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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]].


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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&prime;''(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:

{f}=s{f}f(0)
{f}=s2{f}sf(0)f(0)

One by induction can prove that

{f(n)}=sn{f}i=1nsnif(i1)(0)

Now we consider the following differential equation:

i=0naif(i)(t)=ϕ(t)

with given initial conditions

f(i)(0)=ci

Using the linearity of the Laplace transform it is equivalent to rewrite the equation as

i=0nai{f(i)(t)}={ϕ(t)}

obtaining

{f(t)}i=0naisii=1nj=1iaisijf(j1)(0)={ϕ(t)}

Solving the equation for {f(t)} and substituting f(i)(0) with ci one obtains

{f(t)}={ϕ(t)}+i=1nj=1iaisijcj1i=0naisi

The solution for f(t) is obtained by applying the inverse Laplace transform to {f(t)}.

Note that if the initial conditions are all zero, i.e.

f(i)(0)=ci=0i{0,1,2,...n}

then the formula simplifies to

f(t)=1{{ϕ(t)}i=0naisi}

An example

We want to solve

f(t)+4f(t)=sin(2t)

with initial conditions f(0) = 0 and f′(0)=0.

We note that

ϕ(t)=sin(2t)

and we get

{ϕ(t)}=2s2+4

The equation is then equivalent to

s2{f(t)}sf(0)f(0)+4{f(t)}={ϕ(t)}

We deduce

{f(t)}=2(s2+4)2

Now we apply the Laplace inverse transform to get

f(t)=18sin(2t)t4cos(2t)

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