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Constaint problem: Minimize T(y)=\int_0^1\left(y'^2+x^2\right)\,dx\, s.t. K(y)=\int_0^1y^2\,dx=2\,.

Using the method of Langrange multipliers,

\int_0^1\left(y'^2+x^2+\lambda y^2\right)\,dx = J(y)\,

If the Euler-Lagrange equations are satisfied then this integral has an extremum.

\frac{\partial}{\partial y}\left[y'^2+x^2+\lambda  y^2\right]=\frac{d}{dx}\frac{\partial}{\partial y'}\left[y'^2+x^2+\lambda  y^2\right]\,

2\lambda y = 2y''\,

y''=\lambda y\, so y(x) = c_1\cos\sqrt{\lambda}x + c_2\sin\sqrt{\lambda}x\,

y(0)=0\, gives c_1=0\,

y(x)=c_2\sin\sqrt{\lambda}x\,

y(1)=0\, gives 0=c_2\sin\sqrt{\lambda}\, so \sqrt{\lambda}=n\pi, n=1,2,...\,

The solution is

y(x)=c_2 \sin n\pi x\,

2=\int_0^1y^2\,dx = c_2^2\int_0^1\sin^2(n\pi x)\,dx\,

2=\frac{c_2^2}{2}\int_0^1(1-\cos 2n\pi x)\,dx = \frac{c_2^2}{2}\,

c_2=\pm 2\,

Calculus of Variations

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