Index exceeds matrix dimensions when solving for ODE system.

2017 9 27
1 个回答
回答已采纳
更新时间：2017 9 27
5 次查看（30 天）

0 个投票

Hi. Im trying to solve this system of equation. But it gives me this error. What does it mean and how to solve this? Thank you.
function dphidz = Pphiode(z,phi,Ro,Eta,Lambda)
Co=2.0-Ro-Ro^2; % Coriolis number
dphidz = [phi(2)
Ro.*(phi(1)^2+phi(2).*phi(5)-phi(3)^2+1)-Co*(phi(3)-1.0)
phi(4)
Ro*(2.0*phi(1)*phi(3) + phi(4)*phi(5))+Co*phi(1)
-2.0*phi(1)];
end
Error:
Trial>> Pphiode(20,2,-1,1,1)
Index exceeds matrix dimensions.
Error in Pphiode (line 3)
dphidz = [phi(2)
Let say
z=12, phi=1, R0=-1, Eta=0, Lambda=1

请先登录，再回答此问题。

 采纳的回答

KSSV
KSSV 2017-9-27

1 个投票

The input
phi = 1 ;
should be a vector with elements 5. Try with:
phi = rand(5,1) ;

2 个评论
显示 无 隐藏 无

nanana sehaa
nanana sehaa 2017-9-27
And how about this? Should i enter rand(5,1) for both phi0 and phiinf?
function bc = Pphibc (phi0,phiinf,Ro,Eta,Lambda)
bc = [phi0(1)-Lambda*phi0(2)
phi0(3)-Eta*phi0(4)
phi0(5)
phiinf(1)
phiinf(3)+1];
end

请先登录，再进行评论。

更多回答（0 个）

类别

在 帮助中心 和 File Exchange 中查找有关 Ordinary Differential Equations 的更多信息

标签

Community Treasure Hunt

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

Start Hunting!

Translated by