Partial Differential Equations for Engineering Math 312, Fall 2012

Transcription

1 Partial Differential Equations for Engineering Math 312, Fall 2012 Jens Lorenz July 17, 2012 Contents Department of Mathematics and Statistics, UNM, Albuquerque, NM Second Order ODEs with Constant Coefficients General Solution Initial Value Problems Boundary Value Problems Eigenvalue Problems Second Order Cauchy Euler Equations General Solution The 1D Wave Equation The One Way Wave Equation Derivation of the 1D Wave Equation Solution of the Cauchy Problem for the 1D Wave Equation Remarks on the Wave Equation in 3D and 2D The Heat Equation Derivation Initial and Boundary Conditions The Heat Kernel Solution in 1D BVPs for Laplace s Equation Problem Problem Problem Sine Series Expansions Separation of Variables for the Heat Equation Separating the Time Dependence An Initial Boundary Value Problem for the 1D Heat Equation. 22 1

2 7 The Wave Equation and Helmholtz s Equation for a Circle Separating the Time Dependence The Laplacian in Polar Coordinates Separating the Helmholtz s Equation in Polar Coordinates

3 1 Second Order ODEs with Constant Coefficients Read Section 1.1 of Snider. 1.1 General Solution Consider the differential operator L defined by Ly(x) = ay (x) + by (x) + cy(x) where a, b, c are real or complex constants and a 0. We want to determine the general solution of the homogeneous equation Using the ansatz Ly = 0. y(x) = e px, where p is a parameter, one obtains the characteristic equation The solutions are ap 2 + bp + c = 0. p 1,2 = b 2a ± 1 b 2a 2 4ac. Case 1: b 2 4ac 0 In this case, p 1 p 2 and the general solution of the equation Ly = 0 is y(x) = c 1 e p 1x + c 2 e p 2x. Here c 1 and c 2 are arbitrary constants. Case 2: b 2 4ac = 0 In this case, p 1 = p 2 = p = b 2a. The general solution of Ly = 0 is Here c 1 and c 2 are arbitrary constants. y(x) = c 1 e px + c 2 xe px. Remarks: If p = α + iβ with real α, β, then e px = e αx e iβx = e αx (cos βx + i sin βx). Thus, the real part of p determines the exponential behavior of the function e px and the imaginary part of p determines the wavelength of the oscillatory behavior of e px. Note that the wavelength of the function cos βx is L = 2π/ β. One calls β the wave number of the oscillation cos βx. Roughly speaking, β is the number of waves of the oscillation if x varies from 0 to 2π. If the independent variable is time, we write 3

4 e pt = e αt e iβt = e αt (cos βt + i sin βt). In this case, the oscillation cos βt has period T = 2π/ β and one calls β the frequency of the oscillation. 1.2 Initial Value Problems Consider the equation y (t) = λy(t), where λ is a real constant. We want to determine the solution satisfying the initial conditions y(0) = b 0, y (0) = b 1. Case 1: λ = ω 2 > 0, ω > 0. The general solution is The initial conditions require y(t) = c 1 e ωt + c 2 e ωt. y(0) = c 1 + c 2 = b 0, y (0) = c 1 ω c 2 ω = b 1. These are two conditions for the constants c 1 and c 2. We write these in matrix form as ( ) ( ) ( ) 1 1 c1 b0 =. ω ω Note that the determinant of the matrix is One finds that c 2 deta = 2ω 0. c 1 = 1 2 b ω b 1, c 2 = 1 2 b 0 1 2ω b 1. Thus, the solution of the initial value problem is b 1 e ωt + e ωt y(t) = b 0 2 Recall the definitions + b 1 e ωt e ωt ω 2. Using these notations, cosh τ = 1 2 (eτ + e τ ), sinh τ = 1 2 (eτ e τ ). y(t) = b 0 cosh(ωt) + b 1 ω sinh(ωt). 4

