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Chapter 4 Boundary Value Problems, Characteristic Functions and Representations 1 Boundary Value Problems and Characteristic Functions If a solution to an ordinary differential equation must satisfy certain specified conditions at two or more boundaries then one has a boundary value problem (BVP). A BVP is homogeneous if the differential condition and the boundary conditions are all homogeneous. An expression is homogeneoeus if when satisfied by a function F it is also satisfied by cF where c is a constant. Recall that non-trivial solutions to a second order linear homogeneous ODE can be expressed as a linear combinations of linearly independent functions with two constants. These must be determined from the boundary conditions resulting in an expression in terms of only a single constant. In many cases, the linearly independent solutions depend on a parameter λ which produces non-trivial solutions only for certain specific values λ1 , λ2 , .... Problems of this kind are called characteristic-value or eigenvalue problems. The specific values of λ are called characteristic values or eigenvalues and the solutions are called characteristic functions or eigenfunctions. 2 The Rotating String A taut string (length l, density ρ and tension T ) rotates about its length at angular velocity ω. What is the shape y(x) of the rotating string? A force balance on an element of string yields (assuming small deflections) d2 y + λy = 0 dx2 where λ = ρω 2 /T . Since the ends are fixed the boundary conditions are y(0) = y(l) = 0. The most general solution of the differential equation is √ y(x) = C sin( λx) = Cφ(x) 1 √ where φ(x) = sin( λx). √ In order to satisfy the condition y(l) = 0 it is necessary that λl = nπ where n = 1, 2, ... producing the infinite set λn = n2 π 2 l2 n = 1, 2, ... Therefore, there is an infinite set of φ(x)’s given by q φn (x) = sin( λn x) = sin( nπ x) l n = 1, 2, ... are the characteristic functions of the problem and λn are the characteristic values corresponding to specific, physically meaningful critical values of the angular velocity nπ ωn = l s T ρ n = 1, 2, ... associated with characteristic deflection modes of the string. A general expression capable of accounting for various types of boundary conditions in the above problem is T Lyb0 = αLyb0 = −yb kL where yb is the displacement at the string boundary and k is the spring constant of an elastic spring which may be attached to the the end of the string (i.e. a yielding support). Specifically, if α = 0 (i.e. infinite restoring force) one gets yb = 0, the fixed end condition, while if α = ∞ (i.e. zero restoring force) one obtains yb0 = 0, the free end condition. 3 The Rotating Shaft Consider now a uniform shaft (length l, density ρ, Young’s modulus E and moment of inertia I) rotating about its axis with angular velocity ω. From beam bending theory, the deflexion of the shaft y(x) satisfies d4 y − λy = 0 dx4 where λ = ρω 2 /EI. This has the general solution y(x) = c1 sinh(λ1/4 x) + c2 cosh(λ1/4 x) + c3 sin(λ1/4 x) + c4 cos(λ1/4 x) If the shaft is regarded as hinged at both ends, the appropriate boundary conditions are y(0) = y 00 (0) = 0 2 which gives c2 = c4 = 0, and y(l) = y 00 (l) = 0 which gives the systems of equations c1 sinh(µ) + c3 sin(µ) = 0 and c1 sinh(µ) − c3 sin(µ) = 0 where µ = λ1/4 l. From the basic theory of systems of linear algebraic equations, compatibility