Applied partial differential equations with fourier series and boundary value problems 5th edition rPublished on Feb 8, 2018 Download at: http://testbanklive.com/download/applied-partial-differential-equations-with-fourier-series-and-boundary-value-problems-5th-edition-richa... nokia  Solutions Manual for Applied Partial Differential Equations with Fourier Series and Boundary Value Problems 5th Edition by Richard Haberman Full clear download (no formatting errors) at: http://downloadlink.org/p/solutions-manual-for-applied-partial-differential- equations-with-fourier-series-and-boundary-value-problems-5th-edition-by- richard-haberman/ 1.2.9 (d) Circular cross section means that P = 2πr, A = πr2 , and thus P / A = 2 /r , where r is the radius. Also γ = 0. 1.2.9 (e) u(x, t) = u(t) implies that du 2 h cρ dt = − r u . The solution of this first-order linear differential equation with constan t coefficients, which satisfies the initial condition u(0) = u0 , is u(t) = u0 exp · 2h ¸ − cρr t . Section 1.3 1.3.2 ∂u/∂x is continuous if K 0 ( x 0 −) = K 0 ( x 0 +), that is, if the conductivity is con tin uous. Section 1.4 1.4.1 (a) Equilibrium satisfies (1.4.14), d2 u/dx 2 = 0, whose general solution is (1.4.17), u = c1 + c 2 x. The boundary condition u(0) = 0 implies c1 = 0 and u(L) = T implies c2 = T /L so that u = T x/L . 1.4.1 (d) Equilibrium satisfies (1.4.14), d2 u/dx 2 = 0, whose general solution (1.4.17), u = c1 + c 2 x. F rom the boundary conditions, u(0) = T yields T = c1 and du/dx(L) = α yields α = c 2 . Thus u = T + α x . 1.4.1 (f ) In equilibrium, (1.2.9) becomes d2 u/dx 2 = −Q/K 0 = −x 2 , whose general solution (by in tegrating t wice) isu = − x 4 / 12 + c1 + c 2 x. The boundary condition u(0) = T yields c1 = T , while du/dx(L) = 0 yields c2 = L3 / 3. Thus u = − x 4 / 12 + L3 x/ 3 + T . 1.4.1 (h) Equilibrium satisfies d2 u/dx 2 = 0. One integration yields du/dx = c 2 , the second in tegration yields the general solution u = c1 + c 2 x . x = 0 : c2 −(c1 − T ) = 0 x = L : c2 = α and thus c1 = T + α . Therefore, u = (T + α) + αx = T + α(x + 1). 1.4.7 (a) For equilibrium: d2 u dx 2 x 2 = −1 implies u = − 2 + c 1 x + c2 and du dx = −x + c 1 . From the boundary conditions du (0) = 1 and du (L) = β, c1 = 1 and −L + c1 = β which is consisten t dx dx 2 only if β + L = 1. Ifβ = 1 − L, there is an equilibrium solution (u = − x + x + c 2 ). Ifβ = 1 − L , |
Applied partial differential equations haberman 5th edition pdf download
Related Posts
Copyright © 2024 en.frojeostern Inc.
