Numerics: Wronskian and linear independence

[member 428835]

Hi PF!

I'm solving a differential eigen-value problem in weak form, so I have trial functions. If the Wronskian of trial functions is small but not zero, is linear independence an issue? I have analytic trial functions but am numerically integrating.

 

[jvicens]

Hi there,

Linear independence is definitely something to consider when solving a differential eigen-value problem in weak form. The Wronskian of trial functions can give you an indication of the linear independence of these functions. If the Wronskian is small but not zero, it could mean that your trial functions are approximately linearly dependent. This could potentially lead to inaccuracies in your numerical integration and affect the accuracy of your solution.

To ensure the accuracy of your solution, it is important to choose trial functions that are linearly independent. This can be achieved by using a larger set of trial functions or by using a different set of functions altogether. Additionally, you may also want to consider checking the condition number of your resulting matrix system, as a high condition number can also indicate issues with linear independence.

I hope this helps and good luck with your research!