1 2D Second Order Equations: Separation of Variables


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1 Chpter 12 PDEs in Rectngles 1 2D Second Order Equtions: Seprtion of Vribles 1. A second order liner prtil differentil eqution in two vribles x nd y is A 2 u x + B 2 u 2 x y + C 2 u y + D u 2 x + E u + F u = G. (1) y 2. If G = 0 we sy the problem is homogeneous otherwise it is nonhomogeneous. 3. A Solution is function u(x, y) tht hs the required differentibility nd stisfies the eqution. Unlike the cse of ODEs the ide of generl solution is not very cler. We will look for prticulr solutions. 4. A very useful method for looking for solutions is the Method of Seprtion of Vribles in which we look for solution in the form u(x, y) = ψ(y). 5. (Principle of Superposition) Since the problem is liner, if we find severl solutions, sy {u j (x, y)} n j=1 then for ll constnts {c j } n j=1 is lso solution. u(x, y) = n c j u j (x, y) j=1 6. Equtions in the form (1) cn be clssified s one three types of equtions by () Hyperbolic if B 2 4AC > 0 (b) Prbolic if B 2 4AC = 0 (c) Elliptic if B 2 4AC < 0 7. Min Clssicl Exmples: (replcing y by t in the hyperbolic nd prbolic cses) () Hyperbolic Wve eqution 2 u t (x, t) = 2 u 2 2 (x, t) x2 (b) Prbolic Het eqution u t (x, t) = k 2 u (x, t) x2 (c) Elliptic Lplce s eqution 2 u x (x, y) + 2 u (x, y) = 0 2 y2 8. Exmples of seprtion of vribles for min exmples: () For 2 u t 2 = 2 u 2 x seek u(x, t) = ψ(t) ψ (t) 2 2 ψ(t) = ϕ (x) (b) For u t = k 2 u x seek u(x, t) = ψ(t) ψ (t) 2 kψ(t) = ϕ (x) (c) For 2 u x + 2 u 2 y = 0 seek u(x, y) = ψ(y) ϕ (x) 2 + ψ (y) ψ(y) = λ where λ is constnt. = λ where λ is constnt. = λ where λ is constnt. 1
2 9. Boundry Conditions re given on the physicl boundry of the sptil domin, i.e., t the ends of heted rod or vibrting string. The most common BCs re () Dirichlet Homogeneous BC t x = is u(, t) = 0, Nonhomogeneous BC is u(, t) = γ(t) (b) Neumnn Homogeneous BC t x = is u x (, t) = 0, Nonhomogeneous BC is u x (, t) = γ(t) (c) Robin Homogeneous BC t x = is u x (, t) ± ku(, t) = 0, Nonomogeneous BC is u x (, t) ± ku(, t) = γ(t) where k > 0 is constnt nd (±) is determined by which end of the rod or string. 2 Regulr SturmLiouville problem One of the most importnt ides in functionl nlysis is contined in the following discussion which is very much relted to the ide of eigenvlues nd n orthonorml bsis of eigenvectors. Consider the boundry vlue problem. ϕ (x) q(x) = λ, α 1 ϕ () + α 2 ϕ() = 0, (2) β 1 ϕ (b) + β 2 ϕ(b) = 0 This is n eigenvlue problem which is referred to s Regulr SturmLiouville problem. Theorem 2.1. The problem (2) hs infinitely mny eigenpirs {λ n, ϕ n (x)} which stisfy the following properties: 1. The eigenvlues re ll simple, i.e. they re eigenvlues of multiplicity one which mens tht λ j λ k for j k. 2. The eigenvlues re ll rel nd ll but finite number re negtive. If α 1 + α 2 0, β 1 + β 2 0 then ll (but possibly one) of the eigenvlues re less thn or equl to zero. 3. If we order the eigenvlues in decresing order by λ n < λ n 1 < < λ 2 < λ 1, nd λ n s n. 4. The eigenfunctions re ll rel nd ϕ n (x) hs exctly (n 1) zeros in the intervl (, b). 5. The eigenfunctions form Complete Orthonorml Set in the following sense { b 0, n m ϕ n (x)ϕ m (x) dx = δ n,m = 1, n = m nd if f(x) is piecewise smooth (P C (1) (, b) in my nottion in clss), then (f(x+) + f(x )) = c n ϕ n (x), < x < b, where c n = f(x)ϕ n (x) dx. 2 ( 1/2 N For ny f such tht f = f 2 (x) dx) < we hve f(x) N c n ϕ n (x) 0, i.e., f(x) = c n ϕ n (x) in the sense of L 2 (, b). 2
3 3 1D Het Eqution: Eigenvlues nd Eigenvectors Our first PDE is the het eqution on finite rod x b. u t (x, t) = ku xx (x, t), < x < b, t > 0 u(x, 0) = f(x) There re three min types of boundry conditions imposed t the ends of the rod. The two min conditions re u(, t) = 0, u(b, t) = 0 Dirichlet Conditions u x (, t) = 0, u x (b, t) = 0 Neumnn Conditions α 1 u x (, t) + α 2 u(, t) = 0, β 1 u x (b, t) + β 1 u(b, t) = 0 Robin Conditions We cn lso hve ny combintion of these conditions, i.e., we could hve Dirichlet condition t x = nd Neumnn condition t x = b. Notice tht Dirichlet nd Neumnn BCs re specil cses of the Robin BCs. There