Subject: Applied Mathematics 4

Topic: Calculus of Variation

Difficulty: Medium

Consider $ I = \int (2xy - y^2 - (y')^2) \,\, dx $, y(0) = 0, y(1) = 0

Here, $ \bar{F} = 2xy - y^2 - (y')^2 $

We have to extremist the integral I

$ \bar{y}(x) = C_0 + C_1x + C_2x^2 \\ y(0) = 0; C_0 = 0 \\ y(1) = 0; C_1 + C_2 = 0; C_2 = -C_1 \\ \bar{y}(x) = C_1x + C_2x^2 = C_1x - C_1x^2 = C_1(x)(1-x) \\ \bar{y'}(x) = C_1 + 2C_2x = C_1 - 2C_1x = C_1(1 - 2x) $

Putting the values in I,

$ I = \int_0^1[2x \bar{y} - (\bar{y})^2 - (\bar{y'})^2] \,\, dx \\ = \int_0^1 [2x[C_1(x)(1-x)] - C_1^2(x)^2(1-x)^2 - [ C_1(1 - 2x)]^2] \,\, dx \\ = \int_0^1 [2C_1 x^2 (1 - x) - C_1^2(x)^2(1 - x)^2 - C_1^2 (1 - 2x)^2] \,\, dx \\ = \int_0^1 [2C_1(x^2 - x^3) - C_1^2(x)^2(1 -2x + x^2) - C_1(1 - 4x + 4x^2)] \,\, dx \\ = \int_0^1 [2C_1(x^2 - x^3) - C_1^2[x^2(1 - 2x + x^2) + (1 - 4x + 4x^2)]] \,\, dx \\ = \int_0^1 [2C_1(x^2 - x^3) - C_1^2[x^2 - 2x^3 + x^4 + 1 - 4x + 4x^2]] \,\, dx \\ = \int_0^1 [2C_1(x^2 - x^3) - C_1^2[x^4 - 2x^3 + 5x^2 - 4x + 1]] \,\, dx \\ = C_1 \int_0^1 [2(x^2 - x^3) - C_1[x^4 - 2x^3 + 5x^2 - 4x + 1]] \,\, dx \\ = C_1 [2[\frac{x^3}{3} - \frac{x^4}{4}] - C_1[\frac{x^5}{5} - 2 \frac{x^4}{4} + 5 \frac{x^3}{3} -4 \frac{x^2}{2} + x]]_0^1 \\ = C_1 [2[\frac{1}{3} - \frac{1}{4}] - C_1[\frac{1}{5} - \frac{1}{2} + \frac{5}{3} - 2 + 1]] \\ = C_1 [2[\frac{1}{12} - C_1[\frac{28}{15} - \frac{5}{2} + 1]] \\ = C_1 [\frac{1}{6} - C_1 \frac{56 - 75 + 30}{30}] \\ = C_1 [\frac{1}{6} - C_1[\frac{11}{30}]] = C_1[\frac{1}{6} - \frac{11 C_1}{30}] \\ = [\frac{C_1}{6} - \frac{11 C_1^2}{30}] \\ \frac{dI}{dC_1} = [\frac{1}{6} - \frac{22 C_1}{30}] $

For stationary values,

$ \frac{dI}{dC_1} = 0 \\ [\frac{1}{6} - \frac{22 C_1}{30}] = 0 \\ \therefore C_1 = \frac{5}{22} \\ \bar{y} (x) = \frac{5}{22} (x)(1-x) $