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:

<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 can prove by induction 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. Template:Isbn