Ordinary Differential Equations II

Transcription

1 Ordinary Differential Equations II February Separation of variables Wave eq. (PDE) 2 u t (t, x) = 2 u 2 c2 (t, x), x2 c > 0 constant. Describes small vibrations in a homogeneous string. u(t, x) = displacement of the string at position x and time t. Assume fixed at endpoints: u(t, 0) = u(t, 1) = 0 (length 1 at rest). Look for a solution of the form u(t, x) = w(t)y(x). Gives w (t)y(x) = c 2 w(t)y (x) w (t) c 2 w(t) = y (x) y(x). LHS independent of x, RHS indep. of t, so both must be constant = λ. Gives and Thus, w = λc 2 w(t) y (x) = λy(x), y(0) = y(1) = 0. w(t) = c 1 cos(c λt) + c 2 sin(c λt), y(x) = c 3 cos( λx) + c 4 sin( λx). Boundary condition y(0) = 0 gives c 3 = 0. Boundary condition y(1) = 0 gives c 4 sin( λ) = 0. Nontrivial solution λ = n 2 π 2, n = 1, 2, 3,.... Then, u(t, x) = (c 1 cos(cnπt) + c 2 sin(cnπt)) sin(nπx). Normally one also has initial conditions u(0, x) = u(x), u (0, x) = v(x). Can try t to solve this by making an Ansatz in the form of a series u(t, x) = (c 1,n cos(cnπt) + c 2,n sin(cnπt)) sin(nπx). n=1 1

2 Evaluating at t = 0, we formally obtain u(x) = v(x) = c 1,n sin(nπx), n=1 cnπc 2,n sin(nπx). n=1 Can any function be written like this? Fourier analysis the series for u(x) is uniformly convergent if e.g. u C 1 ([0, 1], C) with u(0) = u(1) = 0. With some extra conditions on u(x) and v(x), one can show that the series for u(t, x) converges uniformly to a C 2 solution of the wave eq. which satisfies the initial conditions and boundary conditions (see pp in Teschl). The numbers λ n can be seen as eigenvalues and the functions y n (x) = sin(nπx) as eigenvectors, or eigenfunctions, for the linear operator A := d2 dx 2. Recall that a linear map A on an n-dim. inner product space V is symmetric (or Hermitian) if Au, v = u, Av u, v V. If A is symmetric, then ON basis of eigenvectors {u j } n j=1 (spectral theorem). Every vector u V can be written u = c j u j j=1 where u j, u k = { 1, j = k, 0, j k and c j = u j, u. A = d2 dx 2 is formally a symmetric operator with respect to the inner product f, g = 1 0 f (x)g(x) dx = 1 0 f(x)g(x) dx, ( * = complex conjugation), as we ll show more generally later. Moreover, the eigenfunctions are orthogonal: sin(nπx), sin(mπx) = 0 if n m. The norm of sin(nπx) is 1/ 2. Setting u n (x) = 2 sin(nπx), n = 1, 2,... we thus get an ON basis of eigenfunctions. Goal: generalize this to eigenvalue problems of the form (p(x)y (x)) + q(x)y(x) = λr(x)y(x), a x b, with separated boundary conditions b 11 y(a) + b 12 y (a) = b 21 y(b) + b 22 y (b) = 0. This is called a Sturm-Liouville problem. Can be written Ay = λy with A = 1 ( d r(x) dx p(x) d ) dx + q(x). 2

