Laplace Transform to ODEs

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:

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \mathcal{L}\{f'\}=s\mathcal{L}\{f\}-f(0)}

${\mathcal {L}}\{f''\}=s^{2}{\mathcal {L}}\{f\}-sf(0)-f'(0)$

One can prove by induction that

${\mathcal {L}}\{f^{(n)}\}=s^{n}{\mathcal {L}}\{f\}-\sum _{i=1}^{n}s^{n-i}f^{(i-1)}(0)$

Now we consider the following differential equation:

$\sum _{i=0}^{n}a_{i}f^{(i)}(t)=\phi (t)$

with given initial conditions

$f^{(i)}(0)=c_{i}$

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

$\sum _{i=0}^{n}a_{i}{\mathcal {L}}\{f^{(i)}(t)\}={\mathcal {L}}\{\phi (t)\}$

obtaining

${\mathcal {L}}\{f(t)\}\sum _{i=0}^{n}a_{i}s^{i}-\sum _{i=1}^{n}\sum _{j=1}^{i}a_{i}s^{i-j}f^{(j-1)}(0)={\mathcal {L}}\{\phi (t)\}$

Solving the equation for ${\mathcal {L}}\{f(t)\}$ and substituting $f^{(i)}(0)$ with $c_{i}$ one obtains

${\mathcal {L}}\{f(t)\}={\frac {{\mathcal {L}}\{\phi (t)\}+\sum _{i=1}^{n}\sum _{j=1}^{i}a_{i}s^{i-j}c_{j-1}}{\sum _{i=0}^{n}a_{i}s^{i}}}$

The solution for f(t) is obtained by applying the inverse Laplace transform to ${\mathcal {L}}\{f(t)\}.$

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

$f^{(i)}(0)=c_{i}=0\quad \forall i\in \{0,1,2,...\ n\}$

then the formula simplifies to

$f(t)={\mathcal {L}}^{-1}\left\{{{\mathcal {L}}\{\phi (t)\} \over \sum _{i=0}^{n}a_{i}s^{i}}\right\}$

An example

We want to solve

$f''(t)+4f(t)=\sin(2t)$

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

We note that

$\phi (t)=\sin(2t)$

and we get

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \mathcal{L}\{\phi(t)\}=\frac{2}{s^2+4}}

The equation is then equivalent to

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle s^2\mathcal{L}\{f(t)\}-sf(0)-f'(0)+4\mathcal{L}\{f(t)\}=\mathcal{L}\{\phi(t)\}}

We deduce

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \mathcal{L}\{f(t)\}=\frac{2}{(s^2+4)^2}}

Now we apply the Laplace inverse transform to get

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle f(t)=\frac{1}{8}\sin(2t)-\frac{t}{4}\cos(2t)}

Licensing

Content obtained and/or adapted from:

• Laplace transform applied to ODEs, Wikipedia under a CC BY-SA license