- #1

skrat

- 748

- 8

## Homework Statement

Determine minimum value of integral ##B(y)=\int_{0}^{2}({y}')^2dx## for function ##y\in C^1(\mathbb{R})## and ##y(0)=y(2)=0## and ##\int_{0}^{2}y^2dx=4##

## Homework Equations

## The Attempt at a Solution

IF I am not mistaken, the idea is to first find the function ##y## for given conditions and than calculate ##B(y)##.

My idea:

Since ##y\in C^1(\mathbb{R})## I simply started with polynomial ##y(x)=ax^2+bx+c##. Of course ##y(0)=0## so immediately ##c=0## than ##y(2)=0=4a+2b=0## so ##b=-2a##.

Third condition says ##\int_{0}^{2}y^2dx=\int_{0}^{2}(ax^2+bx)^2dx=4##

##\frac{32a^2}{5}+\frac{ab16}{2}+8b^2=4## using ##b=-2a## gives me ##a=\sqrt{\frac{5}{28}}## and ##b=-\sqrt{\frac{5}{7}}##.

So using ##y(x)=\sqrt{\frac{5}{28}}x^2-\sqrt{\frac{5}{7}}x## the ##B(y)=\int_{0}^{2}({y}')^2dx=\frac{5}{3}##.

Or is it not?