Chapter 7, Sections 7.1, 7.2, 7.3, 7.4, 7.5
Topics, Definitions, Theorems
Section 7.1 Laplace Transforms and Inverse Transforms
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The Laplace Transform solves differential equations. It should be
advertised as the METHOD of QUADRATURE for Nth ORDER ODE and SYSTEMS of
DIFFERENTIAL EQUATIONS.
DEF. Laplace transform. Direct Laplace transform. Laplace integral.
DEF. Improper integral on (0,infinity). Converge. Diverge.
Assume always in Laplace theory that an input satisfies f(t)=0 for
t<0, and f(t) is defined by a t-expression for t>=0.
Assume always in Laplace theory that a Laplace integral is an
s-expression defined for all large s.
EXAMPLE 1. Laplace integral of f(t)=1.
ANSWER: L(1)=1/s
EXAMPLE 2. Laplace integral of f(t)=exp(at).
ANSWER: L(exp(at))=1/(s-a)
DEF. Generalized factorial function. Gamma function
Gamma(x)=int(G(t,x),t=0..infinity)
where
G(t,x)=t^(x-1) exp(-t), for x>0 and 00, Gamma(1)=1, Gamma(n+1)=n! for
positive integers n. Gamma(1/2)=sqrt(Pi) [page 579]
EXAMPLE 3. Laplace integral of f(t)=t^a, a > -1 real.
ANSWER: L(t^a)=Gamma(a+1)/s^(a+1)
LEMMA. L(t)=1/s^2, L(t^2)=2/s^3, L(t^3)=6/s^4
PROOF. See the lecture slides.
THEOREM 1. Linearity of the Laplace transform. L(af+bg)=aL(f)+bL(g)
EXAMPLE 4. Laplace integral of f(t)=t^(n/2), L(3t^2+4t^(3/2))
ANSWER: L(3t^2+4t^(3/2))=6/s^3+3 sqrt(Pi)/s^(5/2)
The basic important results are for exp(at), cos(kt), sin(kt).
EXAMPLE 5a. Compute L(cosh(kt)) and L(sinh(kt))
ANSWER: s/(s^2-k^2) and k/(s^2-k^2), resp.
EXAMPLE 5b. Compute L(cos(kt)) and L(sin(kt))
ANSWER: s/(s^2+k^2) and k/(s^2+k^2), resp.
EXAMPLE 6. Compute L(3e^(2t) + 2 sin^2(3t))
ANSWER: Use identity cos(2 theta)=1 - 2 sin^2(theta)
L(3e^(2t) + 2 sin^2(3t))=3/(s-2) + 1/s - s/(s^2+36)
INVERSE TRANSFORM
DEF. L^(-1)(F(s))=f(t) means L(f(t))=F(s). Makes sense because of
Lerch's theorem, Theorem 3 below.
CANCELLATION LAW. If L(f(t))=L(g(t)), then f(t)=g(t). The hypothesis
requires "for large s" and the conclusion "at points of continuity".
EXAMPLE 7. Find the inverse transform of
7a. 1/s^3 7b. 1/(s+2) 7c. 2/(s^2+9)
ANSWERS: (a) t^2/2, (b) exp(-2t), (c) (2/3) sin(3t)
WORDS
"Find the function f(t) whose transform is 1/s^3"
"Find the inverse transform of 1/s^3"
"Read the Laplace table backwards"
FIGURE 7.1.2 Short table of transforms
Transform table too long. Too many extras. It is enough to have
entries 1,2,3,5,6,7. Then a second table, of lessor-used transforms.
The words "forward table" and "backward table" summarize what
readers have to learn, in order to use the Laplace table.
Page 581.
DEF. Unit step u(t)
The table entry for u(t-a) should say a>=0.
MISSING: the unit pulse definition u(t-a)-u(t-b). Applications in
7.1 exercises 7, 8, 9, 37, 38, 40, 41, 42.
Page 581
PIECEWISE CONTINUOUS FUNCTIONS
The precise mathematical definition with limits is needed.
