Subject: Applied Mathematics 4

Topic: Calculus of Variation

Difficulty: Medium

We have to extremist,

$ I = \int_{0}^{1} xy+\frac{1}{2} (y')^2 \,\, dx $

where, $ F = (xy+\frac{1}{2} (y')^2) $

Assume the trial solution,

$ \bar{y}(x) = C_0 + C_1x + C_2x^2 $

By data, $ \bar{y} (0) = 0 \hspace{0.25cm}\therefore C_0 = 0 \\ \bar{y}(1) = 0; \hspace{0.25cm} C_1 + C_2 = 0 \hspace{0.25cm} \therefore C_2 = C_1 $

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

$ F = x \bar{y}(x) + \frac{1}{2}[\bar{y'}x]^2 \\ = x[C_1x(1-x)] + \frac{1}{2}[C_1(1-2x)]^2 \\ = C_1[x^2(1-x)] + \frac{1}{2}[C_1^2(1 - 4x + 4x^2)] \\ = C_1(x^2-x^3) + \frac{1}{2}[C_1^2(4x^2 - 4x - 1)] $

Now,

$ I = \int_0^1 [x \bar{y}(x) + \frac{1}{2}[\bar{y'}x]^2] \,\, dx \\ = \int_0^1 C_1(x^2-x^3) + \frac{1}{2}[C_1^2(4x^2 - 4x + 1)] \,\, dx \\ = C_1 \int_0^1 (x^2-x^3) + \frac{1}{2}[C_1(4x^2 - 4x + 1)] \,\, dx \\ = C_1 [(\frac{x^3}{3} - \frac{x^4}{4}) +\frac{1}{2} C_1 (\frac{4x^3}{3} - \frac{4x^2}{2} + x) ]_0^1 \\ = C_1 [(\frac{1}{3} - \frac{1}{4}) + \frac{1}{2} C_1 (\frac{4}{3} - 2 + 1)] \\ = C_1 [\frac{1}{12} + \frac{1}{2} C_1 (\frac{4}{3} - 1)] \\ = C_1 [\frac{1}{12} + \frac{1}{2} C_1 (\frac{1}{3})] \\ = \frac{C_1}{12} + \frac{C_1^2}{6} $

As we know, $ \frac{dI}{dC_1} = 0 $

$ \therefore \frac{1}{12} + \frac{2C_1}{6} = 0 \implies C_1 \frac{1}{3} = - \frac{1}{12} \\ \therefore C_1 = - \frac{1}{4} \\ \therefore \bar{y}(x) = - \frac{1}{4}x(1 - x) = \frac{1}{4}[x(x-1)]$