(PDE) u t k(u xx + u yy ) = 0 (x, y) in Ω, t > 0, (BC) u(x, y, t) = 0 (x, y) on Γ, t > 0, (IC) u(x, y, 0) = f(x, y) (x, y) in Ω.


 Bruce Richard
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1 Seprtion of Vriles for Higher Dimensionl Het Eqution 1. Het Eqution nd Eigenfunctions of the Lplcin: An 2D Exmple Ojective: Let Ω e plnr region with oundry curve Γ. Consider het conduction in Ω with fixed oundry temperture on Γ: (PDE u t k(u xx + u yy = 0 (x, y in Ω, t > 0, (BC u(x, y, t = 0 (x, y on Γ, t > 0, (IC u(x, y, 0 = f(x, y (x, y in Ω. (IDEA: Seprtion of Vriles We look for solutions of (PDE(BC whose sptil structure of the temprture distriution remins invrint with time. In these solutions only the mplitudes of the temprture distriution my chnge with time. Mthemticlly, we try to clssify ll nontrivil solutions of the following form: u(x, y, t = T(tΦ(x, y 0. (THE EQUATIONS FOR PRODUCT SOLUTIONS Plugging u = T(tΦ(x, y in (PDE(BC we see: There is constnt α such tht ( T (t kαt(t = 0 t > 0; ( Φ xx + Φ yy αφ = 0 (x, y in Ω, ( Φ = 0 (x, y on Γ. We look for solutions of (*(**(*** tht re 0. Here, (* nd (** re deduced from (PDE, nd (*** follows from (BC. Eqution ( is esy to solve. The min issue now is to solve the eigenvlue prolem ( (. (EIGENVALUES AND EIGENFUNCTIONS Now focus on the oundry vlue prolem (**(*** for Φ(x, y. The solution structure of this prolem depends on the prmeter vlue α. It cn e shown tht for most choices of α, (**(*** only hs the trivil solution Φ(x, y 0. The specil vlues of α dmiting nontrivil solutions re clled eigenvlues of (**(***, nd in tht cse, the corresponding nontrivil solutions Φ(x, y 0 re clled eigenfunctions. Unfortuntely, for generl domin Ω, it s impossile to write down the eigenvlues nd eigenfunctions of ( ( in explicit form. We do, however, hve mthemticl theorem out this eigenvlue prolem: THEOREM. (i The eigenvlues of ( ( form sequence of negtive numers {α n } such tht 0 > α 1 > α 2 α 3 α 4, lim n α n =. (ii The corresponding eigenfunctions Φ n (x, y re smooth functions in Ω. (iii Φ 1 > 0 does not chnge sign in Ω. (iv Ω Φ m(x, yφ n (x, ydxdy = 0 for m n. Let α = α n e one of the eigenvlues. We now solve eqution (*, which ecomes T (t kα n T(t = 0 t > 0. 1
2 The solutions re T(t = Constnt e kαnt. (CLASSIFICATION OF PRODUCT SOLUTIONS Thus, product solution of (PDE(BC must e constnt multiple of u n (x, t = e kαnt Φ n (x, y, n = 1, 2, 3,. (FROM PRODUCT SOLUTIONS TO GENERAL SOLUTIONS The generl solutions of (PDE(BC re liner comintions of the product solutions: u(x, t = n e kαnt Φ n (x, y, n re constnts. (SOLUTION FORMULA FOR THE INITBDRY VAL PROBLEM To otin the solution of the initiloundry vlue prolem (PDE(BC(IC, we now need to use (IC to determine the vlues of n. Let t = 0: (FOURIER f(x, y = n Φ n (x, y. The coefficients n cn e expressed in terms of f(x, y. The derivtion is similr to the 1D cse. Detils follow: Multiply oth sides of (FOURIER y Φ m (x, y nd integrte over Ω: Ω f(x, yφ m (x, ydxdy = n = m Ω Ω Φ n (x, yφ m (x, ydxdy Φ m (x, y 2 dxdy, y Property (iv of the ove theorem. This finlly gives the solution of the initiloundry vlue prolem (PDE(BC(IC: u(x, t = n e kαnt Φ n (x, y, n = f(x, yφ Ω n(x, ydxdy Φ Ω n(x, y 2 dxdy n = 1, 2, 3,. 2
3 2. The Cse of Rectngulr Ω Ojective: Solve the initiloundry vlue prolem (PDE(BC(IC, for the cse Ω is the rectngle {0 < x <, 0 < y < }: (PDE u t k(u xx + u yy = 0 0 < x <, 0 < y <, t > 0, { u(x, 0, t = 0, u(x,, t = 0 0 < x <, t > 0, (BC u(0, y, t = 0, u(, y, t = 0 0 < y <, t > 0, (IC u(x, y, 0 = f(x, y 0 < x <, 0 < y <. THE EIGENVALUE PROBLEM ( ( cn e solved y further seprtion of vriles: Thnks to the specil symmetry of the rectngulr domin, we cn otin explicit expressions for the eigenvlues nd eigenfunctions. As mtter of fct, the eigenfunctions of the product form: Φ(x, y = X(xY (y will produce