Engineering use is primarily in terms of superposition of pulses
multiplied by shape functions.
f(x) = sum of terms v[n] for n=1 to N
v[n]= continuous function times a pulse
DEF. unit pulse = (t,a,b) --> u(t-a)-u(t-b), for a0, a>0
Should be a>=0. See also Table 7.1.2, where a>=0 is missing.
DEF. Function of exponential order.
Page 583
THEOREM 2. Existence of L(f(t)).
COROLLARY. Limit of F(s) as s --> infinity.
Page 584
THEOREM 3. Lerch's theorem, L(f)=L(g) implies f=g.
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Page 587
7.1 APPLICATION: CAS
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7.2 Transformation of Initial Value Problems
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Page 588
THEOREM 1. If f is continuous and piecewise smooth and of exponential
order, then L(f')=sL(f)-f(0).
The alternate definition of piecewise smooth is that f(x) is a
continuous function expressed as the sum of terms g(x) times a unit
pulse, where g(x) is smooth on -infinity to infinity. Could the
alternate wording appear, along with a displayed equation?
FIGURE 7.2.1. Discontinuities of f'
Page 589
COROLLARY. Transforms of Higher Derivatives
EXAMPLE 1. Solve x''-x'-6x=0, x(0)=2, x'(0)=-1
ANSWER: x(t)=(1/5)(3 exp(3t) + 7 exp(-2t))
DETAILS. Transform by (4) and (5) rules, isolate X(s) left. Solve the
partial fraction problem. Use the Laplace table backwards. Apply
Lerch's theorem or take the inverse transform, to find x(t).
Page 590
REMARK. Partial fraction methods.
Missing is Heaviside's cover-up method.
EXAMPLE 2. Solve x'' + 4x = sin(3t), x(0)=0, x'(0)=0
ANSWER: x(t)=(1/10)(3 sin(2t) - 2 sin(3t))
DETAILS. Uses equation (5) and the shortcut for zero initial
conditions. Then X(s) is isolated left to give a partial fraction
problem with quadratic factors.
Shortcut: Using (4) and (5) in the case of zero conditions is much easier.
Answer checks along the route to isolating X(s) left include
looking for the characteristic equation: a preliminary sanity check.
FIGURE 7.2.4. Flow chart for solving a DE by Laplace.
Page 592
LINEAR SYSTEMS
EXAMPLE 3. Solve the system 2x''=-6x + 2y, y''=2x - 2y + 40 sin(3t),
x(0)=x'(0)=y(0)=y'(0)=0.
ANSWER: x(t)=5 sin(t) - 4 sin(2t) + sin(3t),
y(t)=10 sin(t) + 4 sin(2t) - 6 sin(3t)
DETAILS.
Transform the system, get a 2x2 system for X(s), Y(s). Solve by
Cramer's rule and expand in partial fractions. Find the inverse
transforms of X(s), Y(s) from the Laplace table.
Page 593
TRANSFORM PERSPECTIVE.
==> The background is sections 5.1 to 5.4. Please alter the first
sentence of the subsection to insert the back-reference to section
5.4.
EXAMPLE 3.5. Transform of a forced damped oscillator.
X(s) = F(s)/Z(s) + I(s)/Z(s),
1/Z(s)=transfer function
EXAMPLE 4. Compute L(t exp(at)) and verify it equals 1/(s-a)^2.
EXAMPLE 5. Compute L(t sin(kt))
DETAILS. Use (5), f(t)=t sin(kt) and the facts f(0)=f'(0)=0.
ANSWER: F(s)=2ks/(s^2+k^2)^2
THEOREM 2. Transforms of Integrals
Let f be piecewise continuous of exponential order. Then
L(integral of f on [0,t]) = F(s)/s
EXAMPLE 6. Find the inverse transform of G(s)=(1/s^2)/(s-a)
ANSWER: (1/a^2)(exp(at)-at-1)
DETAILS: Use the integral theorem twice.
Page 596
Proof of theorem 1.
Page 597
Extension of theorem 1.
Page 598
EXAMPLE 7. Let f(t)=1+floor(t), the unit staircase. Apply the extension
to theorem 1, to obtain the identity L(f)=(1/s)/(1-exp(-s)).
