Warning: Results may be inaccurate because of an ill-conditioned Jacobian whose reciprocal condition number is 3.23252e-20.

49 ビュー (過去 30 日間)
Kushagra Saurabh
Kushagra Saurabh 2024 年 11 月 11 日 7:02
編集済み: Torsten 2024 年 11 月 11 日 11:39
Hey!
I was trying to solve these equations and was getting an error as : Warning: Results may be inaccurate because of an ill-conditioned Jacobian whose reciprocal condition number is 3.23252e-20. > In bvp4c (line 288) in compressiblechannel (line 15). How should I resolve it. here's my code and equations.
Thanks for your help in advance.
code:
close all
clear all
G = -1;
%Na = 0.86;
gamma = 1.4;
Pr = 0.71;
M = 0.1;
xstart = -1;
xend = 1;
nx = 51;
x = linspace(xstart,xend,nx);
solinit = bvpinit(x, [0;0;1;0]);
sol = bvp4c(@(y,z)bvpfcn(y,z,G, Pr, M, gamma), @bcfcn, solinit);
Warning: Results may be inaccurate because of an ill-conditioned Jacobian whose reciprocal condition number is 3.23252e-20.
% Extract variables from solution
U = sol.y(1,:); % U from solution
dUdy = sol.y(2,:); % dU/dy from solution
T0 = sol.y(3,:); % T0 from solution
dT0dy = sol.y(4,:); % dT0/dy from solution
y = sol.x; % y values (independent variable)
%Calculate d^2U/dy^2 using the equation from bvpfcn
% d2Udy2 = (dUdy .* dT0dy) + G * exp(T0);
% Plot d^2U/dy^2 vs y
figure(3)
% plot(d2Udy2, y);
% hold on
plot(diff(dUdy), y(2:end));
xlabel('d^2U/dy^2');
ylabel('y');
title('Plot of d^2U/dy^2 vs y');
figure(1)
plot(sol.y(1,:), sol.x)
xlabel ('U')
ylabel ('y')
figure(2)
plot(sol.y(3,:), sol.x)
xlabel ('T0')
ylabel ('y')
function dzdy = bvpfcn(y,z,G,Pr,M,gamma )
U = z(1);
dUdy = z(2);
T0 = z(3);
dT0dy = z(4);
dzdy = zeros(4,1);
dzdy(1) = dUdy;
dzdy(2) = -dUdy*dT0dy*0.76*(T0^ -1) + G/(T0^0.76);
dzdy(3) = dT0dy;
dzdy(4) = -Pr*M^2*(gamma-1)*((T0^0.76)*(dUdy^2));
end
function res = bcfcn(za,zb)
res = [za(1);zb(1);za(3);zb(3)-1];
end
equation:
where mu0 = T0^ 0.76 and boundary conditions are: U = 0 at y= +1 and y = -1 and T0 = 0 at y =-1 and T0 = 1 at y = +1
  1 件のコメント
Torsten
Torsten 2024 年 11 月 11 日 11:24
編集済み: Torsten 2024 年 11 月 11 日 11:39
It seems mu0 = exp(-T0) was a better choice than mu0 = T0^0.76 for "bvp4c".
Since you divide by T0^0.76, T0 mustn't become zero anywhere. But you set T0 = 0 at y = -1. This might be a problem although it's only in a boundary point.
Another problem is that T0 should not become negative in the course of the iterations because T0^0.76 would produce a complex number.

サインインしてコメントする。

回答 (0 件)

カテゴリ

Help Center および File ExchangeBoundary Value Problems 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