MA 201: Partial Differential Equations Lecture - 12

Transcription

1 Two dimensionl Lplce Eqution MA 201: Prtil Differentil Equtions Lecture - 12 The Lplce Eqution (the cnonicl elliptic eqution)

2 Two dimensionl Lplce Eqution Two dimensionl Lplce Eqution 2 u = u xx + u yy = 0. (1) We use seprtion of vrible method to solve it. Assume solution: Substituting in (1) we get X Y + Y X = 0, where k is constnt. u(x, y) = X (x)y (y). (2) i.e., X X = Y Y = k, Cse: k = p 2 > 0: Then X p 2 X = 0, Y + p 2 Y = 0, which give X = c 1 e px + c 2 e px, Y = c 3 cos py + c 4 sin py, i.e., u(x, y) = (c 1 e px + c 2 e px )(c 3 cos py + c 4 sin py). (3)

3 Two dimensionl Lplce Eqution Two dimensionl Lplce Eqution: generl solutions Cse: k = 0: Then X = 0, Y = 0, which give solution u(x, y) = (c 5 x + c 6 )(c 7 y + c 8 ). (4) Cse: k = p 2 < 0: Then X + p 2 X = 0, Y p 2 Y = 0, which give u(x, y) = (c 9 cos px + c 10 sin px)(c 11 e py + c 12 e py ). (5) Note: To get nontrivil solution u(x, y) neither of X (x) nd Y (y) should be identiclly zero.

4 Two dimensionl Lplce Eqution Dirichlet Problem for rectngle Consider the domin D : 0 x, 0 y b. PDE: u xx + u yy = 0, BCs: u(0, y) = u(, y) = u(x, b) = 0, u(x, 0) = f (x). We seek for vible solutions from (3)-(5). Consider (3): u(x, y) = (c 1 e px + c 2 e px )(c 3 cos py + c 4 sin py). The condition u(0, y) = 0 gives c 1 + c 2 = 0. (6) Agin, u(, y) = 0 gives c 1 e p + c 2 e p = 0. (7) However, (6) nd (7) imply c 1 = c 2 = 0, yielding trivil solution.

5 Two dimensionl Lplce Eqution Dirichlet Problem for rectngle Next, consider solution (4): u(x, y) = (c 5 x + c 6 )(c 7 y + c 8 ). The BCs: u(0, y) = u(, y) = 0 yields only the zero solution. Finlly, consider solution (5): u(x, y) = (c 9 cos px + c 10 sin px)(c 11 e py + c 12 e py ). The BC u(0, y) = 0 gives c 9 = 0, nd the BC u(, y) = 0 gives c 10 sin p = 0. Thus, for nontrivil solution (c 10 0) we hve sin p = 0. This gives eigenvlues p n = nπ, n = 1, 2, 3,...

6 Two dimensionl Lplce Eqution Dirichlet Problem for rectngle The corresponding eigenfunctions re ( nπx ) [n u n (x, y) = sin exp(nπy/) + b n exp( nπy/) ]. Using BC: u(x, b) = 0, we get n exp( nπb ) + b n exp( nπb ) = 0, i.e., b n = n exp(nπb/) exp( nπb/). Thus, u n (x, y) = 2 n sin(nπx/) exp( nπb/) = [ exp nπ(y b) ] exp nπ(y b) 2 2 n exp( nπb/) sin(nπx/) sinh ( nπ(y b)/ ) = A n sin(nπx/) sinh ( nπ(y b)/ ), where A n = (2 n )/ ( exp( nπb/) ).

7 Two dimensionl Lplce Eqution Using superposition we get possible nontrivil solution u(x, y) = A n sin(nπx/) sinh ( nπ(y b)/ ). (8) n=1 Now, the BC: u(x, 0) = f (x) gives ( f (x) = A n sinh ( nπb/ )) sin(nπx/) n=1 hlf rnge sine series in the intervl [0, ]. Thus, the Fourier coefficients: A n sinh ( nπb/ ) = 2 A n = 0 2 sinh ( nπb/ ) f (x) sin(nπx/)dx, (8) nd (9) give the required solution. 0 i.e., f (x) sin(nπx/)dx. (9)

8 Two dimensionl Lplce Eqution Exmple. A thin rectngulr homogeneous thermlly conducting plte lies in the xy-plne defined by 0 x, 0 y b. The edge y = 0 is held t temperture Tx(x ), where T is constnt, while the remining edge re held t 0. The other fces re insulted nd no internl sources nd sinks re present. Find the stedy stte temperture inside the plte. Solution. The stedy stte temperture stisfy 2 u = 0. Hence the problem is: PDE: u xx + u yy = 0. BC: u(0, y) = 0, u(, y) = 0, u(x, b) = 0, u(x, 0) = Tx(x ) = f (x).

