Pardon the obvious question, but is there a reason you don't simply rewrite the expression to avoid generating NaN/Inf in the first place? Replacing Inf with 0 also seems like a strange thing to do! Andrew On 3 March 2017 at 21:15, Francis Poulin <fpoulin@uwaterloo.ca> wrote:
Hello,
I am doing a stability calculation for different profiles.
The test case that I have used is Gaussian, where the velocity and gradient of PV is defined using something like this,
Ub = Function(V).interpolate(Uj*exp(-(x[0]-Ly/2)**2)) dQb = Function(V).interpolate(-2.*Uj*exp(-(x[0]-Ly/2)**2)*(2.*(x[0]-Ly/2)**2 - 1.))
I have compared this with the results of using a spectral method and the results agree very well.
I am now considering a less trivial example using something called a Bump function. Yes, technical name is Bump. Thanks to previous help I have figured out how to define it piecewise (thanks again).
Ub = Function(V).interpolate(conditional(le(abs(x[0]-Ly/2.), 1.0), exp(-1./(1. -(x[0]-Ly/2)**2)), 0.0)) dQb = Function(V).interpolate(conditional(le(abs(x[0]-Ly/2.), 1.0), -2.*exp(-1./(1. -(x[0]-Ly/2)**2))*(3.*(x[0]-Ly/2.)**4 - 1)/((x[0]-Ly/2.)**2-1)**4, 0.0))
However, when I try this I get zero growth rate, which does not agree with the spectral method.
Actually, in the spectral method we need to zero out the Nans and Infs in expressions, and that works out nicely. Is there an easy way to convert Nans and Infs to zero in interpolate?
I tried using both a Krylov method and Lapack. The first gives zero, which I don't believe, and the second fails, I presume because I am dividing by zero somewhere.
Any help would be greatly appreciated!
Cheers, Francis
------------------ Francis Poulin Associate Professor Department of Applied Mathematics University of Waterloo
email: fpoulin@uwaterloo.ca Web: https://uwaterloo.ca/poulin-research-group/ Telephone: +1 519 888 4567 x32637 <(519)%20888-4567>
Good question. I agree it's probably better to avoid evaluating a function that divides by zero. In the expression I have a conditional expression that uses less than or equal to something. It occurs to me that if I use only less than (but not equal to) that should do the trick. I would guess that I should change le to lt but not sure. Do you think this is what I should try? Cheers, Francis ------------------ Francis Poulin Associate Professor Department of Applied Mathematics University of Waterloo email: fpoulin@uwaterloo.ca Web: https://uwaterloo.ca/poulin-research-group/ Telephone: +1 519 888 4567 x32637 ________________________________ From: firedrake-bounces@imperial.ac.uk [firedrake-bounces@imperial.ac.uk] on behalf of Andrew McRae [A.T.T.McRae@bath.ac.uk] Sent: Sunday, March 05, 2017 11:15 AM To: firedrake@imperial.ac.uk Subject: Re: [firedrake] Nans and infs Pardon the obvious question, but is there a reason you don't simply rewrite the expression to avoid generating NaN/Inf in the first place? Replacing Inf with 0 also seems like a strange thing to do! Andrew On 3 March 2017 at 21:15, Francis Poulin <fpoulin@uwaterloo.ca<mailto:fpoulin@uwaterloo.ca>> wrote: Hello, I am doing a stability calculation for different profiles. The test case that I have used is Gaussian, where the velocity and gradient of PV is defined using something like this, Ub = Function(V).interpolate(Uj*exp(-(x[0]-Ly/2)**2)) dQb = Function(V).interpolate(-2.*Uj*exp(-(x[0]-Ly/2)**2)*(2.*(x[0]-Ly/2)**2 - 1.)) I have compared this with the results of using a spectral method and the results agree very well. I am now considering a less trivial example using something called a Bump function. Yes, technical name is Bump. Thanks to previous help I have figured out how to define it piecewise (thanks again). Ub = Function(V).interpolate(conditional(le(abs(x[0]-Ly/2.), 1.0), exp(-1./(1. -(x[0]-Ly/2)**2)), 0.0)) dQb = Function(V).interpolate(conditional(le(abs(x[0]-Ly/2.), 1.0), -2.*exp(-1./(1. -(x[0]-Ly/2)**2))*(3.*(x[0]-Ly/2.)**4 - 1)/((x[0]-Ly/2.)**2-1)**4, 0.0)) However, when I try this I get zero growth rate, which does not agree with the spectral method. Actually, in the spectral method we need to zero out the Nans and Infs in expressions, and that works out nicely. Is there an easy way to convert Nans and Infs to zero in interpolate? I tried using both a Krylov method and Lapack. The first gives zero, which I don't believe, and the second fails, I presume because I am dividing by zero somewhere. Any help would be greatly appreciated! Cheers, Francis ------------------ Francis Poulin Associate Professor Department of Applied Mathematics University of Waterloo email: fpoulin@uwaterloo.ca<mailto:fpoulin@uwaterloo.ca> Web: https://uwaterloo.ca/poulin-research-group/ Telephone: +1 519 888 4567 x32637<tel:(519)%20888-4567>
