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I need to find the maximum mean value of the function
F=\int\left|\alpha^{''}(f)\right|^{2}df
I thought i could get an answer finding a constant K to normalize the function so as
K\int\left|\alpha^{''}(f)\right|^{2}df=1
The boundaries could be f_{1} and f_{2}
as \alpha=1-\left|R\right|^{2} i tried to find first the second derivative like this:
\frac{d\alpha}{df}=-2\left|R\right|\frac{d\left|R\right|}{df}
\frac{d^{2}\alpha}{df}=-2\left[\frac{d\left|R\right|}{df}*\frac{d\left|R\right|}{df}+\left|R\right|\frac{d^{2}\left|R\right|}{df}\right]
\frac{d^{2}\alpha}{df}=-2\left[\left|\frac{d\left|R\right|}{df}\right|^{2}+\left|R\right|\frac{d^{2}\left|R\right|}{df}\right]
but know, replacing that into the integral seems harder.
so how can i find the normalization constant and solve the integral?
F=\int\left|\alpha^{''}(f)\right|^{2}df
I thought i could get an answer finding a constant K to normalize the function so as
K\int\left|\alpha^{''}(f)\right|^{2}df=1
The boundaries could be f_{1} and f_{2}
as \alpha=1-\left|R\right|^{2} i tried to find first the second derivative like this:
\frac{d\alpha}{df}=-2\left|R\right|\frac{d\left|R\right|}{df}
\frac{d^{2}\alpha}{df}=-2\left[\frac{d\left|R\right|}{df}*\frac{d\left|R\right|}{df}+\left|R\right|\frac{d^{2}\left|R\right|}{df}\right]
\frac{d^{2}\alpha}{df}=-2\left[\left|\frac{d\left|R\right|}{df}\right|^{2}+\left|R\right|\frac{d^{2}\left|R\right|}{df}\right]
but know, replacing that into the integral seems harder.
so how can i find the normalization constant and solve the integral?