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

${\mathcal {L}}\left\{f(t)\right\}=\int _{0}^{\infty }e^{-st}f(t)\,dt$

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

$\lim _{b\to \infty }e^{-sb}f(b)=0$

Examples

Consider the following initial value problem

$y'+y=e^{at}$

where $a$ is constant, subject to

$y(0)=0$

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

$\int _{0}^{\infty }e^{-st}(y'+y)\,dt=\int _{0}^{\infty }e^{-st}e^{at}\,dt$

Or

$\int _{0}^{\infty }e^{-st}y'\,dt+\int _{0}^{\infty }e^{-st}y\,dt=\int _{0}^{\infty }e^{-st+at}\,dt$

The integral on the right hand side is

$\int _{0}^{\infty }e^{-st+at}\,dt$

$=\lim _{b\to \infty }\int _{0}^{b}e^{-st+at}\,dt$

$=\lim _{b\to \infty }\left[{\frac {1}{a-s}}e^{(a-s)t}-{\frac {1}{a-s}}e^{(a-s)(0)}\right]$

$=\lim _{b\to \infty }\left[{\frac {1}{a-s}}e^{(a-s)t}-{\frac {1}{a-s}}\right]$

If $s>a$,

$\lim _{b\to \infty }\left[{\frac {1}{a-s}}e^{(a-s)t}-{\frac {1}{a-s}}\right]={\frac {1}{a-s}}$

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

$\int uv'\,dt=uv-\int u'v\,dt$

Substitution into the left side will get

$\int _{0}^{\infty }e^{-st}y'\,dt+\int _{0}^{\infty }e^{-st}y'\,dt$

$=e^{-st}y+s\int _{0}^{\infty }e^{-st}y\,dt+\int _{0}^{\infty }e^{-st}y\,dt$

$=\lim _{b\to \infty }e^{-sb}y(b)-e^{-st}y(0)+(s+1)\int _{0}^{\infty }e^{-st}y\,dt$

For the transformation to converge,

$\lim _{b\to \infty }e^{-sb}y(b)=0$

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

$(s+1)\int _{0}^{\infty }e^{-st}y\,dt$

Equating the two sides of the equation:

$(s+1)\int _{0}^{\infty }e^{-st}y\,dt={\frac {1}{a-s}}$

$\int _{0}^{\infty }e^{-st}y\,dt={\frac {1}{(s+1)(a-s)}}$

If $a\neq 1$, we can use partial fractions to changet the right side into

$\int _{0}^{\infty }e^{-st}y\,dt={\frac {1}{a+1}}\left({\frac {1}{a-s}}-{\frac {1}{s+1}}\right)$

and the solution is obtained by noticing:

$\int _{0}^{\infty }e^{-st+at}\,dt={\frac {1}{a-s}}$

Substituting the inverse transform we have

$\int _{0}^{\infty }e^{-st}y\,dt={\frac {1}{a+1}}\left(\int _{0}^{\infty }e^{-st}e^{at}\,dt-\int _{0}^{\infty }e^{-st}e^{-t}\,dt\right)$

$\int _{0}^{\infty }e^{-st}y\,dt=\int _{0}^{\infty }e^{-st}\left[{\frac {1}{a+1}}\left(e^{at}-e^{-t}\right)\right]\,dt$

Which leads to the answer

$y={\frac {1}{a+1}}\left(e^{at}-e^{-t}\right)$

References

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