is it possible to determine the values of the input parameters ca, cb, cc and cd for which the cc value at the output would be maximal?

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Bosnian Kingdom 2019-4-24

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评论： Bosnian Kingdom 2019-5-6

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is it possible to determine the values of the input parameters ca, cb, cc and cd for which the cc value at the output would be maximal?

    dca/dt=-theta(1) * ca-theta(2).cA;
    dcb/dt= theta(1).*cA+theta(4).*cc*theta(3).*cb-theta(5).*cd
    dcc/dt= theta(2).*cA+theta(3).*cb*theta(4).*cCtheta(6).*cd
    dd/dt= theta(5).*cb*theta(6).*cd

Values for all thetas are known.

Thanks in advance.

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Walter Roberson 2019-5-2

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That NaN is sure to cause problems.

There is a solution to the system of

ca(t) = 0, cb(t) = 0, cc(t) = cc0, cd(t) = 0

where cc0 is the initial condition for cc. WIth you having provided 0 as the initial conditions for ca, cb, cd, then you would possibly be hitting this situation.

There appears to be another solution when cb(0) is non-zero:

ca(t) = 0, cb(t) = cb0 * exp(theta3*theta4*cc0*t), cc(t) = cc0, cd(t) = 0

When cb(0) = 0, then there is a solution with a non-zero ca with a simple form, and with cd(t) = 0, and with a complicated cb expression, and with

cc(t) = (-exp(-(theta1+theta2)*t) * ca0 * theta2 + (ca0+cc0) * theta2 + cc0 * theta1) / (theta1 + theta2)

That system is linear in cc0, so you would maximize it when cc(0) is infinity. (The system has cc0*(theta1+theta2) in the numerator, and has (theta1+theta2) in the denominator, so if theta1+theta2 is negative, the two negatives would cancel out as cc0 reached infinity, and so the maxima is at cc(0) -> infinity rather than at cc(0) -> infinity * sign(theta1+theta2) )

The complicated expression for cb does not have a closed form (except perhaps at very specific reationships between the angles), but that does not matter since you just want to know for cb, which does have a simple relationship in this case.

When cc(0) = 0, then there is an additional fairly messy solution that cannot be expressed in closed form, and which is difficult to to maximize. Oddly, even though the expression involves cb0 in non-trivial ways, the derivative is 0 with respect to cb0, suggesting that perhaps the complicated expression could be simplified... just maybe to a fairly simple expression ??

Walter Roberson 2019-5-5

I went through the scenarios in

https://www.mathworks.com/matlabcentral/answers/458387-is-it-possible-to-determine-the-values-of-the-input-parameters-ca-cb-cc-and-cd-for-which-the-cc#comment_701095

If you start with ca 0, cb 0, cd 0, then the output for cc will be the same as your input for cc. Therefore you can get any output you want for cc by using that same input for cc. This is the same as Torsten told you, that to maximize cc then start with cc being +infinity.

This is the "generate1" scenario. You start with all 0 and you get the same cc output as input. Nothing ever changes.

The "generate2" scenario starts with a non-zero cb, and has a cb that changes over time. However, the cc stays the same as input, so to maximize cc, start with cc being infinity.

The "generate3" scenario does have a cc that changes. But look at it: it has terms that are independent of time, and it has a term which is -ca0*theta2*exp(-(theta1+theta2)*t) / (theta1+theta2) .

If theta1+theta2 is positive, then -(theta1+theta2)*t gets more negative as time increases, and exp() of a more negative term decreases towards 0, so -ca0*theta2 times the term will get closer and closer to 0 (even if ca0 or theta2 are negative.) Thus in that situation, yes, cc would increase, but it would increase towards a constant upper bound determined by the terms that are independent of time.

If theta1+theta2 is negative, then -(theta1+theta2)*t gets more positve as time increases, and exp() of a more positive term increases rapidly, and the negative of that grows rapidly smaller. But with the division by (theta1+theta2) and (theta1+theta2) being negative, that would give you a term that would grow rapidly larger in time. But theta1+theta2 would have to be negative, and your input theta are all positive.

If you have to maximize cc, and you want one of the situations where cc is not constant, then you are down to two possibilities:

The generate3 scenario where cc starts as small as it can ever get and grows towards a constant upper bound as the exponential contribution of it decreases until you can ignore it; or
The formulation with ca(0) non-zero, cb(0) non-zero, cc(0) non-zero, cd(0) irrelevant, in which case the situation is too complicated to analyze analytically without a lot of work. ca(t), cb(t), cc(t) would all have to be expressed in terms of functions of integrals.

Bosnian Kingdom 2019-5-6

How can I accept you answer? You helped me. I used generate3. Thank you.

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