is lso more generl problem involving two extr terms tht correspond to het conduction nd convection. u t (x, t) = k ( u xx (x, t) 2u(x, t) x + bu(x, t) ), 0 < x < l, t > 0 u(x, 0) = f(x). Let us consider the het eqution on x b. u t (x, t) = ku xx (x, t), < x < b, t > 0 α 1 u x (, t) + α 2 u(, t) = 0, β 1 u x (b, t) + β 2 u(b, t) = 0 u(x, 0) = Applying seprtion of vribles we seek simple solutions in the form This gives u(x, t) = ψ(t). ψ (t) kψ(t) = ϕ (x) nd since the left side is independent of x nd the right side is independent of t, it follows tht the expression must be constnt: ψ (t) kψ(t) = ϕ (x) = λ We seek to find ll possible constnts λ nd the corresponding nonzero functions ϕ nd ψ. The eqution ψ kλψ = 0 hs generl solution ψ(t) = Ce kλt (3) where C is n rbitrry constnt. 3
4 We lso obtin ϕ λϕ = 0 Furthermore, the boundry conditions give (α 1 ϕ () α 2 ϕ())ψ(t) = 0, (β 1 ϕ (b) + β 2 ϕ(b))ψ(t) = 0 for ll t. Since ψ(t) is not identiclly zero we obtin the desired eigenvlue problem ϕ (x) λ = 0, α 1 ϕ () + α 2 ϕ() = 0, β 1 ϕ (b) + β 2 ϕ (b) = 0. By the SturmLiouville Theorem there re infinitely mny eigenpirs {λ n, ϕ n (x)} with the {ϕ n } re orthonorml, i.e., { b 0, n m ϕ n, ϕ m = ϕ n (x)ϕ m (x) dx = δ n,m = 1, n = m. We seek solution to het problem in the form u(x, t) = c n e λnt ϕ n (x). (4) We need two things: 1. When t = 0 we find vlues for {c n } so tht given by f(x) = u(x, 0) = c n = c n ϕ n (x) f(x)ϕ n (x) dx. 2. Next we need for the infinite sum in (7) to represent solution to the eqution. This is formlly true since u t (x, t) = de λnt c n ϕ n (x) dt = c n λ n e λnt ϕ n (x) = c n e λnt d2 ϕ n dx (x) = 2 u (x, t) 2 x2 A rigorous proof in the cse f is smooth or if we use convergence in L 2 (, b) nd study wek solutions ( topic beyond the scope of this clss). 4
5 4 1D Wve Eqution: Eigenvlues nd Eigenvectors The one dimensionl wve eqution modeling the displcement of vibrting string of length l = (b ) covering the intervl < x < b is u tt (x, t) = c 2 u xx (x, t), < x < b, t > 0 (5) u(x, 0) = f(x) u t (x, 0) = g(x) Once gin we hve the sme three min types of boundry conditions imposed t the ends of the string: Dirichlet Conditions, Neumnn Conditions, nd Robin Conditions. These re ll specil cses of the generl BCs α 1 ϕ () + α 2 ϕ() = 0, β 1 ϕ (b) + β 2 ϕ(b) = 0. Let us consider the het eqution on x b. u tt (x, t) = c 2 u xx (x, t), < x < b, t > 0 α 1 u x (, t) + α 2 u(, t) = 0, β 1 u x (b, t) + β 2 u(b, t) = 0 u(x, 0) = f(x), u t (x, 0) = g(x) Applying seprtion of vribles we seek simple solutions in the form u(x, t) = ψ(t). This gives ψ (t) c 2 ψ(t) = ϕ (x) nd since the left side is independent of x nd the right side is independent of t, it follows tht the expression must be constnt: ψ (t) c 2 ψ(t) = ϕ (x) = λ We seek to find ll possible constnts λ nd the corresponding nonzero functions ϕ nd ψ. In the x vrible we hve ϕ λϕ = 0 Furthermore, the boundry conditions give (α 1 ϕ() + α 2 ϕ())ψ(t) = 0, (β 1 ϕ(b) + β 2 ϕ(b))ψ(t) = 0 for ll t. Since ψ(t) is not identiclly zero we obtin the desired eigenvlue problem ϕ (x) λ = 0, α 1 ϕ () + α 2 ϕ() = 0, β 1 ϕ (b) + β 2 ϕ(b) = 0. 5
6 By the SturmLiouville Theorem there re infinitely mny eigenpirs {λ n, ϕ n (x)} with the eigenfunctions forming complete orthonorml set. In most prcticl problems we hve λ n = µ 2 n so the problem for ψ becomes ψ + c 2 µ 2 nψ = 0 hs generl solution ψ n (t) = n cos(cµ n t) + b n sin(cµ n t) (6) where n nd b n re rbitrry constnts. At lest for continuous initil conditions ϕ we obtin solution to wve eqution in the form where u(x, t) = ( n cos(cµ n t) + b n sin(cµ n t)) ϕ n (x). (7) 1. f(x) = u(x, 0) = 2. g(x) = u t (x, 0) = n ϕ n (x) with n = f(x)ϕ n (x) dx. (cµ n )b n ϕ n (x) with b n = (cµ n ) 1 g(x)ϕ n (x) dx. 3. Next we need for the infinite sum in (7) to represent solution to the eqution. This is formlly true since 2 u t (x, t) = d 2 c 2 n dt ( 2 n cos(cµ n t) + b n sin(cµ n t)) ϕ n (x) = = c n ( cµ n ) 2 ( n cos(cµ n t) + b n sin(cµ n t)) ϕ n (x) c n (c 2 λ n ) ( n cos(cµ n t) + b n sin(cµ n t)) ϕ n (x) = c 2 = c 2 2 u (x, t) x2 c n ( n cos(cµ n t) + b n sin(cµ n t)) d2 ϕ n dx 2 (x) 6
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