Solve a nonlinear equation with constrains
1 次查看(过去 30 天)
显示 更早的评论
Hello,
clear
p0=1000e6;
t0 = 1e-6;
td = 1e-6;
t = t0 + td;
c = 5e6;
a = @(r)log(r)./(t0*(r-1.0));
b = @(r)a(r).*r;
func= @(r) p0*((exp(-a*t) - exp(-b*t))/(exp(-a*t0) - exp(-b*t0)))-c;
r=1.5;
roots = fzero(func,r)
my equation is the func.
![](https://www.mathworks.com/matlabcentral/answers/uploaded_files/1232672/image.png)
where a and b are,
![](https://www.mathworks.com/matlabcentral/answers/uploaded_files/1232677/image.png)
1 个评论
Bora Eryilmaz
2022-12-15
编辑:Bora Eryilmaz
2022-12-15
Your function "p(t)" (func in your code) is not a function of time since you are assigning a fixed scalar value to "t" in your code. So, func() is a function of r, with a fixed "t". So what you are really solving here is p(r) = 0 given fixed values for t, t0, a, b, c.
You will need to reformulate your problem.
采纳的回答
Torsten
2022-12-15
p0 = 1000e6;
t0 = 1e-6;
td = 1e-6;
t = t0 + td;
c = 7.5e8;
a = @(r)log(r)./(t0*(r-1.0));
b = @(r)a(r).*r;
func= @(r) p0*((exp(-a(r)*t) - exp(-b(r)*t))./(exp(-a(r)*t0) - exp(-b(r)*t0))) - c;
r = 0.001:0.1:10;
plot(r,func(r))
root1 = fzero(func,[0.001 0.75])
root2 = fzero(func,[1.1 2.5])
2 个评论
更多回答(0 个)
另请参阅
类别
在 Help Center 和 File Exchange 中查找有关 Calculus 的更多信息
Community Treasure Hunt
Find the treasures in MATLAB Central and discover how the community can help you!
Start Hunting!