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

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \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

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \lim_{b\to\infty} e^{-sb}f(b)=0}

Examples

Consider the following initial value problem

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle y'+y=e^{at}}

where Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle 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

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

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

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

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle = \lim_{b\to\infty}\left[\frac{1}{a-s} e^{(a-s)t} - \frac{1}{a-s}\right]}

If Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle s>a} ,

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \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,

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \int uv'\,dt=uv-\int u'v\,dt}

Substitution into the left side will get

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \int_0^{\infty}e^{-st}y'\,dt + \int_0^{\infty}e^{-st}y'\,dt }

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle =e^{-st}y + s\int_0^{\infty}e^{-st}y\,dt + \int_0^{\infty}e^{-st}y\,dt }

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle =\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,

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \lim_{b\to\infty}e^{-sb}y(b)=0}

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

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle (s+1) \int_0^{\infty}e^{-st}y\,dt }

Equating the two sides of the equation:

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle (s+1) \int_0^{\infty}e^{-st}y\,dt =\frac{1}{a-s}}

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \int_0^{\infty}e^{-st}y\,dt =\frac{1}{(s+1)(a-s)}}

If Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle a \ne 1} , we can use partial fractions to changet the right side into

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \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:

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \int_0^{\infty}e^{-st+at} \,dt=\frac{1}{a-s}}

Substituting the inverse transform we have

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \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) }

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \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

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle 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