Heat & Wave Equation in a Rectangle. Section 12.8
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1 Het & Wve Eqution in Rectngle Section Het Eqution in Rectngle In this section we re concerned with ppliction of the method of seprtion of vriles pplied to the het eqution in two sptil dimensions. In prticulr we will consider prolems in rectngle. Thus we consider u t (x, y, t = k (u xx (x, y, t + u yy (x, y, t, t >, (x, y [, ] [, ], (1.1 u(, y, t =, u(, y, t =, u(x,, t =, u(x,, t = u(x, y, = f(x, y u(x, y = X(xY (yt (t. Sustituting into (.1 nd dividing oth sides y kx(xy (yt (t gives T (t kt (t = Y (y Y (y + X (x X(x Since the left side is independent of x, y nd the right side is independent of t, it follows tht the expression must e constnt: T (t kt (t = Y (y Y (y + X (x X(x = λ. We seek to find ll possile constnts λ nd the corresponding nonzero functions T, X nd Y. We otin T (t kλt (t =, 1
2 nd X (x X(x = λ Y (y Y (y. But since the left hnd side depends only on x nd the right hnd side only on y, we conclude tht there is constnt α X αx =. On the other hnd we could lso write so there exists constnt β so tht Y (y Y (y = λ X (x X(x Y βy =. Thus we hve X αx =, Y βy =, T (t kλt (t = nd λ = α + β. Furthermore, the oundry conditions give X(Y (y =, X(Y (y =, for ll y. Since Y (y is not identiclly zero we otin the desired eigenvlue prolem X (x αx(x =, X( =, X( =. (1. We hve solved this prolem mny times nd we hve α = µ so tht X(x = c 1 cos(µx + c sin(µx. Applying the oundry conditions we hve = X( = c 1 c 1 = = X( = c sin(µ. From this we conclude sin(µ = which implies nd therefore ( nπ α n = µ n = µ = nπ, Xn (x = sin(µ nx, n = 1,,. (1.3 Now from the oundry condition X(xY ( =, X(xY ( = for ll x.
3 This gives the prolem Y (y βy (y =, Y ( =, Y ( =. (1.4 This is the sme s the prolem (. so we otin eigenvlues nd eigenfunctions β m = ν m = ( mπ, Ym (y = sin(ν my, n = 1,,. (1.5 So we otin eigenvlues of the min prolem given y ( (nπ ( mπ λ n,m = + (1.6 nd corresponding eigenfunctions ϕ n,m (x, y = sin(µ n x sin(ν m y. We lso find the solution to T (t kλ n,m T (t = is given y T (t = e kλn,mt. So we look for u s n infinite sum u(x, y, t = c n,m e kλn,mt sin sin ( mπy. (1.7 The only prolem remining is to somehow pick the constnts c n,m so tht the initil condition u(x, y, = f(x, y is stisfied, i.e., with f(x, y = u(x, y, = ( c n,m = for n = 1,,, m = 1,,. c n,m ϕ n,m (x, y. (1.8 ( mπy f(x, y sin sin dx dy Exmple 1.1 (Dirichlet BCs. To simplify the prolem it we set = 1 nd = 1. Nmely we consider u t (x, y, t = k (u xx (x, y, t + u yy (x, y, t, t >, (x, y [, 1] [, 1] (1.9 u(, y, t =, u(1, y, t =, u(x,, t =, u(x, 1, t = u(x, y, = x(1 xy(1 y 3
