Separation of variables in two dimensions. Overview of method: Consider linear, homogeneous equation for u(v 1, v 2 )

Transcription

1 Separation of variables in two dimensions Overview of method: Consider linear, homogeneous equation for u(v 1, v 2 )

2 Separation of variables in two dimensions Overview of method: Consider linear, homogeneous equation for u(v 1, v 2 ) Domain (v 1, v 2 ) [a, b] [c, d] (rectangles, disks, wedges, annuli)

3 Separation of variables in two dimensions Overview of method: Consider linear, homogeneous equation for u(v 1, v 2 ) Domain (v 1, v 2 ) [a, b] [c, d] (rectangles, disks, wedges, annuli) Only linear, homogeneous equations and homogeneous boundary conditions at v 1 = a, v 1 = b

4 Separation of variables in two dimensions Overview of method: Consider linear, homogeneous equation for u(v 1, v 2 ) Domain (v 1, v 2 ) [a, b] [c, d] (rectangles, disks, wedges, annuli) Only linear, homogeneous equations and homogeneous boundary conditions at v 1 = a, v 1 = b ook for separated solutions u = f (v 1 )g(v 2 )

5 Separation of variables in two dimensions Overview of method: Consider linear, homogeneous equation for u(v 1, v 2 ) Domain (v 1, v 2 ) [a, b] [c, d] (rectangles, disks, wedges, annuli) Only linear, homogeneous equations and homogeneous boundary conditions at v 1 = a, v 1 = b ook for separated solutions u = f (v 1 )g(v 2 ) Superpositions of separated solutions will give entire solution.

6 Separation of variables in two dimensions Overview of method: Consider linear, homogeneous equation for u(v 1, v 2 ) Domain (v 1, v 2 ) [a, b] [c, d] (rectangles, disks, wedges, annuli) Only linear, homogeneous equations and homogeneous boundary conditions at v 1 = a, v 1 = b ook for separated solutions u = f (v 1 )g(v 2 ) Superpositions of separated solutions will give entire solution. Other boundary/initial conditions will determine coefficients of superposition.

7 The separation principle Suppose we have independent variables x 1, x 2,..., x n and functions f 1 (x 1 ), f 2 (x 2 ),..., f n (x n ) of each variables separately.

8 The separation principle Suppose we have independent variables x 1, x 2,..., x n and functions f 1 (x 1 ), f 2 (x 2 ),..., f n (x n ) of each variables separately. Suppose f 1 (x 1 ) + f 2 (x 2 ) f n (x n ) = 0, for all (x 1, x 2,..., x n ) Ω Taking partial derivatives gives f i (x i) = 0.

9 The separation principle Suppose we have independent variables x 1, x 2,..., x n and functions f 1 (x 1 ), f 2 (x 2 ),..., f n (x n ) of each variables separately. Suppose f 1 (x 1 ) + f 2 (x 2 ) f n (x n ) = 0, for all (x 1, x 2,..., x n ) Ω Taking partial derivatives gives f i (x i) = 0. It follows each function is a constant: f i (x i ) = λ i ; these are called separation constants.

10 Example: the wave equation u tt = c 2 u xx, u(0, t) = 0 = u(, t), u(x, 0) = φ(x), u t (x, 0) = ψ(x) Note domain is (x, t) [0, ] [0, ).

11 Example: the wave equation u tt = c 2 u xx, u(0, t) = 0 = u(, t), u(x, 0) = φ(x), u t (x, 0) = ψ(x) Note domain is (x, t) [0, ] [0, ). Insert u(x, t) = X(x)T (t) into equation in (3) gives XT = c 2 TX.

