# using fzero to solve nonlinear equation

25 ビュー (過去 30 日間)
Jordan Rosales 2021 年 4 月 10 日
コメント済み: Walter Roberson 2021 年 4 月 10 日
I have the nonlinear equation
I need to use fzero to solve for s at different times, t.
In my script, I defined known parameters, created a function handle for the nonlinear equation set to zero, created a loop to calculate concentration at each time point, and within the loop used the fzero function. For my estimation I used 8, as a graph created earlier in the problem seems to approach 0 around 10. As shown in the command window there seems to be missing inputs within my function. However, I have checked thoroughly for missing inputs and even utilized the function outside of fzero and it worked fine. I assume the issue is then with fzero, but I am not sure what it is. Any advice is welcome!

サインインしてコメントする。

### 回答 (1 件)

Walter Roberson 2021 年 4 月 10 日
The function handle you pass to fzero() or to fsolve() must accept only one parameter.
Your s0, vmax, and k are all known constants, and do not need to be passed to the function; when you create the anonymous function, MATLAB will pull them out of the current workspace.
Your code suggests that you are trying to find s that satisfies the expression, but your text description indicates you need to find t ?
##### 4 件のコメント2 件の古いコメントを表示2 件の古いコメントを非表示
Jordan Rosales 2021 年 4 月 10 日
what is 'T' in this solution?
Walter Roberson 2021 年 4 月 10 日
T is a dummy variable in an anonymous function. arrayfun() in that form will take each value of t and substitute one value at a time for T into fzero(@(s) km*log(s0/s) + (s0-s) - vmax*T, 8) and collect all of the results.
If you prefer to loop then
tvals = 0:0.1:10;
numt = numel(tvals);
s = zeros(size(tvals));
for tidx = 1 : numt
t = tvals(tidx);
s(tidx) = fzero(@(S) km*log(s0/S) + (s0-S) - vmax*t, 8);
end

サインインしてコメントする。

### カテゴリ

Help Center および File ExchangeOptimization についてさらに検索

### Community Treasure Hunt

Find the treasures in MATLAB Central and discover how the community can help you!

Start Hunting!

Translated by