5 Case 2: λ = ω 2 < 0, ω > 0. The general solution is The initial conditions require y(t) = c 1 e iωt + c 2 e iωt. y(0) = c 1 + c 2 = b 0, y (0) = c 1 iω c 2 iω = b 1. These are two conditions for the constants c 1 and c 2. We write these in matrix form as ( ) ( ) ( ) 1 1 c1 b0 =. iω iω Note that the determinant of the matrix is One finds that c 2 deta = 2iω 0. c 1 = 1 2 b iω b 1, c 2 = 1 2 b 0 1 2iω b 1. Thus, the solution of the initial value problem is b 1 Here y(t) = b 0 e iωt + e iωt 2 + b 1 e iωt e iωt ω 2i. One obtains: e iωt + e iωt 2 = cos ωt, e iωt e iωt 2i = sin ωt. y(t) = b 0 cos(ωt) + b 1 ω sin(ωt). Case 3: λ = 0. In this case, the general solution is y(t) = c 1 + c 2 t. The solution of the initial value problem is 1.3 Boundary Value Problems Consider the equation y(t) = b 0 + b 1 t. y (x) = λy(x), 0 x π, where λ is a real constant. We want to determine the solution satisfying the boundary conditions 5

6 y(0) = b 0, y(π) = b 1. Case 1: λ = ω 2 > 0, ω > 0. The general solution is The boundary conditions require y(x) = c 1 e ωx + c 2 e ωx. y(0) = c 1 + c 2 = b 0, y(π) = c 1 e ωπ + c 2 e ωπ = b 1. These are two conditions for the constants c 1 and c 2. We write these in matrix form as ( ) ( ) ( ) 1 1 c1 b0 =. e ωπ e ωπ We see that the determinant of the matrix is c 2 deta = e ωπ e ωπ 0. Therefore, the boundary value problem has a unique solution if λ > 0. Case 2: λ = ω 2 < 0, ω > 0. The general solution is The boundary conditions require y(t) = c 1 e iωt + c 2 e iωt. y(0) = c 1 + c 2 = b 0, y(π) = c 1 e iωπ + c 2 e iωπ = b 1. These are two conditions for the constants c 1 and c 2. We write these in matrix form as ( 1 1 e iωπ e iωπ ) ( c1 We see that the determinant of the matrix is c 2 ) = b 1 ( b0 b 1 deta = e iωπ e iωπ = 2i sin(ωπ). We see that the boundary value problem has a unique solution if and only if the positive number ω is not an integer. In case that ω = n {1, 2,...} we have deta = 0. Then, for most boundary data b 0, b 1, the above matrix equation has no solution and the boundary value problem has no solution. If it happens that the matrix equation is solvable, then the solution of the equation is not unique and the solution of the boundary value problem is not unique. Case 3: λ = 0. In this case, the general solution is y(t) = c 1 + c 2 t. One finds that the boundary value problem is uniquely solvable. 6 ).

7 1.4 Eigenvalue Problems It turns out that the exceptional case of Case 2 of the previous section is of great interest. We can phrase the case as follows: Under what conditions on λ does the boundary value problem y (x) = λy(x) for 0 x π, y(0) = y(π) = 0, (1.1) have a nontrivial solution? In general, one must consider complex λ. One obtains that the boundary value problem has a nontrivial solution if and only if λ = λ n = n 2, n = 1, 2,... The numbers λ n = n 2 are called the eigenvalues of the eigenvalue problem (1.1) and the functions are the corresponding eigenfunctions. y n = sin(nx), n = 1, 2,... 7

8 2 Second Order Cauchy Euler Equations Read Section 1.1 of Snider. 2.1 General Solution An equation of the form Ly(x) ax 2 y (x) + bxy (x) + cy(x) = 0, x > 0, where a, b, c are real or complex constants and a 0, is called a Cauchy Euler equation. To determine the general solution one uses the ansatz y(x) = x r where r is a parameter. One obtains the characteristic equation ar(r 1) + br + c = 0. If there are two distinct solutions, r 1 and r 2, then the general solution of Ly = 0 is y(x) = c 1 x r 1 + c 2 x r 2. If r 1 = r 2 = r, then the general solution is Example 1: x 2 y 2y = 0 The characteristics equation is with solutions The general solution is Example 2: x 2 y + xy + y = 0 The characteristics equation is with solutions The general solution is y(x) = c 1 x r + c 2 x r ln x. r(r 1) 2 = 0 r 1 = 2, r 2 = 1. y(x) = c 1 x 2 + c 2 x. r(r 1) + r + 1 = 0 r 1 = i, r 2 = i. 8

