How to remove the straight line coming at the x-axis at x = -0.11532 in my code

Hi all,

In my code, I want to remove the straight line coming at the x-axis at x = -0.11532.

I attached my code below,

B = 1e-4;
sigma_on = 0.45;
x_on = 0.06;
sigma_p = 4e-5;
A = 1e-10;
sigma_off = 0.013;
x_off = 0.4;
G_m = 0.025;
a = 7.2e-6;
b = 4.7;
beta = 500;
rho = 1e-3;
v_m = 1;
k=50;

t = -1:0.001:1;

for x = 1:length(t)

      G(x) = G_m*t(x)+a*exp(b*sqrt(v_m/(1+exp(-v_m/rho))-v_m/(1+exp(v_m/rho))))*(1-t(x));

      f1(x) = A*sinh(v_m/sigma_off)*exp(-(x_off^2/t(x)^2))*exp(1/(1+beta*G(x)*v_m^2))*(1/(1+exp(k*v_m)));

      f2(x) = B*sinh(v_m/sigma_on)*exp(-(t(x)^2/x_on^2))*exp(G(x)*v_m^2/sigma_p)*(1/(1+exp(-k*v_m)));

      f(x) = f1(x) + f2(x);

end

semilogy(t,f);
xlabel('x','fontsize', 20);
ylabel('y','fontsize', 20);

Could someone help me.

Answer 1

[Walter Roberson]

You cannot do much about it. Your expression for f1 includes

exp(1/(1+beta*G(x)*v_m^2))

That value can be arbitrarily high if (1+beta*G(x)*v_m^2) approaches 0; with your beta = 500 and v_m = 1, that is the condition that G(x) approximately equal -1/500 .

You can go to the line above,

        G(x) = G_m*t(x)+a*exp(b*sqrt(v_m/(1+exp(-v_m/rho))-v_m/(1+exp(v_m/rho))))*(1-t(x));

and construct

        -1/500 == G_m*T+a*exp(b*sqrt(v_m/(1+exp(-v_m/rho))-v_m/(1+exp(v_m/rho))))*(1-T);

and solve for T. You get a result of T about -0.115316250006499 . You are incrementing by 0.001 so when your T becomes -0.115 you have a near singularity .

If your equations are correct then the only way to avoid the singularity is not to calculate near it.

[vetri veeran]

Dear Walter,

Thanks for your answer. Could you please tell me how to avoid the singularity appearing at x = -0.1153162 in my code. I want to remove that straight line anyway.

[Walter Roberson]

Considering that you have 1/(1+beta*G(x)*v_m^2) and the denominator switches between positive and negative, what kind of output were you hoping for?

[vetri veeran]

Thank you very much Walter Robertson