Math 241: More heat equation/laplace equation

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1 Math 241: More heat equation/aplace equation D. DeTurck University of Pennsylvania September 27, 2012 D. DeTurck Math C: Heat/aplace equations 1 / 13

2 Another example Another heat equation problem: u t = 1 2 u xx, u(0, t) = 0, u x (, t) = 0, u(x, 0) = 2x x 2 for t > 0 and 0 x. D. DeTurck Math C: Heat/aplace equations 2 / 13

3 Another example Another heat equation problem: u t = 1 2 u xx, u(0, t) = 0, u x (, t) = 0, u(x, 0) = 2x x 2 for t > 0 and 0 x. To match the boundary conditions this time, we ll assume that we can express u as u(x, t) = n=0 b n e (n+ 1 2 )2 π 2 t/4 2 sin (n )πx and see if we can figure out what the constants b n should be we know that the boundary conditions are automatically satisfied, and perhaps we can choose the b n s so that 2x x 2 = n=0 b n sin (n )πx. D. DeTurck Math C: Heat/aplace equations 2 / 13

4 Integral facts We re trying to find b n s so that 2x x 2 = n=0 b n sin (n )πx. D. DeTurck Math C: Heat/aplace equations 3 / 13

5 Integral facts We re trying to find b n s so that 2x x 2 = We ll use two basic facts: If n m then n=0 b n sin (n )πx. ˆ 0 If n = m then sin (n )πx sin (m )πx dx = 0. ˆ 0 sin (n )πx sin (m )πx dx = ˆ 0 sin 2 (n )πx dx = 2. D. DeTurck Math C: Heat/aplace equations 3 / 13

6 Finding the coefficients We re still trying to find b n s so that 2x x 2 = b n sin (n )πx. n=0 Motivated by the facts on the previous slide, we multiply both sides by sin (m 1 )πx 2 and integrate both sides from 0 to : ˆ 0 (2x x 2 ) sin (m )πx dx ˆ ( = b n sin (n )πx ) 0 n=0 ˆ = b n sin (n )πx n=0 = b m 2 0 sin (m )πx sin (m )πx K D. DeTurck Math C: Heat/aplace equations 4 / 13 dx dx

7 Integration by parts It s an exercise in integration by parts to show that ˆ 0 (2x x 2 ) sin (m )πx dx = 8 3 (m )3 π 3 D. DeTurck Math C: Heat/aplace equations 5 / 13

8 Integration by parts It s an exercise in integration by parts to show that ˆ 0 (2x x 2 ) sin (m )πx dx = 8 3 (m )3 π 3 Therefore, b m = 32 2 (2m + 1) 3 π 3 D. DeTurck Math C: Heat/aplace equations 5 / 13

9 Integration by parts It s an exercise in integration by parts to show that ˆ 0 (2x x 2 ) sin (m )πx dx = 8 3 (m )3 π 3 Therefore, b m = 32 2 (2m + 1) 3 π 3 So we arrive at a candidate for the solution: u(x, t) = n= (2n + 1) 3 π 3 e (n+ 1 2 )2 π 2 t/2 2 sin (n )πx D. DeTurck Math C: Heat/aplace equations 5 / 13

10 Validating the solution The series u(x, t) = n= (2n + 1) 3 π 3 e (n+ 1 2 )2 π 2 t/2 2 sin (n )πx converges for t 0, and certainly satisfies the boundary conditions. What about the initial condition u(x, 0) = 2x x 2? D. DeTurck Math C: Heat/aplace equations 6 / 13

11 Validating the solution The series u(x, t) = n= (2n + 1) 3 π 3 e (n+ 1 2 )2 π 2 t/2 2 sin (n )πx converges for t 0, and certainly satisfies the boundary conditions. What about the initial condition u(x, 0) = 2x x 2? Well,. u(x, 0) = n= (2n + 1) 3 π 3 sin (n )πx D. DeTurck Math C: Heat/aplace equations 6 / 13

12 Graphical evidence Use = 5.Red graph: 10x x 2, Blue graph: sum One term: Three terms: D. DeTurck Math C: Heat/aplace equations 7 / 13

13 Plotting the solution Here is a plot of the sum of of the first three terms of the solution: D. DeTurck Math C: Heat/aplace equations 8 / 13

14 aplace equation on a rectangle The two-dimensional aplace equation is u xx + u yy = 0. Solutions of it represent equilibrium temperature (squirrel, etc) distributions, so we think of both of the independent variables as space variables. D. DeTurck Math C: Heat/aplace equations 9 / 13

15 aplace equation on a rectangle The two-dimensional aplace equation is u xx + u yy = 0. Solutions of it represent equilibrium temperature (squirrel, etc) distributions, so we think of both of the independent variables as space variables. We ll solve the equation on a bounded region (at least at first), and it s appropriate to specify the values of u on the boundary (Dirichlet boundary conditions), or the values of the normal derivative of u at the boundary (Neumann conditions), or some mixture of the two. D. DeTurck Math C: Heat/aplace equations 9 / 13

