Laplace transforms
These are a neat way of solving constant coefficient DE’s and also are used in process control.
The Laplace transform is an operator L(⋅) that maps a function f (t ) , defined for t ≥ 0 , onto
another function F (s )
∞
L( f (t )) = F (s ) ≡ ∫ e − st f (t ) dt
(1)
0
Equation (1) is well defined provided that both the following conditions hold:
• The function f (t ) is piecewise continuous (or smoother)
• The function f (t ) is of exponential order as t → ∞
Definition: The function f (t ) is of exponential order as t → ∞ if there exist constants: a, A, K ,
such that:
f (t ) ≤ Ke at ,
t≥A
(In simple terms, if f (t ) doesn’t grow any faster than exponential.)
Laplace transform of a derivative?
L( f ′(t )) = − f (0 ) + sL( f (t ))
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What about if f (t ) has jumps in it?
Suppose that f ′(t ) is piecewise continuous and of exponential order as t → ∞ and that f (t ) is of
exponential order. What is L( f ′(t )) ?
Let {t j }nj = 0 be the points of discontinuity of f ′(t ) , with t0 = 0 and tn = A . Formally, we have that:
A
L( f ′(t )) = lim ∫ e
A→ ∞
− st
0
f ′(t ) dt = lim
n tj
− st
e
f ′(t ) dt
∑
∫
A→∞
j =1t j −1
We integrate by parts on each interval:
tj
⎞
n ⎛
n
n tj
tj
st
st
−
−
st
st
−
−
j
j
−
1
L( f ′) = lim ∑ ⎜ e f t j −1 + s ∫ e f dt ⎟ = lim ∑ e
f (t j ) − e
f (t j −1 ) + s lim ∑ ∫ e − st f (t ) dt
⎟ A→ ∞ j =1
A→ ∞ j =1⎜
A → ∞ j =1
t j −1
t j −1
⎝
⎠
[
= − f (0 ) + lim e
A→∞
[
]
− sA
]
A
f ( A) + s lim ∫ e − st f (t ) dt
A→ ∞
0
Therefore the expression (2) is valid even with jumps in f (t )
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Higher derivatives?
These can be handled iteratively:
L( f ′′(t )) = − f ′(0) + sL( f ′(t )) = − f ′(0) − sf (0 ) + s 2 L( f (t ))
L( f ′′′(t )) = − f ′′(0 ) + sL( f ′′(t )) = − f ′′(0) − sf ′(0) + s 2 L( f ′(t )) = − f ′′(0 ) − sf ′(0 ) − s 2 f (0) + s 3 L( f (t ))
etc…
Theorem 1: Suppose that f , f ′, f ′′.... f ( n −1) are continuous and that f ( n ) is piecewise
continuous, on any interval [0, A]. Suppose also that each of f , f ′, f ′′.... f ( n −1) , f (n ) is of
exponential order as t → ∞ . Then
( ) = s L( f (t )) − f
L f
(n )
n
(n −1)
(0) − sf
(n − 2 )
(0).... − s
n −1
n −1
f (0 ) = s L( f (t )) − ∑ s j f (n −1− j ) (0 )
n
j =0
Linearity of the Laplace transform
Note that the Laplace transform is a linear operator, i.e. suppose that f (t ), g (t ) have Laplace
transforms F (s ), G (s ) . Then
L(af (t ) + bg (t )) = aF (s ) + bG (s )
for any constants a and b . Why is this true?
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Solving linear IVPs with constant coefficients
Example 1: Use Laplace transforms to solve the following IVP:
y′ + ay = 0
y (0 ) = y0
Solution: Take the transform of the equation
Therefore, the Laplace transform of the solutionY (s ) satisfies: Y (s ) =
Recall that if f (t ) = e − at then F (s ) =
y0
s+a
1
. We immediately recognise that the inverse is:
s+a
y (t ) = y0e − at
which we know is the solution of the IVP, (from our first lecture).
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Points to note:
1. We reduce the differential equation to an algebraic equation.
2. After solving the algebraic equation, we are left with the problem of how to invert the Laplace
transform to recover y (t ) .
Example 2: Use Laplace transforms to solve the following IVP:
y′′ − 9 y = 0
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y(0) = 1, y′(0) = −1
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Inverting Laplace transforms
• There is a 1-1 correspondence between f (t ) and F (s ) , i.e. we can meaningfully work in s-space
and then invert to t-space.
• General formula exists but is not covered in this course.
• Tables exist for a wide range of common functions (see p. 319 in Boyce & DiPrima, 8th ed).