Page 598
EXERCISES 7.2
Exercises 11 to 16 require background reading from section 7.1,
especially Picard's theorem on page 403.
Exercise 16 should have a footnote to 4.1-26 (brine tanks) and
compartment analysis section 5.3.
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7.3 Translation and Partial Fractions
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DEF. A partial fraction is a constant divided by a polynomial with
exactly one root (real or complex).
DEF. A quadratic partial fraction is a linear polynomial divided by a
polynomial with exactly one pair of complex conjugate roots.
Page 600
The total number of terms in the complete partial fraction
expansion produced by (2) and (3) can be predicted from the degree
of Q(x).
"The degree of Q counts the number of terms in the partial
fraction expansion of P(x)/Q(x)."
Page 601
THEOREM 1. Translation on the s-axis [First shifting theorem]
If L(f) exists then so does L(exp(at)f(t)) and L(exp(at)f(t))=F(s-a).
TABLE page 601
Equations (6), (7), (8) are important.
EXAMPLE 1. Spring-mass system with damping, no forcing.
Solve (1/2)x'' + 3x' + 17x=0, x(0)=3, x'(0)=-1.
ANSWER: x(t)=exp(3t)(3 cos(5t) + 2 sin(5t))
DETAILS.
Transform using (4) and (5) from 7.2.
Isolate X(s) left, expand right side s-fraction into partial
fractions with quadratic factors. Find inverse transforms, report
the answer x(t).
Page 602
EXAMPLE 2. Find the inverse transform of R=(s^2+1)/(s^3-2s^2-8s)
ANSWER: -1/8 + (5/12)exp(-2t) + (17/24)exp(4t)
DETAILS.
Partial fractions with simple roots.
EXAMPLE 3. Solve y''+4y'+4y=t^2, y(0)=y'(0)=0
ANSWER: y(t)=t^2/4 - t/2 + 3/8 - t exp(-2t)/4 - 3 exp(-2t)/8
DETAILS.
Transform using (4), (5) in 7.2, isolate Y(s) left, with
right side fraction R = (2/s^3)/(s+2)^2. Degree of the denominator
is 5, need 5 partial fractions in the expansion, which involves
repeated roots.
The technique is longer than sampling the fraction-free equation with 5
samples (s=0,s=2, and 3 invented samples) to obtain a 5x5 linear
algebraic system of equations for A to E.
Page 604
EXAMPLE 4. Spring-mass system with forcing term.
Solve x''+6x'+34x = 30 sin(2t), x(0)=x'(0)=0.
ANSWER: x(t) = (5/29)(-2 cos(2t) + 5 sin(2t)) [periodic steady-state]
+(2/29)exp(-3t)(5 cos(5t) - 2 sin(5t)) [transient]
DETAILS.
Transform using (4), (5) from 7.2, isolate X(s) left and on the
right side the fraction R = (60/(s^2+4)) / ((s+3)^2+25). Expand R
into 2 quadratic partial fractions, with constants A, B, C, D.
Clear fractions. Substitute s^2+1=0 to get a 2x2 system for A.B.
Solve A=-10/29, B=50/29. Substitute (s+3)^2+25=0 to get a 2x2
system for C,D. Solve for C=D=10/29.
Take inverse transforms.
Missing detail in the example. Does not show how to arrange the
fractions in order to use the Laplace table. Most would use the
shifting theorem first on the fraction with (s+3)^2+25.
Page 606
RESONANCE and REPEATED QUADRATIC FACTORS
Formulas (16), (17) for fractions with the square of a quadratic in the
denominator.
EXAMPLE 5. Undamped forced oscillations, spring-mass system.
Solve x'' + omega0^2 x = F0 sin(omega0 t), x(0)=x'(0)=0
ANSWER:
x(t) = (1/2)(F0/omega0^2)(sin(omega0 t) - omega0 t cos(omega0 t))
DETAILS. Transform using (5) in 7.2, isolate X(s) left, and on the
right side is fraction R=F0 omega0 /(s^2+omega0^2)^2. Apply (17).
