I would like to know how can I put some conditions on a function $d(x)$ when solving this optimalization problem:
Given a continuous function $p:[a,b]\rightarrow\mathbb{R}$ on $[a,b]$ find a function $d:[a,b]\rightarrow\mathbb{R}_0^+$ where $d(a)=d(b)=0$ such that the function $d(x)$ minimizes the following functional: $$\int_a^b F(x,d,d')dx=\int_a^b\sqrt[]{1+[p'(x)+d'(x)]^2}dx$$ To find $d(x)$ I tried to apply the Euler-Lagrange equation: $$\frac{\partial F}{\partial d}-\frac{d}{dx}\frac{\partial F}{\partial d'}=0$$ Hence $$0-\frac{d}{dx}\frac{p'+d'}{\sqrt[]{1+[p'+d']^2}}=0$$ $$p'+d'=c\sqrt[]{1+[p'+d']^2}$$ $$(p'+d')^2=c^2+c^2(p'+d')^2$$ $$p'+d'=c_1$$ $$d(x)=c_1x+c_0-p(x)$$ Applying our boundary conditions we get: $$d(x)=\left(\frac{p(a)-p(b)}{a-b}\right)(x-a)+p(a)-p(x)$$ But becouse $d:[a,b]\rightarrow\mathbb{R}_0^+$ this can only be true if $p(x)\le\left(\frac{p(a)-p(b)}{a-b}\right)(x-a)+p(a)$ on $[a,b]$. My question is that if there is some more general solution with no further restrictions (also gives the correct answer if $p(x)\gt c_1x+c_0$)
Thanks for any kind of help with this problem.
In this answer, we'll give a geometric description of OP's variational problem.
Problem. There is given a continuous function profile $$[a,b]~\ni~ x~~\mapsto~~ p(x)~\in~\mathbb{R}.$$ We want to minimize the arclength of the continuous$^1$ function profile $$[a,b]~\ni~ x~~\mapsto~~ f(x)~\in~\mathbb{R}.$$ We are allowed to increase (but not decrease!) the function value $$ f(a)~=~p(a)~~\wedge~~f(b)~=~p(b)~~\wedge~~ \forall x~\in~]a,b[:~~f(x)~\geq~p(x) .$$
Solution: The optimal function profile $f_{\ast}$ is then given by (the boundary of) the convex upward hull of the $p$-profile.
Examples:
If $p$ is convex upward, then $f_{\ast}=p$.
If $p$ is concave upward, then $f_{\ast}$ is the straight line through the endpoints $(a,p(a))$ and $(b,p(b))$.
--
$^1$ Even though OP does not explicitly say so we assume that $f$ must be continuous. Else the problem is trivial: Just pick a constant majorant function $f$ on the open interval $]a,b[$.