Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

I Heaviside's Operational Calculus and Laplace Transform

  1. Mar 7, 2016 #1
    Hello everyone,

    I was studing Heaviside's operators for solving ODE, which I strongly recommend to have a look because it helps a lot when the differential equations have "exotic" inhomogeneous terms, but it is a method that works and you do not know exactly why.

    Some biographies tell that Carson or Browmich proved that Heaviside's operators are equivalent to Laplace Transform but I have not found any work which explain this fact.

    By my own I've found a kind of relation but I suppose it does not mean nothing or is a "mathematical herecy"

    Well, let a general differential equation of constant coefficients:
    $$ay'(x)+by(x)=F(x)$$ $$y'(x)+(b/a)y(x)=(1/a)F(x)$$
    We can simplify the expression ##(b/a)=-p, f(x)=F(x)/a##
    $$y'(x)-py(x)=f(x)$$
    The differential equation obtained can be resolved by multipliying by an integrating factor. In this case it is ##e^{-px}##
    $$e^{-px}y'(x)-pe^{-px}y(x)=e^{-px}f(x)$$
    So:
    $$\frac{d}{dx}(e^{-px}y(x))=e^{-px}f(x)$$

    $$y(x)=e^{px}(\int f(x)e^{-px}\, dx +c1)=e^{px} \int f(x)e^{-px}\, dx +e^{px} c1$$

    By means of Heaviside Operators we can express the differential equation:
    $$(D-p)(y)=f(x)$$
    Which is the ##(D-p)## operator applied on y. Let see what is the meaning of the inverse operator:
    $$y=\frac{1}{(D-p)}f(x)=(D-p)^{-1}f(x)=e^{px}(\int f(x)e^{-px}\, dx +c1)$$

    So, the inverse operator can be interpreted as this Integral:
    $$(D-p)^{-1}(...):=e^{px}(\int(...) e^{-px}\, dx +c1)$$

    Now, we will solve the equation by means of Laplace Transform:

    $$\mathcal{L}[y'(x)-py(x)]=\mathcal{L}[f(x)]$$
    $$s\mathcal{L}[y(x)]-y(0)-p\mathcal{L}[y(x)]=\mathcal{L}[f(x)]$$
    $$(s-p)\mathcal{L}[y(x)]-y(0)=\mathcal{L}[f(x)]$$
    $$\mathcal{L}[y(x)]=\frac{1}{(s-p)}\mathcal{L}[f(x)]+\frac{1}{(s-p)}y(0)=\frac{1}{(s-p)}\int_{0}^{\infty} f(x)e^{-px}\, dx +\frac{1}{(s-p)}y(0)$$
    $$\mathcal{L}[y(x)]=\mathcal{L}[e^{px}]\mathcal{L}[f(x)]+\mathcal{L}[e^{px}]y(0)$$
    Then by solving the Transform of ##f(x)## we can find the form of ##\mathcal{L}[y(x)]## and then ##y##

    Compiling the results, we have:
    $$y=\frac{1}{(D-p)}f(x)=e^{px}(\int f(x)e^{-px}\, dx +c1)$$
    $$\mathcal{L}[y(x)]=\frac{1}{(s-p)}\int_{0}^{\infty} f(x)e^{-px}\, dx +\frac{1}{(s-p)}y(0)$$

    These equations does not represent the same concept. The first one is the solution of the differential equation and the second one is the transform of that solution but there is a similarity between them so (maybe) this let us think that there is a close relation between this Operators and the Laplace Transform. This would suppose that the Heaviside Operators only works when Laplace Transform converges, among other things. What do you think?

    The method of the Heaviside Operators for ODE of nth order is similar. We would have an operator of nth order and we would have to factorize it (finding its roots) and then solving n linear equations.

    Maybe my exposition was pedantic, but I wanted to explain the problem clearly.

    Thanks for your help.

    Julano
     
  2. jcsd
  3. Mar 12, 2016 #2
    Thanks for the post! This is an automated courtesy bump. Sorry you aren't generating responses at the moment. Do you have any further information, come to any new conclusions or is it possible to reword the post?
     
  4. Mar 25, 2016 #3

    Twigg

    User Avatar
    Gold Member

    I completely forgot about this! My bad!

    That's the tricky bit. Like you said, your relationship shows that the Heaviside operational solution can be derived from the Laplace transform when the Laplace transform converges. Here's a direct proof:

    Given:
    ##\mathcal{L}[y] =(\frac{1}{s - p}) \int_{0}^{\infty} f(x)e^{-px}dx + (\frac{1}{s - p})y_{0}##
    ##(s - p)\mathcal{L}[y] + y_{0} = \int_{0}^{\infty} f(x)e^{-px}dx##
    ##s\mathcal{L}[y] - y_{0} - p\mathcal{L}[y] = \int_{0}^{\infty} f(x)e^{-px}dx ##
    ##\mathcal{L}[Dy] - \mathcal{L}[py] = \int_{0}^{\infty} f(x)e^{-px}dx##
    ##\mathcal{L}[(D-p)y] = \int_{0}^{\infty} f(x)e^{-px}dx##
    ##(D-p)y = \mathcal{L^{-1}}[\int_{0}^{\infty} f(x)e^{-px}dx]##

    This allows you to construct the Heaviside D operator from the Laplace transform if it converges. However, that's not the only way to get such an operator. Generally, D is a linear operator on analytic functions. You can write it recursively for real analytic functions as an upper triangular shift matrix in the basis of powers of x with a factor of n+1. This is the approach found in Boas.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted