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

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

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

• 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