Why it is not making plot for different values of "M"?

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syms x
alpha = -0.1;
sigma = 0.1;
eps = -0.1;
e = 0.2;
a = 2;
lambda = 2;
psi = 10;
figure
M_list = [4, 6, 8, 10];
for i = 1:numel(M_list)
M = M_list(i);
hbar = @(x) a - a.*x + x;
A1 = eps + alpha^3 + (3 * sigma^2 * alpha);
B1 = @(x) (-3 * lambda * M) * ((hbar(x).^2) + (2 .* hbar(x) .* alpha) + (sigma^2) + (alpha^2));
a1 = @(x) tanh(M .* hbar(x));
b1 = @(x) 1 - ((tanh(M .* hbar(x))).^2);
c1 = (M * alpha) - ((M^3 * A1)/3);
d1 = 2 * (M^2) * (1 + lambda);
C1 = @(x) a1(x) + (b1(x) .* c1);
D1 = @(x) d1 .* ((hbar(x).^3) + (3 .* (hbar(x).^2) .* alpha) + (3 .* hbar(x) .* (alpha)^2) + (3 .* hbar(x) .* (sigma)^2) + eps + (3 * alpha * (sigma^2)) + (alpha^3));
f1 = @(x) B1(x) + (3 * lambda .* C1(x) .* hbar(x)) + (3 * lambda .* C1(x) .* alpha) + (D1(x) .* C1(x));
f2 = @(x) 12 * (M^2) * (1 + lambda) .* C1(x);
f3 = psi * (e^3);
f4 = (1 + lambda) *180 * ((1 - e)^2); % 180 is not given in paper
f5 = 1/(2 + lambda);
F = @(x) ((f5 .* f1(x))./f2(x)) + (f3/f4);
q1 = @(x) hbar(x) ./ (2 .* F(x));
Q1 = integral(q1,0,1);
q2 = @(x) 1./(F(x));
Q2 = integral(q2,0,1);
Q = Q1/Q2;
p1 = @(x) (1./F(x)) .* ((0.5 .* hbar(x)) - Q);
P = @(x) integral(p1,0,x);
fplot(P, [0 1])
ylim([0 1])
set(gca, 'ytick', 0:0.1:1);
set(gca, 'xtick', 0:0.2:1);
xlabel('x')
ylabel('P(x)')
end
Warning: Reached the limit on the maximum number of intervals in use. Approximate bound on error is 1.6e+02. The integral may not exist, or it may be difficult to approximate numerically to the requested accuracy.
Warning: Function behaves unexpectedly on array inputs. To improve performance, properly vectorize your function to return an output with the same size and shape as the input arguments.
Warning: Reached the limit on the maximum number of intervals in use. Approximate bound on error is 8.0e+02. The integral may not exist, or it may be difficult to approximate numerically to the requested accuracy.
Warning: Function behaves unexpectedly on array inputs. To improve performance, properly vectorize your function to return an output with the same size and shape as the input arguments.
Warning: Reached the limit on the maximum number of intervals in use. Approximate bound on error is 1.7e+02. The integral may not exist, or it may be difficult to approximate numerically to the requested accuracy.
Warning: Function behaves unexpectedly on array inputs. To improve performance, properly vectorize your function to return an output with the same size and shape as the input arguments.
Warning: Reached the limit on the maximum number of intervals in use. Approximate bound on error is 2.1e+02. The integral may not exist, or it may be difficult to approximate numerically to the requested accuracy.
Warning: Function behaves unexpectedly on array inputs. To improve performance, properly vectorize your function to return an output with the same size and shape as the input arguments.

Accepted Answer

KSSV
KSSV on 8 Jul 2022
Edited: KSSV on 8 Jul 2022
You have to use hold on.
syms x
warning off
alpha = -0.1;
sigma = 0.1;
eps = -0.1;
e = 0.2;
a = 2;
lambda = 2;
psi = 10;
M_list = [4, 6, 8, 10];
figure
hold on
for i = 1:numel(M_list)
M = M_list(i);
hbar = @(x) a - a.*x + x;
A1 = eps + alpha^3 + (3 * sigma^2 * alpha);
B1 = @(x) (-3 * lambda * M) * ((hbar(x).^2) + (2 .* hbar(x) .* alpha) + (sigma^2) + (alpha^2));
a1 = @(x) tanh(M .* hbar(x));
b1 = @(x) 1 - ((tanh(M .* hbar(x))).^2);
c1 = (M * alpha) - ((M^3 * A1)/3);
d1 = 2 * (M^2) * (1 + lambda);
C1 = @(x) a1(x) + (b1(x) .* c1);
D1 = @(x) d1 .* ((hbar(x).^3) + (3 .* (hbar(x).^2) .* alpha) + (3 .* hbar(x) .* (alpha)^2) + (3 .* hbar(x) .* (sigma)^2) + eps + (3 * alpha * (sigma^2)) + (alpha^3));
f1 = @(x) B1(x) + (3 * lambda .* C1(x) .* hbar(x)) + (3 * lambda .* C1(x) .* alpha) + (D1(x) .* C1(x));
f2 = @(x) 12 * (M^2) * (1 + lambda) .* C1(x);
f3 = psi * (e^3);
f4 = (1 + lambda) *180 * ((1 - e)^2); % 180 is not given in paper
f5 = 1/(2 + lambda);
F = @(x) ((f5 .* f1(x))./f2(x)) + (f3/f4);
q1 = @(x) hbar(x) ./ (2 .* F(x));
Q1 = integral(q1,0,1);
q2 = @(x) 1./(F(x));
Q2 = integral(q2,0,1);
Q = Q1/Q2;
p1 = @(x) (1./F(x)) .* ((0.5 .* hbar(x)) - Q);
P = @(x) integral(p1,0,x);
fplot(P, [0 1])
end
legend(num2str(M_list'))
ylim([0 1])
set(gca, 'ytick', 0:0.1:1);
set(gca, 'xtick', 0:0.2:1);
xlabel('x')
ylabel('P(x)')

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