Page 529 - Fluid-Structure Interactions Slender Structure and Axial Flow (Volume 1)

SOME OF THE BASIC METHODS OF NONLINEAR DYNAMICS 499

where yc E ker L and ys E range L*, and to define two orthogonal operators, P onto the
range of L, and the complementary operator I - P, where I is the identity operator, such
that (Golubitsky & Schaeffer 1985)
(F. 84a)

(F.84b)
in which y is replaced by (F.83). We can then apply the implicit function theorem to find
ys in (F.84a) as a function of y, and u, since the mapping Pf(y, + ys, u,) acting onto the
range of L is regular. This means that it is possible to express ys as

Ys = h(Y,, 4). (F.85)

Equation (F.85) is valid only locally, in a neighbourhood of (0, u,) with tangency prop-
erties similar to (F.17), i.e.

=
hfy,, 4) 0, hdy,, u,) = 0, (F.86)
which means that h is at least second order in yc, h = 61~~)~. obtain the bifurcation
To
equations, we introduce (F.83) into (F.84b) and use (F.85) to find

(1 - P)f(Yc + We, uc), ue) = 0. (F.87)
Usually, h(y,, u,) cannot be found analytically in closed form, but it can be obtained using
Taylor series in terms of y, and p = u - u,, to any desired order. Note that to determine
(F.87) to a certain order n, we need to find h(y,, u,) only to order n - 1, since h is a
nonlinear function of y,. In fact, in many cases, e.g. under certain symmetry conditions,
it is not necessary to solve (F.84a) to find the bifurcation equation to order n.
To show the reader how such an operation can be carried out, the following simple
nondimensional system is considered (Troger & Steindl 1991):

(F.88)

It corresponds to the buckling of a rod under a compressive load u = P/EZ. We assume
that the rod is simply supported at both ends, i.e. $'(O) = @'(l) = 0, and we want to
find the bifurcation equation for the solution $ = 0. The linear operator L is defined by
taking the (FrCchet) derivative of (F.88) at $ = 0:

(F.89)

with boundary conditions ~'(0) = ~'(1) = 0. Solving the eigenvalue problem Lx = 0
leads to the well-known critical parameter u, = n2, with the corresponding eigenfunction
B cos(n6). Consequently, kerL = span{cos nt} and since L* = L, dimkerL =
dimkerL* = 1. The range of L is given by the function g(x) orthogonal to the range
of L*, i.e. Jd g(6) cos(nt)d( = 0.
We can now decompose the image space: to find the projection P onto the range of
L, we recall that for any function f(c), the projection (I - P)f((-) must be in kerL*,