Solve an inequality involving a function of time

Good afternoon everyone, I am looking for solving an inequality involving a function of time as follow:

a * u(t) + b < alpha

with a, b and alpha are constant variables; and u(t) a function of time t

I want to find the set of u(t) that respect the above inequality.

In order to solve this problem, I run the following script:

 syms t a b alpha;
u(t) = sym('u(t)');
eqn = a*u(t) + b < alpha;
eqn = rewrite(eqn,'log');
[solx, param, cond] = solve(eqn, u(t), 'ReturnConditions',1);
clc
disp('solx')
pretty(solx)
disp('param')
pretty(param)
disp('cond')

Unfortunately, it does not solve the inequality by conducting to an empty structure (solx, param and cond). I tried to replace u(t) by t in order to follow a similar code provided by the Mathworks help solve function and it works well even it is not my problem ... Many thanks for your help,

Regards,

Mathias

Answer 1

[Walter Roberson]

Is there a particular reason you are doing the rewrite to log?

With current versions you cannot use sym('u(t)') and should instead use

syms u(t) a b alpha

Are you expecting an answer different from u(t) < (alpha - b)/a ?

MATLAB is sometimes weak about reasoning about functions, and weak about reasoning about inequalities. The work-around is to convert into an equality and use isolate()

syms u(t) a b alpha
syms du positive
eqn = a*u(t) + b + du == alpha
solu = expand(isolate(eqn, u(t)))

This gives

u(t) == alpha/a - b/a - du/a

Then, knowing that du is an arbitrary positive quantity, then in the case where a is positive, you can interpret this as

u(t) < alpha/a - b/a

but if a is negative you need to change to >

[mathias dupont]

Hi,

Thanks for your answer.

1) I use the rewrite to log function because I have seen in this forum it as a trick to help Matlab to solve this problem ... maybe it is not useful.

2) Actually no I am not expected another solution. However, I would like able to specify another more complex contraints as alpha*t. As a result, I am expected a solution involving time t. I guess it will work with this solution right ?

3) I don't really understand why do you use "du" ?

Thanks,

Mathias

[Walter Roberson]

For finite real values, the formula

 A < B

can only be true if there is some positive C such that

 A + C = B

or in other words

A = B - C

for some positive C.

Here I called C "du" for lack of better name.

 a*u(t) + b + du == alpha

and

a*u(t) + b == alpha - du

are equivalent for finite real values. du is the amount by which alpha exceeds a*u(t) + b since alpha is known to be greater than a*u(t) + b

When reasoning about inequalities, it can often help a lot to introduce auxiliary variables with assumptions on them that are used to express the exact amount of the difference. Especially when you have multiple inequalities in a system: then the requirements that the auxiliary variables be positive (or non-negative in the case if <= or >= ) can generally be used to resolve to intervals where the system is true.

[mathias dupont]

Ok thanks for your explanations. To complete them, I recommend for future users the link as follow. Course

I have another question because I guess I do not use the right function to solve my problem ... If I add now new constraints as follow

a*u'(t) < B

with a, B constant variables and u'(t) the derivative function of u(t) according to the time t.

Which function I need to use according to you to solve this problem please ?

Thanks for your help,

Mathias

[Walter Roberson]

syms u(t) a b B alpha
du = diff(u);
syms delta1 delta2 positive  %inbalance in inequalities
syms C      %arbitrary constant of integration to balance integral of du
eqn = a*u(t) + b == alpha - delta1;
eqn2 = a*du == B - delta2;
sol1 = expand(isolate(eqn, u(t)));
sol2 = expand(isolate(eqn2, u(t)));
sol3 = int(lhs(sol2),t) == int(rhs(sol2),t) + C;

At this point, sol1 and sol3 are both expressions for u(t). Now what?

>> expand(isolate(rhs(sol1)==rhs(sol3),delta1))
ans =
delta1 == alpha - b - C*a - B*t + delta2*t

delta1 is required to be positive. alpha - b - C*a are constants and are whatever they are. We can isolate down to delta1 = D + (delta2-B)*t for constant D = alpha - b - C*a . If positive and negative times are permitted, then with delta2 restricted to positive, then if delta2 > B then (delta2-B) would be positive, but there will be a sufficiently negative time at which D + (delta2-B)*t will go negative. That puts a constraint on delta1 for this case. If delta2 < B then (delta2-B) would be negative, and there would be a sufficiently positive time at which D + (delta2-B)*t would go negative; that puts a constraint on delta1 for this case. Let t = 0 to get a baseline delta1.

... I don't know where to go from here. This isn't really talking about the possible forms for u(t), this is talking about the inequalities.

Perhaps it would be more useful to consider delta1(t) and delta2(t) -- to look at how the inequalities would change over time. So for example instead of

u(t) == alpha/a - b/a - delta1/a

to work with

u(t) == alpha/a - b/a - delta1(t)/a

??

But I do not know what you are looking for?

Where you hoping for something like MATLAB solving and saying "Oh, to get that to work, u(t) has to be this form involving exp(heaviside(a - t/B))" ??