3 Example. Consider a string with varying density. The equation for small vibrations is then given by the variable wave eq. 2 u t (t, x) = 2 c2 (x) 2 u (t, x). x2 Separation of variables leads to the eigenvalue problem y = λc 2 (x)y, so that p(x) 1, q(x) 0 and r(x) = c 2 (x). We first need to discuss some general properties of inner product spaces. Inner product spaces H complex vector space. A map, : H H C is called an inner product if (1) f, f > 0, f 0 (positive definiteness), (2) f, g = g, f (Hermitian symmetry), (3) f, α 1 g 1 + α 2 g 2 = α 1 f, g 1 + α 2 f, g 2 (linearity in 2nd argument). Remarks: (3) 0, 0 = 0. (2) and (3) α 1 f 1 + α 2 f 2, g = α 1 f 1, g + α 2 f 2, g. An inner product is linear in the 2nd argument and conjugate linear in the 1st. This is called sesquilinearity (generalization of bilinearity). Usual convention in physics; in mathematics usually the opposite. We define the norm corresponding to, by Examples. = f, f. C n with a, b = n j=1 a jb j and a = n j=1 a j 2. l 2 = {u = {u j } j=1 : j=1 u j 2 < } with u, v = j=1 u jv j and u = j=1 u j 2 C([a, b], C) with f, g = b a f(x) g(x) dx and f = b a f(x) 2 dx. If H is complete with the norm coming from the inner product (a Banach space), it s called a Hilbert space. We won t require this. 3

4 We write f g if f, g = 0. Recall that f + g 2 = f 2 + g 2, f g. We also have the Cauchy-Schwarz inequality (Thm 5.2) f, g f g. This yields the triangle inequality for, showing that it s indeed a norm: f + g 2 = f + g, f + g = f, f + f, g + g, f + g, g f f g + g 2 = ( f + g ) 2. A set of vectors {u j } is called an orthonormal set if { 1, j = k, u j, u k = 0, j k. We can define the orthogonal projection on an ON set using the following lemma. Lemma. Suppose {u j } n is an ON set and let V be the span of {u j } n. Every f H can be be written f = f + f, f = u j, f u j, with f f. Moreover, u j, f = 0 for all j and (1) f 2 = u j, f 2 + f 2. Every ˆf in V satisfies f ˆf f with equality iff ˆf = f. That is, f is the unique vector in V closest to f. Proof. u j, f = u j, f f = u j, f u j, f = u j, f u j, u k, f u k = u j, f k=0 u k, f u j, u k = u j, f u j, f = 0. k=0 In particular, f, f = 0 since f is in V. Thus, f 2 = f 2 + f 2 4

5 with f 2 = u j, f u j, u k, f u k = k=0 Fix a vector ˆf = n α ju j in V. Then u j, f u k, f u j, u k = k=0 u j, f 2. f ˆf 2 = f + f ˆf 2 = f 2 + f ˆf 2, so that with equality iff ˆf = f. f ˆf f (1) Bessel s inequality: u j, f 2 f 2 with equality iff f is in the span of {u j } n. Assume that H is infinite-dimensional. An orthonormal set {u j }, is called an orthonormal basis if f 2 = u j, f 2 f H. Set If {u j } is an ON set then (1) Hence, f n f in H, i.e. iff {u j } is an ON basis. f n = u j, f u j. f f n 2 = f 2 f = u j, f 2. u j, f u j A linear operator on H is a simply a linear map A: H H. It s useful to extend this definition so that A only is defined on some subspace D(A) of H (with values in H). The space D(A) is called the domain of A. A is called symmetric if Af, g = f, Ag, f, g D(A) (the book also requires that D(A) is dense in H, but it s not important). 5

6 Example. The operator A = d2 can be be considered as a linear operator on dx 2 H = C([a, b], C) with D(A) = {f C 2 ([a, b], C): f(a) = f(b) = 0}. Our previous calculation shows that A is symmetric. λ C is called an eigenvector if nonzero u D(A), s.t. Au = λu. Theorem. Let A be symmetric. Then all eigenvalues are real and eigenvectors corresponding to different eigenvalues are orthogonal. Proof. Suppose that u is an eigenvector corresponding to λ. Then λ u, u = u, Au = Au, u = λ u, u λ = λ since u, u = u 2 0. That is, λ is real. If Au j = λ j u j, j = 1, 2, λ 1 λ 2, then λ 2 u 1, u 2 = u 1, Au 2 = Au 1, u 2 = λ 1 u 1, u 2 since λ 1 R. Since λ 1 λ 2 we obtain u 1, u 2 = 0. 6