ll eigenvlues. Although, in generl there might e eigenfunctions of nonproduct forms, those nonproduct eigenfunctions re liner comiintions of product eigenfunctions. In this sense, we only need to solve ( ( for product functions Φ(x, y = X(xY (y. This will seprte the 2D eigenvlue prolem ( ( to two 1D eigenvlue prolems which we know how to solve. An ALTERNATIVE APPROACH is to strt from the eginnning with product solutions of the form: u(x, y, t = T(tX(xY (y 0. Plugging u = T(tX(xY (y in (PDE(BC we see: there re constnts λ nd µ such tht ( T (t + k(λ + µt(t = 0 t > 0; ( x X (x + λx(x = 0 (0 < x <, X(0 = X( = 0; ( y Y (y + µy (y = 0 (0 < y <, Y (0 = Y ( = 0. We know how to solve 1D eigenvlue prolem ( x : If nd only if λ is one of the following vlues λ m = (mπ/ 2, m = 1, 2, prolem ( x hs nontrivil solutions which re constnt multiples of X m (x = sin(mπx/. Another 1D eigenvlue prolem ( y is solved similrly: If nd only if µ is one of the following vlues µ n = (nπ/ 2, n = 1, 2, prolem ( y hs nontrivil solutions which re constnt multiples of Y n (y = sin(nπy/. For λ = λ m nd µ = µ n, we now solve eqution (*, which ecomes T (t + k(λ m + µ n T(t = 0 t > 0. 3
4 The solutions re T(t = Constnt e k(λm+µnt = Constnt e k[(mπ/2 +(nπ/ 2 ]t. (CLASSIFICATION OF PRODUCT SOLUTIONS Thus, (triplefctor product solution of (PDE(BC must e constnt multiple of u m,n (x, y, t = sin(mπx/ sin(nπy/e k[(mπ/2 +(nπ/ 2 ]t, m, n = 1, 2, 3,. (SOLUTION FORMULA FOR THE INITBDRY VAL PROBLEM The solution of the initiloundry vlue prolem (PDE(BC(IC is given y: u(x, t = m,n = m=1 = 4 m,n sin(mπx/ sin(nπy/e k[(mπ/2 +(nπ/ 2 ]t, f(x, y sin(mπx/ sin(nπy/dxdy {sin(mπx/ sin(nπy/}2 dxdy f(x, y sin sin dxdy. 4
5 EXERCISES [1] Consider the 2D het eqution in rectngle, with top nd ottom sides insulted, nd left nd right oundry temperture fixed t 0: (1 u t u xx u yy = 0 0 < x <, 0 < y <, t > 0, (2i u y (x, 0, t = 0, u y (x,, t = 0 0 x, 0 y, t > 0, (2ii u(0, y, t = 0, u(, y, t = 0 0 y, t > 0, (3 u(x, y, 0 = f(x, y 0 < x <, 0 < y <. ( Find ll nontrivil solutions u(x, y, t to (1(2 of the form u(x, y, t = X(xY (yt(t. ( Find the Fourier series solution formul for (1(2(3 with the generl initil dt f(x, y. (c Find the Fourier series formul in the cse f(x, y = xy. [2] Consider the 3D het eqution in 3D ox with insulted oundry conditions: (4 u t u xx u yy u zz = 0 0 < x <, 0 < y <, 0 < z < c, t > 0, (5 u x (0, y, z, t = u x (, y, z, t = u y (x, 0, z, t = u y (x,, z, t = u z (x, y, 0, t = u z (x, y, c, t = 0 0 < x <, 0 < y <, 0 < z < c, t > 0, (6 u(x, y, z, 0 = f(x, y, z 0 < x <, 0 < y <, 0 < z < c. ( Find ll nontrivil solutions u(x, y, z, t to (4(5 of the form u(x, y, z, t = X(xY (yz(zt(t. ( Find the Fourier series solution formul for (4(5(6 with the generl initil dt f(x, y, z. ANSWERS: [1] ( u(x, y, t = C sin e [(mπ/2 +(nπ/ 2 ]t, m is ny positive integer, n is ny nonnegtive integer, nd C is ny nonzero constnt. 5
6 ( u(x, y, t = m,0 = 2 m,n = 4 (c u(x, y, t = + m=1 m=1 n=0 m=1 m,n sin f(x, y sin f(x, y sin e [(mπ/2 +(nπ/ 2 ]t, dxdy (m 1 ( 1 m+1 sin e (mπ/2 t mπ 4( 1 m {1 ( 1 n } mn 2 π 3 sin dxdy (m 1, n 1 e [(mπ/2 +(nπ/ 2 ]t ( kπz [2] ( u(x, y, z, t = C c m, n, k re ny nonnegtive integers nd C is ny nonzero constnt. ( kπz ( u(x, y, z, t = m,n,k c 0,0,0 = 1 m,0,0 = 2 0,n,0 = 2 0,0,k = 2 m,n,0 = 4 0,n,k = 4 m,0,k = 4 m,n,k = 8 m=0 n=0 k=0 z=0 f(x, y, zdxdydz f(x, y, z f(x, y, z ( kπz f(x, y, z c f(x, y, z ( kπz f(x, y, z c ( kπz f(x, y, z c f(x, y, z (m 1, n 1, k 1 e [(mπ/2 +(nπ/ 2 +(kπ/c 2 ]t, dxdydz (m 1 dxdydz (n 1 dxdydz (k 1 e [(mπ/2 +(nπ/ 2 +(kπ/c 2 ]t, dxdydz (m 1, n 1 dxdydz (n 1, k 1 dxdydz (m 1, k 1 ( kπz c dxdydz 6
u(x, y, t) = T(t)Φ(x, y) 0. (THE EQUATIONS FOR PRODUCT SOLUTIONS) Plugging u = T(t)Φ(x, y) in (PDE)(BC) we see: There is a constant λ such that
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