Elzaki and Sumudu transforms

The Sumudu transform was proposed by G.K. Watugala in 1993:
\[ S\left[ f(t) \right] (\nu ) = \frac{1}{\nu} \int_0^{\infty} f(t)\,e^{-t/\nu} \,{\text d}t . \]
A sufficient condition for the existence of the Sumudu transform of a function f is its exponential order: \( \left\vert f(t) \right\vert \le M\,e^{k\,t} , \) for some positive constants M and k. The Sumudu transformation is related to the Laplace transform via formula:
\[ S\left[ f(t) \right] (\nu ) = \frac{1}{\nu}\,{\cal L} \left[ f(t) \right] \left( \frac{1}{\nu} \right) = \int_0^{\infty} f(t\nu )\,e^{-t} \,{\text d}t, \]
and to Elzaki transform
\[ S\left[ f(t) \right] (\nu ) = \nu^2 E\left[ f(t) \right] (\nu ) , \]
where the Elzaki transformation is
\[ E\left[ f(t) \right] (\nu ) = \nu\,\int_0^{\infty} f(t)\,e^{-t/\nu} \,{\text d}t = \nu^2 \int_0^{\infty} f(t\nu )\,e^{-t} \,{\text d}t . \]
The inverse Elzaki transform:
\[ E^{-1} \left[ F(\nu ) \right] (t ) = \frac{1}{2\pi{\bf j}}\,\int_{a-{\bf j}\infty}^{a+{\bf j}\infty} F\left( \frac{1}{\nu} \right)\,e^{t\nu} \,\nu \,{\text d}\nu = \sum \,\mbox{residues of } \left[ F\left( \frac{1}{\nu} \right)\,e^{t\nu} \,\nu \right] . \]
The inverse Sumudu transform:
\[ S^{-1} \left[ F(\nu ) \right] (t ) = \frac{1}{2\pi{\bf j}}\,\int_{a-{\bf j}\infty}^{a+{\bf j}\infty} \,{\text d}t . \]

The ELzaki and Laplace transforms exhibit a duality relation expressed as follows

\[ E \left[ f(t) \right] (\nu ) = \nu\,{\cal L}\left[ f(t) \right] \left( \frac{1}{\nu} \right) \qquad\mbox{and} \qquad {\cal L}\left[ f(t) \right] \left( \lambda \right) = \lambda \, E \left[ f(t) \right] \left( \frac{1}{\lambda} \right) . \]
The main properties of these transformations are
  1. Convolution property:
    \begin{align*} S \left[ f(t) * g(t) \right] (\nu ) &= \nu\,S \left[ f(t) \right] (\nu ) \, \nu\,S \left[ g(t) \right] (\nu ) = \nu^2 S \left[ f(t) \right] S \left[ g(t) \right] ; \\ E \left[ f(t) * g(t) \right] (\nu ) &= \frac{1}{\nu}\,E \left[ f(t) \right] E \left[ g(t) \right] \end{align*}
    where \( (f*g)(t) = \int_0^t f(\tau )\,g(t-\tau )\,{\text d}\tau = (g*f)(t) \) is the convolution of two functions.
  2. Differentiation property:
    \begin{align*} S \left[ f' (t) \right] (\nu ) &= \frac{1}{\nu}\,S \left[ f (t) \right] (\nu ) - \frac{1}{\nu}\,f(+0) , \\ S \left[ f'' (t) \right] (\nu ) &= \frac{1}{\nu^2}\,S \left[ f (t) \right] (\nu ) - \frac{1}{\nu^2}\,f(+0) - \frac{1}{\nu}\,f'(+0) , \\ E \left[ f' (t) \right] (\nu ) &= \frac{1}{\nu} \,E \left[ f (t) \right] (\nu ) - \nu\,f(+0) , \\ E \left[ f'' (t) \right] (\nu ) &= \frac{1}{\nu^2} \,E \left[ f (t) \right] (\nu ) - \nu\,f'(+0) -f(+0) , \\ S \left[ t\,f' (t) \right] (\nu ) &= \\ S \left[ t^2 f' (t) \right] (\nu ) &= \\ E \left[ t\,f' (t) \right] (\nu ) &= \nu^2 \frac{\text d}{{\text d}\nu} \left[ \frac{1}{\nu}\, E \left[ f (t) \right] (\nu ) -\nu\,f(0) \right] - E \left[ f (t) \right] (\nu ) + \nu^2 f(0) , \\ E \left[ t^2 f' (t) \right] (\nu ) &= \nu^4 \frac{{\text d}^2}{{\text d}\nu^2} \left[ \frac{1}{\nu}\, E \left[ f (t) \right] (\nu ) - \nu\,f(0) \right] , \\ S \left[ t\,f'' (t) \right] (\nu ) &= \\ S \left[ t^2 f'' (t) \right] (\nu ) &= \\ E \left[ t\,f'' (t) \right] (\nu ) &= \nu^2 \frac{\text d}{{\text d}\nu} \left[ \frac{1}{\nu^2}\, E \left[ f (t) \right] (\nu ) -\nu\,f'(0) - f(0) \right] - \frac{1}{\nu}\,E \left[ f (t) \right] (\nu ) + \nu\, f(0) + \nu^2 f' (0) , \\ E \left[ t^2 f'' (t) \right] (\nu ) &= \nu^4 \frac{{\text d}^2}{{\text d}\nu^2} \left[ \frac{1}{\nu^2}\, E \left[ f (t) \right] (\nu ) - f(0) - \nu\,f' (0) \right] . \end{align*}

