College Math Teaching

November 12, 2013

Why I teach multiple methods for the inverse Laplace Transform.

I’ll demonstrate with a couple of examples:

y''+4y = sin(2t), y(0) = y'(0) = 0

If we use the Laplace transform, we obtain: (s^2+4)Y = \frac{2}{s^2+4} which leads to Y = \frac{2}{(s^2+4)^2} . Now we’ve covered how to do this without convolutions. But the convolution integral is much easier: write Y = \frac{2}{(s^2+4)^2} = \frac{1}{2} \frac{2}{s^2+4}\frac{2}{s^2+4} which means that y = \frac{1}{2}(sin(2t)*sin(2t)) = \frac{1}{2}\int^t_0 sin(2u)sin(2t-2u)du = -\frac{1}{4}tcos(2t) + \frac{1}{8}sin(2t) .

Note: if the integral went too fast for you and you don’t want to use a calculator, use sin(2t-2u) = sin(2t)cos(2u) - cos(2t)sin(2u) and the integral becomes \frac{1}{2}\int^t_0 sin(2t)cos(2u)sin(2u) -cos(2t)sin^2(2u)du =

\frac{1}{2} (sin(2t))\frac{1}{4}sin^2(2u)|^t_0 - cos(2t)(\frac{1}{4})( t - \frac{1}{4}sin(4u)|^t_0 =

\frac{1}{8}sin^3(2t) - \frac{1}{4}tcos(2t) +\frac{1}{16}sin(4t)cos(2t) =

\frac{1}{8}(sin^3(2t) +sin(2t)cos^2(2t))-\frac{1}{4}tcos(2t)

= \frac{1}{8}sin(2t)(sin^2(2t) + cos^2(2t))-\frac{1}{4}tcos(2t) = -\frac{1}{4}tcos(2t) + \frac{1}{8}sin(2t)

Now if we had instead: y''+4y = sin(t), y(0)=0, y'(0) = 0

The Laplace transform of the equation becomes (s^2+4)Y = \frac{1}{s^2+1} and hence Y = \frac{1}{(s^2+1)(s^2+4)} . One could use the convolution method but partial fractions works easily: one can use the calculator (“algebra” plus “expand”) or:

\frac{A+Bs}{s^2+4} + \frac{C + Ds}{s^2+1} =\frac{1}{(s^2+4)(s^2+1)} . Get a common denominator and match numerators:

(A+Bs)(s^2+1) + (C+Ds)(s^2+4)  = 1 . One can use several methods to resolve this: here we will use s = i to see (C + Di)(3) = 1 which means that D = 0 and C = \frac{1}{3} . Now use s = 2i so obtain (A + 2iB)(-3) = 1 which means that B = 0, A = -\frac{1}{3} so Y = \frac{1}{3} (\frac{1}{s^2+1} - \frac{1}{s^2+4} so y = \frac{1}{3} (sin(t) - \frac{1}{2} sin(2t)) = \frac{1}{3}sin(t) -\frac{1}{6}sin(2t)

So, sometimes the convolution leads us to the answer quicker than other techniques and sometimes other techniques are easier.

Of course, the convolution method has utility beyond the Laplace transform setting.

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2 Comments »

  1. the above are very important as say prof dr mircea orasanu and prof horia orasanu

    Comment by perringu — February 7, 2018 @ 4:06 am

  2. also these we see as say prof dr mircea orasanu and prof horia orasanu as followed
    LAGRANGIAN AND FOURIER TRANSFORM
    Author Horia orasanu
    ABSTRACT
    In each of these cases the lone nonhomogeneous boundary condition will take the place of the initial condition in the heat equation problems that we solved a couple of sections ago. We will apply separation of variables to the each problem and find a product solution that will satisfy the differential equation and the three homogeneous boundary conditions
    1 INTRODUCTION
    Using the Principle of Superposition we’ll find a solution to the problem and then apply the final boundary condition to determine the value of the constant(s) that are left in the problem. The process is nearly identical in many ways to what we did when we were solving the heat equation.Note that in this case, unlike the heat equation we must solve the boundary value problem first. Without knowing what is there is no way that we can solve the first differential equation here with only one boundary condition since the sign of will affect the solution. and thus appear
    THESIS

    Fondul problemei. LAGRANGIAN si DEFINIREA CONSTRANGERILOR
    Aici noi consideram ca

    Astfel ca problema de mai sus se reduce la o problema importanta avalorilor proprii de
    Forma

    /t = [p ( x) u]-q ( x ) u

    cand sunt indeplinite conditiile la limita si initiale de forma

    u( )=

    aici conditiile initiale si la limita reprezinta legaturile cinematice ale vitezei care sunt ec
    hivalente cu legaturile neolonome ale miscarilor enuntate mai sus ,deci cele care fac part
    e din dinamica sistemelor de puncte materiale ,sau sisteme de particule fluide. Deci astfel ca in acest caz determinant al curburii de tip neolonomic este presiunea pe lini
    E sau suprafata ,unde potentialul vitezelor are o forma data,precum in cazul miscarilor

    Punctelor singular

    In acest caz ca element determinant al curburii de tip neolonomic este presiunea pe linie
    Sau suprafata ,unde potentialul vitezelor are o forma da

    Si deci rezultatele numerice pentru schema si problemele neolonomice in cazul ptobleme
    Lor cu cavitati pot fi comparate cu existent solutiilor din punct de vedere fizic
    We use cookies to enhance your experience on our website. By continuing to use our website, you are agreeing to our use of cookies. You can change your cookie settings at any time. Find out more
    Let us now assume that the fractal curve is immersed in a three-dimensional space and that X of components Xi (i = 1, 2, 3) is the position vector of a point on the curve. Let us also consider a function f(X, t) and the following series expansion up to the second order:
    df=f(Xi+dXi,t+dt)−f(Xi,dt)=(∂∂XidXi+∂∂tdt)f(Xi,t)+12(∂∂XidXi+∂∂tdt)2f(Xi,t)

    Comment by perringu — February 7, 2018 @ 4:20 am


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