Asked by pxg882
on 29 Oct 2012

Hi all, I'm running the following code:

function M = nonNewtonian(~)

M = bvpinit(linspace(0,10,301),@VKinit); sol = bvp4c(@VK,@VKbc,M);

figure; hold all; plot(sol.x,sol.y(2,:)); plot(sol.x,sol.y(4,:)); hold off; xlabel('\zeta') xlabel('\zeta') hleg = legend('F\prime','G\prime',... 'Location','NorthEast'); %#ok<NASGU>

figure; hold all; plot(sol.x,sol.y(1,:)); plot(sol.x,sol.y(3,:)); plot(sol.x,(-1)*sol.y(5,:)); hold off; xlabel('\zeta') hleg = legend('F','G','-H',... 'Location','East'); %#ok<NASGU>

function yprime = VK(x,y)

n=1;

yprime = [ y(2) n^(-1)*((y(2)^(2)+y(4)^(2))^((n-1)/2))^(-1)*((y(1)^(2)-y(3)^(2)+(y(5)+((1-n)/(n+1))*y(1)*x)*y(2))*(1+(n-1)*(y(2)^(2)+y(4)^2)^(-1)*y(4)^(2))-(n-1)*y(2)*y(4)*(y(2)^(2)+y(4)^(2))^(-1)*(2*y(1)*y(3)+(y(5)+((1-n)/(n+1))*y(1)*x)*y(4))) y(4) n^(-1)*((y(2)^(2)+y(4)^(2))^((n-1)/2))^(-1)*((2*y(1)*y(3)+(y(5)+((1-n)/(n+1))*y(1)*x)*y(4))*(1+(n-1)*(y(2)^(2)+y(4)^2)^(-1)*y(2)^(2))-(n-1)*y(2)*y(4)*(y(2)^(2)+y(4)^(2))^(-1)*(y(1)^(2)-y(3)^(2)+(y(5)+((1-n)/(n+1))*y(1)*x)*y(2))) -2*y(1)-(1-n)/(n+1)*x*y(2)];

function res = VKbc(ya,yb)

res = [ya(1);ya(3)-1;ya(5);yb(2)-(yb(5)*yb(1));yb(4)-(yb(5)*yb(3))];

function yinit = VKinit(~)

yinit = [0;0;1;0;0];

but receive the following error message:

??? Error using ==> bvp4c at 252 Unable to solve the collocation equations -- a singular Jacobian encountered.

Error in ==> nonNewtonian at 4 sol = bvp4c(@VK,@VKbc,M);

I'm using n=1 as a test case here. I know the solutions to this system for n=1 but would like to look into the solutions when n is not equal to one. Can anyone explain why I am getting this error?

Any help anyone could give would be greatly appreciated. Thanks!

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Answer by Star Strider
on 29 Oct 2012

Edited by Star Strider
on 30 Oct 2012

The most likely reason you got the error was that the your boundary values are too large.

I suggest you experiment with:

N = 1; M = bvpinit(linspace(0,N,301),@VKinit);

and increase `N` until you again get the error. A value of `N` that is `<10` may not be what you want, but it may be what you have to live with, given finite machine precision. (I did not try to run your code because I cannot determine where the rows of your `yprime` matrix end. It always helps to format your code. Highlight your code and use the `‘{} Code’` button at the top of the text window to format it.)

Answer by pxg882
on 30 Oct 2012

Thanks for taking the time to look at this. I didn't realise how horribly my code had been displayed on here.

Changing the boundary values is useful you were right. The problem was in the initial conditions I was using.

Thanks for the advice.

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