Extremum problems for eigenvalues of discrete Laplace operators

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$Math 664 Homework #2: Solutions 1. Let Î© = R, F = {A â R : either A ...$

4 REN GUO where s 2 ≥ 0,s 3 ≥ 0 and s 2 +s 3 = π. Then lim t→∞ a 1 (t) = ∞ and a 2 (t)+a 3 (t) > 0 for t ∈ [0, ∞). Hence lim (a 1(t) + a 2 (t) + a 3 (t)) = ∞. t→∞ If θ i + θ j < π, then cot θ i > cot(π − θ j ) = −cot θ j . Therefore a i + a j > 0. Hence 2f = (a 1 + a 2 ) + (a 2 + a 3 ) + (a 3 + a 1 ) > 0. Thus f has the absolute minimum. But the absolute minimum can not be achieved at a point in the boundary of Ω 3 . It must be achieved at the unique critical point ( π 3 , π 3 , π 3 ). This shows that a 1 + a 2 + a 3 ≥ √ 3 and the equality holds if and only if θ 1 = θ 2 = θ 3 = π 3 . 2.3. Since λ 2 = a 1 + a 2 + a 3 + √ (a 1 + a 2 + a 3 ) 2 − 3, we have λ 2 ≥ √ 3 and the equality holds if and only if θ 1 = θ 2 = θ 3 = π 3 . Since λ 1 λ 2 = 3, we have λ 1 ≤ √ 3 and the equality holds if and only if θ 1 = θ 2 = θ 3 = π 3 . 3. quadrilaterals 3.1. The vertices of a cyclic quadrilateral decompose its circumcircle into four arcs. We assume the radius of the circumcircle is 1 and the lengths of the four arcs are 2θ 1 ,2θ 2 ,2θ 3 ,2θ 4 . Let a i := cot θ i for i = 1,...,4. The condition θ 1 +θ 2 +θ 3 +θ 4 = π implies (3) a 1 a 2 a 3 + a 1 a 2 a 4 + a 1 a 3 a 4 + a 2 a 3 a 4 = a 1 + a 2 + a 3 + a 4 . There are two ways to decompose a cyclic quadrilateral in to a union of two triangles. The two ways produce the same discrete Laplace operator: L 4 = ⎛ ⎜ ⎝ The characteristic polynomial of L 4 is P 4 (x) = x 4 − 2(a 1 + a 2 + a 3 + a 4 )x 3 = x 4 − 2(a 1 + a 2 + a 3 + a 4 )x 3 by the equation (3). ⎞ a 1 + a 4 −a 1 0 −a 4 −a 1 a 1 + a 2 −a 2 0 ⎟ 0 −a 2 a 2 + a 3 −a 3 ⎠ . −a 4 0 −a 3 a 3 + a 4 + (3(a 1 a 2 + a 2 a 3 + a 3 a 4 + a 4 a 1 ) + 4(a 1 a 3 + a 2 a 4 ))x 2 − 4(a 1 a 2 a 3 + a 1 a 2 a 4 + a 1 a 3 a 4 + a 2 a 3 a 4 )x + (3(a 1 a 2 + a 2 a 3 + a 3 a 4 + a 4 a 1 ) + 4(a 1 a 3 + a 2 a 4 ))x 2 − 4(a 1 + a 2 + a 3 + a 4 )x,
EXTREMUM PROBLEMS FOR EIGENVALUES OF DISCRETE LAPLACE OPERATORS 5 3.2. By the similar argument in the case of triangles, we can show that a 1 +a 2 +a 3 + a 4 has the unique critical point at ( π 4 , π 4 , π 4 , π 4 ). And we have 2(a 1 +a 2 +a 3 +a 4 ) = (a 1 + a 2 ) + (a 1 + a 3 ) + (a 3 + a 4 ) + (a 4 + a 1 ) > 0. Next, we investigate the behavior of the function a 1 + a 2 + a 3 + a 4 when the variable (θ 1 ,θ 2 ,θ 3 ,θ 4 ) approaches the boundary of the domain Ω 4 = {(θ 1 ,θ 2 ,θ 3 ,θ 4 ) | θ 1 + θ 2 + θ 3 + θ 4 = π,θ i > 0,i = 1,2,3,4}. Let (θ 1 (t),θ 2 (t),θ 3 (t),θ 4 (t)), t ∈ [0, ∞), be a path in the domain Ω 4 . Let a i (t) = cot θ 1 (t) for i = 1,2,3,4. Without loss of generality, we assume lim (θ 1(t),θ 2 (t),θ 3 (t),θ 4 (t)) = (0,s 2 ,s 3 ,s 4 ) t→∞ where s i ≥ 0 for i = 2,3,4 and s 2 + s 3 + s 4 = π. And we can assume that s 2 < π 2 and s 3 < π 2 . Then lim t→∞ a 1 (t) = ∞, a 2 (t) > 0 when t is sufficiently large and a 3 (t) + a 4 (t) > 0 for any t ∈ [0, ∞). Hence lim (a 1(t) + a 2 (t) + a 3 (t) + a 4 (t)) = ∞ t→∞ . Therefore a 1 + a 2 + a 3 + a 4 achieves its absolute minimum at the unique critical point ( π 4 , π 4 , π 4 , π 4 ). Hence a 1 + a 2 + a 3 + a 4 ≥ 4 and the equality holds if and only if θ 1 = θ 2 = θ 3 = θ 4 = π 4 . Therefore λ 1 + λ 2 + λ 3 ≥ 8, λ 1 λ 2 λ 3 ≥ 16 and the equality holds if and only if θ 1 = θ 2 = θ 3 = θ 4 = π 4 . 3.3. To verify the statement about λ 1 λ 2 + λ 2 λ 3 + λ 3 λ 1 , by the formula of the characteristic polynomial P 4 (x), it is enough to show 3(a 1 a 2 + a 2 a 3 + a 3 a 4 + a 4 a 1 ) + 4(a 1 a 3 + a 2 a 4 ) ≥ 20 and the equality holds if and only if θ 1 = θ 2 = θ 3 = θ 4 = π 4 . In fact, consider g := 3(a 1 a 2 +a 2 a 3 +a 3 a 4 +a 4 a 1 )+4(a 1 a 3 +a 2 a 4 ) as a function defined on the domain Ω 4 . To find the absolute minimum of g, we apply the method of Lagrange multiplier. Let G = 3(a 1 a 2 + a 2 a 3 + a 3 a 4 + a 4 a 1 ) + 4(a 1 a 3 + a 2 a 4 ) + y(θ 1 + θ 2 + θ 3 + θ 4 − π). Then 0 = ∂G ∂θ 1 = −(3a 2 + 3a 4 + 4a 3 )(1 + a 2 1) + y, 0 = ∂G ∂θ 2 = −(3a 1 + 3a 3 + 4a 4 )(1 + a 2 2) + y, 0 = ∂G ∂θ 3 = −(3a 2 + 3a 4 + 4a 1 )(1 + a 2 3) + y, 0 = ∂G ∂θ 4 = −(3a 3 + 3a 1 + 4a 2 )(1 + a 2 4) + y, 0 = ∂G ∂y = θ 1 + θ 2 + θ 3 + θ 4 − π. The first and the third equation above imply that (3a 2 + 3a 4 + 4a 3 )(1 + a 2 1) = (3a 2 + 3a 4 + 4a 1 )(1 + a 2 3)
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Extremum problems for eigenvalues of discrete Laplace operators

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