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18.034 - Honors Differential Equations, Spring 2007 Dr. Vera Mikyoung Hur LT2. APPLICATIONS OF THE LAPLACE TRANSFORM Convolution. Consider a linear system in which the effect at the present time t of a stimulus f (t1 )dt1 at the past time t1 is proportional to the stimulus. On physical grounds we assume that the proportionality constant depends only on the elapsed time t − t1 and hence is of the form g(t − t1 ). The effect at the present time t is therefore f (t1 )g(t − t1 )dt1 . Since the system is linear, the response to the history can be obtained by adding these effects and we are lead to the integral � t (1) w(t) = f (t1 )g(t − t1 )dt1 . 0 (The lower limit 0 means that the process is assumed to have started at time t = 0.) The expression (1) is called the convolution of f and g. It gives the response at the present time t as a weighted superposition over the inputs at times t1 � t. The weighting factor g(t − t1 ) characterizes the system and f (t1 ) characterizes the history of the input. The function w in (1) is denoted by f ∗ g. The following theorem show that convolution in the t-domain corresponds to multiplication in the s-domain. Theorem 1. If f, g ∈ E then f ∗ g ∈ E and L[f ∗ g] = L[f ]L[g]. Proof. I leave the proof of f ∗ g ∈ E to the reader. By deﬁning f (t) = 0 and g(t) = 0 for all negative t, we write � ∞ �� ∞ � L[f ∗ g(t)] = f (t1 )g(t − t1 )dt1 e−st dt −∞ −∞ � ∞ �� ∞ � −st = g(t − t1 )e dt f (t1 )dt1 . −∞ −∞ We set t − t1 = t2 , t = t1 + t + 2, and the assertion follows. � The convolution plays a prominent role in the study of heat conduction, wave motion, plastic ﬂow and creep, and in other branches of mathematical physics. It is also encountered in sociology, economics, ecology, and genetics in which an effect at a given time t1 induces a delayed response at a later time t. Example 2. Consider the problem y �� + ω 2 y = ω 2 f (t), y(0) = y � (0) = 0, where ω is a constant and f ∈ E. Taking the Laplace transform yields that ω2 Ly = L[f ], s2 + ω2 or L[y] = L[f ]L[g] = L[f ∗ g], 1 where g(t) = ω sin ωt. By uniqueness, then y = f ∗ g, or more explicitly � t y(t) = ω f (t1 ) sin ω(t − t1 )dt1 . 0 The unit impulse function. We consider the response of a system to a narrow spike function (an impulse) that acts for a very short time but produces a large effect, e.g., a point charge or an electron point. To formulate the idea of an impulse, let a be a small positive constant and let δa (t) be the function deﬁned as � 1 for 0 � t < a, (2) δa (t) = a 0 elsewhere. Note that the area under the curve is the unity while the function is nonzero only on a small interval. Its Laplace transform is 1 − e−sa � a 1 −st L[δa (t)] = e dt = . 0− a sa It is a conceptual aid to introduce an expression δ(t) that describes the effect of δa (t) as a → 0 and to say that L[δ(t)] = 1. The symbol δ(t) is called the unit impulse function or the Dirac delta distribution. is loosely thought of as a function on the real line which is zero everywhere except at the origin, δ(t) = 0 for x �= 0, which is subject to � ∞ δ(t)dt = 1. −∞ Let me say at the outset that δ(t) is not truly a function. Rigorous treatment of the Dirac delta requires the theory of distributions∗. Example 3 (The unit impulse response). We consider the following RLC- circuit problem with the rest initial conditions (D2 + 2D + 2)x = f (t), x(0) = x� (0) = 0. The unit impulse response is the solution x(t) to a unit impulse function δ(t), that is, the solution to (D2 + 2D + 2)x = δ(t), x(0) = x� (0) = 0. Taking the Laplace transform and using that L[δ(t)] = 1, we obtain in succession 1 L[x] = , x(t) = u(t)e−t sin t. (s + 1)2 + 1 It is immediate to see that the unit impulse input has a lasting effect. We note that the solution x is continuous for all t and it satisﬁes the differential equation for � t = 0. At t = 0, however, it satisﬁes neither the differential equation nor the initial conditions, and it is not even differentiable. Indeed, x� (0−) = 0 and x� (0+) = 1. The unit impulse thus produces a jump of magnitude 1 in x� (t) at t = 0. ∗ A distribution is characterized, not by giving its value δ(t) at each t, but by giving its value δφ on a suitable class of functions φ. Distributions were introduced in mid 1930s by Sergei Sobolev, and independently, in late 1940s by Laurent Schwartz. 2 Test functions. The basis of the theory of δ(t) rests on the fact that δ(t) is not a function whose value is deﬁned for each t, but is deﬁned via the action of δ(t) on other functions φ(t), which is called test functions. In the general theory of distributions, test functions are assumed to have derivatives of all orders and have support on a ﬁnite interval. Here we require that φ(t) is contin uously differentiable as many times as are needed. The deﬁning characteristic of the Dirac delta is � ∞ (3) δ(t)φ(t)dt = φ(0). −∞ If δ(t) were an ordinary function and if the integral in (3) were a ordinary Riemann integral, then a change of variable would give � ∞ � ∞ δ(t − c)φ(t)dt = δ(t)φ(t + c)dt. −∞ −∞ By deﬁnition in (3), the right side is φ(c). This is now taken as the deﬁnition of the left side, that is, � ∞ δ(t − c)φ(t)dt = φ(c). −∞ Similarly, we deﬁned the derivative of δ(t) by � ∞ � ∞ � δ (t)φ(t)dt = δ(t)φ� (t)dt. −∞ −∞ Approximate identity. The delta function can be viewed as the limit of a sequence of functions δa (t) in (2). The term approximate identity has a particular meaning in harmonic analysis, in relation to a limiting sequence to an identity element for the convolution operation. Another kind of nascent delta functions is the normal distribution 1 2 2 δa (t) = √ e−x /a . a π The transfer function. Let p(D) be a linear differential operator with constant coefﬁcient and with the characteristic polynomial p(s). The DE p(D)x = Aest could be solved, in general, by trying the solution x = Best . Under the assumption that p(s) �= 0, 1 the result is p(s)Best = Aest , or B = p(s) A. The function W (s) = 1/P (s) that transforms the input amplitude A into the output amplitude B is called the transfer function. The major property of a transfer function is given by (output)=(transfer function)(input). Suppose now that we characterize the input f and output x by their Laplace transforms F and X. If f ∈ E the rest solution of the equation p(D)x = f satisﬁes p(s)Lx = Lf or X = W (s)F, where W (s) = 1/p(s) is the transfer function as deﬁned above. This allows an arbitrary input f . Note that the passage from F to X involves merely a multiplication, while the passage from f to x requires the solution of a differential equation. This is one of the advantages of working in the s-domain. 3 The problem of identiﬁcation. Here we put in a known input, measure the output, and try to ﬁnd the differential equation. This is a common view in bio-mathematics and in mathematical medicine. After measuring the response to a known regime of drug dosage, for example, one can try to discover the mechanism by which that response was produced. It should be said at the outset that the most one can hope for is to ﬁnd the coefﬁcients in the operator p(D), and if many choices of system parameters give the same p(D) then the situation cannot be further disentangled by mere study of the input-output relation. As an illustration, the vibrations of a frictionless mass-spring system are characterized by two parameters m and k, the mass and the stiffness constant. However, even complete knowledge of the differential equation mx�� + kx = kf (t) yields only the ratio k/m, not k and m separately. The differential operator p(D) is determined by its characteristic polynomial p(s) and succes sively p(s) is determined by the transfer function W (s) = 1/p(s). The input function f (t) ∈ E is said to be nontrivial if its transform F (s) is not identically zero. The output associated with any nontrivial inpput sufﬁces for the complete determination of the transfer function W (s). For example, the inputs f1 (t) = u(t) and f2 (t) = δ(t) yields, respectively, W (s) = sX1 (s) and W (s) = X2 (s). The solutions corresponding to X1 and X2 are called, respectively, the unit step response and the unit impulse response. Note that the second equation states that the transfer function is the unit impulse response. Lumped-parameter circuits. The voltage V across a capacitor is related to the charge Q by V = Q/C where 1/C is the elastance. Since the current is I = dQ/dt an integration gives � t V = Q/C = S Idt + Q(0)/C. 0 The initial charge Q(0) is typically 0 and this condition is assumed here. If the capacitor is connected in series with a resistor of resistance R, an inductor of inductance L and a source of voltage V (t) satisfy � t dI L + RI + S Idt = V (t). dt 0 The transform of this with I(0) = 0 is (sL + R + s−1 S)L[I] = L[V ]. The conditions Q(0) = 0 and I(0) = 0 are used here. For a circuit with n independent loops a calculation of the same sort can be carried out for each loop. The result after transformation is a system of the form n � (sLij + Rij + s−1 Sij )L[Ij ] = L[Vi ], j=1 where Ij is the current and Vj is the impressed voltage associated with the jth loop. The resistences Rij and elastances Sij are 0 for i �= j but the inductances Lij need not have this property because of the possibility of mutual inductance between branches. In the matrix form, the above is written V = ZI, I =YV 4 where Z = sL + R + s−1 S and Y = (sL + R + s−1 S)−1 . In the ﬁrst case, I is considered to be input, V is output, and the transfer function Z is an impedance. In the second case V is considered to be input, I is output, and Y is an admittance. Circuits that are characterized by numbers Rij , Lij Sij as above are said to be lumped parameter, to distinguish them from distributed parameter systems such as a waveguide or a submarine cable. R EFERENCES [1] Garret Birkhoff & Gian-Carlo Rota, Ordinary Differential Equations, 4th ed., New York: Wiley, 1978. [2] Ray Redheffer & Dan Port Differential Equations, theory and applications, Jones and Barlett publishers, Boston, 1991. Problems. 1. If f is continuous at t = c, where c > 0, show that lim u(t − c)f (t) = 0, lim h(t − c)f (t) = f (c). t→c− t→c+ Hence u(t − c)f (t) is continuous at c if and only if f (c) = 0. (Hint: Continuity at c means limt→c f (t) = f (c). ) 2. If x = 0 for t < 0 and x� (t) = δ(t) for t > 0 then the Laplace transform suggests that sLx = 1. Assuming this, conclude that x agrees with the unit step function u(t) except perhaps at t = 0. In that sense δ(t) = u� (t). 5