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Examples of Eigenvalue Problems Leif Mejlbro Download free books at Leif Mejlbro Examples of Eigenvalue Problems Calculus 4c-2 Download free eBooks at bookboon.com Examples of Eigenvalue Problems – Calculus 4c-2 © 2008 Leif Mejlbro & Ventus Publishing ApS ISBN 978-87-7681-381-9 Download free eBooks at bookboon.com Examples of Eigenvalue Problems Contents Contents Introduction Initial and boundary value problems Eigenvalue problems 14 Nontypical eigenvalue problems 67 www.sylvania.com We not reinvent the wheel we reinvent light Fascinating lighting offers an ininite spectrum of possibilities: Innovative technologies and new markets provide both opportunities and challenges An environment in which your expertise is in high demand Enjoy the supportive working atmosphere within our global group and beneit from international career paths Implement sustainable ideas in close cooperation with other specialists and contribute to inluencing our future Come and join us in reinventing light every day Light is OSRAM Download free eBooks at bookboon.com Click on the ad to read more Examples of Eigenvalue Problems Introduction Introduction Here we present a collection of examples of eigenvalue problems The reader is also referred to Calculus 4b as well as to Calculus 3c-2 It should no longer be necessary rigourously to use the ADIC-model, described in Calculus 1c and Calculus 2c, because we now assume that the reader can this himself Even if I have tried to be careful about this text, it is impossible to avoid errors, in particular in the first edition It is my hope that the reader will show some understanding of my situation Leif Mejlbro 20th May 2008 Download free eBooks at bookboon.com Examples of Eigenvalue Problems Initial and boundary value problems Initial and boundary value problems Example 1.1 Solve the following eigenvalue problem y ′′ + λy = 0, x ∈ [0, L], y(0) = y ′ (0) = This is a pure initial value problem y(0) = and y ′ (0) = 0, hence the solution is unique Obviously, the zero solution is the only solutions Example 1.2 Prove that the boundary value problem d2 y dy + 2y = 0, +2 dx dx2 x ∈ [0, π], y(0) = 1, y(π) = −e−π , has infinitely many solutions and find these Sketch the graphs of some of these solution The characteristic polynomial R2 + 2R + = (R + 1)2 + has the roots R = −1 ± i The complete solution is given by y(x) = c1 e−x cos x + c2 e−x sin x, x ∈ [0, π], c1 , c2 ∈ R 0.8 0.6 0.4 0.2 0.5 1.5 x 2.5 Download free eBooks at bookboon.com Examples of Eigenvalue Problems Initial and boundary value problems It follows from the boundary values that y(0) = c1 = og y(π) = −c1 e−π = −eπ We get in both cases that c1 = 1, and we have no requirement on c2 ∈ R The complete solution of the boundary value problem is y(x) = e−x cos x + ce−x sin x, x ∈ [0, π], c ∈ R arbitrær 360° thinking Discover the truth at www.deloitte.ca/careers Download free eBooks at bookboon.com © Deloitte & Touche LLP and affiliated entities Click on the ad to read more Examples of Eigenvalue Problems Initial and boundary value problems Example 1.3 For a loaded column at equilibrium, one can as a mathematical model for a (small) bending y(x) in a convenient coordinate system use the following linear boundary value problem, EI d2 y + P y = −P e, dx2 x ∈ [0, L], y(0) = 0, y ′ (L) = Here, E, I, L, P and e are given positive constants For convenience we write P/(EI) = k 1) Find the solution of the boundary value