12 Example: the wave equation u tt = c 2 u xx, u(0, t) = 0 = u(, t), u(x, 0) = φ(x), u t (x, 0) = ψ(x) Note domain is (x, t) [0, ] [0, ). Insert u(x, t) = X(x)T (t) into equation in (3) gives XT = c 2 TX. Dividing separates variables; each side equals a constant T c 2 T = X X = λ,

13 Example: the wave equation u tt = c 2 u xx, u(0, t) = 0 = u(, t), u(x, 0) = φ(x), u t (x, 0) = ψ(x) Note domain is (x, t) [0, ] [0, ). Insert u(x, t) = X(x)T (t) into equation in (3) gives XT = c 2 TX. Dividing separates variables; each side equals a constant T c 2 T = X X = λ, Inserting u(x, t) = X(x)T (t) into homogeneous boundary conditions X(0) = 0 = X(). Get two ODEs: X + λx = 0, X(0) = 0 = X(), T + c 2 λt = 0.

14 Example: the wave equation cont. Solution of eigenvalue problem: X n = sin ( nπx ) ( nπ, λ n = ) 2, n = 1, 2, 3,...

15 Example: the wave equation cont. Solution of eigenvalue problem: X n = sin ( nπx ) ( nπ, λ n = ) 2, n = 1, 2, 3,... Solutions for time equation are T = sin(c λt) or T = cos(c λt), so for each n get two solutions sin(cnπt/), cos(cnπt/)

16 Example: the wave equation cont. Solution of eigenvalue problem: X n = sin ( nπx ) ( nπ, λ n = ) 2, n = 1, 2, 3,... Solutions for time equation are T = sin(c λt) or T = cos(c λt), so for each n get two solutions sin(cnπt/), cos(cnπt/) Therefore all possible separated solutions are sin(cnπt/) sin(nπx/), cos(cnπt/) sin(nπx/), n = 1, 2, 3,....

17 Example: the wave equation cont. Superposition of separated solutions: u(x, t) = ( nπx ) [A n cos(nπct/) + B n sin(nπct/)] sin. n=1

18 Example: the wave equation cont. Superposition of separated solutions: u(x, t) = ( nπx ) [A n cos(nπct/) + B n sin(nπct/)] sin. n=1 Invoking the initial conditions φ(x) = ( nπx ) A n sin, ψ(x) = n=1 n=1 ( nπc ) ( nπx ) B n sin. These are orthogonal expansions, so coefficients are found by taking inner products with each eigenfunctions X n = sin(nπx/): A n = φ, X n X n, X n, B n = ( ) ψ, Xn nπc X n, X n.

19 Example: the wave equation cont. Meaning of solution: Solution is composed of standing waves

20 Example: the wave equation cont. Meaning of solution: Solution is composed of standing waves Often most important feature: characteristic frequencies ω n = nπc, n = 1, 2, 3,... These special frequencies form basis for sound waves, atomic spectra, elastic vibrations, etc.

21 Example: the wave equation cont. Meaning of solution: Solution is composed of standing waves Often most important feature: characteristic frequencies ω n = nπc, n = 1, 2, 3,... These special frequencies form basis for sound waves, atomic spectra, elastic vibrations, etc. Notice longer strings have smaller frequencies.

22 Example 2: the diffusion equation u t = Du xx, u(0, t) = 0 = u(, t), u(x, 0) = φ(x)

23 Example 2: the diffusion equation u t = Du xx, u(0, t) = 0 = u(, t), u(x, 0) = φ(x) Insert u(x, t) = X(x)T (t) into equation gives T /(DT ) = X /X = λ

24 Example 2: the diffusion equation u t = Du xx, u(0, t) = 0 = u(, t), u(x, 0) = φ(x) Insert u(x, t) = X(x)T (t) into equation gives T /(DT ) = X /X = λ Same eigenvalue problem but now equation for T is T = DλT, whose solutions are T = exp( Dλt).

