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Session 1 Introduction 1 Introduction Complex systems in engineering and science are often fruitfully investigated by means of mathematical models. These models are typically formulated as differential or integral equations that must be solved subject to specific boundary and initial conditions. The methods of advanced calculus are key to the analysis and solution of the model equations. The mathematical problems resulting from modeling can be deterministic or stochastic and they can also be linear or non-linear. In this course we will focus on models leading to linear deterministic mathematical problems. These problems are fairly well understood, useful closed form solutions exist and they constitute an excellent foundation for subsequent study of more sophisticated models. For the sake of motivation, a few selected examples of mathematical models of systems of engineering interest are now described. The solution of two-dimensional problems in elastic media subject to constant body forces is given by the integral of the following differential equation ∂4φ ∂ 4φ ∂ 4φ + 2 + =0 ∂x4 ∂x2 ∂y 2 ∂y 4 where φ(x, y) is the stress function, expressable as φ = Re[z ∗ ψ(z) + χ(z)] where z∗ = x − iy and ψ and χ are suitably chosen analytic functions. Thin straight fins of triangular profile, length l and thickness b at the base are used in heat exchangers to enhance heat flow. An energy balance in the fin produces the following expression for the fin excess temperature over the surrounding environment θ as a function of distance from its tip x θ 1 dθ d2 θ − β =0 + dx2 x dx x 1 where β = 2hl , k is the thermal conductivity of the fin material and h is the heat transfer kb coefficient. The above equation is a modified Bessel equation and has the solution √ I0 (2 βx) √ θ = θb I0 (2 βl) where I0 (z) is the modified Bessel function of the first kind, of order 0 and θb is the temperature at the base of the fin. The velocity distribution v inside an initially motionless fluid produced by wall suddenly set in motion can be shown to be given by x v(x, t) = V [1 − erf ( √ )] 2 νt where V is the velocity of the wall and ν is the kinematic viscosity of the fluid. If the motion of an incompressible fluid is simply a pure rotation about a given axis fixed in space at a rate of w revolutions per unit time, the velocity vector V of the fluid at any given point satisfies the following equations ∇·V =0 and ∇ × V = 2w Now, if the fluid does not rotate at all (i.e. w = 0) one talks about irrotational flow for which a velocity potential ϕ exists such that V = ∇ϕ Since the fluid is incompressible, the above yields Laplace’s equation for the velocity potential ∇2 ϕ = 0 Solutions of this equation subject to specific boundary conditions represent the behavior of a large number of incompressible irrotational flows. The shape h(r) of an axisymmetric pendant drop is approximately described by the following differential equation d2 h 1 dh − (h0 − h) = 0 + dr2 r dr where h0 is a length parameter. The solution of the above equation is √ h(r) = h0 [J0 (r) − J0 ( Bo)] 2 0 where J0 (z) is the Bessel function of the first kind, of order 0 and Bo = ρgR is the Bond σ number of the drop with ρ being the density, σ the surface tension and R0 the drop footprint radius. The vibrations of an elastic string of length l fixed at its ends and released from an initial deflection y0 (x) are described by the solution of the wave equation 2 ∂ 2y 2∂ y = c ∂t2 ∂x2 where y is the string deflection and c2 = Tρ where T is the tension and ρ the string density. A solution to the above equation can be obtained using the method of separation of variables and is y(x, t) = ∞ X An cos( n=1 nπ nπ ct) sin( x) l l with An = 2 l Z 0 t y0 (ξ) sin( nπ ξ)dξ l which is a linear combination of an infinite number of standing waves. The process of uni-directional solidification of molten metal in a mold can be modeled using the heat equation. In the solidified portion, this is ∂Ts ∂ 2 Ts = αs 2 ∂t ∂x in the liquid portion (convection neglected) ∂ 2 Tl ∂Tl = αl 