FIGURE 7.3.4. Graphic of x(t) and envelope curves.
EXAMPLE 6. Solve (D^4+2D^2+1)y=4 t exp(t), y(0)=y'(0)=y''(0)=y'''(0)=0
ANSWER: y(t) = (t-2) exp(t) + (t+1) sin(t) + 2 cos(t)
DETAILS.
Transform by (5), (7) in 7.2, use zero initial data shortcut.
Isolate Y(s) left, with right side fraction R=(4/(s-1)^2)/(s^2+1)2
The denominator of R has degree 6, the partial fraction expansion
has 6 constants A to F, with 2 partial fractions and two quadratic
partial fractions. Clear fractions. Determine A=1 using s=1. Insert
5 samples s=0, -1, 2, -2, 3 to get a 5x5 system for B to F. Solve
it for B=-2, C=2, D=0, E=2, F=1. Use Laplace tables to finish.
Page 608
EXERCISES 7.3
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7.3 APPLICATION: Damping and Resonance
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A good reference for maple commands. CAS with maple is used in a long maple lab involving a
variety of engineering functions, periodic extensions and solutions of
initial value problems.
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7.4 Derivatives, Integrals and Products of Transforms
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Page 610
EXAMPLE 1a. Show that L(cos(t) sin(t) does not equal L(cos(t))
L(sin(t)).
DEF. Convolution f*g
For completeness, an exercise can be added which relates
the convolution of two functions f, g on -infinity to infinity to
f*g. This uses the assumption that f, g are zero for t<0, in Laplace
theory.
EXERCISE. Define =integral of f(x-t)g(t) over t=-inf to inf.
Assume f and g are zero for t<0. Show that =f*g.
This verifies that the convolution defined in Fourier
theory agrees with f*g, provided f=g=0 for t<0.
Page 611
EXAMPLE 1. Verify cos(t)*sin(t) = (1/2) t sin(t)
THEOREM 1. CONVOLUTION PROPERTY
Let f, g be piecewise continuous and of exponential order.
Then f*g is defined and L(f*g) =L(f)L(g)
Page 612
EXAMPLE 2. Find sin(2t) * exp(t) and use the answer to compute the
inverse transform of the fraction R=(2/(s^2+4))/(s-1).
ANSWER:
L(sin(2t)=2/(s^2+4)
L(exp(t))=1/(s-1)
L(sin(2t)*exp(t))=(2/5) exp(t) - (1/5) sin(2t) - (2/5) cos(2t)
inverse laplace of R = t-expression on previous line
Missing in the example are the two equations L(sin(2t))=2/(s^2+4)
and L(exp(t))=1/(s-1), essential to understanding the details.
The purpose of the example is to compute the Laplace
inverse of the fraction R.
THEOREM 2. Differentiation of Transforms [s-diff theorem]
If f is piecewise continuous and of exponential order, then
L(-t f(t)) = F'(s)
L((-t)^n f(t)) = (d/ds)^n F(s)
EXAMPLE 3. Find L(t^2 sin(kt))
ANSWER: (6ks^2-2k^3)/(s^2+k^2)^2
DETAILS.
Apply Theorem 2 to get F(s)=(-1)^2 (d/ds)^2 (k/(s2+k^2). Rest is
calculus.
EXAMPLE 4. Find the inverse transform of F(s)=arctan(1/s)
ANSWER: f(t)=sin(t)/t
DETAILS.
Apply Theorem 2, L((-t)f(t)) = (d/ds) F(s) to get L(-t f(t)) =
-1/(s^2+1), which equals L(-sin(t)). Then the cancellation law
implies -t f(t) = -sin(t).
EXAMPLE 5. Find X(s)=L(x(t)) by Laplace's method for Bessel's equation
t x'' + x' + t x = 0 with x(0)=1 and x'(0)=0.
ANSWER: X=1/sqrt(s^2+1).
DETAILS.
Laplace's method gives (s^2+1)X'(s)+X(s)=0, which is a separable
differential equation with solution X=C/sqrt(s^2+1). C=1 obtained in
an exercise.
Page 614.
