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

Question

Kushagra Saurabh 2024-11-11，7:02

0
链接

此问题的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/2165333-warning-results-may-be-inaccurate-because-of-an-ill-conditioned-jacobian-whose-reciprocal-condition

编辑： Torsten 2024-11-11，11:39

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 个评论
显示 -1更早的评论隐藏 -1更早的评论

Torsten 2024-11-11，11:24

编辑：Torsten 2024-11-11，11:39

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.

请先登录，再进行评论。

请先登录，再回答此问题。

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

1 个评论
显示 -1更早的评论隐藏 -1更早的评论

回答（0 个）

另请参阅

类别

标签

Community Treasure Hunt

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

1 个评论 显示 -1更早的评论隐藏 -1更早的评论

回答（0 个）

另请参阅

类别

标签

Community Treasure Hunt

1 个评论
显示 -1更早的评论隐藏 -1更早的评论