Optimization with a differential term

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I am trying to find V_L(t) that can maximize the following function

P = {I_s*sin(wt) - [C*V_L'(t) + V_L/R]} * V_L

Any thoughts? Thank you very much!

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David Goodmanson 2018-7-1

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Hi Stephan,

Calculus of variations is certainly involved here, but to use it you have to satisfy certain conditions including specifications at the end points, and it matters whether a maximum or minimum is sought. For example for two fixed points in the plane, there is a minimum path length between them but no maximum path length.

For the problem as stated,

V = N*V0*exp(-N*w*t)

is a perfectly acceptable function to test for a maximum of the integral. Here N is a dimensionless constant and V0 is a fixed reference voltage such as 1V. Then:

syms I_s V V0 R C t w N
V = N*V0*exp(-N*w*t)
u1 = int(I_s*sin(w*t)*V,t,0,pi/w)          % first term in integral
u2 = int(-C*V*diff(V,t) -V^2/R,t,0,pi/w)   % second two terms in integral
% result
u1 = I_s*N*V0*(1/(w*(N^2 + 1)) + exp(-pi*N)/(w*(N^2 + 1)))
u2 = N*V0^2*(C*N*R*w - 1)*(1 - exp(-2*pi*N))/(2*R*w)

(u2 is the same as what Matlab provides except I multiplied through by a leading minus sign).

u1 is positive and contributes to the maximum. For large N, u2 is positive and arbitrarily large. So there is no maximum value for this problem.

Stephan 2018-7-3

编辑：Stephan 2018-7-3

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Hi,

did you already update your matlab? Maybe this helps for the problem wirh your symbolic toolbox:

https://de.mathworks.com/matlabcentral/answers/404882-how-can-i-fix-this-freezing-problem-when-use-syms

Back to your question: I do not have much experience solving such problems. I just read something about it because I consider it an exciting topic. After Torsten was able to clarify the question, I understood that your functional is:

E = int (P, 0, pi / w)

For this functional, the function V_L is searched, which gives the functional an extreme value (really maximum?). So I chose that approach and set that expression equal to zero. But due to lack of experience or missunderstanding, it is possible that I am wrong here.

In fact, if i use functionalDerivative with P, set this equal to zero and then isolate V_L, i get the result you have calculated by hand. So i think you are right to use f instead of F for the calculation. Another day when I'm a bit wiser in the evening than I got up in the morning... ;-)

Best regards

Stephan

Walter Roberson 2018-7-3

If the symbolic engine was generally working but you got that internal error, then I suspect you ran out of memory.

Joseph Liu 2018-7-4

Thanks guys!

@ Stephan Jung: everything got fixed! It is indeed an interesting problem to me. I used to solve optimal point instead of curve, surface or even higher dimension.

Another question to everyone: if I assume V_L is some constant (time invariant) that optimizes E. Does Euler-Lagrange Equation still hold? Because the solution is then V_L = I_s*R_p/2*sin(wt), which is time variant and contradictory to the assumption. At this point, we need to to do functional derivative on E instead of P. Any idea?

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David Goodmanson 2018-7-5

编辑：David Goodmanson 2018-7-7

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Hi Joseph,

( for constant V situation see [2] )

[1] ------------------

This is not an ordinary Euler-Lagrange situation. Often there would be a term involving (dV/dt)^2 so there is a high price to be paid for having a function with a large derivative. Here you only have a term proportional to V dV/dt. The resulting contribution to the functional depends only on the boundary values at t=0 and t=pi/w, and nothing more, i.e. not on the shape.

Boundary conditions (bc’s) specify boundary values, but if there are no bc’s the boundary values are free to come out like they come out.

If you do not specify bc's, then in general you are not justified in using Euler-Lagrange to find Fmax, as you did. Certainly that’s true for any functional that contains dV/dt as this one does.

As I mentioned in a previous comment on this thread, the original problem as stated does not have a maximum. For a function V(t) shown there, the functional can become arbitrarily large. That is very much related to the fact that you are not specifying bc’s for V at the end points. It does appear that you have the maximum F under restricted conditions though.