4 In this cse we otin eigenvlues λ n,m = π (n + m, α n = π n, β m = π m, n, m = 1,,. The corresponding eigenfunctions re given y X n (x = sin(nπx, Y m (y = sin(mπy. Our solution is given y u(x, y, t = c n,m e kλn,mt sin(nπx sin(mπy. The coefficients c n,m re otined from We hve c n,m = 1 1 x(1 xy(1 y = c n,m sin(nπx sin(mπy. x(1 xy(1 y sin(nπx sin(mπy dx dy = 8(( 1n 1(( 1 m 1 n 3 m 3 π 6. Tht is u(x, y, t = 16 π 6 u(x, y, t = c n,m e kλn,mt sin (nπx sin (mπy. (1.1 (( 1 n 1(( 1 m 1e kλn,mt n 3 m 3 sin (nπx sin (mπy. (1.11 Exmple 1. (Mixed Dirichlet nd Neumnn BCs. To simplify the prolem it gin set = 1 nd = 1. Nmely we consider u t (x, y, t = k (u xx (x, y, t + u yy (x, y, t, t >, (x, y [, 1] [, 1] (1.1 u(, y, t =, u(1, y, t =, u y (x,, t =, u y (x, 1, t = u(x, y, = x(1 xy In this cse we otin eigenvlues nd λ n, = π n, Y (y = 1, λ n,m = π (n + m, α n = π n, β m = π m, n, m = 1,,, with corresponding eigenfunctions re given y X n (x = sin(nπx, Y m (y = cos(mπy. 4
5 Our solution is given y u(x, y, t = c n, e kλ n,t sin(nπx + c n,m e kλn,mt sin(nπx cos(mπy. Setting t = we otin x(1 xy = c n, sin(nπx + c n,m sin(nπx cos(mπy. This doule Fourier series is evluted gin using orthogonlity reltions. We hve c n,m = 1 1 x(1 xy sin(nπx cos(mπy dx dy = 8(( 1n 1(( 1 m 1 n 3 m π 5. Finlly we otin the coefficients c n, from c n, = 1 1 x(1 xy sin(nπx dx dy = (( 1 n 1 n 3 π 3. (1 ( 1 n u(x, y, t = e kλn,t sin(nπx n 3 π 3 16 (( 1 n 1(( 1 m 1e kλn,mt sin(nπx cos(mπy. π 5 n 3 m Wve Eqution in Higher Dimensions In this section we re concerned with ppliction of the method of seprtion of vriles pplied to the wve eqution in two dimensionl rectngle. Thus we consider u tt (x, y, t = c (u xx (x, y, t + u yy (x, y, t, t >, (x, y [, ] [, ], (.1 u(, y, t =, u(, y, t =, u(x,, t =, u(x,, t = u(x, y, = f(x, y, u t (x, y, = g(x, y 5
6 u(x, y = X(xY (yt (t. Sustituting into (.1 nd dividing oth sides y X(xY (y gives T (t c T (t = Y (y Y (y + X (x X(x Since the left side is independent of x, y nd the right side is independent of t, it follows tht the expression must e constnt: T (t c T (t = Y (y Y (y + X (x X(x = λ. We seek to find ll possile constnts λ nd the corresponding nonzero functions T, X nd Y. We otin X (x X(x = λ Y (y T (t c λt (t =. Y (y Thus we conclude tht there is constnt α On the other hnd we could lso write so there exists constnt β so tht Furthermore, the oundry conditions give X αx =. Y (y Y (y = λ X (x X(x Y βy =. X(Y (y =, X(Y (y = for ll y. Since Y (y is not identiclly zero we otin the desired eigenvlue prolem X (x αx(x =, X( =, X( =. (. We hve solved this prolem mny times nd we hve α = µ so tht X(x = c 1 cos(µx + c sin(µx. Applying the oundry conditions we hve = X( = c 1 c 1 = = X( = c sin(µ. From this we conclude sin(µ = which implies µ = nπ 6