9 y(x) = c 1 x i + c 2 x i. Since the equation has only real coefficients, it is desirable to obtain a general solution in real form. We have x = e ln x, thus and x i = e i ln x = cos(ln x) + i sin(ln x) One obtains x i i ln x = e = cos(ln x) i sin(ln x) ( ) ( ) y(x) = c 1 cos(ln x) + i sin(ln x) + c 2 cos(ln x) i sin(ln x) = (c 1 + c 2 ) cos(ln x) + i(c 1 c 2 ) sin(ln x) Thus, we can write the general solution as y(x) = d 1 cos(ln x) + d 2 sin(ln x) where d 1 and d 2 are arbitrary constants. Note that the solution oscillates rapidly as x 0. 9

10 3 The 1D Wave Equation Read Section 4.3 of Snider. 3.1 The One Way Wave Equation The equation u t + cu x = 0 is known as the one way wave equation. Here c is a real constant. If we give the initial condition where f C 1, then u(x, 0) = f(x), x R, u(x, t) = f(x ct). Let x 0 be any fixed x value. If one considers the straight line in the (x, t) plane consisting of all points (x, t) with then, on this line, x ct = x 0, u(x, t) = u(x 0 + ct, t) = f(x 0 ). This says that the solution u(x, t) carries the value f(x 0 ) along the line x ct = x 0. In other words, the solution u(x, t) describes the movement of the initial function f(x) with speed c. 3.2 Derivation of the 1D Wave Equation Consider a string which in equilibrium occupies the interval 0 x L of the x axis. Let u(x, t) describe the displacement of the point at x at time t. If ρ is the density of the 1D string then the mass in the interval [x, x + x] is We will use Newton s law, M = ρ x. M u tt (x, t) = force, to derive an equation for u(x, t). Let T denote the tension in the string. The tension vectors are, to leading order, if u x (x, t) is small: One obtains the vertical force T(x, t) = T ( i u x (x, t)j) T(x + x, t) = T (i + u x (x + x, t)j) 10

11 ( ) T u x (x + x, t) u x (x, t) T u xx (x, t) x. This leads to ρu tt (x, t) = T u xx (x, t), or u tt (x, t) = c 2 u xx (x, t), c 2 = T/ρ. We have derived the 1D wave equation for a vibrating string. 3.3 Solution of the Cauchy Problem for the 1D Wave Equation We can write the wave equation as Here ( 2 t 2 c2 2 ) x 2 u(x, t) = 0. 2 t 2 c2 2 ( x 2 = t + c )( x t c ). x Therefore, if u(x, t) has the form u(x, t) = f(x ct) + g(x + ct), with smooth functions f and g, then u(x, t) solves the wave equation. We see that u(x, t) consists of a wave f(x ct) moving with speed c to the right (assuming c > 0) and of a wave g(x + ct) moving with speed c to the left. The Cauchy problem for the 1D wave equation is the following: Given smooth functions F (x) and G(x), find the solution u(x, t) of the wave equation u tt = c 2 u xx satisfying the initial conditions If u(x, 0) = F (x), u t (x, 0) = G(x) for x R. then the initial conditions require u(x, t) = f(x ct) + g(x + ct) u t (x, t) = cf (x ct) + cg (x + ct) f(x) + g(x) = F (x) cf(x) + cg(x) = In matrix form one obtains x 0 G(s) ds + C =: H(x) 11

12 Therefore, ( 1 1 c c ) ( f(x) g(x) ) = ( F (x) H(x) ). f(x) = 1 2 F (x) 1 2c H(x), g(x) = 1 2 F (x) + 1 2c H(x). For the solution of the Cauchy problem one obtains u(x, t) = f(x ct) + g(x + ct) The solution formula, = 1 2 F (x ct) 1 2c H(x ct) F (x + ct) + 1 H(x + ct) 2c = 1 ( ) F (x ct) + F (x + ct) + 1 x+ct G(s) ds 2 2c u(x, t) = 1 2 x ct ( ) F (x ct) + F (x + ct) + 1 2c x+ct x ct G(s) ds, is called d Alembert s formula. It is one of the rare cases of an explicit and rather simple general solution of a PDE. One can use the formula to discuss the domain of dependence and the domain of influence for the 1D wave equation. The discussion shows that disturbances travel at speed c in positive and negative direction. For F data (prescribing u at time zero), c is the sharp speed of propagation. For G data (prescribing u t at time zero), c is the maximal speed of propagation. 3.4 Remarks on the Wave Equation in 3D and 2D 12

