„Analízis (MSc) típusfeladatok” változatai közötti eltérés
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<big>3)</big> <small>[2016V1]</small> Keressük meg az extremális függvényt az <math>I(y) = \int_0^1 y(2-y') dx,~y(0) = 1,~ y(1) = 2</math> operátorra vonatkozóan a <math>J(y) = \int_0^1 y'^2 = \frac{13}{3}</math> feltétel mellett! | |||
{{Rejtett | |||
|mutatott=Megoldás: | |||
|szöveg= | |||
<math>F = y(2-y') - \lambda y'^2</math> | |||
Erre alkalmazzuk az Euler-Lagrange egyenletet: | |||
<math>2-y' - \frac{d}{dx}(-y - 2\lambda y') = 2-y' + y' + 2\lambda y'' = 2 + 2\lambda y'' = 0</math> | |||
<math>y'' = \frac{-1}{\lambda}</math> | |||
<math>\frac{dy'}{dx} = \frac{-1}{\lambda}</math> | |||
<math>\int dy' = \int \frac{-1}{\lambda} dx</math> | |||
<math>y' = \frac{-x}{\lambda} + c_1</math> | |||
<math>\frac{dy}{dx} = \frac{-x}{\lambda} + c_1</math> | |||
<math>\int dy = \int \frac{-x}{\lambda} + c_1 dx</math> | |||
<math>y = \frac{-x^2}{2 \lambda} + c_1 x + c_2</math> | |||
Használjuk fel a kezdeti feltételeket! | |||
<math>y(0) = c_2 = 1</math> | |||
<math>y(1) = \frac{-1}{2 \lambda} + c_1 + 1 = 2</math> | |||
<math>c1 = 1 + \frac{1}{2 \lambda}</math> | |||
A <math>\lambda</math>-hoz ki kell számolni J(y)-t. | |||
<math>y = \frac{-x^2}{2 \lambda} + x + \frac{x}{2 \lambda} + 1</math> | |||
<math>y' = \frac{-x}{\lambda} + 1 + \frac{1}{2 \lambda}</math> | |||
<math>y'^2 = \frac{x^2}{\lambda^2} - \frac{2x}{\lambda} + 1 - \frac{2x}{2\lambda^2} + \frac{2}{2\lambda} + \frac{1}{4 \lambda^2} = \frac{1}{\lambda^2} \left( x^2 - 2x\lambda + \lambda^2 - x + \lambda + \frac{1}{4} \right)</math> | |||
<math>\int_0^1 y'^2 = \frac{1}{\lambda^2} \left[ \frac{x^3}{3} - \lambda x^2 + \lambda^2 x - \frac{x^2}{2} + \lambda x + \frac{x}{4} \right]_0^1 = \frac{1}{\lambda^2} \left( \frac{1}{3} - \lambda + \lambda^2 - \frac{1}{2} + \lambda + \frac{1}{4} \right) = 1 + \frac{1}{12\lambda^2} = \frac{13}{3}</math> | |||
<math>\lambda^2 = \frac{3}{120} = \frac{1}{40}</math> | |||
<math>\lambda = \pm \frac{1}{\sqrt{40}}</math> | |||
Visszaírva y-ba: | |||
<math>y(x) = \mp \sqrt{10} x^2 + (1\pm\sqrt{10}) x + 1</math> | |||
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