Jtdylktuy

The following problem is derived from a Finite element problem.

We have a beam which is subject tot a constant axial force. With out loss of generality the beam can have a variable section area $A(x)$. The beam cannot exceed a certain strain value, that is, it cannot be shorten by more than a given percentage. We want the beam to be with the smallest possible volume.

I have no idea how to approach the following optimization problem. Let $Volume=int_a^bA(x)dx$. The constraint is given by $int_a^b frac{1}{A(x)}dxleq const$.

Hence:
$$min_{A} {Volume} \S.T. int_a^b frac{1}{A(x)}dxleq const$$

• How do I approach such a problem if at all possible?


• If not, is there a way to solve this, given the relaxation of $A(x)$ being a polynomial function lower than the 4th (inertia momentum) degree?

This is not a general approach, but it will lead you to the final answer quickly.

Let's say $int_a^b frac{dx}{A} le K$ for some given constant $K$, then by Cauchy Schwarz

$$K int_a^b A(x) dx ge int_a^b frac{dx}{A(x)} int_a^b A(x) dx stackrel{CS}{ge} left(int_a^b dxright)^2 = (b-a)^2$$
This leads to following lower bound for the integral $$int_a^b A(x) dx ge frac{(b-a)^2}{K}$$
Since this lower bound is achieved by the constant function $A(x) = frac{b-a}{K}$, the minimum volume is $frac{(b-a)^2}{K}$.

For a general approach, you need to set this up as a calculus of variation problem with constraint.
The constrainted Lagrange has the form

$$mathcal{L}(x) = A(x) + lambdaleft(frac{1}{A(x)} - kright)$$
for unknown constant $lambda$ and $k$.

You proceed to solve the associated Euler-Lagrange equation

$$frac{partial mathcal{L}}{partial A} - frac{d}{dx}left(frac{partial mathcal{L}}{partial A'}right) = 0$$
In this case, the EL-equation tell us

$$1 - frac{lambda}{A(x)^2} = 0$$
This means the solution which minimize the functional $int_a^b A(x) dx$ subject to constraint $int_a^b left( frac{1}{A(x)} - kright) dx = 0$ is probably a constant function. Finally, you need to verify by suitable choice of constant value of $A(x)$, you will get the actual minimum.

To make this precise, it probably take one or two chapters from a book. I will just stop here.

First I will show that any minimum value of the integral you are considering can be attained by a constant function $A(x).$ Then, once you assume $A(x)$ is constant, the answer falls into your lap.

For the sake of simplicity, I will assume that we are integrating over the interval $[0,1]$ instead of $[a,b]$. I will also assume that $A(x)le 1$ for all $xin [0,1]$, which I think is a reasonable assumption because "beams" are typically longer than they are thick.

So, to restate the question, we want to find a function $A(x)$ that minimizes
$$I=int_0^1 A(x)dx$$
with the constraint that
$$int_0^1 frac{dx}{A(x)}le C$$
where $C$ is some given constant.

To solve this problem, we will make use of the following fact, which can be verified using integration by substitution: if $f$ is an integrable function on $[0,1]$, then
$$int_0^1 f(x)dx=int_0^1 frac{f(frac{x}{2})+f(frac{x+1}{2})}{2}dx$$
Now suppose that $A_m(x)$ is a "well-behaved" function that minimizes $I$. Then, by the above identity, $I$ takes on the same value if we use the function
$$frac{A_m(frac{x}{2})+A_m(frac{x+1}{2})}{2}$$
instead. Further, notice that
$$A_m(x)le 1 implies frac{2}{A_m(frac{x}{2})+A_m(frac{x+1}{2})}le frac{1}{2} bigg(frac{1}{A_m(frac{x}{2})}+frac{1}{A_m(frac{x+1}{2})}bigg)$$
Which implies that
$$int_0^1 frac{2dx}{A_m(frac{x}{2})+A_m(frac{x+1}{2})}le int_0^1 frac{dx}{A_m(x)}le C$$
This is all that we need! This demonstrates that if $A_m(x)$ is a "minimizing function," then the function
$$frac{A_m(frac{x}{2})+A_m(frac{x+1}{2})}{2}$$
is also a "minimizing function." If we repeat the same process with this function, we see that
$$frac{A_m(frac{x}{4})+A_m(frac{x+1}{4})+A_m(frac{x+2}{4})+A_m(frac{x+3}{4})}{4}$$
is also a "minimizing function." In fact, repeating this process infinitely many times shows that the constant function function
$$int_0^1 A_m(t)dt$$
is also a "minimizing function." This proves that for any function minimizing the given integral, there exists a constant function that also minimizes it; for this reason, you may assume that your function $A(x)$ is constant.

So let $A(x)=A$ be constant. Then we have
$$int_0^1 frac{dx}{A}=frac{1}{A}le C$$
and so $Age 1/C$, implying that $Ige 1/C$ and $I=1/C$ is the minimum value of your integral. Done!