requires the vanishing of the determinant of coefficients, i.e. −2 sinh(µ) sin(µ) = 0 This can only happen if µ = 0 or if sin(µ) = 0. Nontrivial solutions result only in the second case, that is if µn = nπ n = 1, 2, ... Substituting into the first equation of the system above and recalling the properties of the sinh function necessarily requires that c1 = 0 so that the eigenfunctions (deflection modes) are nπx ) n = 1, 2, ... φn (x) = sin( l the eigenvalues are λn = n4 π 4 l4 n = 1, 2, ... and the critical speeds are n2 π 2 ωn = 2 l 4 s EI ρ n = 1, 2, ... Buckling of Axially Loaded Columns Consider now an uniform column, axially loaded at one end (length l, load P , density ρ, Young’s modulus E and moment of inertia I). The deflection y(x) must satisfy the equation d2 y d4 y + λ =0 dx4 dx2 3 where λ = P/EI. The general solution of this is of the form y(x) = c1 sin(λ1/2 x) + c2 cos(λ1/2 x) + c3 x + c4 If both ends of the column are hinged the boundary conditions are y(0) = y 00 (0) = 0 which gives c2 = c4 = 0 and y(l) = y 00 (l) = 0 which gives non-trivial solutions only if λn = n2 π 2 l2 n = 1, 2, ... with c3 = 0. The characteristic functions (deflection modes) in this case are then φn (x) = sin( nπx ) l n = 1, 2, ... The critical buckling loads are Pn = n2 π 2 EI l2 n = 1, 2, ... The unbent column is then stable only if P < P1 = π 2 EI l2 which is called the Euler load and the fundamental buckling mode y(x) = C sin(πx/l) occurs when P = P1 . 5 Orthogonality of Characteristic Functions Two arbitrary functions φm (x) and φn (x) for x ∈ [a, b] for which Z b a r(x)φm (x)φn (x)dx = 0 are called orthogonal with respect to the weight function r(x). If all pairs of functions in a set are orthogonal, the set itself is called orthogonal. The sets of characteristic functions corresponding to a wide variety of BVP encountered in applications are orthogonal with respect to various specific weighing functions. 4 Consider the generic BVP for x ∈ [a, b] defined by the following ODE in standard form dy d [p(x) ] + [q(x) + λr(x)]y = Ly + λr(x)y = 0 dx dx for which two eigenfunctions are φ1 (x) and φ2 (x). The characteristic functions must satisfy dφ1 d [p(x) ] + [q(x) + λ1 r(x)]φ1 = 0 dx dx and dφ2 d [p(x) ] + [q(x) + λ2 r(x)]φ2 = 0 dx dx Multiplying the first of these by φ2 , the second one by φ1 substracting and integrating results in (λ2 − λ1 ) Z Z b rφ1 φ2 dx = a b a [φ2 d dφ1 d dφ2 (p ) − φ1 (p )]dx dx dx dx dx Integrating the RHS by parts gives Z (λ2 − λ1 ) b a rφ1 φ2 dx = [p(x){(φ2 dφ1 dφ2 b ) − (φ1 )}] dx dx a Therefore, if the boundary conditions associated with the original problem are of the form y = 0, or y 0 = 0, or y + γy 0 = 0, at each end point the RHS term equals zero. The term can also be zero if p(x) = 0 when x = a or x = b; or if p(b) = p(a) and y(b) = y(a) and y 0 (b) = y 0 (a). Finally, the RHS is also zero if the solutions are required to be periodic. In all these cases one has then Z b a r(x)φ1 (x)φ2 (x)dx = 0 λ2 6= λ1 Boundary value problems of the above types are known as Sturm-Liouville problems. Moreover, if p(x) > 0, r(x) > 0, q(x) ≤ 0 and if γ 6= 0 but is negative if the condition is on the left boundary and positive otherwise, one has then a proper Sturm-Liouville problem. Proper S-L problems are interesting because all their associated eigenvalues are real and ≥ 0 and their eigenfunctions are real too. Furthermore, the eigenvalues constitute an infinite discrete set. Finally, the integral Z Cn = b a r(x)[φn (x)]2 dx is called the norm of φn with respect to r and if the coefficient in φn is chosen so that Cn = 1, φn is considered normalized with respect to r. A set of normalized orthogonal functions is called orthonormal. 