First-Order Relaxation - Direct Solution and Laplace-Transform Preview

(Example)

LT01E1: First-Order Relaxation

Direct Solution and Laplace-Transform Preview

This companion example follows one simple first-order system through two solution paths.

The purpose is not yet to teach a transform table. The purpose is to see, in one concrete problem, why the transform-domain representation is attractive.

Consider the first-order relaxation model

dy- τdt + y = K, y(0) = y0,

(1)

where

τ > 0 is the time constant,
K is the steady forcing level,
y₀ is the initial value.

This equation describes the same mathematical structure that appears in first-order thermal models, simple electrical circuits, mixing problems, and many other relaxation processes.

Figure 1. A generic first-order relaxation system. The state begins at y₀ and approaches the steady value K with characteristic time scale τ.

1 What should the answer look like before we solve anything?

At steady state, the derivative vanishes:

dy- dt = 0.

(2)

Equation (1) then gives

yss = K.

(3)

So the solution should approach K as time becomes large.

At t = 0, the solution must satisfy

y(0) = y0.

(4)

We therefore expect a transient that connects y₀ to K.

2 Direct time-domain solution

Divide Equation (1) by τ:

dy-+ 1-y = K-. dt τ τ

(5)

The integrating factor is

μ(t) = et∕τ.

(6)

Multiply Equation (5) by this factor:

dy 1 K et∕τ---+ --et∕τy = --et∕τ. dt τ τ

(7)

The left side is one derivative:

( ) -d et∕τy = K-et∕τ. dt τ

(8)

Integrate from 0 to t:

∫ t∕τ K-- t u∕τ e y(t) − y0 = τ e du. 0

(9)

The integral is

∫ t ( ) eu∕τ du = τ et∕τ − 1 . 0

(10)

Therefore

( ) et∕τy(t) − y0 = K et∕τ − 1 .

(11)

Rearrange:

et∕τy(t) = Ket ∕τ + y0 − K.

(12)

Finally,

y(t) = K + (y − K )e−t∕τ. 0

(13)

Figure 2. Direct time-domain route. Normalize the ODE, choose an integrating factor, combine the left side into one derivative, integrate, apply the initial value, and simplify.

3 Interpret the direct solution

Equation (13) separates naturally into

y(t) = steady part + transient part.

(14)

The steady part is

K.

(15)

The transient is

− t∕τ (y0 − K )e .

(16)

At t = 0,

y(0) = K + (y0 − K ) = y0.

(17)

As t →∞,

e−t∕τ → 0,

(18)

y(t) → K.

(19)

The time constant controls how rapidly the transient decays.

t = τ,

(20)

the transient has been reduced by the factor

−1 e ≈ 0.368.

(21)

Figure 3. First-order relaxation for several initial values with the same K and τ. Every trajectory approaches the same steady value, while the exponential transient carries the memory of the initial condition.

4 Now solve the same problem using a Laplace-transform preview

The detailed transform rules will be derived later in the LT series. Here we use two results only to compare the structure of the calculation.

Let

Y (s) = ℒ {y(t)}.

(22)

The derivative property is previewed as

{ } dy ℒ --- = sY (s) − y0. dt

(23)

The transform of the constant forcing is

ℒ{K } = K-. s

(24)

Apply the transform to

′ τ y + y = K.

(25)

Using linearity,

K τ [sY (s) − y0] + Y (s) =--. s

(26)

Collect Y (s):

(τs + 1)Y (s) = K--+ τy . s 0

(27)

Solve algebraically:

K ∕s + τy0 Y (s) = ----------. τs + 1

(28)

The differential equation has disappeared. The unknown transform Y (s) is now found by ordinary algebra.

Figure 4. Transform-domain route. The derivative becomes sY − y₀, the initial value appears immediately, and the ODE reduces to an algebraic equation for Y (s).

5 Rewrite the transform so the time-domain pieces are visible

Starting from Equation (28), use

( ) 1 τ s + 1 = τ s + -- . τ

(29)

Then

Y (s) = ---K----- + --τy0-. s(τs + 1) τs + 1

(30)

A particularly useful rearrangement is

K y0 − K Y (s) = s--+ s-+-1∕τ-.

(31)

To verify this, combine the right side:

+	=
	= .	(32)

Multiplying numerator and denominator by τ gives

τsy0-+-K-- s(τs + 1),

(33)

which matches Equation (28).

Figure 5. Algebraic decomposition of Y (s). The transform separates into a steady term K∕s and a transient term with denominator s + 1∕τ.

6 Return to the time domain

The two inverse-transform pairs used here are

1-← → 1, s

(34)

and

--1-- ← → e−at. s + a

(35)

Therefore Equation (31) gives

−t∕τ y(t) = K + (y0 − K )e .

(36)

This is exactly the direct result in Equation (13).

7 Compare the two routes

The direct route required

normalizing the ODE,
constructing the integrating factor,
recognizing a product derivative,
integrating,
inserting the initial condition,
and simplifying.

The transform route required

transforming the derivative and constant forcing,
solving an algebraic equation for Y (s),
rewriting the result into recognizable pieces,
and taking an inverse transform.

For this very simple problem, the direct method is not inferior. In fact, it is probably shorter.

The advantage of the transform method becomes clearer when the forcing is delayed, piecewise, impulsive, or when several initial conditions and higher derivatives appear.

Figure 6. Two routes to the same solution. The direct method emphasizes integration in time; the Laplace route reorganizes the problem into algebra and then returns to time.

8 A numerical example

Let

τ = 2 s, K = 10, y0 = 4.

(37)

Then

y(t) = 10 − 6e− t∕2.

(38)

At t = 0,

y(0) = 4.

(39)

At one time constant,

y(2)	= 10 − 6e⁻¹	(40)
	≈ 7.79.	(41)

At two time constants,

y(4)	= 10 − 6e⁻²	(42)
	≈ 9.19.	(43)

At three time constants,

y(6)	= 10 − 6e⁻³	(44)
	≈ 9.70.	(45)

The system approaches K = 10 exponentially.

9 What this example is meant to teach

This example should not yet be memorized as a collection of transform rules.

Its purpose is to make three structural ideas concrete:

the initial condition enters the transformed equation through the derivative rule,
the differential equation becomes algebraic in Y (s),
and the steady and transient parts become recognizable algebraic pieces in the transform domain.

The remaining LT articles will derive the transform rules that were only previewed here.

10 Self-check questions

What is the steady-state value of y in Equation (1)?
What role does τ play in the time-domain solution?
Why does the direct solution naturally contain the factor e^−t∕τ?
Where does y₀ appear in the transformed equation?
Why is (τs + 1)Y (s) = K∕s + τy₀ easier to manipulate than the original differential equation?
What two time-domain pieces correspond to the two terms in Equation (31)?
Why is the direct method still competitive for this simple example?
What kinds of forcing functions would make the transform method more attractive?

References

[1] E. Kreyszig, Advanced Engineering Mathematics, Wiley.

[2] R. V. Churchill, Operational Mathematics, McGraw-Hill.

"First-Order Relaxation - Direct Solution and Laplace-Transform Preview" is owned by bloftin.
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