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.
First consider the following property of the Laplace transform:
![{\displaystyle {\mathcal {L}}\{f'\}=s{\mathcal {L}}\{f\}-f(0)}](https://wikimedia.org/api/rest_v1/media/math/render/svg/38c4778e0226d35ab990383602960a029c80af87)
![{\displaystyle {\mathcal {L}}\{f''\}=s^{2}{\mathcal {L}}\{f\}-sf(0)-f'(0)}](https://wikimedia.org/api/rest_v1/media/math/render/svg/864209046b4627ad57e2e380695e107c8f922d17)
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)}](https://wikimedia.org/api/rest_v1/media/math/render/svg/0ec78bb43e6fc1df7618075c4080cf7998d3524a)
Now we consider the following differential equation:
![{\displaystyle \sum _{i=0}^{n}a_{i}f^{(i)}(t)=\phi (t)}](https://wikimedia.org/api/rest_v1/media/math/render/svg/31d6f6b1b629b123eeb8e6ed4883fa25ce68e337)
with given initial conditions
![{\displaystyle f^{(i)}(0)=c_{i}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/fb5e84e0a4868e792e7c81795b2d4fd69656aaa9)
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)\}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/8021ebd9e4c0f56a8fa2024416e7a38cc862e6ea)
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)\}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/3967dd75c59758605c672a1ada0a95207d560afe)
Solving the equation for
and substituting
with
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}}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/9afee08e8dd3015de43b631c95bd3874f7b8999f)
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.
![{\displaystyle f^{(i)}(0)=c_{i}=0\quad \forall i\in \{0,1,2,...\ n\}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/5d9deced0bee2f122793751b0e970aa54dc03ff2)
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\}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/9ab37a095ec0f07502a1b8d43b7679b59d346a78)
An example
We want to solve
![{\displaystyle f''(t)+4f(t)=\sin(2t)}](https://wikimedia.org/api/rest_v1/media/math/render/svg/7108de2c6ab151cd8129e2c610069e6419e23953)
with initial conditions f(0) = 0 and f′(0)=0.
We note that
![{\displaystyle \phi (t)=\sin(2t)}](https://wikimedia.org/api/rest_v1/media/math/render/svg/5aa8b3a81124fbd190c768f1909fac24087e8f55)
and we get
![{\displaystyle {\mathcal {L}}\{\phi (t)\}={\frac {2}{s^{2}+4}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/6a281ca6ea972cb29aa64da883a645a06ca33c6f)
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)\}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/0745e1fe37e32503e9d1146f5e76e92bd0262861)
We deduce
![{\displaystyle {\mathcal {L}}\{f(t)\}={\frac {2}{(s^{2}+4)^{2}}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/a2d01da41841d414a0aa084e23fe9b12bc74a358)
Now we apply the Laplace inverse transform to get
![{\displaystyle f(t)={\frac {1}{8}}\sin(2t)-{\frac {t}{4}}\cos(2t)}](https://wikimedia.org/api/rest_v1/media/math/render/svg/87fc0df5945678fb828bb87086d5cb89efdd27ad)
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