Error tolerance in matlab

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I'm using R2022a: I received this when I tried solve of mixed PDEs using finite difference method.
Warning: Failure at t=6.122922e-01. Unable to meet integration tolerances without reducing the step size below the
smallest value allowed (1.776357e-15) at time t.
  3 Comments
University Glasgow
University Glasgow on 6 Oct 2022
Hmmmm, Okay, thank you.

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Accepted Answer

Star Strider
Star Strider on 6 Oct 2022
Singularities occur when the denominator of a fraction approaches zero, or for certrain transcendental functions (e.g. exp, tan) when they have arguments that cause them to approach ±Inf, and (similar to exp) a value is raised to a power and the value is high enough to create a singularity.
Look for those occurrences. Plotting the function over the intended independent variable range defining the integration is the easiest way to find them.
I looked at your code, however I can’t follow what you’re doing. The initial conditions vector ‘u0’ is a 198-element column vector, indicating to me that you’re integrating 198 separate differential equations.
I discourage the use of global variables, since they create problems and make the code extremely difficult to troubleshoot. It would be better to pass them as additional parameters, as described in Passing Extra Parameters.
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More Answers (1)

Torsten
Torsten on 6 Oct 2022
The ode integrators try to guarantee a predefined error tolerance when solving your system of ordinary differential equations. If they cannot succeed (even by reducing the stepsize dt to a very small value), they give up.
Usual reasons are errors in the implementation, singularities in the solutions for the differential equations etc.
  4 Comments
University Glasgow
University Glasgow on 6 Oct 2022
I don't understand what you are saying. Find attached the code
Torsten
Torsten on 6 Oct 2022
I mean integrate up to t=0.6 and write out "rhsode" to inspect its values.
Further plot the solution up to 0.6 to see if it looks as expected.

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