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

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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;
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);
xlabel('x','fontsize', 20);
ylabel('y','fontsize', 20);
Could someone help me.

Accepted Answer

Walter Roberson
Walter Roberson on 21 Apr 2017
You cannot do much about it. Your expression for f1 includes
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.
Walter Roberson
Walter Roberson on 21 Apr 2017
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?

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