problem 2) Prove that y(L) → ∞ for P → EI π2 , no matter how small the fixed constant e is 4L2 3) Sketch y(L) as a function of kl = π P · L, ≤ kL < EI 1) By a division with EI > the equation is transferred into the inhomogeneous equation d2 y + k y = −k e, dx2 k2 = P > EI (a) First find the complete solution The characteristic equation R2 + k = 0, i.e R = ±ik, [NB k > 0] provides us with the following solution of the corresponding homogeneous equation c1 cos kx + c2 sin kx, c1 , c2 are arbitrary We guess a particular solution as the constant y = −e Since the equation is linear, the complete solution is y = −e + c1 cos(kx) + c2 sin(kx), x ∈ [0, L], c1 , c2 arbitrary NB Unfortunately e is a constant which has nothing to with the usual mathematical constant 2, 718 (b) Insert into the boundary conditions We get y(0) = = −e + c1 , dvs c1 = e, and y ′ (L) = = −c1 k sin(kL) + c2 k cos(kL), hence [because k > 0] c2 cos(kL) = e · (kL) π + pπ, then the left hand side is 0, and the right hand side is ±e Therefore we π not have any solution for kL = + pπ, p ∈ N0 If kL = Download free eBooks at bookboon.com Examples of Eigenvalue Problems If kL = Initial and boundary value problems π + pπ, p ∈ N0 , then cos(kL) = 0, so c2 = e · tan(kL) By insertion of c1 = e and c2 = e · tan(kL) we get the solution y = −e + e cos(kx) + e tan(kL) sin(kx) (cos(kL) · cos(kx) + sin(kL) · sin(kx)) − = e cos(kL) cos(k(L − x)) −1 , x ∈ [0, L] = e cos(kL) 2) If P → EI kL → π2 P π → from below, then k = EI 4L · from below, so L2 π − By insertion of x = L we get y(L) = e −1 cos(kL) →∞ for P → EI π2 − 4L2 3) The function yk (L) = e −1 cos(kL) = e{sec(kL) − 1} (secant = 1/cosine) is easily sketched on a figure 0.2 0.4 0.6 x 0.8 1.2 1.4 Legend Funktionen $y_k(L)$ Download free eBooks at bookboon.com Examples of Eigenvalue Problems Initial and boundary value problems Example 1.4 Consider the boundary value problem y ′′ + 3y = 0, x ∈ [0, π], y(0) = y(π) = Set up the linear system of equations Bc = z, and check it √ The characteristic polynomial R2 + has the two simple roots R = ±i 3, so the complete solution is √ √ y = c1 cos( 3x) + c2 sin( 3x), x ∈ [0, π], c1 , c2 , arbitrary It follows from the boundary conditions, ⎧ +0 · c2 = y(0) = 0, ⎨ c1 ⎩ √ √ c1 · cos( 3π) +c2 · sin( 3π) = y(π) = The matrix equation is Bc = √ √ cos( 3π) sin( 3π) c1 c2 0 = Since √ det B = sin( 3π) = 0, the solution c1 = c2 = is unique end the zero solution is the only solution Example 1.5 Consider the boundary value problem y ′′ + 4y = 0, x ∈ [0, π], y(0) = y(π) = Set up the linear system of equations Bc = z, and check it is Since the characteristic polynomial R2 +4 has the two simple roots R = ±2i, the complete solution y = c1 cos(2x) + c2 sin(2x), x ∈ [0, π], c1 , c2 arbitary It follows from the boundary conditions that c1 + · c2 = y(0) = 0, c1 + · c2 = y(π) = The matrix equation becomes Bc = 1 c1 c2 = 0 , 10 Download free eBooks at bookboon.com ... eBooks at bookboon.com Examples of Eigenvalue Problems Contents Contents Introduction Initial and boundary value problems Eigenvalue problems 14 Nontypical eigenvalue problems 67 www.sylvania.com... bookboon.com Click on the ad to read more Examples of Eigenvalue Problems Introduction Introduction Here we present a collection of examples of eigenvalue problems The reader is also referred to...Leif Mejlbro Examples of Eigenvalue Problems Calculus 4c-2 Download free eBooks at bookboon.com Examples of Eigenvalue Problems – Calculus 4c-2 © 2008 Leif Mejlbro