 2023
close all;clc
tspan = [0 1];
y0(1)=2;
y0(2)=1;
y0(3)=3;
source = 0;
tol = 1e-3;%Критерии допуска для определения наилучшей стоимости Мой критерий остановки - Tol= 1e-3.
error = 1.0;%Начальная ошибка запуска
while error > tol%ЗАПУСТИТЕ ЦИКЛ WHILE
sol = ode45(@(t,y) ode2(t,y,source), [0 1], y0);% ВЫЗЫВАЕМ СОЛВЕР BVP4C:
t = linspace(0,1,100);
source_new = trapz(t,y(3,:));
error = abs(source_new - source);
source = source_new;
end
figure(1);
subplot(3,1,1)
plot(sol.x,sol.y(1,:),'color','r','Linewidth',1.2);
grid on
hold on
xlabel('\bf t'); ylabel('$$y$$','interpreter','latex','fontsize',16);
subplot(3,1,2)
plot(sol.x,sol.y(2,:),'color','r','Linewidth',1.2);
grid on
hold on
xlabel('\bf t'); ylabel('$$\dot{y}$$','interpreter','latex','fontsize',16);
subplot(3,1,3)
plot(sol.x,sol.y(3,:),'color','r','Linewidth',1.2);
grid on
hold on
xlabel('\bf t'); ylabel('$$y"$$','interpreter','latex','fontsize',16);
function dy = ode2(~,y,source)%y''+y'=1+int^1_0(y'*x)dx;y(0) = 2;y(1) = 4;
m=0.05;
dy = zeros(3,1); % создает нулевой вектор-столбец
dy(1)=y(2);
dy(2)=y(3);
dy(3)=-3/m*y(3)+1/(m.^2)*(1-2*y(2)+4.*source);
end
Undefined function or variable 'y'.
Error in IDTinitial6 (line 14)
source_new = trapz(t,y(3,:));

Sign in to answer this question.

Accepted Answer

Walter Roberson
Walter Roberson on 26 May 2023
Later you keep referring to sol.y so perhaps the reference to y should instead be to sol.y
  7 Comments
Show 5 older comments
Torsten
Torsten on 29 May 2023
Edited: Torsten on 29 May 2023
... which is y2(1) - y2(0) = y2(1) - 1.
But this also doesn't seem to help - the iteration diverges.
See my symbolic solution and the boundary value solution below.

Sign in to comment.

More Answers (1)

Torsten
Torsten on 29 May 2023
Edited: Torsten on 29 May 2023
Ran in:
syms t y(t)
syms s real
m = 0.05;
Dy = diff(y,t);
D2y = diff(y,t,2);
D3y = diff(y,t,3);
eqn = D3y -( -3/m*D2y + 1/m^2*(1-2*Dy+4*s)) ;
sol_ode = dsolve(eqn)
sol_ode = 
vars = symvar(sol_ode)
vars = 
cond1 = subs(sol_ode,t,0) == 2;
cond2 = subs(diff(sol_ode,t),t,0) == 1;
cond3 = subs(diff(sol_ode,t,2),t,0) == 3;
cond4 = subs(diff(sol_ode,t),t,1) - subs(diff(sol_ode,t),t,0) == s;
sol_cond = solve([cond1,cond2,cond3,cond4])
sol_cond = struct with fields:
C1: 1663/800 C2: -(6*exp(20) + 7*exp(40))/(800*(exp(40) - 4*exp(20) + 2)) C3: (17*exp(40) + 6)/(400*(exp(40) - 4*exp(20) + 2)) s: (10*exp(40) - 23*exp(20) + 13)/(20*(exp(40) - 4*exp(20) + 2))
double(sol_cond.s)
ans = 0.5000
sol_ode = simplify(subs(sol_ode,[vars(1),vars(2),vars(3),vars(4)],[sol_cond.C1,sol_cond.C2,sol_cond.C3,sol_cond.s]))
sol_ode = 
figure(1)
fplot(sol_ode,[0 1])
figure(2)
fplot(diff(sol_ode,t),[0 1])
figure(3)
fplot(diff(sol_ode,t,2),[0 1])
simplify(diff(sol_ode,t,3) -( -3/m*diff(sol_ode,t,2) + 1/m^2*(1-2*diff(sol_ode,t)+4*(subs(diff(sol_ode,t),t,1)-subs(diff(sol_ode,t),t,0)))))
ans = 
0
  2 Comments
Torsten
Torsten on 29 May 2023
Ran in:
A different method to solve the problem:
y0(1)=2;
y0(2)=1;
y0(3)=3;
source = 0.0;
solinit = bvpinit(linspace(0,1,100),y0,source);
sol = bvp4c(@mat4ode, @mat4bc, solinit);
sol = struct with fields:
solver: 'bvp4c' x: [0 0.0101 0.0202 0.0303 0.0404 0.0505 0.0606 0.0707 0.0808 0.0909 0.1010 0.1111 0.1212 0.1313 0.1414 0.1515 0.1616 0.1717 0.1818 0.1919 0.2020 0.2121 0.2222 0.2323 0.2424 0.2525 … ] y: [3×100 double] yp: [3×100 double] parameters: 0.5000 stats: [1×1 struct]
sol.parameters
ans = 0.5000
figure(1)
plot(sol.x,sol.y(1,:))
figure(2)
plot(sol.x,sol.y(2,:))
figure(3)
plot(sol.x,sol.y(3,:))
function dydx = mat4ode(x,y,source) % equation being solved
m = 0.05;
dydx = [y(2);y(3);-3/m*y(3)+1/(m.^2)*(1-2*y(2)+4.*source)];
end
function res = mat4bc(ya,yb,source) % boundary conditions
res = [ya(1)-2.0;ya(2)-1.0;ya(3)-3.0;yb(2)-1.0-source];
end

Sign in to comment.

Categories

Find more on Linear Algebra in Help Center and File Exchange

Tags

Community Treasure Hunt

Find the treasures in MATLAB Central and discover how the community can help you!

Start Hunting!