I am trying to solve an optimization problem: $$\begin{aligned}&\max_{G(v)}\int_p^1 \frac{G(v)}{v} dv\\ s.t.& \int_0^1(1+\log u)G'(u)du=-\overline v<0\\ &\int_0^x(1-\frac{G(v)}{v})dv\leq\int_0^xH(v)dv, \text{ for any } x\in[0,1]\\ &\frac{G(v)-v*G'(v)}{v^2}>0, \text{ for any }v\in(0,1]\\ &G(1)=0 \text{ and } G'(p)=0, \end{aligned} $$ where $p\in[0,1]$ and $H(v)$ is a weakly increasing function on $[0,1]$ with $H(0)=0$ and $H(1)=1$.
(Actually, this problem is a second version of the problem in my another question Functional Extremum Problem with inhomgeneous boundary condition, just simply let $G(v)=v(1-F(v))$ to avoid inhomgeneous boundary condition.)
Now with the constraint I describe above, the new objective function should be $$\begin{aligned}&\int_p^1 \left[\frac{G(v)}{v}+\lambda(1+\log v)G'(v)+\mu(v)\frac{G(v)-v*G'(v)}{v^2}\right]dv\\+&\int_0^1\gamma(x)\int_0^x(1-\frac{G(v)}{v})dvdx\\ =&\int_p^1 \left[\frac{G(v)}{v}+\lambda(1+\log v)G'(v)+\mu(v)\frac{G(v)-v*G'(v)}{v^2}+\int_v^1\gamma(x)dx\times (1-\frac{G(v)}{v})\right]dv, \end{aligned}$$ where $\lambda,\gamma(\cdot),\mu(\cdot)$ are the Lagrange Multiplier (Function), and the equality is from the exchange of integral sequence.
By the Calculus of Variations, the marginal condition is $$\frac{1}{v}+\lambda\cdot(-\frac{1}{v})+\int_v^1\gamma(x)dx\cdot(-\frac{1}{v})+\left[\mu(v)\cdot(\frac{1}{v^2}-\frac{1}{v^2})+\mu'(v)\cdot\frac{1}{v}\right]=0,$$ or, $$1-\lambda-\int_v^1\gamma(x)dx+\mu'(v)=0.$$ Neither $G(v)$ nor $G'(v)$ disappears in the condition! I know that this is the similar case when we use Lagrange Multiplier in linear programming problem. But in linear programming, there is other ways. In my case, I don't know how to solve this problem at all without the Lagrange method.