Calculus of variation with inequality constraints

Question

Find the function $y$ which maximizes the functional $$J[y] = \int_0^1 g(x) y(x) dx$$ subject to $0 \leq y(x) \leq 1$ for all $x\in [0,1]$ and $$\int_0^1 y(x) dx = k$$ where $g$ is a strictly increasing function.

I know that I can take care of the isoperimetric constraint quite easily using the Lagrangian

$$K[y] = \int_0^1 (g(x) y(x) + \lambda y(x)) dx$$

I also know that I can take care of constraints of the form $y(x) \leq 1$ using a substitution such as $u^2(x) = 1 - y(x)\geq 0$ to get

$$K[u] = \int_0^1 \left( g(x) \left( 1 - u^2(x) \right) + \lambda \left( 1 - u^2(x) \right) \right) dx$$

However, I am quite at a loss with a constraint of the form $0 \leq y(x) \leq 1$, i.e., when two inequalities are involved at the same time. How can I take care of this?

why not taking $y = z^2$ For $y \geq 0$ ? then you have $u^2 = 1- z^2$ for the $y \leq 1$ — Red shoes, Sep 28 '21 at 04:54

score 5 · Accepted Answer · edited Jul 09 '18 at 14:54

5

Suppose that on two infinitesimal intervals $A = [a, a+dx]$ and $B = [b, b+dx]$, we have $a < b$, $y > \epsilon$ on $A$ and $y< 1 - \epsilon$ on $B$. Let $C = [0,1] - A - B$. Since $g$ is strictly increasing, $\int_A g(x) \, dx \le \int_B g(x) \, dx$. Define $$y_1(x) =\begin{cases} y(x) - \epsilon & x\in A\\y(x) + \epsilon & x\in B\\y(x) & x\in C\end{cases}$$

Then, $$\begin{align}\int_0^1 g(x)y_1(x) \, dx &= \int_A g(x)y_1(x) \, dx + \int_B g(x)y_1(x) \, dx +\int_C g(x)y_1(x) \, dx\\ &= \int_A g(x)(y(x)-\epsilon) \, dx + \int_B g(x)(y(x)+\epsilon) \, dx +\int_C g(x)y(x) \, dx\\ &= \int_0^1 g(x)y(x) \, dx + \epsilon\left(\int_B g(x) \, dx - \int_Ag(x) \, dx\right)\\ & \ge \int_0^1 g(x)y(x) \, dx\end{align}$$ Thus, raising the value of $y$ for higher values of $x$ at the expense of the value of $y$ for lower values of $x$ increases $J$. Taking this to the extreme while still satisfying $0 \le y \le 1$ and $\int_0^1 y \, dx = k$ gives $$y(x)=\begin{cases} 0 & x < 1-k\\1&x \ge 1 - k\end{cases}.$$

edited Jul 09 '18 at 14:54

Jon Warneke

5,017

answered Jan 15 '16 at 21:13

Paul Sinclair

45,932

1

That's great and I see why it works in this case. However, I'd still be interested in solving this kind of problem using calculus of variation as described above. Any advice there? – sami Jan 20 '16 at 18:48
3

@shomi - Late, but I'll provide an answer anyway. First, this actually is a calculus of variations argument - just not the form you were looking for. As for that, you could have set $u^2 = y(1-y)$, which is only non-negative for $y\in[0,1]$. – Paul Sinclair Mar 30 '16 at 18:31
1

Thanks a lot for following up! That's such a simple variation of what I did for $y(x) \leq 1$, it's embarrassing I did not see it. – sami Mar 30 '16 at 19:16
2

@PaulSinclair I tried doing this problem using the methods outlined in the question and your suggested substitution $u^2 = y(1-y)$, but I don't get anything useful from the Euler-Lagrange equation in that case. I must be doing something wrong -- could you walk me through how to arrive at your solution via Euler-Lagrange and your substitution? – Jon Warneke Jul 16 '18 at 15:06

score 4 · Answer 2 · edited Dec 19 '17 at 15:02

4

I think its better to write your constraint in this form: $0 \leq y(x) \leq 1 \implies -\frac{1}{2} \leq y(x)-\frac{1}{2} \leq \frac{1}{2} \implies \left( y(x)-\frac{1}{2} \right)^2 \leq \frac{1}{4} \implies y(x)^2-y(x)<0$. Now you can use your knowledge to solve the problem. Good luck.

edited Dec 19 '17 at 15:02

Jon Warneke

5,017

answered May 24 '16 at 08:00

Mostafa

121

Calculus of variation with inequality constraints

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