9 Two dimensionl Lplce Eqution The solution is where u(x, y) = A n sin(nπx/) sinh ( nπ(y b)/ ). n=1 A n sinh ( nπb/ ) = 2 Thus, u(x, y) = n=1 0 f (x) sin(nπx/)dx = 2T x(x ) sin(nπx/)dx 0 = = 42 T n 3 π 3 (cos nπ 1) = 42 T n 3 π 3 (( 1)n 1) [ (( 1) n 1 ) 4 2 T nπb ] n 3 cosech sin nπx π3 nπ(y b) sinh.

10 Two dimensionl Lplce Eqution Non-homogeneous BC t more thn one side In generl the Dirichlet BVP will be like u xx + u yy = 0, 0 x, 0 y b, u(0, y) = g 1 (y), u(, y) = g 2 (y), 0 y b, u(x, 0) = f 1 (x), u(x, b) = f 2 (x), 0 x. where f 1 (x), f 2 (x), g 1 (y), g 2 (y) re given functions. Though the eqution is liner nd homogenous, the BCs re not. Here, the BVP cn be be split into four BVPs ech contining one non-homogenous BC. The solution u of the problem is given by u = u 1 + u 2 + u 3 + u 4, 0 x, 0 y b, where u i re solutions of the following BVPs.

11 Two dimensionl Lplce Eqution BVP I nd BVP II: u 1,xx + u 1,yy = 0; u 2,xx + u 2,yy = 0, 0 x, 0 y b; u 1 (0, y) = 0, 0 y b; u 2 (0, y) = 0, 0 y b; u 1 (, y) = 0, 0 y b; u 2 (, y) = 0, 0 y b; u 1 (x, 0) = f 1 (x), 0 x ; u 2 (x, 0) = 0, 0 x ; u 1 (x, b) = 0, 0 x ; u 2 (x, b) = f 2 (x), 0 x ; BVP III nd BVP IV: u 3,xx + u 3,yy = 0; u 4,xx + u 4,yy = 0, 0 x, 0 y b; u 3 (0, y) = g 1 (y), 0 y b; u 4 (0, y) = 0, 0 y b; u 3 (, y) = 0, 0 y b; u 4 (, y) = g 2 (y), 0 y b; u 3 (x, 0) = 0, 0 x ; u 4 (x, 0) = 0, 0 x ; u 3 (x, b) = 0, 0 x ; u 4 (x, b) = 0, 0 x.

12 Two dimensionl Lplce Eqution Lplce Eqution in rectngle with Neumnn conditions Consider the domin D : 0 x, 0 y b. PDE: u xx + u yy = 0, BCs: u x (0, y) = u x (, y) = u y (x, 0) = 0, u y (x, b) = f (x). We seek for vible solutions from (the k < 0 cse): u(x, y) = (c 1 cos px + c 2 sin px)(c 3 e py + c 4 e py ). (10) We hve u x (x, y) = ( c 1 p sin px + c 2 p cos px)(c 3 e py + c 4 e py ) u y (x, y) = (c 1 cos px + c 2 sin px)(c 3 pe py c 4 pe py ) The BC: u x (0, y) = 0 gives c 2 p(c 3 e py + c 4 e py ) = 0, i.e., c 2 = 0. The BC: u y (x, 0) = 0 gives c 3 p c 4 p = 0, i.e., c 3 = c 4.

13 Two dimensionl Lplce Eqution Dirichlet Problem for rectngle The BC: u x (, y) = 0 gives c 1 p sin p = 0 nd so (with c 1 0) the eigenvlues p n = nπ, n = 0, 1, 2,... The corresponding eigenfunctions re ( nπx ) [bn u n (x, y) = n cos exp(nπy/) + b n exp( nπy/) ]. ( nπx ) = A n cos cosh nπy (putting A n = 2 n b n ) Thus, using superposition we get ( nπx ) u(x, y) = A n cos cosh nπy. n=0 Finlly, the BC: u y (x, b) = f (x) gives ( nπx ) nπ f (x) = A n cos n=0 sinh nπb.

14 Two dimensionl Lplce Eqution i.e., i.e., f (x) = A n nπ n=0 sinh nπb [ A n nπ = 2 i.e., A n = 2 1 nπ sinh nπb With these A n, the solution is u(x, y) = n=0 nπb ] sinh cos 0 0 f (x) cos f (x) cos ( nπx ), ( nπx ) dx. ( nπx ) A n cos cosh nπy. ( nπx ) dx. (11)