Using interp1 within vpasolve
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Adam Zieser
on 13 Jul 2021
Commented: Walter Roberson
on 14 Jul 2021
I'm having difficulty with the following situation trying to numerically solve an equation (with vpasolve, or something similar) that uses interp1-interpolated data within the solve. Consider:
with implementation:
syms Z;
vpasolve(Z.^4 == ( muL .* interp1(X,GH,Z) - muH .* interp1(X,GL,Z) )./( muL .* interp1(X,FH,Z) ...
- muH .* interp1(X,FL,Z) ), Z);
where and are constants, so that the μ-values are constant, and F and G are 2D matrices squeezed to 1D. These 1D matrices are known on a grid Z = 5:15. Then, I'm attempting to solve this equation for Z, which presumably has real-numbered values as solution. I assumed vpasolve worked iteratively, with an initial guessed query of Z, such that interp1 with (syms Z) is valid to use inside vpasolve. Of course, this doesn't work.
Is there any way to work this? I've simplified things for ease of explanation, but suffice it to say that just doing a polynomial fit on F and G instead of using interp1 is not on the table: the data in the F and G grids need to only be interpolated for non-integer values of Z. Here's some data, just to make it easier to run if needed.
muH = 0.178125;
muL = 0.142500;
GL = [0.2684 0.2690 0.2696 0.2704 0.2714 0.2724 0.2734 0.2746 0.2756 0.2768 0.2781];
GH = [0.3258 0.3280 0.3307 0.3342 0.3381 0.3431 0.3474 0.3520 0.3568 0.3618 0.3670];
FL = 1E-6 .* [0.4234 0.4180 0.4100 0.4017 0.3932 0.3840 0.3766 0.3698 0.3628 0.3554 0.3480];
FH = 1E-6 .* [0.4234 0.4180 0.4100 0.4017 0.3932 0.3840 0.3766 0.3698 0.3628 0.3554 0.3480];
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Accepted Answer
Walter Roberson
on 14 Jul 2021
Turn it into a numeric system
z0 = 1;
muH = 0.178125;
muL = 0.142500;
GL = [0.2684 0.2690 0.2696 0.2704 0.2714 0.2724 0.2734 0.2746 0.2756 0.2768 0.2781];
GH = [0.3258 0.3280 0.3307 0.3342 0.3381 0.3431 0.3474 0.3520 0.3568 0.3618 0.3670];
FL = 1E-6 .* [0.4234 0.4180 0.4100 0.4017 0.3932 0.3840 0.3766 0.3698 0.3628 0.3554 0.3480];
FH = 1E-6 .* [0.4234 0.4180 0.4100 0.4017 0.3932 0.3840 0.3766 0.3698 0.3628 0.3554 0.3480];
best_z = fsolve(@(Z) Z.^4 - (( muL .* interp1(X,GH,Z,'linear','extrap') - muH .* interp1(X,GL,Z,'linear','extrap') )./( muL .* interp1(X,FH,Z,'linear','extrap') ...
- muH .* interp1(X,FL,Z,'linear','extrap') )), z0);
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