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Heat & Wave Equation in a Rectangle Section 12.8 1 Heat

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Heat & Wave Equation in a Rectangle Section 12.8 1 Heat Heat & Wave Equation in a Rectangle Section 12.8 1 Heat Equation in a Rectangle In this section we are concerned with application of the method of separation of variables ap- plied to the heat equation in two spatial dimensions. In particular we will consider problems in a rectangle. Thus we consider ut(x; y; t) = k (uxx(x; y; t) + uyy(x; y; t)) ; t > 0; (x; y) 2 [0; a] × [0; b]; (1.1) u(0; y; t) = 0; u(a; y; t) = 0; u(x; 0; t) = 0; u(x; b; t) = 0 u(x; y; 0) = f(x; y) u(x; y) = X(x)Y (y)T (t): Substituting into (2.1) and dividing both sides by kX(x)Y (y)T (t) gives T 0(t) Y 00(y) X00(x) = + kT (t) Y (y) X(x) Since the left side is independent of x, y and the right side is independent of t, it follows that the expression must be a constant: T 0(t) Y 00(y) X00(x) = + = λ. kT (t) Y (y) X(x) We seek to find all possible constants λ and the corresponding nonzero functions T , X and Y . We obtain T 0(t) − kλT (t) = 0; 1 and X00(x) Y 00(y) = λ − : X(x) Y (y) But since the left hand side depends only on x and the right hand side only on y, we conclude that there is a constant α X00 − αX = 0: On the other hand we could also write Y 00(y) X00(x) = λ − Y (y) X(x) so there exists a constant β so that Y 00 − βY = 0: Thus we have X00 − αX = 0;Y 00 − βY = 0;T 0(t) − kλT (t) = 0 and λ = α + β: Furthermore, the boundary conditions give X(0)Y (y) = 0;X(a)Y (y) = 0; for all y: Since Y (y) is not identically zero we obtain the desired eigenvalue problem X00(x) − αX(x) = 0;X(0) = 0;X(a) = 0: (1.2) We have solved this problem many times and we have α = −µ2 so that X(x) = c1 cos(µx) + c2 sin(µx): Applying the boundary conditions we have 0 = X(0) = c1 ) c1 = 0 0 = X(a) = c2 sin(µa): From this we conclude sin(µa) = 0 which implies nπ µ = a and therefore r nπ 2 2 α = −µ2 = − ;X (x) = sin(µ x); n = 1; 2; ··· : (1.3) n n a n a n Now from the boundary condition X(x)Y (0) = 0;X(x)Y (b) = 0 for all x: 2 This gives the problem Y 00(y) − βY (y) = 0;Y (0) = 0;Y (b) = 0: (1.4) This is the same as the problem (2.2) so we obtain eigenvalues and eigenfunctions r mπ 2 2 β = −ν2 = − ;Y (y) = sin(ν y); n = 1; 2; ··· : (1.5) m m b m b m So we obtain eigenvalues of the main problem given by nπ 2 mπ 2 λ = − + (1.6) n;m a b and corresponding eigenfunctions 2 'n;m(x; y) = p sin(µnx) sin(νmy). ab 0 We also find the solution to T (t) − kλn;mT (t) = 0 is given by T (t) = ekλn;mt: So we look for u as an infinite sum 1 2 X nπx mπy u(x; y; t) = p c ekλn;mt sin sin : (1.7) n;m a b ab n;m=1 The only problem remaining is to somehow pick the constants cn;m so that the initial condi- tion u(x; y; 0) = f(x; y) is satisfied, i.e., 1 X f(x; y) = u(x; y; 0) = cn;m'n;m(x; y): (1.8) n;m=1 with 2 Z b Z a nπx mπy cn;m = p f(x; y) sin sin dx dy ab 0 0 a b for n = 1; 2; ··· , m = 1; 2; ··· . Example 1.1 (Dirichlet BCs). To simplify the problem a bit we set a = 1 and b = 1. Namely we consider ut(x; y; t) = k (uxx(x; y; t) + uyy(x; y; t)) ; t > 0; (x; y) 2 [0; 1] × [0; 1] (1.9) u(0; y; t) = 0; u(1; y; t) = 0; u(x; 0; t) = 0; u(x; 1; t) = 0 u(x; y; 0) = x(1 − x)y(1 − y) 3 In this case we obtain eigenvalues 2 2 2 2 2 2 2 λn;m = −π (n + m ); αn = −π n ; βm = −π m ; n; m = 1; 2; ··· : The corresponding eigenfunctions are given by p p Xn(x) = 2 sin(nπx);Ym(y) = 2 sin(mπy): Our solution is given by 1 X kλn;mt u(x; y; t) = 2 cn;me sin(nπx) sin(mπy): n=1 The coefficients cn;m are obtained from 1 X x(1 − x)y(1 − y) = 2 cn;m sin(nπx) sin(mπy): n;m=1 We have Z 1 Z 1 8((−1)n − 1)((−1)m − 1) cn;m = 2 x(1 − x)y(1 − y) sin(nπx) sin(mπy) dx dy = 3 3 6 : 0 0 n m π 1 X kλn;mt u(x; y; t) = 2 cn;me sin (nπx) sin (mπy) : (1.10) n;m=1 That is 1 16 X ((−1)n − 1)((−1)m − 1)ekλn;mt u(x; y; t) = sin (nπx) sin (mπy) : (1.11) π6 n3m3 n;m=1 Example 1.2 (Mixed Dirichlet and Neumann BCs). To simplify the problem a bit again set a = 1 and b = 1. Namely we consider ut(x; y; t) = k (uxx(x; y; t) + uyy(x; y; t)) ; t > 0; (x; y) 2 [0; 1] × [0; 1] (1.12) u(0; y; t) = 0; u(1; y; t) = 0; uy(x; 0; t) = 0; uy(x; 1; t) = 0 u(x; y; 0) = x(1 − x)y In this case we obtain eigenvalues 2 2 λn;0 = −π n ;Y0(y) = 1; and 2 2 2 2 2 2 2 λn;m = −π (n + m ); αn = −π n ; βm = −π m ; n; m = 1; 2; ··· ; with corresponding eigenfunctions are given by p p Xn(x) = 2 sin(nπx);Ym(y) = 2 cos(mπy): 4 Our solution is given by 1 p 1 p p X kλn;0t X kλn;mt u(x; y; t) = cn;0e 2 sin(nπx) + cn;me 2 sin(nπx) 2 cos(mπy): n=1 n;m=1 Setting t = 0 we obtain 1 1 X p X p p x(1 − x)y = cn;0 2 sin(nπx) + cn;m 2 sin(nπx) 2 cos(mπy): n=1 n;m=1 This double Fourier series is evaluated again using orthogonality relations. We have Z 1 Z 1 −8((−1)n − 1)((−1)m − 1) cn;m = 2 x(1 − x)y sin(nπx) cos(mπy) dx dy = 3 2 5 : 0 0 n m π Finally we obtain the coefficients cn;0 from p p Z 1 Z 1 − 2((−1)n − 1) cn;0 = 2 x(1 − x)y sin(nπx) dx dy = 3 3 : 0 0 n π 1 X (1 − (−1)n) u(x; y; t) = 2 ekλn;0t sin(nπx) n3π3 n=1 1 16 X ((−1)n − 1)((−1)m − 1)ekλn;mt − sin(nπx) cos(mπy): π5 n3m2 n;m=1 2 Wave Equation in Higher Dimensions In this section we are concerned with application of the method of separation of variables applied to the wave equation in a two dimensional rectangle. Thus we consider 2 utt(x; y; t) = c (uxx(x; y; t) + uyy(x; y; t)) ; t > 0; (x; y) 2 [0; a] × [0; b]; (2.1) u(0; y; t) = 0; u(a; y; t) = 0; u(x; 0; t) = 0; u(x; b; t) = 0 u(x; y; 0) = f(x; y); ut(x; y; 0) = g(x; y) 5 u(x; y) = X(x)Y (y)T (t): Substituting into (2.1) and dividing both sides by X(x)Y (y) gives T 00(t) Y 00(y) X00(x) = + c2T (t) Y (y) X(x) Since the left side is independent of x, y and the right side is independent of t, it follows that the expression must be a constant: T 00(t) Y 00(y) X00(x) = + = λ. c2T (t) Y (y) X(x) We seek to find all possible constants λ and the corresponding nonzero functions T , X and Y . We obtain X00(x) Y 00(y) = λ − T 00(t) − c2λT (t) = 0: X(x) Y (y) Thus we conclude that there is a constant α X00 − αX = 0: On the other hand we could also write Y 00(y) X00(x) = λ − Y (y) X(x) so there exists a constant β so that Y 00 − βY = 0: Furthermore, the boundary conditions give X(0)Y (y) = 0;X(a)Y (y) = 0 for all y: Since Y (y) is not identically zero we obtain the desired eigenvalue problem X00(x) − αX(x) = 0;X(0) = 0;X(a) = 0: (2.2) We have solved this problem many times and we have α = −µ2 so that X(x) = c1 cos(µx) + c2 sin(µx): Applying the boundary conditions we have 0 = X(0) = c1 ) c1 = 0 0 = X(a) = c2 sin(µa): From this we conclude sin(µa) = 0 which implies nπ µ = a 6 and therefore r nπ 2 2 α = −µ2 = − ;X (x) = sin(µ x); n = 1; 2; ··· :: (2.3) n n a n a n Now from the boundary condition X(x)Y (0) = 0;X(x)Y (b) = 0 for all x: This gives the problem Y 00(y) − βY (y) = 0;Y (0) = 0;Y (b) = 0: (2.4) This is the same as the problem (2.2) so we obtain eigenvalues and eigenfunctions r mπ 2 2 β = −ν2 = − ;Y (y) = sin(ν y); n = 1; 2; ··· :: (2.5) m m b m b m So we obtain eigenvalues of the main problem given by nπ 2 mπ 2 λ = − + (2.6) n;m a b and corresponding eigenfunctions 2 'n;m(x; y) = p sin(µnx) sin(νmy): ab 00 2 We also find the solution to T (t) − c λn;mT (t) = 0 is given by Tn;m(t) = [an;m cos(c!n;mt) + bn;m sin(c!n;mt)] where we have defined r nπ 2 mπ 2 ! = + : n;m a b So we look for u as an infinite sum 1 2 X nπx mπy u(x; y; t) = p [a cos(c! t) + b sin(c! t)] sin sin : (2.7) n;m n;m n;m n;m a b ab n;m=1 We have left to find the constants an;m and bn;m so that the initial condition u(x; y; 0) = f(x; y) and ut(x; y; 0) = g(x; y) are satisfied, i.e., 1 2 X nπx mπy f(x; y) = u(x; y; 0) = p a sin sin : (2.8) n;m a b ab n;m=1 7 Thus we conclude that 2 Z b Z a nπx mπy an;m = p f(x; y) sin sin dx dy ab 0 0 a b for n = 1; 2; ··· , m = 1; 2; ··· .
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