2 ∂t ∂x and at the solid-liquid interface X ks dX ∂Ts ∂Tl − kl =L ∂x ∂x dt Here, T is temperature, α thermal diffusivity, k thermal conductivity, and L is the latent heat of solidification per unit volume. Closed form solutions exist for the semi-infite case but numerical methods are required in most situations. The self-propagating high temperature synthesis (SHS) process has been modeled by the expression ρCp ∂ ∂T Q ∂T = (k ) + Hr K0 (1 − φ)n exp(− ) ∂t ∂x ∂x RT 3 where Hr is the heat released by the synthesis reaction, K0 is a reaction rate constant, φ is the fraction reacted and Q is the activation energy of the reaction. Again, although a few closed form solutions are available, most cases of interest require numerical computation. The electric current j and the electric field E inside a current carrying medium subject to an electric potential gradient satisfy the following simplified form of Maxwell’s equations ∇·j=0 and E= j κ where κ is the electric conductivity. The potential φ in turn obeys Laplace’s equation ∇2 φ = 0 The solution of Laplace’s equation for the potential subject to specifically stated boundary conditions allows determination of the current and electric field distributions inside the conductor. A popular model for the price f of a stock option is the Black-Scholes equation ∂f 1 ∂2f ∂f + rS + σ 2 S 2 2 = rf ∂t ∂S 2 ∂S where r is the risk-free interest rate and S is the stock price, commonly represented by an Ito process of the form √ dS = µdt + σ dt S Here µ is the expected rate of return of the stock, σ 2 is the variance rate of the proportional change in the stock price and is a random drawing from a standarized normal distribution. Many more examples could be given but the above should suffice to show that in order to obtain a quantitative understanding of real world problems one must know the basic tools and techniques of advanced calculus. 2 Variables: Real and Complex Real variables x are quantities who adopt values from the set of real numbers while complex variables z adopt √ values from the set of complex numbers x + iy where x and y are real variables and i = −1. x is called the real part and y the imaginary part of the complex number x + iy. Real numbers can be represented as points along the real axis while complex numbers must √be represented using a (complex) plane. Useful related concepts are the modulus |z| = x2 + y 2 and the complex conjugate z∗ = x − iy. 4 If z1 = x1 + iy1 and z2 = x2 + iy2 then the basic algebraic operations are: z1 ± z2 = (x1 ± x2 ) + i(y1 ± y2 ) z1 z2 = (x1 + iy1 )(x2 + iy2 ) z2 /z1 = x1 x2 + y1 y2 x1 y2 − x2 y1 +i 2 2 x1 + y1 x21 + y12 Introducing polar coordinates with radius r and amplitude θ then x = r cos θ and y = r sin θ and a useful alternative representation of z is z = x + iy = r(cos θ + i sin θ) = reiθ The factor eiθ represents a rotation of the real vector r through an angle θ in the complex plane. 3 Elementary Functions Functions whose arguments are real variables are called functions of a real variable, f (x), while functions whose arguments are complex variables are called functions of a complex variable, f (z). In general, since z = x + iy then f (z) = u(x, y) + iv(x, y). For instance, the integral power function (n = integer) is f (z) = z n = (x + iy)n = rn (cos θ + i sin θ)n = rn einθ = rn (cos nθ + i sin nθ) The polynomial is just a linear combination of the above, i.e. N X f (z) = An z n n=0 The power series about a is f (z) = ∞ X An (z − a)n n=0 which converges whenever |z − a| < 1 1 = L limn→∞ | AAn+1 | n The circular and hyperbolic functions are sin z = eiz − e−iz 2i 5 cos z = eiz + e−iz 2 sinh z = ez − e−z 2 ez + e−z cosh z = 2 If the complex variable z = ew where w is complex, then the complex logarithm is w = Logz = log |z| + iθ Note that this function is multivalued having infinitely many branches and a single branch point at z = 0. The principal value of θ is restricted to 0 ≤ θP < 2π. 4 Analytic Functions of a Complex Variable The derivative of a function of a complex variable is defined as f (z + ∆z) − f (z) df = f 0 (z) = lim ∆z→0 