16 Uniqueness proof #1 One uniqueness proof: suppose u = 0 on the boundary of the region (so u could arise as the difference between two solutions of the Dirichlet problem). D. DeTurck Math C: Heat/aplace equations 10 / 13

17 Uniqueness proof #1 One uniqueness proof: suppose u = 0 on the boundary of the region (so u could arise as the difference between two solutions of the Dirichlet problem). Since (u u) = u u + ( u) ( u), the divergence theorem tells us: ˆˆ ˆˆ u 2 da = u u n ds u 2 u da. R R R D. DeTurck Math C: Heat/aplace equations 10 / 13

18 Uniqueness proof #1 One uniqueness proof: suppose u = 0 on the boundary of the region (so u could arise as the difference between two solutions of the Dirichlet problem). Since (u u) = u u + ( u) ( u), the divergence theorem tells us: ˆˆ ˆˆ u 2 da = u u n ds u 2 u da. R R R But the right side is zero because u = 0 on R (the boundary of R) and because 2 = 0 throughout R. D. DeTurck Math C: Heat/aplace equations 10 / 13

19 Uniqueness proof #1 One uniqueness proof: suppose u = 0 on the boundary of the region (so u could arise as the difference between two solutions of the Dirichlet problem). Since (u u) = u u + ( u) ( u), the divergence theorem tells us: ˆˆ ˆˆ u 2 da = u u n ds u 2 u da. R R R But the right side is zero because u = 0 on R (the boundary of R) and because 2 = 0 throughout R. So we conclude u is constant, and thus zero since u = 0 on the boundary. D. DeTurck Math C: Heat/aplace equations 10 / 13

20 Case 1: R is a rectangle We ll first solve the Dirichlet problem on a rectangle, say 0 x and 0 y M. D. DeTurck Math C: Heat/aplace equations 11 / 13

21 Case 1: R is a rectangle We ll first solve the Dirichlet problem on a rectangle, say 0 x and 0 y M. et s see what we get by separation of variables: et u(x, y) = X (x)y (y), then X Y + XY = 0, or X X = Y Y = λ a constant because X /X is a function of x alone and Y /Y is a function of y alone. D. DeTurck Math C: Heat/aplace equations 11 / 13

22 Case 1: R is a rectangle We ll first solve the Dirichlet problem on a rectangle, say 0 x and 0 y M. et s see what we get by separation of variables: et u(x, y) = X (x)y (y), then X Y + XY = 0, or X X = Y Y = λ a constant because X /X is a function of x alone and Y /Y is a function of y alone. The boundary data might look like: u(x, 0) = f (x), u(x, M) = h(x), u(, y) = g(y) u(0, y) = k(y) but that s too much to deal with all at once. D. DeTurck Math C: Heat/aplace equations 11 / 13

23 One side at a time You can add together four solutions to problems with boundary data like: u(x, 0) = 0, u(, y) = 0 u(x, M) = 0, u(0, y) = k(y) D. DeTurck Math C: Heat/aplace equations 12 / 13

24 One side at a time You can add together four solutions to problems with boundary data like: u(x, 0) = 0, u(, y) = 0 u(x, M) = 0, u(0, y) = k(y) Since u is zero for two values of y, we ll use Y + α 2 Y = 0, so X α 2 X = 0. This gives ( nπy ) Y = sin n = 1, 2, 3... M and α = nπ/m. D. DeTurck Math C: Heat/aplace equations 12 / 13

25 One side at a time You can add together four solutions to problems with boundary data like: u(x, 0) = 0, u(, y) = 0 u(x, M) = 0, u(0, y) = k(y) Since u is zero for two values of y, we ll use Y + α 2 Y = 0, so X α 2 X = 0. This gives ( nπy ) Y = sin n = 1, 2, 3... M and α = nπ/m. A useful way to write the corresponding X solutions is ( nπ ) X = sinh M ( x) so that X () = 0. D. DeTurck Math C: Heat/aplace equations 12 / 13

26 Fourier series So we now have u(x, y) = n=1 and we have to match: u(0, y) = k(y) = ( nπ ) ( nπy ) b n sinh M ( x) sin. M b n sinh n=1 ( ) nπ sin M ( nπy ). M D. DeTurck Math C: Heat/aplace equations 13 / 13

27 Fourier series So we now have u(x, y) = n=1 and we have to match: u(0, y) = k(y) = ( nπ ) ( nπy ) b n sinh M ( x) sin. M b n sinh n=1 ( ) nπ sin M ( nπy ). M Thus b n = 2 M sinh ( ) nπ M ˆ M 0 k(y) sin ( nπy ) dy M D. DeTurck Math C: Heat/aplace equations 13 / 13

28 aplace on a disk Next up is to solve the aplace equation on a disk with boundary values prescribed on the circle that bounds the disk. We ll use polar coordinates for this, so a typical problem might be: 2 u = 1 ( r u ) u r r r r 2 θ 2 = 0 on the disk of radius R = 3 centered at the origin, with boundary condition { 1 0 θ π u(3, θ) = sin 2 θ π < θ < 2π D. DeTurck Math C: Heat/aplace equations 14 / 13

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