A First Course on Complex Functions by G. Jameson

By G. Jameson

This ebook will be of curiosity to undergraduates taking classes on complicated features.

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Example text

1) produces y 1+y2− yy 2 1+y2 = c1 . Reordering and another integration yields x = c1 1 y 2 − c21 dy. Hyperbolic substitution enables the integration as x = c1 cosh−1 ( y ) + c2 . c1 Finally the solution curve generating the minimal surface of revolution between the two points is y = c1 cosh( x − c2 ), c1 where the integration constants are resolved with the boundary conditions as y0 = c1 cosh( x0 − c2 ), c1 44 Applied calculus of variations for engineers and y1 = c1 cosh( x1 − c2 ). 2 where the meridian curves are catenary curves.

Yn , y1 , y2 , . . , yn )dx with a pair of boundary conditions given for all functions: yi (x0 ) = yi,0 and yi (x1 ) = yi,1 for each i = 1, 2, . . , n. The alternative solutions are: Yi (x) = yi (x) + i ηi (x); i = 1, . . , n with all the arbitrary auxiliary functions obeying the conditions: ηi (x0 ) = ηi (x1 ) = 0. The variational problem becomes x1 I( 1 , . . , n) = f (x, . . , yi + i ηi , . . , yi + i ηi , . )dx, x0 whose derivative with respect to the auxiliary variables is ∂I = ∂ i x1 x0 ∂f dx = 0.

The simplest case is that of two independent variables, and this will be the vehicle to introduce the process. The problem is of the form y1 x1 I(z) = f (x, y, z, zx, zy )dxdy = extremum. y0 x0 Here the derivatives are zx = ∂z ∂x and ∂z . ∂y The alternative solution is also a function of two variables zy = Z(x, y) = z(x, y) + η(x, y). 40 Applied calculus of variations for engineers The now familiar process emerges as y1 x1 I( ) = f (x, y, Z, Zx , Zy )dxdy = extremum. y0 x0 The extremum is obtained via the derivative ∂I = ∂ y1 y0 x1 x0 ∂f dxdy.

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