 

Example: The following formulas follow from the corresponding Laplace transformations:

\begin{align*} S \left[ H(t) \right] (\nu ) &= 1, \\ E \left[ H(t-a) \right] (\nu ) &= \nu^2 e^{-a/\nu} , \\ S \left[ \delta (t) \right] (\nu ) &= \frac{1}{\nu} , \\ E \left[ \delta (t-a) \right] (\nu ) &= \nu\, e^{-a/\nu} , \\ S \left[ \sin t\,H(t) \right] (\nu ) &= \frac{\nu}{1 + \nu^2} , \\ E \left[ \sin (at)\,H(t) \right] (\nu ) &= \frac{a\,\nu^3}{1+ a^2 \nu^2} , \\ S \left[ \cos t\,H(t) \right] (\nu ) &= \\ E \left[ \cos (at)\,H(t) \right] (\nu ) &= \frac{\nu^2}{1+ a^2 \nu^2} , \\ S \left[ \sinh t\,H(t) \right] (\nu ) &= \frac{\nu}{1 - \nu^2} , \\ E \left[ \sinh (at)\,H(t) \right] (\nu ) &= \frac{a\,\nu^3}{1- a^2 \nu^2} , \\ S \left[ \cos t\,H(t) \right] (\nu ) &= \\ E \left[ \cosh (at)\,H(t) \right] (\nu ) &= \frac{\nu^2}{1- a^2 \nu^2} , \\ S \left[ t^n \,H(t) \right] (\nu ) &= \\ E \left[ t^n \,H(t) \right] (\nu ) &= n! \nu^{n+2} , \\ S \left[ J_0 (at) \,H(t) \right] (\nu ) &= \\ E \left[ J_0 (at) \,H(t) \right] (\nu ) &= \frac{\nu^2}{\sqrt{1 + a^2 \nu^2}} . \end{align*}
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Example: Consider the Bessel equation subject to the initial condition

\[ t\,y'' + y' + a^2 t\,y =0, \qquad y(0) =1. \]
Application of the Elzaki transform yields
\[ \nu^2 \frac{\text d}{{\text d}\nu} \left[ \frac{1}{\nu^2}\,E[y] -1 -\nu y'(0) \right] + \frac{1}{\nu}\,E[y] -\nu + a^2 \left[ \nu^2 \frac{\text d}{{\text d}\nu}\,E[y] - \nu\,E[y] \right] =0 . \]
Let us denote by T(ν) the Elzaki transform of the unknown function y(t). The above equation can be rewritten as
\[ T' (\nu ) - \frac{2}{\nu}\, T(\nu ) + a^2 \nu^2 T' (\nu ) - a^2 \nu\,T(\nu ) =0 \]
or separating of variables,
\[ \frac{T' (\nu )}{T(\nu )} = \frac{2}{\nu} - \frac{a^2 \nu}{1 + a^2 \nu^2} . \]
Integrating both sides, we get
\[ T(\nu ) = \frac{C\,\nu^2}{\sqrt{1 + a^2 \nu^2}} , \]
where C is a constant of integration. Inventing the Elzaki transform, we obtain
\[ y(t) = E^{-} \left[ T(\nu ) \right] (t) = C\,J_0 \left( at \right) . \]
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Example: Consider the initial value problem for the Euler equation

\[ t^2 y'' + 4t\,y' + 2\,y = 6\,t^2 , \qquad y(0) = y' (0) =0. \]
First we apply the Sumudu transform to this equation to find
\[ \nu^2 S'' + 4\nu\,S' + 2\,S = 12\,\nu^2 , \]
which is the same differential equation. We conclude that the Sumudu transform does not provide any advantage.

Now if the Elzaki transform to the given equation is applied, it yields

\[ \nu^ \frac{{\text d}^2}{{\text d} \nu^2} \left[ \frac{E[y](\nu )}{\nu^2} \right] + 4\nu^2\, \frac{{\text d}}{{\text d} \nu} \left[ \frac{E[y](\nu )}{\nu} \right] + 2\,E[y] = 12\,\nu^2 , \]
The general solution to the latter is
\[ E[y] = \nu^4 + c_1 \nu + c_2 . \]
Using the initial conditions, we find the values of constant c1 and c2 to be both zeroes. By using the inverse Elzaki transform we find the solution in the form: \( y(t) = t^2 /2 . \)    ■

 

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