MATLAB: How to increase the number of mesh points that bvp5c uses

bvp5cMATLABmesh points

I'm trying to solve a 2nd order differential equation using bvp5c. It can get moderately stiff for certain parameters.

I get the following warning.

Warning: Unable to meet the tolerance without using more than 5000 mesh points. The last mesh of 892 points and the solution are available in the output argument. The maximum error is 0.00120499, while requested accuracy is 1e-07. > In bvp5c at 339

How can I increase the number of mesh points beyond 5000?

Thanks.

Best Answer

Please take the time to read the documentation of bvp5c. There you find the link to the function to set the options http://www.mathworks.com/help/matlab/ref/bvpset.html, where the parameter "NMax" is explained:

 Maximum number of mesh points allowed when solving the BVP, ...

Of course this message has the strong meta-message, that it is recommended to read the help and doc information before asking the forum. Matlab's docs are very useful and complete, and they have a much higher quality than the manuals of other software manufacturers.

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MATLAB: Problem with too many inputs into bvp5c

You should not use additional inputs at all, even some integrators still support this ancient method. See http://www.mathworks.com/matlabcentral/answers/1971: Use anonymous functions to provide parameters. This is explained in the documentation also: https://www.mathworks.com/help/matlab/ref/bvp5c.html:

 Parameterizing Functions explains how to provide additional 
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with a link to: https://www.mathworks.com/help/matlab/math/parameterizing-functions.html.

MATLAB: Though the linspace has 100 points but it takes 131 points for soly(2,:) in bvp4c

Hi,

see the documentation for bvpinit:

x is a vector that specifies an initial mesh. If you want to solve the BVP on [a,b], then specify x(1) as a and x(end) as b. The solver adapts this mesh to the solution, so a guess like xb=nlinspace(a,b,10) often suffices.

Note that you give x to the bvpinit function - not to the solver. So the solver decided to make 133 nodes to get a good solution.

Best regards

Stephan

Best Answer

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