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The [[Laplace transform]] is a powerful [[integral transform]] used to switch a function from the [[time domain]] to the [[Laplace transform#s-Domain equivalent circuits and impedances|s-domain]]. The Laplace transform can be used in some cases to solve [[linear differential equation]]s with given [[Initial value problem|initial conditions]]. | |||
First consider the following property of the Laplace transform: | |||
:<math>\mathcal{L}\{f'\}=s\mathcal{L}\{f\}-f(0)</math> | |||
:<math>\mathcal{L}\{f''\}=s^2\mathcal{L}\{f\}-sf(0)-f'(0)</math> | |||
One by [[Mathematical induction|induction]] can prove that | |||
:<math>\mathcal{L}\{f^{(n)}\}=s^n\mathcal{L}\{f\}-\sum_{i=1}^{n}s^{n-i}f^{(i-1)}(0)</math> | |||
Now we consider the following differential equation: | |||
:<math>\sum_{i=0}^{n}a_if^{(i)}(t)=\phi(t)</math> | |||
with given initial conditions | |||
:<math>f^{(i)}(0)=c_i</math> | |||
Using the [[linearity]] of the Laplace transform it is equivalent to rewrite the equation as | |||
:<math>\sum_{i=0}^{n}a_i\mathcal{L}\{f^{(i)}(t)\}=\mathcal{L}\{\phi(t)\}</math> | |||
obtaining | |||
:<math>\mathcal{L}\{f(t)\}\sum_{i=0}^{n}a_is^i-\sum_{i=1}^{n}\sum_{j=1}^{i}a_is^{i-j}f^{(j-1)}(0)=\mathcal{L}\{\phi(t)\}</math> | |||
Solving the equation for <math> \mathcal{L}\{f(t)\}</math> and substituting <math>f^{(i)}(0)</math> with <math>c_i</math> one obtains | |||
:<math>\mathcal{L}\{f(t)\}=\frac{\mathcal{L}\{\phi(t)\}+\sum_{i=1}^{n}\sum_{j=1}^{i}a_is^{i-j}c_{j-1}}{\sum_{i=0}^{n}a_is^i}</math> | |||
The solution for ''f''(''t'') is obtained by applying the [[inverse Laplace transform]] to <math>\mathcal{L}\{f(t)\}.</math> | |||
Note that if the initial conditions are all zero, i.e. | |||
:<math>f^{(i)}(0)=c_i=0\quad\forall i\in\{0,1,2,...\ n\}</math> | |||
then the formula simplifies to | |||
:<math>f(t)=\mathcal{L}^{-1}\left\{{\mathcal{L}\{\phi(t)\}\over\sum_{i=0}^{n}a_is^i}\right\}</math> | |||
==An example== | |||
We want to solve | |||
:<math>f''(t)+4f(t)=\sin(2t)</math> | |||
with initial conditions ''f''(0) = 0 and ''f′''(0)=0. | |||
We note that | |||
:<math>\phi(t)=\sin(2t)</math> | |||
and we get | |||
:<math>\mathcal{L}\{\phi(t)\}=\frac{2}{s^2+4}</math> | |||
The equation is then equivalent to | |||
:<math>s^2\mathcal{L}\{f(t)\}-sf(0)-f'(0)+4\mathcal{L}\{f(t)\}=\mathcal{L}\{\phi(t)\}</math> | |||
We deduce | |||
:<math>\mathcal{L}\{f(t)\}=\frac{2}{(s^2+4)^2}</math> | |||
Now we apply the Laplace inverse transform to get | |||
:<math>f(t)=\frac{1}{8}\sin(2t)-\frac{t}{4}\cos(2t)</math> | |||
==Bibliography== | |||
* A. D. Polyanin, ''Handbook of Linear Partial Differential Equations for Engineers and Scientists'', Chapman & Hall/CRC Press, Boca Raton, 2002. ISBN 1-58488-299-9 | |||
[[Category:Integral transforms]] | |||
[[Category:Differential equations]] | |||
[[Category:Differential calculus]] | |||
[[Category:Ordinary differential equations|Ordinary differential equations]] |
Revision as of 17:41, 31 January 2014
The Laplace transform is a powerful integral transform used to switch a function from the time domain to the s-domain. The Laplace transform can be used in some cases to solve linear differential equations with given initial conditions.
First consider the following property of the Laplace transform:
One by induction can prove that
Now we consider the following differential equation:
with given initial conditions
Using the linearity of the Laplace transform it is equivalent to rewrite the equation as
obtaining
Solving the equation for and substituting with one obtains
The solution for f(t) is obtained by applying the inverse Laplace transform to
Note that if the initial conditions are all zero, i.e.
then the formula simplifies to
An example
We want to solve
with initial conditions f(0) = 0 and f′(0)=0.
We note that
and we get
The equation is then equivalent to
We deduce
Now we apply the Laplace inverse transform to get
Bibliography
- A. D. Polyanin, Handbook of Linear Partial Differential Equations for Engineers and Scientists, Chapman & Hall/CRC Press, Boca Raton, 2002. ISBN 1-58488-299-9