25 Example 2: the diffusion equation u t = Du xx, u(0, t) = 0 = u(, t), u(x, 0) = φ(x) Insert u(x, t) = X(x)T (t) into equation gives T /(DT ) = X /X = λ Same eigenvalue problem but now equation for T is T = DλT, whose solutions are T = exp( Dλt). Separated solutions are therefore exp( D(nπ/) 2 t) sin(nπx/)

26 Example 2: the diffusion equation u t = Du xx, u(0, t) = 0 = u(, t), u(x, 0) = φ(x) Insert u(x, t) = X(x)T (t) into equation gives T /(DT ) = X /X = λ Same eigenvalue problem but now equation for T is T = DλT, whose solutions are T = exp( Dλt). Separated solutions are therefore exp( D(nπ/) 2 t) sin(nπx/) Superposition of separated solutions is ( nπx ) u(x, t) = A n exp( D(nπ/) 2 t) sin. n=1

27 Example 2: the diffusion equation u t = Du xx, u(0, t) = 0 = u(, t), u(x, 0) = φ(x) Insert u(x, t) = X(x)T (t) into equation gives T /(DT ) = X /X = λ Same eigenvalue problem but now equation for T is T = DλT, whose solutions are T = exp( Dλt). Separated solutions are therefore exp( D(nπ/) 2 t) sin(nπx/) Superposition of separated solutions is ( nπx ) u(x, t) = A n exp( D(nπ/) 2 t) sin. n=1 Invoking the initial condition φ(x) = A n sin n=1 ( nπx ), A n = φ, X n X n, X n,

28 Example 3: the aplace equation u xx +u yy = 0, u(0, y) = 0 = u(, y), u(x, 0) = h(x), u(x, H) = g(x)

29 Example 3: the aplace equation u xx +u yy = 0, u(0, y) = 0 = u(, y), u(x, 0) = h(x), u(x, H) = g(x) Separating u = X(x)Y (y) leads to Y /Y = X /X = λ.

30 Example 3: the aplace equation u xx +u yy = 0, u(0, y) = 0 = u(, y), u(x, 0) = h(x), u(x, H) = g(x) Separating u = X(x)Y (y) leads to Y /Y = X /X = λ. Same eigenvalue problem for X, and Y = λy.

31 Example 3: the aplace equation u xx +u yy = 0, u(0, y) = 0 = u(, y), u(x, 0) = h(x), u(x, H) = g(x) Separating u = X(x)Y (y) leads to Y /Y = X /X = λ. Same eigenvalue problem for X, and Y = λy. Two linearly independent solutions Y = exp( λy) and Y = exp( λy).

32 Example 3: the aplace equation u xx +u yy = 0, u(0, y) = 0 = u(, y), u(x, 0) = h(x), u(x, H) = g(x) Separating u = X(x)Y (y) leads to Y /Y = X /X = λ. Same eigenvalue problem for X, and Y = λy. Two linearly independent solutions Y = exp( λy) and Y = exp( λy). Separated solutions are therefore exp(nπy/) sin(nπx/) and exp( nπy/) sin(nπx/)

33 Example 3: the aplace equation u xx +u yy = 0, u(0, y) = 0 = u(, y), u(x, 0) = h(x), u(x, H) = g(x) Separating u = X(x)Y (y) leads to Y /Y = X /X = λ. Same eigenvalue problem for X, and Y = λy. Two linearly independent solutions Y = exp( λy) and Y = exp( λy). Separated solutions are therefore exp(nπy/) sin(nπx/) and exp( nπy/) sin(nπx/) Superposition of these is u(x, y) = ( nπx ) [A n exp(nπy/)+b n exp( nπy/)] sin. n=1

34 Example 3, cont. Satisfy the inhomogeneous boundary conditions by setting y = 0 and y = H, g(x) = n=1 h(x) = ( nπx ) (A n + B n ) sin n=1 ( nπx ) [A n exp(nπh/) + B n exp( nπh/)] sin.

35 Example 3, cont. Satisfy the inhomogeneous boundary conditions by setting y = 0 and y = H, g(x) = n=1 h(x) = ( nπx ) (A n + B n ) sin n=1 ( nπx ) [A n exp(nπh/) + B n exp( nπh/)] sin. Taking inner products with the eigenfunctions X n = sin(nπx/), one gets a system of two equations for each pair A n, B n A n +B n = h, X n X n, X n, A n exp(nπh/)+b n exp( nπh/) = g, X n X n, X n