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

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Nauryzbay
Nauryzbay am 26 Mai 2023
Kommentiert: Nauryzbay am 30 Mai 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,:));

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Akzeptierte Antwort

Walter Roberson
Walter Roberson am 26 Mai 2023
Later you keep referring to sol.y so perhaps the reference to y should instead be to sol.y
  7 Kommentare
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Torsten
Torsten am 29 Mai 2023
Bearbeitet: Torsten am 29 Mai 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.
Nauryzbay
Nauryzbay am 29 Mai 2023
Thank you very much. I'll try to work on it myself.

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Weitere Antworten (1)

Torsten
Torsten am 29 Mai 2023
Bearbeitet: Torsten am 29 Mai 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 Kommentare
Torsten
Torsten am 29 Mai 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

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