Laplace transform

Laplace transforms are one of the ways of solving linear ordinary differential equations (Linear ODEs) with constant coefficients. This technique allows us to transform a Linear ODE into a linear algebraic equation.

Definition

The unilateral Laplace transform is defined by

<math>\mathcal{L} \left\{f(t)\right\}=\int_0^{\infty} e^{-st} f(t) \,dt </math>

Given the integral converges. A necessary condition for this integral to converge is

<math>\lim_{b\to\infty} e^{-sb}f(b)=0</math>

Examples

Consider the following initial value problem

<math>y'+y=e^{at}</math>

where <math>a</math> is constant, subject to

<math>y(0)=0</math>

To solve this problem using laplace transform, first apply laplace transform to both sides of the equation, obtaining:

<math>\int_0^{\infty}e^{-st} (y'+y) \,dt= \int_0^{\infty}e^{-st} e^{at} \,dt</math>

Or

<math>\int_0^{\infty}e^{-st}y'\,dt + \int_0^{\infty}e^{-st}y\,dt = \int_0^{\infty}e^{-st+at} \,dt</math>

The integral on the right hand side is

<math>\int_0^{\infty}e^{-st+at} \,dt</math>

<math>=\lim_{b\to\infty} \int_0^b e^{-st+at} \,dt</math>

<math>=\lim_{b\to\infty}\left[\frac{1}{a-s} e^{(a-s)t} - \frac{1}{a-s}e^{(a-s)(0)}\right]</math>

<math>= \lim_{b\to\infty}\left[\frac{1}{a-s} e^{(a-s)t} - \frac{1}{a-s}\right]</math>

If <math>s>a</math>,

<math>\lim_{b\to\infty} \left[\frac{1}{a-s}e^{(a-s)t} - \frac{1}{a-s}\right]= \frac{1}{a-s}</math>

For the left side, if we apply integration by parts,

<math>\int uv'\,dt=uv-\int u'v\,dt</math>

Substitution into the left side will get

<math>\int_0^{\infty}e^{-st}y'\,dt + \int_0^{\infty}e^{-st}y'\,dt </math>

<math>=e^{-st}y + s\int_0^{\infty}e^{-st}y\,dt + \int_0^{\infty}e^{-st}y\,dt </math>

<math>=\lim_{b\to\infty}e^{-sb}y(b)-e^{-st}y(0) + (s+1) \int_0^{\infty}e^{-st}y\,dt </math>

For the transformation to converge,

<math>\lim_{b\to\infty}e^{-sb}y(b)=0</math>

Therefore, substituting the initial condition y(0)=0 the left side becomes

<math>(s+1) \int_0^{\infty}e^{-st}y\,dt </math>

Equating the two sides of the equation:

<math>(s+1) \int_0^{\infty}e^{-st}y\,dt =\frac{1}{a-s}</math>

<math>\int_0^{\infty}e^{-st}y\,dt =\frac{1}{(s+1)(a-s)}</math>

If <math>a \ne 1</math>, we can use partial fractions to changet the right side into

<math>\int_0^{\infty}e^{-st}y\,dt =\frac{1}{a+1}\left(\frac{1}{a-s}-\frac{1}{s+1}\right)</math>

and the solution is obtained by noticing:

<math>\int_0^{\infty}e^{-st+at} \,dt=\frac{1}{a-s}</math>

Substituting the inverse transform we have

<math>\int_0^{\infty}e^{-st}y\,dt=\frac{1}{a+1}\left(\int_0^{\infty}e^{-st}e^{at}\,dt-\int_0^{\infty}e^{-st}e^{-t}\,dt \right) </math>

<math>\int_0^{\infty}e^{-st}y\,dt=\int_0^{\infty}e^{-st}\left[\frac{1}{a+1}\left(e^{at}-e^{-t} \right)\right]\,dt </math>

Which leads to the answer

<math>y=\frac{1}{a+1}\left(e^{at}-e^{-t} \right)</math>

References

• D. Lomen and D. Lovelock, Differential Equations Graphics. Model. Data., John Wiley and Sons, Toronto, 1999.
• Laplace transform on Wolfram Mathworld