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. First consider the following property of the Laplace transform:

[math]\displaystyle{ \mathcal{L}\\{f'\\}=s\mathcal{L}\\{f\\}-f(0) }[/math]

[math]\displaystyle{ \mathcal{L}\\{f''\\}=s^2\mathcal{L}\\{f\\}-sf(0)-f'(0) }[/math]

One can prove by induction that

[math]\displaystyle{ \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]\displaystyle{ \sum_{i=0}^{n}a_if^{(i)}(t)=\phi(t) }[/math]

with given initial conditions

[math]\displaystyle{ f^{(i)}(0)=c_i }[/math]

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

[math]\displaystyle{ \sum_{i=0}^{n}a_i\mathcal{L}\\{f^{(i)}(t)\\}=\mathcal{L}\\{\phi(t)\\} }[/math]

obtaining

[math]\displaystyle{ \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]\displaystyle{ \mathcal{L}\\{f(t)\\} }[/math] and substituting [math]\displaystyle{ f^{(i)}(0) }[/math] with [math]\displaystyle{ c_i }[/math] one obtains

[math]\displaystyle{ \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]\displaystyle{ \mathcal{L}\\{f(t)\\}. }[/math]

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

[math]\displaystyle{ f^{(i)}(0)=c_i=0\quad\forall i\in\\{0,1,2,...\ n\\} }[/math]

then the formula simplifies to

[math]\displaystyle{ 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]\displaystyle{ f''(t)+4f(t)=\sin(2t) }[/math]

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

We note that

[math]\displaystyle{ \phi(t)=\sin(2t) }[/math]

and we get

[math]\displaystyle{ \mathcal{L}\\{\phi(t)\\}=\frac{2}{s^2+4} }[/math]

The equation is then equivalent to

[math]\displaystyle{ s^2\mathcal{L}\\{f(t)\\}-sf(0)-f'(0)+4\mathcal{L}\\{f(t)\\}=\mathcal{L}\\{\phi(t)\\} }[/math]

We deduce

[math]\displaystyle{ \mathcal{L}\\{f(t)\\}=\frac{2}{(s^2+4)^2} }[/math]

Now we apply the Laplace inverse transform to get

[math]\displaystyle{ 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

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