• Use tables in combination with general results, to identify how to invert a particular formula
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Example 3: Use Laplace transforms to solve the following IVP:
y′′ − 2 y′ − 2 y = 0
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y(0) = 1, y′(0) = 1
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Properties of Laplace transforms
• Frequently the problem is to invert Y (s ) , the solution to our IVP in s-space.
• We have a limited number of "known" transforms of common functions and their inverses.
• To expand our range of known functions & inverses, we apply a number of general results
Assume that f (t ) , g (t ) , h(t ) have Laplace transforms F (s ) , G (s ) , H (s ) and a , b , c , are positive
constants.
• Linearity:
L(af (t ) + bg (t )) = aF (s ) + bG (s )
Utility: solve IVP in s-space & the solution Y (s ) is of form: Y (s ) = aF (s ) + bG (s ) + cH (s ) + ....
Then the solution in t-space, will be:
y (t ) = af (t ) + bg (t ) + ch(t ) + ....
• Stretching:
L( f (at )) =
1 ⎛s⎞
F⎜ ⎟
a ⎝a⎠
Proof:
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• Differentiation I: (see theorem 1)
L( f ′(t )) = − f (0 ) + sL( f (t )) ,
or applied iteratively:
(
L f
(n )
(t )) = s L( f (t )) − ∑ s j f (n −1− j ) (0) ,
n
n −1
j =0
Utility: turning IVP's into algebraic equations
• Differentiation II: (not the inverse of the above)
(
L (− t )
n
)
dn
f (t ) = n F (s ) = F (n ) (s ).
ds
Proof:
Utility: possibly we can recognise that the solution in s-space, Y (s ) , is a derivative with respect
to s of a function that we know the transform of.
Example: Suppose that Y (s ) = 6 / s 4 , find y (t )
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• Shifting I:
(
)
L e at f (t ) = F (s − a )
Proof:
Utility: we can recognise that the solution in s-space, Y (s ) , is of form F (s − a ) , where we know
the inverse transform of F (s ) .
Example: Suppose that Y (s ) =
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, find y (t )
2
(s − 2) + 25
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• Shifting II: We introduce the unit step function uc (t ):
⎧0 0 ≤ t ≤ c
uc (t ) = ⎨
⎩1 c < t
and note that the Laplace transform of uc (t ) is simply
∞
L(u c (t )) = ∫ u c (t )e
0
− st
∞
dt = ∫ e
− st
c
e −cs
dt =
, s>0
s
Now for a given function f (t ) , suppose we "shift" the function to the right by an amount c,
defining f (t ) = 0 for t < 0 , i.e.
0≤t ≤c
⎧0
g (t ) = ⎨
⎩ f (t − c ) c < t
The function g (t ) is a delayed form of f (t ) . We can also write
g (t ) = uc (t ) f (t − c ) .
The second form of shifting theorem is:
L(uc (t ) f (t − c )) = e − cs F (s )
Proof:
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Utility: we can recognise that the solution in s-space, Y (s ) , is of form F (s )e − cs , where we know
the inverse transform of F (s ) .
e− s + e − 2 s − e −3s − e − 4 s
Example: If Y (s ) =
, find y (t ) .
s
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• Convolution theorem: If H (s ) = F (s )G (s ) then
t
t
0
0
h(t ) = ∫ f (τ )g (t − τ ) dτ = ∫ g (τ ) f (t − τ ) dτ
Proof: (see text for reverse version).
Notation: we denote the convolution of two functions by the symbol f ∗ g , i.e.
t
h(t ) = ( f ∗ g )(t ) = ∫ f (τ )g (t − τ ) dτ
0
Utility: Often we identify the solution in s-space, Y (s ) , as a product of form F (s )G (s ) , where we
know the inverse transforms of F (s ) and G (s ) . This arises with inhomogeneous equations.