Hello Andrew, Quick check shows that it does work nicely using less than (lt). Thanks for encouraging me to write better code! Cheers, Francis ------------------ Francis Poulin Associate Professor Department of Applied Mathematics University of Waterloo email: fpoulin@uwaterloo.ca Web: https://uwaterloo.ca/poulin-research-group/ Telephone: +1 519 888 4567 x32637 ________________________________ From: Francis Poulin Sent: Sunday, March 05, 2017 11:32 AM To: firedrake@imperial.ac.uk Subject: RE: [firedrake] Nans and infs Good question. I agree it's probably better to avoid evaluating a function that divides by zero. In the expression I have a conditional expression that uses less than or equal to something. It occurs to me that if I use only less than (but not equal to) that should do the trick. I would guess that I should change le to lt but not sure. Do you think this is what I should try? Cheers, Francis ------------------ Francis Poulin Associate Professor Department of Applied Mathematics University of Waterloo email: fpoulin@uwaterloo.ca Web: https://uwaterloo.ca/poulin-research-group/ Telephone: +1 519 888 4567 x32637 ________________________________ From: firedrake-bounces@imperial.ac.uk [firedrake-bounces@imperial.ac.uk] on behalf of Andrew McRae [A.T.T.McRae@bath.ac.uk] Sent: Sunday, March 05, 2017 11:15 AM To: firedrake@imperial.ac.uk Subject: Re: [firedrake] Nans and infs Pardon the obvious question, but is there a reason you don't simply rewrite the expression to avoid generating NaN/Inf in the first place? Replacing Inf with 0 also seems like a strange thing to do! Andrew On 3 March 2017 at 21:15, Francis Poulin <fpoulin@uwaterloo.ca<mailto:fpoulin@uwaterloo.ca>> wrote: Hello, I am doing a stability calculation for different profiles. The test case that I have used is Gaussian, where the velocity and gradient of PV is defined using something like this, Ub = Function(V).interpolate(Uj*exp(-(x[0]-Ly/2)**2)) dQb = Function(V).interpolate(-2.*Uj*exp(-(x[0]-Ly/2)**2)*(2.*(x[0]-Ly/2)**2 - 1.)) I have compared this with the results of using a spectral method and the results agree very well. I am now considering a less trivial example using something called a Bump function. Yes, technical name is Bump. Thanks to previous help I have figured out how to define it piecewise (thanks again). Ub = Function(V).interpolate(conditional(le(abs(x[0]-Ly/2.), 1.0), exp(-1./(1. -(x[0]-Ly/2)**2)), 0.0)) dQb = Function(V).interpolate(conditional(le(abs(x[0]-Ly/2.), 1.0), -2.*exp(-1./(1. -(x[0]-Ly/2)**2))*(3.*(x[0]-Ly/2.)**4 - 1)/((x[0]-Ly/2.)**2-1)**4, 0.0)) However, when I try this I get zero growth rate, which does not agree with the spectral method. Actually, in the spectral method we need to zero out the Nans and Infs in expressions, and that works out nicely. Is there an easy way to convert Nans and Infs to zero in interpolate? I tried using both a Krylov method and Lapack. The first gives zero, which I don't believe, and the second fails, I presume because I am dividing by zero somewhere. Any help would be greatly appreciated! Cheers, Francis ------------------ Francis Poulin Associate Professor Department of Applied Mathematics University of Waterloo email: fpoulin@uwaterloo.ca<mailto:fpoulin@uwaterloo.ca> Web: https://uwaterloo.ca/poulin-research-group/ Telephone: +1 519 888 4567 x32637<tel:(519)%20888-4567>
participants (2)
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Andrew McRae
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Francis Poulin