5 Although the above statements referred to the given ODE in standard form they are widely applicable since that the generic ODE a0 (x) d2 y dy + a1 (x) + [a2 (x) + λa3 (x)]y = 0 2 dx dx can always be transformed to the standard form by setting Z a1 dx) a0 p = exp( q= a2 p a0 r= a3 p a0 and Exercise. Investigate the eigenvalues and eigenfunctions associated with the BVP in x ∈ [0, l], y 00 + λy = 0 subject to y(0) = y(l) = 0 6 Characteristic Function Representations Linear combinations of eigenfunctions from a complete orthogonal set constitute useful formal representations of arbitrary functions. This is reminiscent of the representation of arbitrary vectors in Euclidean space by means of linear combination of the complete orthogonal set of unit vectors. Assume that an arbitrary function f (x) can be represented as a linear combination of characteristic functions forming an orthogonal set with respect to a certain weighting function r(x), {φn (x)}∞ 0 f (x) = A0 φ0 (x) + A1 φ1 (x) + A2 φ2 (x) + ... = ∞ X An φn (x) n=0 Multiplication of the above by r(x)φk (x), followed by integration from x = a to x = b and taking into account the orthogonality of the eigenfunctions leads to Rb r(x)f (x)φn (x)dx An = aR b 2 a r(x)[φn (x)] dx 6 which determines the formal series representation of f (x). If the associated Sturm-Liouville problem is proper and p(x), q(x) and r(x) are regular in (a, b), the formal representation given above of any piecewise differentiable function f (x) converges to f (x) inside (a, b) wherever f (x) is continuous and converges to the mean value wherever finite jumps occur. Furthermore, if f (x) is continuous and its derivative is piecewise differentiable, the term by term derivative of the series converges to f 0 (x) wherever the derivative is continuous. Note the great generality of series representation in terms of eigenfunctions in comparison with power series expansions. While Taylor series representations require the existence of derivatives of all orders and even in such case the representation may not exist, this is not the case at all with Fourier series. The characteristic function representation can be operated upon by the operator L = dy d (p(x) ) + q to give dx dx Lf (x) ≈ ∞ X An Lφn (x) = −r(x) n=0 ∞ X λn An φn (x) n=0 and if Lf (x)/r(x) is piecewise differentiable ∞ Lf (x) X = Bn φn (x) r(x) n=0 with Bn = −λn An . Therefore, in these conditions the operator can by applied term by term to the series. 7 Boundary Value Problems involving Nonhomogeneous Equations Consider the nonhomogeneous differential equation Ly + Λry = h(x) and its associated homogeneous problem when h(x) = 0 with corresponding eigenfunctions P φn (x) and eigenvalues λn with the series representation of its solution y = an φn Since P Lφn = −λn rφn , substitution into the original ODE gives r (Λ − λn )an φn = h, i.e. X X h(x) = f (x) = (Λ − λn )an φn (x) = An φ(x) r(x) where h/r is assumed to be piecewise differentiable. The solution of the nonhomogeneous problem is then y(x) = X An φn (x) Λ − λn 7 8 Fourier Series The characteristic functions φn (x) = sin(nπx/l) are eigenfunctions of the problem d2 y + λy = 0, dx2 y(0) = 0, y(l) = 0 with the characteristic numbers λn = n2 π 2 /l2 . If a function f (x) can be represented as f (x) = ∞ X An φn (x) = n=1 ∞ X An sin( n=1 nπx ) l in x ∈ [0, l] then, because of orthogonality, Rl An = f (x) sin( nπx )dx 2Z l nπx l )dx = f (x) sin( Rl nπx 2 l 0 l 0 [sin( l )] dx 0 The series is the Fourier sine series representation of f (x) in terms of individual harmonics and An are called the Fourier coefficients and are equal to twice the average value of f sin(nπx/l) in [0, l]. Since the sign of each harmonic is reversed when x is replaced by −x, the Fourier sine series is a representation of an odd function f (x) (i.e. one for which f (−x) = −f (x)) in the interval [−l, l]. If f (x) is odd and periodic, the series represents f (x) everywhere. If instead the following BVP is considered d2 y + λy = 0, dx2 y 0 (0) = 0, y 0 (l) = 0 with characteristic functions φn (x) = cos( nπx ), l (n = 0, 1, 2, ...) and eigenvalues λn = n2 π 2 /l2 and this is used to produce a (Fourier cosine series) representation of the function f (x) f (x) = A0 + ∞ X An cos( n=1 nπx ) l the corresponding Fourier coefficients are A0 = 1 l Z 0 l f (x)dx and 2Z l nπx )dx, f (x) cos( An = l 0 l 8 (n = 1, 2, 3, ...) The Fourier cosine series is a representation of an even function f (x) (i.e. one for which f (−x) = f (x)) in the interval [−l, l]. If f (x) is even and periodic, the series represents f (x) everywhere. Since all harmonics are even and periodic, they constitute convergent representations to piecewise differentiable even functions f (x) everywhere. Since any function f (x) can be expressed as the sum of an even fe (x) and an odd function fo (x) with fe (x) = A0 + ∞ X An cos( n=1 fo (x) = ∞ X Bn sin( n=1 nπx ) l nπx ) l such that , over the interval [−l, l], f (x) = A0 + ∞ X (An cos( n=1 nπx nπx ) + Bn sin( )) l l where 1 Zl f (x)dx 2l −l A0 = 1 An = l Z l −l f (x) cos( nπx )dx l f (x) sin( nπx )dx l and 1 Bn = l Z l −l The series representation of f (x) is called the complete Fourier sine-cosine series representation. This can be expressed in more compact form by using Euler’s formula: f (x) = ∞ X Ck eikx k=−∞ where Ck = 1 Zl 1 i f (x)e−ikx dx = Ak − Bk 2π −l 2 2 The complete Fourier series representation of f (x) over any interval [a, a + P ] is easily obtained. 9 Fourier series representations of f (x) can be differentiated term by term and the result becomes a convergent representation of f 0 (x) as long as f 0 is piecewise differentiable. Exercise. Obtain Fourier sine series and cosine series representations of f (x) = ex in x ∈ [0, π]. For the sine series An = 2 π Z 0 π ex sin(nx)dx = 2 n (1 − eπ cos(nπ)) π n2 + 1 And for the cosine series 1 A0 = π Z 0 π ex dx = eπ − 1 π 2Zπ x 2 1 (eπ cos(nπ) − 1) An = e cos(nx)dx = 2 π 0 πn +1 9 Fourier-Bessel Series The BVP p2 d dy (x ) + (− + µ2 x)y = 0 dx dx x is of Sturm-Liouville form with p(x) = x, q(x) = −p2 /x, r(x) = x and λ = µ2 . Its general solution is of the form y(x) = Zp (µx) where Zp (µx) = c1 Jp (µx) + c2 J−p (µx) when p is not integer and Zp (µx) = c1 Jp (µx) + c2 Yp (µx) when p is integer. The functions Jp and Yp constitute and orthogonal set of characteristic functions. If at x = 0 y is finite and xy 0 = 0 and at x = l, y(l) = 0, the eigenvalues µn s are the roots of Jp (µn l) = 0. If instead at x = l, y 0 (l) = 0, the µn s are the roots of Jp0 (µn l) = 0. Finally if at x = l, ky(l) + ly 0 (l) = 0 (with k ≥ 