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N sin αn − 2 cos αn Choosing the initial value α1 = 2 we get α2 = 2, 029048 and α3 = 2, 028758 = α4 , hence α = 2, 028758 · 1 L where λ = α2 = 4, 115858 · 1 . L2 A generating eigenfunction is y1 (x) = sin 2, 028758 · x . 14 Consider an axle which is simply supported at its endpoints x = 0 and x = L. The axle is rotating with the constant angular speed ω. For some values of ω, called the critical angular speeds, the axle may rotate in a bent form. The model equation for small bendings of the rotating axle is (5) EI d4 u − ω 2 u = 0, dx4 x ∈ [0, L], where E is the elasticity module of the axle, I is the moment of inertia, and Given the boundary conditions is the mass per length.

2 = c1 + The only solution is c1 = c2 = 0, so λ = 0 cannot be an eigenvalue either. If λ = k 2 > 0, k > 0, then the complete solution is y = c1 cos(kx) + c2 sin(kx) where y = kc2 cos(kx) − kc1 sin(kx). We get by insertion into the boundary conditions that y(0) + y (0) = c1 + kc2 = 0, y 1 2 = cos k 2 c1 + sin k 2 c2 = 0. This system has proper solutions (c1 , c2 ) = (0, 0), if and only if the corresponding determinant is zero, thus 0= 1 k k cos 2 k sin 2 = sin k 2 − k · cos k 2 , k > 0. 5 3 x Since cos tan k = 0 for every solution, this condition is equivalent to the equation 2 k = k.

NNE Pharmaplan is a company in the Novo Group. com 33 Examples of Eigenvalue Problems Eigenvalue problems 3) Let λ = k 2 > 0, k > 0. • The complete solution of d2 y + k2 y = 0 dx2 is y(x) = c1 cos(kx) + c2 sin(kx) where y (x) = −c1 k sin(kx) + c2 k cos(kx). • Insertion into the boundary conditions. It follows from the ﬁrst boundary condition that y(0) = 0 = c1 , so we must necessarily have y(x) = c2 sin(kx) where y (x) = c2 k cos(kx). It follows from the second boundary condition y(1) = y (1) that c2 sin(k) = c2 k cos(k).