Laplace transform applied to differential equations - Alchetron, the free social encyclopedia

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:

L { f ′ } = s L { f } − f ( 0 ) L { f ″ } = s 2 L { f } − s f ( 0 ) − f ′ ( 0 )

One can prove by induction that

L { f ( n ) } = s n L { f } − ∑ i = 1 n s n − i f ( i − 1 ) ( 0 )

Now we consider the following differential equation:

∑ i = 0 n a i f ( i ) ( t ) = ϕ ( 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

∑ i = 0 n a i L { f ( i ) ( t ) } = L { ϕ ( t ) }

obtaining

L { f ( t ) } ∑ i = 0 n a i s i − ∑ i = 1 n ∑ j = 1 i a i s i − j f ( j − 1 ) ( 0 ) = L { ϕ ( t ) }

Solving the equation for L { f ( t ) } and substituting f ( i ) ( 0 ) with c i one obtains

L { f ( t ) } = L { ϕ ( t ) } + ∑ i = 1 n ∑ j = 1 i a i s i − j c j − 1 ∑ i = 0 n a i s i

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

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

f ( i ) ( 0 ) = c i = 0 ∀ i ∈ { 0 , 1 , 2 , . . . n }

then the formula simplifies to

f ( t ) = L − 1 { L { ϕ ( t ) } ∑ i = 0 n a i s i }

An example

We want to solve

f ″ ( t ) + 4 f ( t ) = sin ⁡ ( 2 t )

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

We note that

ϕ ( t ) = sin ⁡ ( 2 t )

and we get

L { ϕ ( t ) } = 2 s 2 + 4

The equation is then equivalent to

s 2 L { f ( t ) } − s f ( 0 ) − f ′ ( 0 ) + 4 L { f ( t ) } = L { ϕ ( t ) }

We deduce

L { f ( t ) } = 2 ( s 2 + 4 ) 2

Now we apply the Laplace inverse transform to get

f ( t ) = 1 8 sin ⁡ ( 2 t ) − t 4 cos ⁡ ( 2 t )

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