0), the µs are the roots of kJp (µn l) + µn lJp0 (µn l) = 0. In the first case the characteristic functions are φn (x) = Jp (µn x) which are orthogonal on [0, l] with respect to r(x) = x. 10 So, the representation f (x) = ∞ X An Jp (µn x) n=1 with An = and Z Cn = 0 l 1 Cn Z l 0 xf (x)Jp (µn x)dx 1 dJp (µn x) |x=l )2 ] [(µ2n l2 − p2 )Jp (µn l)2 + l2 ( 2 2µn dx x[Jp (µn x)]2 dx = is called the Fourier-Bessel series representation of f (x). Therefore, if y(l) = 0, Jp (µn l) = 0 and Z Cn = 0 If y (l) = 0, Jp0 (µn l) l 0 x[Jp (µn x)]2 dx = l2 [Jp+1 (µn l)]2 2 = 0 and Cn = µ2n l2 − p2 [Jp (µn l)]2 2 2µn And if ky(l) + ly 0 (l) = 0, kJp (µn l) + µn lJp0 (µn l) = 0 Cn = µ2n l2 − p2 + k 2 [Jp (µn l)]2 2µ2n l2 Note that in all the above cases the characteristic functions are of the form φn (x) = Jp (µn x) The integral involved in the calculation of An can be readily obtained by numerical methods. Exercise. Determine the Fourier-Bessel series expansion of the function f (x) = 1 for x ∈ [0, l] in terms of J0 . Let 1= ∞ X An J0 (µn x) n=1 where J0 (µn l) = 0. Thus An = with Z Cn = 0 l 1 Zl xJ0 (µn x)dx Cn 0 x[J0 (µn x)]2 dx = 11 l2 [J1 (µn l)]2 2 10 Legendre Series Legendre’s differential equation d dy [(1 − x2 ) ] + p(p + 1)y = 0 dx dx has the form of the differential equation of a Sturm-Liouville problem with p(x) = 1 − x2 , q(x) = 0, r(x) = 1 and λ = p(p + 1). Finite solutions at x = ±1 require p = 0, 1, 2, ... and are φn (x) = Pn (x) where Pn (x) are the Legendre polynomials. Since these are orthogonal, any piecewise differentiable function in the interval (−1, 1) can be represented as ∞ X f (x) = An Pn (x) n=0 where R1 f (x)Pn (x)dx = An = −1 R1 2 −1 [Pn (x)] dx ( 2n+1 R 1 −1 f (x)Pn (x)dx 2 n 2n+1 R 1 (1 − x2 )n d dxf (x) n dx 2n+1 n! −1 Because of the properties of Pn , if f (x) is even, An = 0 if n is odd and viceversa. Any polynomial of degree k can be expressed as a linear combination of the first k + 1 Legendre polynomials. 11 Fourier Integral R Consider the expression 0∞ A(u) sin(ux)du where u is any positive number. Could this be a representation of a well behaved function f (x) for 0 < x < ∞? Assume that is the case, therefore, Z f (x) = ∞ 0 A(u) sin(ux)du Multiplying both sides by sin(u0 x), integrating from 0 to l, taking the limit as l → ∞ and rearranging gives 2 A(u) = π Z 0 ∞ f (x) sin(ux)dx Switching the dummy integration variable from x to t and substituting in the original representation Z ∞ 2Z∞ f (x) = sin(ux) f (t) sin(ut)dtdu π 0 0 12 This is the FourierR sine integral representation of f (x) and is valid in ∀x if f is odd, piecewise differentiable and 0∞ |f (x)|dx exists. Similarly, the Fourier cosine integral representation can be defined as Z Z ∞ 2 ∞ f (x) = cos(ux) f (t) cos(ut)dtdu π 0 0 valid ∀x if f is even. For a function f (x) = fe (x) + fo (x), the complete Fourier integral representation in −∞ < x < ∞ Z f (x) = ∞ 0 [A(u) cos(ux) + B(u) sin(ux)]du with Z ∞ A(u) = 1 π B(u) = 1Z∞ f (t) sin(ut)dt π 0 0 f (t) cos(ut)dt and Substituting and using complex notation f (x) = The integral transform is R∞ −∞ 1 2π Z Z ∞ −∞ eiux ∞ −∞ e−iut f (t)dtdu ¯ of f . The inverse e−ixt f (x)dx is called the Fourier transform f (u) 1 Z ∞ iux ¯ e f (u)du f (x) = 2π −∞ Fourier sine and cosine integral transforms can also be defined as Z fS (x) = ∞ 0 f (x) sin(ux)dx with 2 f (x) = π and Z 0 Z fC (x) = ∞ 0 ∞ fS (u) sin(ux)du f (x) cos(ux)dx with 2Z∞ f (x) = fC (u) cos(ux)du π 0 13