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

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AVINASH SAHU
AVINASH SAHU 2022-7-8
评论： AVINASH SAHU 2022-7-8
Ran in:
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.

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采纳的回答

KSSV
KSSV 2022-7-8
编辑：KSSV 2022-7-8
Ran in:
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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