INTEGRATION OF TRANSFORMS
THEOREM 3. Integration of transforms
Assume f piecewise continuous of exponential order and f(t)/t has a
finite right limit at t=0. Then
L(f(t)/t) = integral of F(s) over [s,infinity)
EXAMPLE 6. Let f(t)=sinh(t). Find L(f(t)/t).
ANSWER: L(f(t)/t)=(1/2)ln|s+1|-(1/2)ln|s-1|
DETAILS.
Check the right limit at t=0 for f(t)/t using L'Hopital's rule.
Apply Theorem 3. Use L(sinh(t)) = (1/2)/(s-1)-(1/2)/(s+1). Then
integrate to get logs. We have to write the difference of logs as
the log of a quotient in order to evaluate at s=infinity (ln(1)=0).
EXAMPLE 7. Find the inverse Laplace transform of F(s)=2s/(s^2-1)^2.
ANSWER: f(t)= t sinh(t).
DETAILS.
Apply Theorem 3, L(f(t)/t)==integral of F(s) over [s,infinity).
The integration of F(s) gives 1/(s^2-1)=L(sinh(t)). Then
L(f(t)/t)=L(sinh(t)), and cancellation implies f(t)/t = sinh(t).
Page 615.
PROOFS OF THE THEOREMS
Try to read pages 615-616. The proofs of theorems 2, 3 are usually done in class.
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Page 617
EXERCISES
Problems 1-34 are recommended. The remaining problems are quite challenging.
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7.5 Periodic and Piecewise Continuous Input Functions
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Page 618
DEF. unit step. Repeated from 7.1.
THEOREM. L(u(t-a)) = exp(-as)/s for a>=0, s>0.
THEOREM 1. Translation on the t-axis [second shift theorem]
(a) If L(f) exists for s>c and a>=0, then for s>c+a
L(u(t-a)f(t-a)) = exp(-as) F(s)
(b) If a>=0 and L(g(t+a)) exists for large s, then
L(u(t-a)g(t)) = exp(-as) L(g(t+a))
Error: Page 618, statement of theorem 1, missing a>=0.
Missing: Page 618, statement of theorem 1, missing part (b) of the second
shifting theorem. Part (b) is suited for finding the transform of
terms like sin(t) u(t-2) whereas part (a) is suited for inverse
transforms of terms like exp(-2s)/(s^2+4).
Parts (a) and (b) have an alternative notation, which suggests the
translations explicitly, but also they are easier to use on a given
problem, due to looking like the calculus u-substitution.
(a) If L(f) exists for s>c and a>=0, then for s>c+a
L(u(x)f(x)|x=t-a) = exp(-as) F(s)
(b) If a>=0 and L(g(x)|x=t+a) exists for large s, then
L(u(t-a)g(t)) = exp(-as) L(g(x)|x=t+a)
Example: L(sin(t)u(t-Pi)) = exp(-Pi s) L(sin(x)|x=t+Pi)
Page 619
Error: Proof of theorem 1. Missing a>=0, used when splitting the integral
over [0,a] and [a,infinity).
EXAMPLE 1. Find the inverse transform of exp(-as)/s^3.
ANSWER: (1/2)(t-a)^2u(t-a) for a>=0
DETAILS.
Using Theorem 1 part (a), exp(-as)L(f(t))= L(u(t-a)f(t-a))
[backwards on purpose] and comparing to exp(-as)/s^3, choose
L(f(t))=1/s^3 or f(t)=t^2/2. Then exp(-as)/s^3 = exp(-as)L(t^2/2)=
L(x^2/2 u(x)|x=t-a), giving the answer x^2/2 u(x)|x=t-a =
(1/2)(t-a)^2 u(t-a).
EXAMPLE 2. Find L(g(t)) if g(t)=t^2u(t-3).
ANSWER: exp(-3s)(2/s^3 + 6/s^2 + 9/s)
DETAILS.
Using theorem 1, part (b), L(t^2 u(t-3))=exp(-3s)L(x^2|x=t+3)=
exp(-3s)L((t+3)^3). Expanding the square and using the Laplace
table gives the result.
Page 620
EXAMPLE 3. Pulse problem.
