sol4 - MAE294B/SIO203B: Methods in Applied Mechanics Winter...

Info iconThis preview shows pages 1–3. Sign up to view the full content.

View Full Document Right Arrow Icon
MAE294B/SIO203B: Methods in Applied Mechanics Winter Quarter 2010 http://maecourses.ucsd.edu/mae294b Solution IV 1 The leading-order solution y 0 satisfies y 0 0 + y 0 x - 1 xy 0 = 0 . This can be solved by separation of variables to yield y 0 = ± 1 + Ax - 2 . For the two boundary conditions here, the solutions are y 0 = 1 and y 0 = 1 + 3 x - 2 , so in both cases the limit for large x is 1. For small x , from the equation one cannot neglect ε with respect to x when x = O ( ε ) . However the magnitude of y may also change, as it clearly does in case 2, so care is needed. The general rescaling x = ε α X , y = ε β Y gives Y X + Y X + ε 1 - α X - X ( X + ε 1 - α ) 2 ε - 2 β Y = 0 . To find the appropriate exponents, we will need the O ( ε ) equation for y = O ( 1 ) : y 0 1 + ± 1 + 1 y 2 0 ² y 1 x = y 0 x 2 - 2 x 2 y 0 , y 1 ( 1 ) = 0 . For the first case, we find y 1 = x - 2 - x - 1 . Hence y = 1 + ε ( x - 2 - x - 1 )+ ··· . The expansion becomes disordered when ξ = ε - 1 / 2 x = O ( 1 ) , at which point y = O ( 1 ) . That is, we need to solve the case with α = 1 / 2 and β = 0, which to leading order gives y ξ + y ξ - 1 ξ y = 0 . We have solved this before and the answer is y = p 1 + B ξ - 2 . Van Dyke’s rule gives H 0 E 1 y = H 0 [ 1 + ε ( x - 2 - x - 1 )] = H 0 [ 1 + ξ - 2 - ε 1 / 2 ξ - 1 ] = 1 + ξ - 2 , E 1 H 0 y = E 1 p 1 + B ε x - 2 = 1 + 1 2 B ε x - 2 . Hence B = 2 and y = p 1 + 2 ξ - 2 . This is still not enough because this solution is no longer valid for X = ε - 1 x = O ( 1 ) . Then, from what we have just found, y = O ( ε - 1 / 2 ) and we have α = 1 and β = - 1 / 2. The governing equation becomes, to leading order and using Y because the scaling has changed, Y X + Y X + 1 = 0 , 1
Background image of page 1

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
which has the solution Y = C ( X + 1 ) - 1 . Match with the solution in the ξ -region (noting that ξ = ε 1 / 2 X ): H 0 E 0 y = H 0 [ p 1 + 2 ε - 1 X - 2 ] = 2 ε - 1 / 2 X - 1 , E 0 H 0 y = E 0 [ ε - 1 / 2 C ( ε - 1 / 2 ξ + 1 ) - 1 ] = C ξ - 1 , so C = 2 and the solution becomes y = p 2 / ε ( X + 1 ) - 1 . Hence y ( 0 ) = p 2 / ε . For the second case, the governing equation for y 1 is less tractable, but shows that y 1 x - 2 again for small x . Hence the variable ξ is again relevant and α = 1 / 2, but this time y = O ( x - 1 ) for small x and so β = - α = - 1 / 2. Then the governing equation becomes, in terms of Y = ε
Background image of page 2
Image of page 3
This is the end of the preview. Sign up to access the rest of the document.

This note was uploaded on 03/16/2010 for the course MAE 294b taught by Professor Young,w during the Winter '08 term at UCSD.

Page1 / 7

sol4 - MAE294B/SIO203B: Methods in Applied Mechanics Winter...

This preview shows document pages 1 - 3. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online