where c 1 , …, c 4 are arbitrary constants of integration. Requiring this solution to
satisfy the boundary conditions (2.7), we obtain from y(0) = y′′(0) = 0 that c 2 =
c 4 = 0, and from y(l) = y′(l) = 0 that:
c 1 sinðc lÞ þ c 3 l ¼ 0
c 1 c cosðclÞ þ c 3 ¼ 0:
ð2:9Þ
The trivial solution c 1 = c 3 = 0 corresponds to the straight equilibrium y = 0.
For nontrivial solutions to exist the determinant of coefficients must vanish, that is,
sin(cl) − clcos(cl) = 0, or:
tan a ¼ a; a cl:
ð2:10Þ
This transcendent equation has an infinity of solutions, as appears from Fig. 2.2.
Using, e.g., Newton-Raphson iteration, the lowest three solutions turns out to be
a 1 = 4.4934095ÁÁÁ, a 2 = 7.7252518ÁÁÁ and a 3 = 10.904121ÁÁÁ. The eigenvalues k i are
then given by:
k i ¼ c
2
i ¼ a i =l
ð
Þ
2 ; i ¼ 1; 1:
ð2:11Þ
With c 2 = c 4 = 0 and c 3 = −c 1 ccos(c l) inserted into (2.8) the solution y
(x) becomes
yðxÞ ¼ c 1 sinðcxÞ À cx cosðclÞ
ð
Þ :
ð2:12Þ
Hence, for each eigenvalue k i an associated eigenfunction (buckling mode) exists
(see Fig. 2.3) :
y i ðxÞ ¼ c 1i sinð
ffiffiffiffi
k i
p
xÞ À
ffiffiffiffi
k i
p
x cosð
ffiffiffiffi
k i
p
lÞ
:
ð2:13Þ
For each eigenfunction y i (x) one constant c 1i remains undetermined. This reflects
the incapability of linear models to capture any post-buckling behavior. Linear
theory predicts buckling amplitudes to grow unbounded when k ! k 1 . As
amplitudes grow, however, nonlinearities come into play that will inevitably limit
Fig. 2.2 Locating solutions of tana = a
2.4 Stability-Related EVPs
55