Difference between revisions of "Laplace transform"
Jump to navigation
Jump to search
(It's a start. Example(s) will be completed later.) |
m (expanding example; will be completed later) |
||
| Line 6: | Line 6: | ||
<!-- | <!-- | ||
| − | == | + | ==Examples== |
===Example 1=== | ===Example 1=== | ||
Consider the following initial value problem | Consider the following initial value problem | ||
| Line 15: | Line 15: | ||
To solve this problem using laplace transform, first apply laplace transform to both sides of the equation, obtaining: | 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} | + | :<math>\int_0^{\infty}e^{-st} (y'+y) \,dt= \int_0^{\infty}e^{-st} e^{at} \,dt</math> |
Or | Or | ||
| − | :<math>\int_0{\infty}e^{-st}y'\,dt + \int_0{\infty}e^{-st}y | + | :<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 | The integral on the right hand side is | ||
| − | :<math>\int_0{\infty}e^{-st+at} \,dt=\ | + | :<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>=e^{-st}y + (s+1) \int_0^{\infty}e^{-st}y\,dt </math> | ||
--> | --> | ||
==References== | ==References== | ||
| − | D. Lomen and D. Lovelock, ''Differential Equations Graphics. Model. Data.'', John Wiley and Sons, Toronto, 1999. | + | *D. Lomen and D. Lovelock, ''Differential Equations Graphics. Model. Data.'', John Wiley and Sons, Toronto, 1999. |
| − | + | *[http://mathworld.wolfram.com/LaplaceTransform.html Laplace transform] on Wolfram Mathworld | |
{{stub}} | {{stub}} | ||
Revision as of 15:10, June 27, 2009
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>
References
- D. Lomen and D. Lovelock, Differential Equations Graphics. Model. Data., John Wiley and Sons, Toronto, 1999.
- Laplace transform on Wolfram Mathworld