7 nd therefore ( nπ α n = µ n =, Xn (x = sin(µ nx, n = 1,,.. (.3 Now from the oundry condition This gives the prolem X(xY ( =, X(xY ( = for ll x. Y (y βy (y =, Y ( =, Y ( =. (.4 This is the sme s the prolem (. so we otin eigenvlues nd eigenfunctions ( mπ β m = νm =, Ym (y = sin(ν my, n = 1,,.. (.5 So we otin eigenvlues of the min prolem given y ( (nπ ( mπ λ n,m = + (.6 nd corresponding eigenfunctions ϕ n,m (x, y = sin(µ n x sin(ν m y. We lso find the solution to T (t c λ n,m T (t = is given y T n,m (t = [ n,m cos(cω n,m t + n,m sin(cω n,m t] where we hve defined ω n,m = (nπ ( mπ. + So we look for u s n infinite sum u(x, y, t = [ n,m cos(cω n,m t + n,m sin(cω n,m t] sin sin ( mπy. (.7 We hve left to find the constnts n,m nd n,m so tht the initil condition u(x, y, = f(x, y nd u t (x, y, = g(x, y re stisfied, i.e., f(x, y = u(x, y, = n,m sin sin ( mπy. (.8 7
8 Thus we conclude tht ( n,m = for n = 1,,, m = 1,,. In similr wy we hve with g(x, y = u t (x, y, = ( n,m = ωn,m for n = 1,,, m = 1,,. ( mπy f(x, y sin sin dx dy cω n,m n,m sin sin ( mπy ( mπy g(x, y sin sin dx dy Exmple.1. In this exmple we set c = 1, = π nd = π. Nmely we consider. (.9 u tt (x, y, t = (u xx (x, y, t + u yy (x, y, t, t >, (x, y [, π] [, π] (.1 u(, y, t =, u(π, y, t =, u(x,, t =, u(x, π, t = u(x, y, = x(π xy(π y, u t (x, y, =. In this cse we otin eigenvlues λ n,m = (n + m, α n = n, β m = m, n, m = 1,,. The corresponding eigenfunctions re given y Our solution is given y u(x, y, t = π where we hve defined X n (x = π sin(nx, Y m(y = π sin(my. [ n,m cos(ω n,m t + n,m sin(ω n,m t] sin(nx sin(my ω n,m = n + m. The coefficients n,m re otined from We hve n,m = π π π x(π xy(π y = π n,m sin(nx sin(my. x(π xy(π y sin(nx sin(my dx dy = 8(( 1n 1(( 1 m 1. n 3 m 3 π 8
9 Since u t (x, y, = g(x, y = we hve n,m =. u(x, y, t = 16 π (( 1 n 1(( 1 m 1 n 3 m 3 e kλn,mt sin (nx sin (my. (.11 Exmple.. In this exmple we set c = 1, = π nd = π. Nmely we consider u tt = (u xx (x, y, t + u yy (x, y, t, t >, (x, y [, π] [, π] (.1 u x (, y, t =, u x (π, y, t =, u(x,, t =, u(x, π, t = u(x, y, = x(π xy, u t (x, y, =. We get eigenvlue prolem in x given y X αx =, X ( =, X (π =. Therefore we hve eigenvlues nd eigenvectors α =, X (x = 1 π, α n = n, X n (x = The eigenvlue prolem in y is given y The corresponding eigenvlues re Y βy =, Y ( =, Y (π =. β m = m, Y m (y = In this cse we otin eigenvlues cos(nx, n = 1,, 3,. π sin(my, m = 1,, 3,. π λ n,m = (n + m, α n = n, β m = m, n, m = 1,,. The corresponding eigenfunctions re given y For this exmple we lso hve eigenvlues ϕ n,m (x, y = cos(nx sin(my. π λ,m = m, X (x = 1 π. Our solution is given y u(x, y, t = 1 π + π m=1 [,m cos(ω,m t +,m sin(ω,m t] sin(my [ n,m cos(ω n,m t +,m sin(ω,m t] cos(nx sin(my. 9
10 Setting t = we otin x(π xy = π n, cos(nx + π n,m cos(nx sin(my. We hve n,m = π π π Finlly we otin the coefficients n, from x(1 xy cos(nx sin(my dx dy = π( 1m (( 1 n + 1. n m,m = π π π Finlly we rrive t the solution x(π xy sin(my dx dy = ( 1 m+1 π 3. 6m u(x, y, t = with ω n,m = n + m. + π ( 1 m+1 cos(ω,m t sin(my 3m m=1 4(( 1 n + 1(( 1 m n m cos(ω n,m t cos(nx sin(my 1
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