[2] ---------------------------------

The constant V problem does not specify bc's. But if you want to use a constant for V, just set V = constant and start integrating. Let

t1 = 0, t2 = pi/w, and I_s --> I, sin(wt) --> s, V_L --> V

The functional is

F = Int{t1,t2} [IsV -CVV' -V^2/R] dt

(here IsV means I_s*sin(wt)*V, hopefully this will not cause confusion)

In the functional, V' = 0 so the term involving C drops out. You can integrate the other two terms to get

F = IV(2/w) - (V^2/R)(pi/w)

which as a function of the scalar parameter V does have a maximum, which is

Fmax1 = (1/pi)(I^2R/w) = .32 (I^2R/w)          <---- Fmax1
with
V(t1) = V(t2) = IR/pi.                               boundary values

This is a perfectly good calculation. No Euler-Lagrange in this case. Boundary values came out like they came out. There is no functional derivative, merely minimizing with respect to a parameter.

[3] ----------------------------

It’s interesting to compare that maximum with the one from the original question, which was to minimize the functional F for an arbitrary function V, also with unspecified bc’s. This can be done without using Euler-Lagrange, and leads to the same result that you found,

V = IRsin(wt)/2

Plugging that into F, the result is

Fmax2 = (pi/8)(I^2R/w) = .39 (I^2R/w) <-- Fmax2

Which is a bit larger than Fmax1.

ADDED

[4] I don't understand what you mean by the statement " Also in the example, it practices on a function F(x,y,y') instead of y' '. "

The term that I said was quite common in the functional is proportional to (V ')^2, and the form of f is still f(x,y,y'). Nothing involves f ' '.

[5a] As far as maximizing the functional

F = Int{t1,t2} [IsV -CVV' -V^2/R] dt,

first consider the term involving C. Its integral is

 -(C/2)V^2 (evaluated at end points)
                = (C/2) [V(t1)^2 - V(t2)^2]     <-- the C term

The C term depends on the value of V at the end points, but not on the shape of V. This is an example of how important boundary values can be. The function V(t) I mentioned in a previous comment (I*R here replaces V0 there) is

V = N*I*R*exp(-N*w*t)

where N is a dimensionless number. For large N, this function is very large at t1 and much smaller at t2, so the C term makes a large contribution. The other two terms in F stay medium sized.

Since F can be arbitrarily large, it has no maximum. (There is no attempt here to maximize F under certain constraints. I only have to demonstrate that for some V(t), F can be arbitrarily large).

[5b]

For the original question, F can be maximized for the subclass of all possible V that meet the following boundary condition:

V(t1) = V(t2) condition 1

Then the C term goes away.

The strategy will be (a) maximize F for all possible functions V. (b) then show that the V that maximizes F obeys condition 1. This is the same as, but easier than, maximizing only those functions V that satisfy condition 1 from the get-go.

For the two remaining terms in F (that are not the C term), completing the square gives

F = Int{t1,t2} (1/R) [ (IR)^2(s^2/4) - (V-(IR/2)s)^2 ]

Since the second term in brackets comes in with a minus sign, it has to be minimized and since it is a squared term it must set to zero for all t. Then

V = IRsin(wt)/2

as you found. This has V(t1) = V(t2) = 0 and meets condition 1. The remaining (first) term gives

Fmax = (pi/8)(I^2R/w)

as previously demonstrated. No Euler-Lagrange in this process.

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Joseph Liu 2018-7-6

Thanks David for the detailed explanation,

[1]

According to example given in https://en.wikipedia.org/wiki/Euler%E2%80%93Lagrange_equation, my equation fits the form of L(t,q(t),q'(t)), which is P(t,V_L(t),V_L'(t)) in my case. Also in the example, it practices on a function F(x,y,y') instead of y''.

[3]

Would you mind to show how to achieve the time variant function without using Euler-Lagrange equation? Are we doing this, diff(int(P,t),V_L) = 0? I was stuck here. Any simple way to solve it?

David Goodmanson 2018-7-7

Hi Joseph, see additions to the answer

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Optimization with a differential term

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