13 4 The Heat Equation Read Section 4.5 of Snider. 4.1 Derivation Let x = (x, y, z) and t denote space and time coordinates and let u(x, t) denote the temperature at the point x at time t. (We assume that all quantities are already dimensionless.) Let q(x, t) denote the heat flux vector. Fourier s law says that q(x, t) = k u(x, t) where k is the thermal conductivity of the material. Consider a rectangular box B located near x with sides of length x, y, z. Denote the unit outward normal on the surface of B by n = n(x). We compute the heat energy H which flows into B in the time interval form t to t + t. The box B has six rectangular surfaces. Let us consider the sides S 1 and S 2 which are perpendicular to i. On S 1 we have n = i and on S 2 we have n = i. One obtains, to leading order, H 1 + H 2 = ( ) q(x + x, y, z, t) i + q(x, y, z, t) i y z t. Using Fourier s law: ( ) H 1 + H 2 = k u x (x + x, y, z, t) u x (x, y, z, t) y z t. Using the approximation we have, to leading order, u x (x + x, y, z, t) u x (x, y, z, t) u xx (x, y, z, t) x H 1 + H 2 = ku xx (x, y, z, t) V t where V is the volume of the box. Adding the heat flux through the other four sides of B yields where H = k 2 u(x, y, z, t) V t 2 u(x, y, z, t) = (u xx + u yy + u zz )(x, y, z, t). Heat flux into or out of B leads to a change of temperature in B. If ρ is the density of the material and c is its specific heat, then 13

14 Dividing by V and t yields ( ) cρ V u(x, y, z, t + t) u(x, y, z, t) H. We have derived the heat equation, cρ u t (x, y, z, t) = k 2 u(x, y, z, t). u t (x, y, z, t) = γ 2 u(x, y, z, t), γ = k cρ. Here the constants k, c, ρ depend on the material. 4.2 Initial and Boundary Conditions Let D denote a region in space with surface S. Let n = n(x) denote the unit outward normal on S. A reasonable IBVP (initial boundary value problem) for the heat equation is the following: Find a function u = u(x, y, z, t) so that u t (x, t) = γ 2 u(x, t) for x D, t 0 ; (4.1) u(x, 0) = F (x) for x D ; (4.2) u(x, t) = G(x) for x S, t 0. (4.3) Here F and G are given functions. Our problem consists of the PDE (4.1), the initial condition (4.2) and the boundary condition (4.3). One calls (4.3) a Dirichlet boundary condition. Instead of (4.3) one also uses a condition kn u(x, t) = H(x) for x S, t 0. (4.4) This condition prescribes the heat flux through S. If H = 0 then no heat flows through the surface S; the body if thermally insulated. One calls (4.4) a Neumann boundary condition. A third boundary condition is Newton s law of cooling: kn u(x, t) = C(u(x, t) F (x)) for x S, t 0. (4.5) Here F is the outside temperature and C is a constant. Condition (4.5) says that the heat flux through S is proportional to the temperature difference between the inside and the outside of S. Formally, for C =, one obtains the Dirichlet condition (4.2). For C = 0 one obtains the homogeneous Neumann condition corresponding to thermal insulation. 14

