How to solve for gamma

Hello, I'm trying to solve for gamma in this equation, I was hoping if someone can help out.

considering everything else is a constant that can be put as an input.

[Torsten]

An analytical expression for gamma does not seem possible.

Use "fzero" or "fsolve" as numerical root finders for

left-hand side(gamma) - right-hand side(gamma) = 0

[smith]

Hello, I tried but It's not solving.

clc
clear all

dbs=5.3e-10;
das=1.01e-9;
cb=4e3;
ca=1e3;
x0 =0;
x = fsolve(@myfun,x0)

function F = myfun(x)

F = (1-erf(sqrt(x/dbs)))-(((cb/ca)*sqrt(dbs/das))*(erf(sqrt(x/das)))*(exp((x/das)-(x/dbs))));


end 

Answer 1

[Torsten]

format long
dbs=5.3e-10;
das=1.01e-9;
cb=4e3;
ca=1e3;
x0 =0;
options = optimset('TolX',1e-14,'TolFun',1e-14);
x = fsolve(@(x)myfun(x,dbs,das,cb,ca),x0,options)

Equation solved.

fsolve completed because the vector of function values is near zero
as measured by the value of the function tolerance, and
the problem appears regular as measured by the gradient.

x = 
     2.980227787591144e-08

myfun(x,dbs,das,cb,ca)

ans = 
    -7.180679523372318e-12

function F = myfun(x,dbs,das,cb,ca)
  F = (1-erf(sqrt(x/dbs)))-(((cb/ca)*sqrt(dbs/das))*(erf(sqrt(x/das)))*(exp((x/das)-(x/dbs))));
end 

[smith]

What does the (-7e-12) resemble exactly ??

[Walter Roberson]

-7e-12 is the value the function actually calculates at x. Ideally you would want it to compute exactly 0, but that is not usually realistic due to floating point round-off (and the fact that you did not use as tight a tolerance as absolutely possible.)

[smith]

is there a way to get x more accurately, because according to the background, x should be smaller in order (e-10)

[Torsten]

Seems your equation has multiple solutions.

A good idea is always to plot the function to have a good guess for the solution.

dbs=5.3e-10;
das=1.01e-9;
cb=4e3;
ca=1e3;
F = @(x)(1-erf(sqrt(x/dbs)))-(((cb/ca)*sqrt(dbs/das))*(erf(sqrt(x/das))).*(exp((x/das)-(x/dbs))));
x = 0:1e-12:2e-10;
plot(x,F(x))

sol = fzero(F,[0 1e-9])

sol = 4.7822e-11

F(sol)

ans = 1.2922e-09

[smith]

Thank you @Torsten!