Index in position 2 exceeds array bounds (Index must not exceed 1) errors?

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JEESHAN ANSARI
JEESHAN ANSARI 2024-1-16
评论： JEESHAN ANSARI 2024-1-19
% Code to solve system of four simultaneous differential equation
clc
N0 = [9.66e17;0;0;0]; % Intital conditions of N1, N2, N3, N4 (9.66e17,0,0,0)
[T, N] = ode15s(@rate_eq,[0 126e-15], N0); % Intital conditions of time from 0 to 126e-15
plot(T,N(:,4),T,N(:,3),T,N(:,2),T,N(:,1)) % Plotting of N1 N2 N3 N4 vs Time
function dN = rate_eq( t,N )
t = 126e-15;
W = [0.2716 0.4243 0.6110 0.8317 1.0862 1.3748 1.6972]*1e9;
PION = [0.7072 0.8874 1.0254 1.2375 1.4143 1.5911 1.7679]*1e11;
dN = zeros(4,7); % Blank array to store N1 N2 N3 N4 from solution for each values of seven values of W and PION
Q = 10e10; % Constant value
A = 10e8; % Constant value
Ppd = 10e10; % Constant value
% W and PION is not constant and its values are calculated in first part of
% code using FOR loop
for j = 1:length(W)
dN(1,j) = (-W(j)*N(1,j)) + ((Q+A)*N(2,j)) + ((W(j)+Q+A)*N(3,j)) + 0; % first equation
dN(2,j) = 0 + (-(Q+A)*N(2,j)) + ((Q+A)*N(3,j)) + 0; % second equation
dN(3,j) = (W(j)*N(1,j)) + 0 - (W(j)+(2*Q)+(2*A)+ Ppd + PION(j))*N(3,j) + 0; % third equation
dN(4,j) = 0 + 0 + PION(j)*N(3,j) + 0; % fourth equation
end
end
%-----------end of code----------------
I want to store values of N1 N2 N3 N4 for each seven different values of W and PION.
I also want to plot N1 N2 N3 N4 vs time for each W and PION.
But I am getting the following errors:
Index in position 2 exceeds array bounds. Index must not exceed 1.
Error in test_soln_rate>rate_eq (line 24)
dN(1,j) = (-W(j)*N(1,j)) + ((Q+A)*N(2,j)) + ((W(j)+Q+A)*N(3,j)) + 0; % first equation
Error in odearguments (line 92)
f0 = ode(t0,y0,args{:}); % ODE15I sets args{1} to yp0.
Error in ode15s (line 148)
odearguments(odeIsFuncHandle, odeTreatAsMFile, solver_name, ode, tspan, y0, options, varargin);
Error in test_soln_rate (line 6)
[T, N] = ode15s(@rate_eq,[0 126e-15], N0); % Intital conditions of time from 0 to 126e-15

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采纳的回答

Alan Stevens
Alan Stevens 2024-1-16
N0 is a 4x1 vector, but you call it with two indices in the for loop! Turn N(1,j) into N(1) and similarly for the other calls to N.
  7 个评论
显示 5更早的评论
Alan Stevens
Alan Stevens 2024-1-19
Ran in:
Like this?
N0 = [9.66e17;0;0;0]; % Intital conditions of N1, N2, N3, N4 (9.66e17,0,0,0)
dt = 126e-15/50;
tspan = 0:dt:126e-15;
W = [0.2716 0.4243 0.6110 0.8317 1.0862 1.3748 1.6972]*1e9;
PION = [0.7072 0.8874 1.0254 1.2375 1.4143 1.5911 1.7679]*1e11;
for j = 1:7
w = W(j); p = PION(j);
[T, N] = ode15s(@(t,N) rate_eq(t,N,w,p),tspan, N0);
N1(:,j) = N(:,1); N2(:,j) = N(:,2); N3(:,j) = N(:,3); N4(:,j) = N(:,4);
end
% Plotting of N1 N2 N3 N4 vs Time
figure
plot(T,N1),grid
xlabel('T'),ylabel('N1')
legend('j=1','j=2','j=3','j=4','j=5','j=6','j=7','Location','southwest')
figure
plot(T,N2),grid
xlabel('T'),ylabel('N2')
legend('j=1','j=2','j=3','j=4','j=5','j=6','j=7','Location','northwest')
figure
plot(T,N3),grid
xlabel('T'),ylabel('N3')
legend('j=1','j=2','j=3','j=4','j=5','j=6','j=7','Location','northwest')
figure
plot(T,N4),grid
xlabel('T'),ylabel('N4')
legend('j=1','j=2','j=3','j=4','j=5','j=6','j=7','Location','northwest')
function dN = rate_eq(~,N,w,p )
Q = 10e10; % Constant value
A = 10e8; % Constant value
Ppd = 10e10; % Constant value
N1 = N(1); N2 = N(2); N3 = N(3); N4 = N(4);
dN1 = (-w*N1) + ((Q+A)*N2) + ((w+Q+A)*N3) + 0; % first equation
dN2 = 0 + (-(Q+A)*N2) + ((Q+A)*N3) + 0; % second equation
dN3 = (w*N1) + 0 - (w+(2*Q)+(2*A)+ Ppd + p)*N3 + 0; % third equation
dN4 = 0 + 0 + p*N3 + 0; % fourth equation
dN = [dN1; dN2; dN3; dN4];
end

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