Laplace transform applied to differential equations
In mathematics, 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.
Approach
[edit ]First consider the following property of the Laplace transform:
- {\displaystyle {\mathcal {L}}\{f'\}=s{\mathcal {L}}\{f\}-f(0)}
- {\displaystyle {\mathcal {L}}\{f''\}=s^{2}{\mathcal {L}}\{f\}-sf(0)-f'(0)}
One can prove by induction that
- {\displaystyle {\mathcal {L}}\{f^{(n)}\}=s^{n}{\mathcal {L}}\{f\}-\sum _{i=1}^{n}s^{n-i}f^{(i-1)}(0)}
Now we consider the following differential equation:
- {\displaystyle \sum _{i=0}^{n}a_{i}f^{(i)}(t)=\phi (t)}
with given initial conditions
- {\displaystyle f^{(i)}(0)=c_{i}}
Using the linearity of the Laplace transform it is equivalent to rewrite the equation as
- {\displaystyle \sum _{i=0}^{n}a_{i}{\mathcal {L}}\{f^{(i)}(t)\}={\mathcal {L}}\{\phi (t)\}}
obtaining
- {\displaystyle {\mathcal {L}}\{f(t)\}\sum _{i=0}^{n}a_{i}s^{i}-\sum _{i=1}^{n}\sum _{j=1}^{i}a_{i}s^{i-j}f^{(j-1)}(0)={\mathcal {L}}\{\phi (t)\}}
Solving the equation for {\displaystyle {\mathcal {L}}\{f(t)\}} and substituting {\displaystyle f^{(i)}(0)} with {\displaystyle c_{i}} one obtains
- {\displaystyle {\mathcal {L}}\{f(t)\}={\frac {{\mathcal {L}}\{\phi (t)\}+\sum _{i=1}^{n}\sum _{j=1}^{i}a_{i}s^{i-j}c_{j-1}}{\sum _{i=0}^{n}a_{i}s^{i}}}}
The solution for f(t) is obtained by applying the inverse Laplace transform to {\displaystyle {\mathcal {L}}\{f(t)\}.}
Note that if the initial conditions are all zero, i.e.
- {\displaystyle f^{(i)}(0)=c_{i}=0\quad \forall i\in \{0,1,2,...\ n\}}
then the formula simplifies to
- {\displaystyle f(t)={\mathcal {L}}^{-1}\left\{{{\mathcal {L}}\{\phi (t)\} \over \sum _{i=0}^{n}a_{i}s^{i}}\right\}}
An example
[edit ]We want to solve
- {\displaystyle f''(t)+4f(t)=\sin(2t)}
with initial conditions f(0) = 0 and f′(0)=0.
We note that
- {\displaystyle \phi (t)=\sin(2t)}
and we get
- {\displaystyle {\mathcal {L}}\{\phi (t)\}={\frac {2}{s^{2}+4}}}
The equation is then equivalent to
- {\displaystyle s^{2}{\mathcal {L}}\{f(t)\}-sf(0)-f'(0)+4{\mathcal {L}}\{f(t)\}={\mathcal {L}}\{\phi (t)\}}
We deduce
- {\displaystyle {\mathcal {L}}\{f(t)\}={\frac {2}{(s^{2}+4)^{2}}}}
Now we apply the Laplace inverse transform to get
- {\displaystyle f(t)={\frac {1}{8}}\sin(2t)-{\frac {t}{4}}\cos(2t)}
Bibliography
[edit ]- 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