dz ∆z If a function f (z) has a finite derivative (regardless the direction of approach) and is single valued at each point in a region it is called analytic and its derivative is continuous. Let z = x + iy and w = f (z) = u(x, y) + iv(x, y) then, if f (z) is analytic the CauchyRiemann equations hold ∂v ∂u = ∂x ∂y ∂v ∂u =− ∂y ∂x A consequence of the analyticity of w is that its real and imaginary parts satisfy Laplace’s equation and are called harmonic functions. ∇2 u = ∇2 v = 0 6 5 Line Integrals R The integral of a function of a real variable ab f (x)dx generalizes to the line integral (over a curve C from z0 to z1 ) of a function of a complex variable f (z) = u + iv as follows Z Z f (z)dz = C C (udx − vdy) + i Z (vdx + udy) C If the curve C can be enclosed in a simple connected region where f (z) is analytic the integral is path independent. Also, if f (z) = dF (z)/dz then Z Z z1 f (z)dz = C z0 dF (z) = F (z1 ) − F (z0 ) Furthermore, if z0 = z1 I f (z)dz = 0 C which is Cauchy’s integral theorem. The positive direction of integration is the one that maintains the enclosed area to the left. Example: Study the line integration of the function f (z) = 1/z. Finally, if M is an upper bound for |f (z)| and L is the length of C | Z C f (z)dz| ≤ M L If C is a curve in the complex plane and C a smaller circular contour (center = α, radius = ) completely inside C I C I f (z) dz = z−α C f (z) dz z−α and in the limit as → 0 1 f (z) = 2πi I C f (α) dα α−z which is Cauchy’s integral formula. 6 Ordinary Differential Equations An ordinary differential equation is an equation relating two variables in terms of derivatives. The linear ODE of order n is a0 (x) dn y dn−1 y dy + a (x) + ... + an−1 (x) + an (x)y = Ly = h(x) 1 n n−1 dx dx dx 7 where L is the linear differential operator given by a0 (x) dn dn−1 d + a (x) + ... + an−1 (x) + an (x) 1 n n−1 dx dx dx and represents the mathematical operation which produces h(x) when applied onto y(x). Linear operators satisfy basic properties such as commutativity and distributivity and they can often be factored. The function y = u(x) which when substituted above makes left and right hand sides equal is called a solution of the differential equation. If h(x) = 0 the differential equation is called homogeneous. 7 Linear Dependence The set of functions u1 (x), u2 (x), ..., un (x) is called linearly independent if none of the functions in the set can be expressed as a linear combination of the others. To check that the given functions are linearly independent it is sufficient to show that their Wronskian determinant u1 u2 .....un du du du n 1 2 ..... dx dx dx W (u1 , u2 , ..., un ) = ................................. n−1 dn−1 u1 dn−1 u2 d un n−1 n−1 ..... n−1 dx dx dx is not equal to zero. 8 Complete Solutions If u1 (x), u2 (x), ..., un (x) are linearly independent solutions of the n-th order linear ODE Ly = 0, the general solution of the equation is yH (x) = n X ck uk (x) k=1 Now, if for the associated non-homogeneous equation Ly = h(x) with h(x) 6= 0, a particular solution is yP (x), then the complete solution becomes y(x) = yH (x) + yP (x) = n X k=1 8 ck uk (x) + yP (x) 9 Linear Differential Equations The standard form of a first order linear differential equation is dy + a1 (x)y = h(x) dx R Introducing the integrating factor p(x) = e a1 (x)dx , the solution is C 1Z phdx + y(x) = p p where C is an integration constant. Specifically, if in addition to the differential equation one is given the initial condition y(x0 ) = y0 , the solution becomes Z x y(x) = x0 p(ξ) p(x0 ) h(ξ)dξ + y0 p(x) p(x) Example. Solve xy 0 + (1 − x)y = xex . The equation dn−1 y dy dn y + an y = Ly = h(x) + a + ... + an−1 1 n n−1 dx dx dx is called the n−th order linear ODE with constant coefficients. Note that the symbol L stands here for L= dn−1 d dn + an + a + ... + an−1 1 n n−1 dx dx dx The general homogeneous solution of the above is yH (x) = n X ck erk x k=1 and the associated characteristic equation is rn + a1 rn−1 + ... + an−1 r + an = 0 which may involve imaginary roots. Also, sometimes roots of the characteristic equation