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Determine the function {\hat  {y}}\in C^{2}[0,1]\, that minimizes the functional J(y)=\int _{0}^{1}\left[y'(x)\right]^{2}dx+[y(1)]^{2},y(0)=1,h(0)=0\,.

First, compute the first variation so that it can be set to zero:

\delta J(y,h)\, ={\frac  {d}{d\varepsilon }}J(y+\varepsilon h)\left.\right|_{{\varepsilon =0}}
={\frac  {d}{d\varepsilon }}\left[\int _{0}^{1}(y^{\prime }(x)+\varepsilon h^{\prime }(x))^{2}\ dx+(y(1)+\varepsilon h(1))^{2}\right]\left.\right|_{{\varepsilon =0}}
=\left[\int _{0}^{1}2(y^{\prime }(x)+\varepsilon h^{\prime }(x))h^{\prime }(x)\ dx+2(y(1)+\varepsilon h(1))h(1)\right]\left.\right|_{{\varepsilon =0}}
=\int _{0}^{1}2y^{\prime }(x)h^{\prime }(x)\ dx+2y(1)h(1)

To get this only in terms of h\,, integrate by parts. Let

u=2y^{\prime }(x) du=2y^{{\prime \prime }}(x)\ dx
dv=h^{\prime }(x)\ dx v=h(x)\,

Thus, setting the first variation to zero

2y^{\prime }(x)h(x)\left.\right|_{0}^{1}-\int _{0}^{1}2y^{{\prime \prime }}(x)h(x)\ dx+2y(1)h(1) =0\,
2y^{\prime }(1)h(1)-\int _{0}^{1}2y^{{\prime \prime }}(x)h(x)\ dx+2y(1)h(1) =0\,

Since h(1)\, is unspecified, suppose h(1)=0\,. Then

-2\int _{0}^{1}y^{{\prime \prime }}(x)h(x)\ dx =0\,

By the fundamental lemma, for all h(x)\,

y^{{\prime \prime }}(x) =0\,
y^{\prime }(x) =c_{1}\,
y(x)\, =c_{1}x+c_{2}\,

From the initial condition, y(0)=1\Longrightarrow c_{2}=1. Now, suppose h(1)=1\,. Then, the first variation gives

2y^{\prime }(1)\cdot 1-(0)+2y(1)\cdot 1 =0\,
2y^{\prime }(1)+2y(1) =0\,

Substituting in gives

2(c_{1})+2(c_{1}+c_{2})\, =0\,
4c_{1}+2(1)\, =0\,
c_{1}\, =-{\frac  {1}{2}}\,

Therefore, the function that minimizes the functional is {\hat  {y}}=-{\frac  {x}{2}}+1.


Main Page : Calculus of Variations