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2) Prove that there does not exist any eigenvalue λ < 1. 3) Prove that the n-th positive eigenvalue is λn = n2 π 2 + 1, and ﬁnd a corresponding eigenfunction. 1) Let λ = 1. Then the characteristic equation is R2 + 2R + 1 = (R + 1)2 = 0. • Since R = −1 is a double root, the complete solution is y(x) = c1 xe−x + c2 e−x . com 35 Examples of Eigenvalue Problems Eigenvalue problems • Insertion into the boundary conditions: It follows from the ﬁrst boundary condition that y(0) = 0 = c2 , hence we shall only look for candidates of the structure y = c1 xe−x .

Since c1 = 0, it follows from the latter boundary condition that u (1) = 0 + c2 Ω2 (1 − r) · 1 = c2 Ω2 (1 − r) = 0. Since 1 − r > 0 and Ω > 0, we must have c2 = 0 and thus c3 = 0, and we only get the zero solution, so we have no eigenvalue when 0 < r < 1. 20 Given the diﬀerential equation y + 2λy + 2λ2 y = 0, 0 ≤ x ≤ π, with y(0) − y (0) = 0 and y(π) − y (π) = 0, and where the parameter λ ∈ R is considered as a possible eigenvalue. 1) Prove that λ = 0 is not an eigenvalue. 2) Find all the eigenvalues and the corresponding eigenfunctions.

17 Consider the eigenvalue problem y + λy = 0, x ∈ [−π, π], y(−π) = y(π), y (−π) = y (−π). 1) Prove that λ = 0 is an eigenvalue and ﬁnd a corresponding eigenfunction. 2) Prove that there are no negative eigenvalues. 3) Find all the positive eigenvalues and prove that each of them has two corresponding linearly independent eigenfunctions. Explain why this is not a counterexample to Sturm’s oscillation theorem. 1) If λ = 0, then the complete solution is y(x) = c1 x + c2 med y (x) = c1 . It follows from the boundary conditions that −c1 π + c2 = c1 π + c2 and c 1 = c1 , hence c1 = 0, while c2 is arbitrary.