15 4.3 The Heat Kernel Solution in 1D Consider the Cauchy problem for the heat equation in one space dimension, The function u t = u xx, u(x, 0) = F (x). (4.6) G(x, t) = 1 4πt e x2 /4t, < x <, t > 0, is called the 1D heat kernel. If F is a bounded continuous function, the solution of (4.6) can be shown to be and u(x, t) = G(x y, t)f (y) dy for t > 0 (4.7) Discussion: An important formula is u(x, 0) = F (x) for t = 0. (4.8) J := e x2 dx = π. We show this by considering J 2 and using polar coordinates: J 2 = = = π = π = π. 2π The formula J = π implies that It is also not difficult to see that 0 e x2 y 2 dxdy e r2 r drdθ 2re r2 dr e s ds Because of (4.9) and (4.10) one often writes G(x, t) dx = 1 for t > 0. (4.9) lim G(x, t) = 0 for x 0. (4.10) t 0+ where δ 0 (x) is the delta function peaked at x = 0. Using calculus, it is not difficult to show: lim G(x, t) = δ 0(x) (4.11) t 0+ 15

16 Lemma 4.1 The heat kernel satisfies G t (x, t) = G xx (x, t) for t > 0. Then, differentiating under the integral sign, it follows that the function u(x, t) = G(x y, t)f (y) dy for t > 0 satisfies the heat equation, u t = u xx, for t > 0. The property (4.11) of the heat kernel G(x, t) makes it plausible that the function u(x, t) defined by (4.7), (4.8) is continuous at t = Derivation of the Heat Kernel via Fourier Transform 16

17 5 BVPs for Laplace s Equation Read Section 5.2 of Snider, in particular Problem 1, 2, 3 on pages Problem 1 Let D denote the square consisting of all points (x, y) with 0 < x < π, 0 < y < π in the (x, y) plane. We want to determine the solution of subject to the boundary conditions and u xx + u yy = 0 in D u(0, y) = u(π, y) = 0 for 0 < y < π (5.1) u(x, 0) = 0, u(x, π) = sin x for 0 < x π. We first ignore the boundary conditions and seek solutions of the pde in separated variables, One obtains u(x, y) = X(x)Y (y). X (x) = λx(x), Y (y) = λy (y) where λ is called the separation constant. Imposing the boundary conditions (5.1) leads to X(0) = X(π) = 0. A nontrivial solution of the ODE X = λx with these boundary conditions is obtained if and only if and then λ = λ n = n 2, n = 1, 2,... X n (x) = c n sin(nx). The corresponding Y equation is Y = n 2 Y with general solution Y n (y) = d 1 e ny + d 2 e ny. Thus, the functions ( u n (x, y) = c n sin(nx) d 1 e ny + d 2 e ny), n = 1, 2,... 17

18 satisfy the pde and the two homogeneous boundary conditions (5.1). boundary condition u(x, 0) = 0 applied to u n (x, y) yields d 2 = d 1. Thus ( u n (x, y) = c n sin(nx) e ny e ny), n = 1, 2,... The satisfies the pde and the three homogeneous boundary conditions. To satisfy the remaining boundary condition u(x, π) = sin x we let n = 1 and ( u(x, y) = c 1 sin(x) e y e y). One obtains the condition The solution of the bvp is 5.2 Problem 2 c 1 (e π e π ) = 1, c 1 = u(x, y) = We want to determine the solution of subject to the boundary conditions and 1 e π e π. 1 e π e π sin(x)(ey e y ). u xx + u yy = 0 in D u(x, 0) = u(x, π) = 0 for 0 < x < π (5.2) u(0, y) = 0, u(π, y) = sin 3 y for 0 < y < π. Proceeding as in Problem 1, one finds that the functions u n (x, y) = (e nx e nx ) sin(ny), n = 1, 2,... satisfy the pde and the three homogeneous boundary conditions. One can take linear combinations of these such functions: Any linear combination u(x, y) = a 1 u 1 (x, y) + a 2 u 2 (x, y) + a 3 u 3 (x, y) +... (5.3) will again satisfy the pde and the three homogeneous boundary conditions. We now try to find a linear combination which satisfies the inhomogeneous boundary condition u(π, y) = sin 3 (y). 18