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Example:
Consider the inhomogeneous 2nd order constant coefficient IVP:
ay′′ + by′ + cy = g (t )
y (0 ) = y0 , y′(0 ) = y0′
Take the Laplace transform of this equation and use theorem 1:
(as
2
)
+ bs + c Y (s ) − (as + b ) y0 − ay0′ = G (s )
Therefore, Y (s ) = Φ(s ) + Ψ (s ) , where:
(as + b ) y0 + ay0′
G (s )
as 2 + bs + c
as 2 + bs + c
We break Φ (s ) into partial fractions & invert: Φ(s ) → φ (t ) . Define the transfer function F (s ) by:
Φ (s ) =
,
Ψ (s ) =
1
as 2 + bs + c
We can find the inverse of F (s ) , say f (t ) , by breaking into partial fractions. The function f (t ) is
called the impulse response of the system. Using the convolution theorem, Ψ (s ) = F (s )G (s )
inverts to:
ψ (t ) = f ∗ g
F (s ) =
and the solution of the inhomogeneous IVP will be:
y (t ) = φ (t ) + ψ (t )
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Notes:
1. The function φ (t ) satisfies the homogeneous IVP:
ay ′′ + by ′ + cy = 0
y (0 ) = y0 , y ′(0 ) = y0′
2. The function ψ (t ) satisfies the inhomogeneous IVP:
ay′′ + by′ + cy = g (t )
y (0 ) = 0, y′(0 ) = 0
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Differential Equations with Piecewise Continuous Data Functions
Consider the IVP:
y′′ + p(t ) y′ + q(t ) y = g (t )
y (0 ) = y0
y′(0 ) = y0′
(3)
Suppose that the data functions p(t ), q(t ) , g (t ) are piecewise continuous on the interval I = (0, β ),
with points of discontinuity t = t j : j = 0,...., n , where t0 = 0 and t n = β .
Questions:
1. Can we find a solution to (3) defined on I = (0, β )?
2. How smooth can a solution be?
3. What happens if p(t ), q(t ) , g (t ) are continuous on I , but are only piecewise differentiable?
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Example 1: Consider the IVP:
1
(u2 (t )(t − 2) − u4 (t )(t − 4)) − u8 (t )
2
y (0 ) = 0
y′(0 ) = 0
y′′ + y =
• Describe the qualitative behaviour of the solution for t ∈ (0,10)
• Solve the IVP using Laplace transforms
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Example 2: Consider the IVP:
17
y′′ + 3 y′ + y = uπ (t ) − u2π (t )
2
y (0 ) = 0
y′(0 ) = 0
• Describe the qualitative behaviour of the solution for t > 0
• Solve the IVP using Laplace transforms
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Example 3: Consider the IVP:
y (iv ) − 16 y = sin t − u2π (t )sin (t − 2π )
y (0 ) = 0 y′(0 ) = 0
y′′(0 ) = 0 y′′′(0 ) = 0
Solve the IVP using Laplace transforms
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Differential Equations with Impulses
Consider the function dτ (t ) :
⎧1
⎪
dτ (t ) = ⎨ 2τ
⎪⎩ 0
Define the function I (τ ) by:
I (τ ) =
∞
τ
−∞
−τ
t ≤τ
(4)
t >τ
∫ dτ (t ) dt = ∫ dτ (t ) dt =
τ
1
∫ 2τ dt = 1
−τ
Now consider what happens to both I (τ ) and dτ (t ) as τ → 0 :
• Straightforwardly, we get: lim I (τ ) = 1
τ →0
• For any fixed t ≠ 0 we have: lim dτ (t ) = 0 , but integral of dτ (t ) remains uniformly equal to 1.
τ →0
This prompts the definition of the Dirac delta function δ (t ):
δ (t ) = lim dτ (t )
(5)
τ →0
which is not really well defined in a conventional sense. The delta function acts like an intense
pulse, of unit strength.
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For differentiable functions f (t ) , the delta function δ (t ) has the following property:
∞
∫ δ (t − t0 ) f (t ) dt = f (t0 )
−∞
Sketch proof:
∞
∞
∫ δ (t − t0 ) f (t ) dt = ∫
−∞
−∞
[lim d (t − t )]f (t ) dt
τ →0
τ
0
∞
= lim ∫ dτ (t − t0 ) f (t ) dt
τ →0
−∞
⎡ 1 t 0 +τ
⎤
= lim ⎢
∫ f (t ) dt ⎥
τ → 0 2τ
⎢⎣ t 0 −τ
⎥⎦
= lim[ f (t0 ) + O(τ )] = f (t0 )
τ →0
Directly from the above, we can derive the Laplace transform of the delta function:
∞
L(δ (t − t 0 )) = ∫ δ (t − t 0 )e
− st
∞
dt = ∫ δ (t − t 0 )e − st dt = e − st0
0
−∞
Now taking t0 → 0
L(δ (t − t0 )) = lim e − st 0 = 1
t0 →0
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Example 1: Consider the IVP:
2 y′′ + y′ + 2 y = δ (t − 5)
y (0) = 0
y′(0) = 0
Describe the qualitative behaviour of the solution for t > 0 . Solve the IVP using Laplace transforms
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Example 2: Consider the IVP:
y′′ + 4 y = δ (t − π ) − δ (t − 2π )
y (0 ) = 0
y′(0 ) = 0
Describe the qualitative behaviour of the solution for t > 0 . Solve the IVP using Laplace transforms
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