are repeated. In this case less than n independent solutions result but the missing solution can be obtained from the condition of the vanishing of the partial derivatives with respect to the repeated roots. Therefore, the part of the homogeneous solution corresponding to an m-fold root r1 is er1 x (c1 + c2 x + c3 x2 + ... + cm xm−1 ) 9 A commonly encountered equation is the n-th order equidimensional linear equation xn n−1 dy dn y y n−1 d + b x + ... + bn−1 x + bn y = f (x) 1 n n−1 dx dx dx The above is easily solved by introducing a new independent variable z such that x = ez which transforms it into a simple linear equation with constant coefficients. Example. Solve x2 y 00 − 2xy 0 + 2y = x2 + 2 An even simpler procedure is available in the case of f (x) = 0. 10 Particular Solutions by Variation of Parameters Suppose that the general solution of the homogeneous equation Ly = dn−1 y dy dn y + a (x) + ... + an−1 (x) + an (x)y = 0 1 n n−1 dx dx dx P has been obtained and it is of the form yH (x) = nk=1 ck uk (x). A particular solution of the associated non-homogeneous equation Ly = h(x) can be obtained by replacing the constant parameters ck is the above solution by certain functions of x satisfying certain conditions, i.e. n X yP (x) = Ck (x)uk (x) k=1 Succesive differentiation and manipulation of yP (x) produces two expressions which are used to obtain the Ck ’s. For instance, for the equation dy d2 y + a1 (x) + a2 (x)y = h(x) 2 dx dx where a1 , a2 and h(x) are continuous in the domain of interest, the solution obtained by the method of variation of parameters is yP (x) = C1 (x)u1 (x) + C2 (x)u2 (x) with C1 (x) = − Z x u2 (ξ)h(ξ) dξ + c1 W (ξ) and Z C2 (x) = x u1 (ξ)h(ξ) dξ + c2 W (ξ) 10 where u1 (x) and u2 (x) are linearly independent solutions of the associated homogeneous equation and u u W (x) = W (u1 (x), u2 (x)) = du11 du22 dx dx is the Wronskian of u1 (x) and u2 (x). 11 Initial and Boundary Value Problems The general solution of an n-th order ODE involves n arbitrary constants which must be determined by n suitably prescribed supplementary conditions. For example, the value of the function and of its first n − 1 derivatives may be prescribed at the single point x = a. If this is the case one talks about an initial value problem and if the coefficients and h(x) are continuous there exists a unique solution n X y(x) = ck uk (x) + yP (x) k=1 On the other hand, if the values of the function and/or certain of the derivatives are prescribed at two points x = a and x = b, othe talks about a boundary value problem. In this case the solution may or may not exist and may or may not be unique. 12 Practice Problems 12.1 Determine the real functions of x and y which are the real and imaginary parts of the following complex functions a) z 2 b) z 3 Answers: a) x2 − y 2 , 2xy b) x3 − 3xy 2 , 3x2 y − y 3 12.2 Express the function z π in the form z π = exp(π log z) 11 √ and find the principal value when z = (1 + i)/ 2 = √12 + i √12 in the form a + ib. Answer: Let z = r exp(iθ) = r(cos(θ) + i sin(θ)) thus z π = rπ exp(iπθ) = rπ (cos(πθ) + i sin(πθ)) = √ rπ (cos(π(θP + 2kπ)) + i sin(π(θP + 2kπ))). Now using z = (1 + i)/ 2 (which implies θP = π4 ) with k = 0 yields the principal value cos(π 2 /4) + i sin(π 2 /4). 12.3 12.4 The following differential equations arise in dealing with the problems noted. Find the general solution of each equation assuming k 6= 0 a) Vibration of a beam. d4 y − k4y = 0 dx4 b) Beam on an elastic foundation. d4 y + 4k 4 y = 0 dx4 c) Bending of a plate. 2 d4 y 2d y − 2k + k4y = 0 dx4 dx2 Answers: a) The characteristic equation is r4 − k4 = 0 with roots k, −k, ik and −ik so the general solution is y = c1 exp(kx) + c2 exp(−kx) + c3 cos(kx) + c4 sin(kx) b) The characteristic equation is r4 + 4k 4 = 0 12 with roots (1 + i)k, (1 − i)k, (−1 + i)k and (−1 − i)k so the general solution is y = exp(kx)(c1 cos(kx) + c2 sin(kx)) + exp(−kx)(c3 cos(kx) + c4 sin(kx)) c) The characteristic equation is r4 − 2k 2 r2 + k 4 = 0 with roots k and −k (repeated), so the general solution is y = exp(kx)(c1 + c2 x) + exp(−kx)(c3 + c4 x) 13