19 We use Euler s identity to show that sin 3 (y) = 3 4 sin(y) 1 sin(3y). (5.4) 4 Proof: Let c = cos y, s = sin y. We have e 3iy = (e iy ) 3 = (c + is) 3 = c 3 + 3c 2 is 3cs 2 is 3 and e 3iy = cos(3y) + i sin(3y). Comparing the imaginary parts yields sin(3y) = 3c 2 s s 3 = 3(1 s 2 )s s 3 = 3s 4s 3 Solving for s 3 = sin 3 (y) we obtain the identity (5.4). Since the boundary function sin 3 (y) is a linear combination of sin y and sin(3y) we seek a solution of the bvp of the form Then u(x, y) will satisfy if u(x, y) = a 1 (e x e x ) sin(y) + a 3 (e 3x e 3x ) sin(3y). u(π, y) = sin 3 y a 1 (e π e π ) = 3 4 and a 3 (e 3π e 3π ) = 1 4. We can solve for a 1 and a 3 and obtain the solution u(x, y) = Problem 3 e x e x e π e π sin(y) 1 4 We want to determine the solution of subject to the boundary conditions u xx + u yy = 0 in D e 3x e 3x e 3π sin(3y). e 3π u(0, y) = u(π, y) = 0 for 0 < y < π (5.5) 19

20 and u(x, π) = 0, u(x, 0) = x(π x) for 0 < x < π. Proceeding as in Problem 1, one finds that the functions u n (x, y) = sin(nx)(e ny e 2nπ e ny ), n = 1, 2,... satisfy the pde and the three homogeneous boundary conditions. We want to have a somewhat more convenient form: Recall that sinh α = 1 2 (eα e α ). Therefore, sinh(n(π y)) = 1 2 ( e n(π y) e n(y π)). It follows that u n (x, y) = e nπ sin(nx)(e n(y π) e n(π y) ) = 2e nπ sin(nx) sinh(n(π y)) Again, one can take linear combinations of these functions and obtain functions u(x, y) = N α n sin(nx) sinh(n(π y)) n=1 which satisfy the pde and the three homogeneous boundary condtions. The inhomogeneous boundary condition becomes u(x, 0) = N α n sinh(nπ) sin(nx) = x(π x). n=1 Since the boundary function x(π x) is not a finite linear combination of the trigonometric functions sin(nx), n = 1, 2,..., we must write the function x(π x) as a sine series, i.e., as an infinite sum. We will show below that the function x(π x) has the sine series representation x(π x) = 8 π To solve the bvp we let u(x, y) = j=0 1 sin((2j + 1)x) for 0 x π. (2j + 1) 3 β j sinh((2j + 1)(π y)) sin((2j + 1)x)) j=0 where the coefficients β j must be determined so that 20

21 u(x, 0) = x(π x) for 0 x π. Comparing with the sine series representation of x(π x) we obtain the conditions β j sinh((2j + 1)π) = 8 π 1, j = 0, 1,... (2j + 1) 3 We solve for β j and obtain the following solution of the bvp: u(x, y) = 8 π 3 sinh((2j + 1)(π y)) (2j + 1) sinh((2j + 1)π) j=0 sin((2j + 1)x)). 5.4 Sine Series Expansions The following orthogonality relations are very important. Lemma 5.1 1) For every integer m we have 2π 0 e imx dx = 2) For all k, n {1, 2,...} we have { 0 for m 0 2π for m = 0 (5.6) π 0 { 0 for k n sin(nx) sin(kx) dx = π 2 for k = n (5.7) Proof: 1) The formula (5.6) is clear for m = 0. If m 0 one should note that e imx 1 is the derivative of im eimx and that e imx equals one for x = 0 and x = 2π. 2) Using Euler s identity we have sin(nx) sin(kx) = 1 (e inx e inx)( e ikx e ikx) 4 = 1 (e i(n+k) + e i(n+k)x e i(k n)x e i(n k)x) 4 Denote β kn = π 0 sin(nx) sin(kx) dx. Then, since the integrand is even and has period 2π: 2β kn = = π π 2π 0 sin(nx) sin(kx) dx sin(nx) sin(kx) dx We now use the above formula for sin(nx) sin(kx) and use (5.6) to obtain (5.7). 21

