Wave Equation on a Two Dimensional Rectangle
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1 Wve Eqution on Two Dimensionl Rectngle In these notes 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 = c 2 (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 u(x, y = X(xY (yt (t. Sustituting into (1 nd dividing oth sides y X(xY (y gives T (t c 2 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 2 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 2 λt (t =. Y (y Thus we conclude tht there is constnt α On the other hnd we could lso write X αx =. Y (y Y (y = λ X (x X(x 1
2 so there exists constnt β so tht Y βy =. 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( =. (2 We hve solved this prolem mny times nd we hve α = µ 2 so tht X(x = c 1 cos(µx + c 2 sin(µx. Applying the oundry conditions we hve = X( = c 1 c 1 = = X( = c 2 sin(µ. From this we conclude sin(µ = which implies nd therefore ( nπ α n = µ 2 n = µ = nπ 2, Xn (x = sin(µ n x, n = 1, 2,.. (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 (2 so we otin eigenvlues nd eigenfunctions β m = ν 2 m = ( mπ 2, Ym (y = sin(ν m y, n = 1, 2,.. (5 So we otin eigenvlues of the min prolem given y ( (nπ 2 ( mπ 2 λ n,m = + (6 nd corresponding eigenfunctions ϕ n,m (x, y = sin(µ n x sin(ν m y. 2
3 We lso find the solution to T (t c 2 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 2 ( mπ 2. + 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. (7 The only prolem remining is to somehow pick 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 At this point we recll our orthogonlity reltions l ( ( jπξ c 2 πξ 2 j = k sin sin dξ = l l l j k.. (8 So we first multiply oth sides of (8 y sin nd integrte in x from to to get ( 2 f(x, y sin dx = n,m sin. Next we multiply this expression y sin nd integrte in x from to to get ( ( f(x, y sin sin dx = n,m. 2 2 n=1 Thus we conclude tht n,m = ( 4 for n = 1, 2,, m = 1, 2,. f(x, y sin sin dx dy g(x, y = u t (x, y, = cω n,m n,m sin 3 sin. (9
4 Just s ove we get ( g(x, y sin sin dx = 2 ( n,m cω n,m. 2 ( 4 n,m = cω n,m for n = 1, 2,, m = 1, 2,. g(x, y sin sin dx dy Generl Methodology The sme methodology cn e pplied for more generl oundry conditions. Consider t >, (x, y [, ] [, ], nd u tt = c 2 (u xx (x, y, t + u yy (x, y, t, (1 γ 1 u(, y, t + γ 2 u x (, y, t =, γ 3 u(, y, t + γ 4 u x (, y, t =, γ 5 u(x,, t + γ 6 u y (x,, t =, γ 7 u(x,, t + γ 8 u y (x,, t = u(x, y, = f(x, y u t (x, y, = g(x, y As usul we seek simple solution in the form u(x, y = X(xY (yt (t. Sustituting into (1 nd dividing oth sides y X(xY (y gives T (t c 2 T (t = Y (y Y (y + X (x X(x As ove we otin three prolems involving constnts λ, α nd β with λ = α + β: X (x αx(x =, γ 1 X( + γ 2 X ( =, γ 3 X( + γ 4 X ( =. (11 After some work we otin n infinite set of negtive eigenvlues nd eigenfunctions α n = µ 2 n, X n (x, n = 1, 2,. (12 4
5 Similrly, Y (y βy (y =, g 5 Y ( + γ 6 Y ( =, γ 7 Y ( + γ 8 Y ( =. (13 Once gin we otin eigenvlues nd eigenfunctions β m = ν 2 m, Y m (y, n = 1, 2,.. (14 We lso find the solution to T (t c 2 λ n,m T (t =. To this end let us gin define ω n,m = λ = (µ n 2 + (ν m 2. Then the solution is given y T n,m (t = [ n,m cos(cω n,m t + n,m sin(cω n,m t]. 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] X n (xy m (y. (15 Finlly we need to pick the constnts n,m nd n,m so tht the initil conditions u(x, y, = f(x, y nd u t (x, y, = g(x, y re stisfied, i.e., f(x, y = u(x, y, = g(x, y = u t (x, y, = n,m X n (xy m (y. (16 n,m cω n,m X n (xy m (y. (17 The generl Sturm-Liouville theory gurntees tht the eigenfunctions corresponding to distinct eigenfunctions re distinct, i.e. { { κ j j = k, X j (xx k (x dx = j k., κ j j = k, Y j (xy k (x dx = (18 j k. for some positive constnts κ j nd κ j. So we otin nd for n = 1, 2,, m = 1, 2,. ( 1 n,m = κ n κ m ( 1 n,m = cω n,m κ n κ m f(x, yx n (xy m (y dx dy (19 5 g(x, yx n (xy m (y dx dy (2
6 Exmple 1. To simplify the prolem it 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] (21 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 2 + m 2, α n = n 2, β m = m 2, n, m = 1, 2,. The corresponding eigenfunctions re given y X n (x = sin(nx, Y m (y = sin(my. Our solution is given y (15 which here hs the form u(x, y, t = [ n,m cos(ω n,m t + n,m sin(ω n,m t] sin(nx sin(my n=1 where we hve defined ω n,m = n 2 + m 2. The coefficients n,m re otined from (18 where in this cse κ n = κ m = π, n, n = 1, 2,. 2 We hve From (19 we hve x(π xy(π y = n,m sin(nx sin(my. n,m = 4 π 2 π π Since u t (x, y, = g(x, y = we hve x(π xy(π y sin(nx sin(my dx dy = 16(( 1n 1(( 1 m 1 n 3 m 3 π 2. n,m =. u(x, y, t = c n,m e kλn,mt sin (nπx sin (mπy. (22 6
7 Exmple 2. To simplify the prolem it we set c = 1, = 1 nd = 1. Nmely we consider u tt = (u xx (x, y, t + u yy (x, y, t, t >, (x, y [, π] [, π] (23 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, α n = n 2, X n (x = cos(nx, n = 1, 2, 3,. The eigenvlue prolem in y is given y Y βy =, Y ( =, Y (π =. The corresponding eigenvlues re β m = m 2, X m (x = sin(mx, m = 1, 2, 3,. In this cse we otin eigenvlues λ n,m = (n 2 + m 2, α n = n 2, β m = m 2, n, m = 1, 2,. The corresponding eigenfunctions re given y X n (x = sin(nπx, Y m (y = sin(mπy. For this exmple we lso hve eigenvlues λ,m = m 2, X (x = 1. Our solution is given y (15 which here hs the form u(x, y, t = [,m cos(ω,m t +,m sin(ω,m t] sin(my m=1 + [ n,m cos(ω n,m t +,m sin(ω,m t] cos(nx sin(my. Setting t = we otin x(π xy = n, cos(nx + n,m sin(nx sin(my. n=1 7
8 We hve n,m = 4 π 2 π π Finlly we otin the coefficients n, from,m = 2 π 2 π x(π xy cos(nx sin(my dx dy = 8(( 1n 1(( 1 m 1 n 2 m 3 π 2. π Finlly we rrive t the solution u(x, y, t = + with ω n,m = n 2 + m 2. m=1 x(π xy sin(my dx dy = 2(( 1m 1 m 3. 2(( 1 m 1 m 3 cos(ω,m t sin(my 8(( 1 n 1(( 1 m 1 n 2 m 3 π 2 cos(ω n,m t cos(nx sin(my 8
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