Find L(f(t)), given f(t) = cos(2t) pulse(t,0,Pi)
ANSWER: s(1-exp(-2 PI s))/(s^2+4)
DETAILS.
Write the pulse as 1-u(t - 2 Pi). Then
f(t)=cos(2t) - cos(2t)u(t - 2 Pi)
Replace cos(2t) by the equivalent expression cos(2t - 2 Pi) in the
second term. Then use the second shifting theorem as follows
L(f(t))=L(cos(2t)) - L(cos(x)u(x)|x=2t-2Pi)
=L(cos(2t)) - exp(-2Pi s)L(cos(2t))
= (1-exp(-2Pi s))(2/(s^2+4))
EXAMPLE 4. Forced harmonic oscillator with pulse force.
Solve x'' + 4x = f(t), x(0)=x'(0)=0, where
f(t)=cos(2t) pulse(t,0,2Pi)
ANSWER: x(t) = (1/4) t sin(2t) for t < 2Pi, f(t) = (Pi/2)sin(2t) for
t>= 2Pi.
DETAILS.
Transform the differential equation by (5) in section 7.2. Isolate
X(s) left and on the right expand L(f(t)) using the answer from
Example 3. Then X(s)=s/(s^2+4)^2 - exp(-2Pi s)(s/(s^2+4)^2). Use
the Laplace table on the inside cover of the book to obtain the
formula L((t/4)sin(2t))=s/(s^2+4)^2. Use the second shifting
theorem on the second term in X(s), as follows:
exp(-2Pi s)(s/(s^2+4)^2) = L((x/4)sin(2x)u(x)|x=t-2Pi)
= L(((t-2Pi)/4)sin(2(t-2Pi))u(t-2Pi))
The last challenge is to write this piecewise defined function as a
normal display.
Page 621
TRANSFORMS of PERIODIC FUNCTIONS
THEOREM 2. Periodic Function Theorem
Let f(t) satisfy f(t+p)=f(t), where p>0 is the period of f, and
assume f is piecewise continuous for t>=0. Then L(f(t)) exists and is
given by the equation
L(f(t)) = P(t)/(1-exp(-ps)),
P(t) = integral of f(t)exp(-st) over t=0 to t=p.
Page 622
Another use of the geometric series.
EXAMPLE 5. Square wave of period p=2a, f(t)=(-1)^floor(t/a).
Verify L(f(t))=(1/s)tanh(as/2).
DETAILS.
Use the periodic function theorem. The main task is to integrate
f(t)exp(-st) on t=0 to t=p. This is done by writing the integral
as the sum of two integrals, the first over [0,a] and second over
[a,2a]. The integrands simplify and integration is possible, giving
integral=(1-exp(-as))^2/s. The rest is a trick, to multiply the
fraction for L(f(t)) by exp(as/2)/exp(as/2), distribute, and then
discover sinh(as/2) and cosh(as/2) in the fraction. Their quotient
is tanh(as/2).
EXAMPLE 6. Triangular wave g(t), period p=2a. Verify the identity
L(g(t))=tanh(as/2)/s^2.
DETAILS.
Because g'(t)=f(t) in Example 5, then the parts formula implies
sL(g)-g(0)=L(g')=L(f)=tanh(as/2)/s.
Evaluate g(0)=0. Then sL(g)=tanh(as/2)/s, which implies the
identity.
EXAMPLE 7. Forced damped spring-mass system.
Let f(t) be the square wave of amplitude 20 and period
p=2Pi. Solve the problem x''+4x'+20x=f(t), x(0)=x'(0)=0.
ANSWER: Too complicated to write here. See Page 624.
DETAILS.
Write F(s)=L(f(t)). Then
F(s)=20/s+(40/s)(series of (-1)^n exp(-n Pi s), n=1..infinity)
Transform the DE into X(s)=F(s)/Z(s), where Z(s)=s^2+4s+20.
Expand F/Z into an infinite series and apply the second shifting
theorem on each term.
EXERCISES
7.5 APPLICATION: Engineering functions
This material is used heavily in the maple lab on Laplace theory.