22 6 Separation of Variables for the Heat Equation Read Section 5.3 of Snider. 6.1 Separating the Time Dependence Consider the heat equation v t (x, t) = 2 v(x, t). Here x is the space variable and t is the time variable. We try to find solutions in separated variables, v(x, t) = u(x)t (t). Substituting into the heat equations yields thus u(x)t (t) = 2 u(x)t (t), T (t) T (t) = u(x) u(x) One obtains that, for some constant λ:. The equation T (t) = λt (t) and 2 u(x) = λu(x). 2 u(x) = λu(x) is called Helmholtz s equation. In many appliactions, boundary conditions for v(x, t) translate into boundary conditions for u(x). In turn, these determine the values of λ for which the Helmholtz equation has a nontrivial solution. 6.2 An Initial Boundary Value Problem for the 1D Heat Equation We want to solve the heat equation subject to the initial condition and boundary conditions v t (x, t) = v xx (x, t), 0 < x < π, t > 0, v(x, 0) = f(x) for 0 < x < π v(0, t) = v(π, t) = 0 for t > 0. Here f(x) is a given function. For example, 22

23 f(x) = x(π x) = 8 π n=1,n odd 1 n 3 sin(nx). Proceeding as above, we first ignore the initial condition and seek solutions of the pde of the form One obtains The boundary conditions for v require v(x, t) = u(x)t (t). T (t) = λt (t) and u (x) = λu(x). u(0) = u(π) = 0. We are lead to an eigenvalue problem for u. Precisely, we want to know for which values of λ does the bvp u = λu, u(0) = u(π) = 0 have a nontrivial solution? We obtain the eigenvalues and eigenfunctions The corresponding time equation has the solution The functions λ n = n 2, u n (x) = sin(nx), n = 1, 2,... T (t) = n 2 T (t) T (t) = e n2t T (0). v n (x, t) = e n2t sin(nx), n = 1, 2,... satisfy the pde v t = v xx and the homogeneous boundary conditions. superposition, we obtain that any function of the form Using v(x, t) = c n e n2t sin(nx) n=1 satisfies the pde and the boundary conditions, assuming rapid convergence of the series which allows to differentiate term by term. The initial condition requires v(x, 0) = c n sin(nx) = f(x), 0 < x < π. n=1 23

24 For example, if the initial condition reads then the solution is v(x, 0) = x(π x) = 8 π n=1,n odd 1 n 3 sin(nx) The first two terms are v(x, t) = 8 π n=1,n odd 1 n 3 e n2t sin(nx). v(x, t) = 8 π e t sin(x) π 27 e 9t sin(3x) +... We see that, as t increases, the solution is essentially v(x, t) 8 π e t sin(x). 24

25 7 The Wave Equation and Helmholtz s Equation for a Circle Read Section 5.3 of Snider. 7.1 Separating the Time Dependence Let D denote the open unit circle, and let D denote its boundary. Consider the wave equation with boundary condition D = {(x, y) : x 2 + y 2 < 1} v tt (x, t) = 2 v(x, t) for x D v(x, t) = 0 for x D. We try to find solutions in separated variables, v(x, t) = u(x)t (t) which satisfy the boundary conditions. yields Substituting into the wave equation u(x)t (t) = 2 u(x)t (t), thus T (t) T (t) = u(x) u(x) One obtains that, for some constant K: The equation T (t) = KT (t) and 2 u(x) = Ku(x). 2 u(x) = Ku(x) is called Helmholtz s equation. We try to find nontrivial solutions satisfying the boundary condition u(x) = 0 on D.. 25

26 7.2 The Laplacian in Polar Coordinates Let u(x, y) denote a function where x, y are Cartesian coordinates in the plane. Denote polar coordinates by r, θ where x = r cos θ, y = r sin θ. Expressing u in polar coordinates leads to the function ũ(r, θ) = u(r cos θ, r sin θ). If f(x, y) = u xx (x, y) + u yy (x, y) is the Laplacian of u, we can also write f in polar coordinates, f(r, θ) = f(r cos θ, r sin θ). The following formula gives the relation between f(r, θ) and ũ(r, θ): f = 1 r (rũ r) r + 1 r 2 ũθθ. It is common to drop the notation. 7.3 Separating the Helmholtz s Equation in Polar Coordinates Consider the equation 1 r (ru r) r + 1 r 2 u θθ + Ku = 0 for a function u = u(r, θ). We try to find solutions of the form u(r, θ) = R(r)Θ(θ). 26