ODE-ProjectODE-Project Solving Initial Value Problems

Section 6.2 Solving Initial Value Problems

Objectives

To understand how the Laplace transform acts on derivatives.
To understand how the Laplace transform can be used to solve initial value problems such as

$\begin{aligned} a y^{″} (t) + b y^{'} (t) + c y (t) & = f (t) \\ y (0) & = y_{0} \\ y^{'} (0) & = y_{0}^{'} \end{aligned}$

even when $f (t)$ is discontinuous.

There is no need for Laplace transforms when solving many initial value problems. If we consider the initial value problem

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\begin{aligned} y^{″} + 7 y^{'} + 10 y & = 0 \\ y (0) & = 1 \\ y^{'} (0) & = 1, \end{aligned}

the methods described in Chapter 4 work quite well. The characteristic polynomial of

y^{″} + 7 y^{'} + 10 y = 0

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r^{2} + 7 r + 10 = (r + 2) (r + 5),

and the general solution of the differential equation must be

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y (t) = c_{1} e^{- 2 t} + c_{2} e^{- 5 t} .

Applying the initial conditions, we find that the solution to our initial value problem is

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y (t) = 2 e^{- 2 t} - e^{- 5 t} .

However, suppose that we have a harmonic oscillator with a discontinuous forcing term,

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\begin{matrix} y^{″} + 2 y^{'} + 5 y = h (t) \\ y (0) = y^{'} (0) = 0, \end{matrix}

where

h (t)

is given by

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h (t) = {\begin{cases} 7 & t < 5 \\ 0 & t \geq 5. \end{cases}

The previous techniques that we described might prove cumbersome to solve such an initial value problem. We can easily imagine such equations arising in physics or engineering.

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Fortunately, Laplace transforms forms behave very nicely with respect to derivatives. We can use Laplace transforms to transform an initial value problem into an algebraic equation. Once the algebraic equation is solved, we can use the inverse transform to obtain the solution to our original initial value problem.

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Subsection 6.2.1 Laplace Transforms of the Derivative

Suppose that we have linear differential equation with constant coefficients

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a y^{″} (t) + b y^{'} (t) + c y (t) = f (t)

and initial conditions

y (0) = y_{0}

and

y^{'} (0) = y_{0}^{'} .

We can take the Laplace transform of both sides to obtain

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a L (y^{″} (t)) + b L (y^{'} (t)) + c L (y (t)) = L (f (t)) .

Notice that we have used the fact that the Laplace transform is a linear operator (Theorem 6.1.3). To proceed further, we need to know the Laplace transform of the derivative of a function.

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Theorem 6.2.1.

Let

y = y (t)

be a piecewise continuous, exponentially bounded function and assume that

y^{'}

is also exponentially bounded. Then for large values of

s

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L (y^{'}) (s) = s L (y) (s) - y (0) = s Y (s) - y (0),

where

Y (s)

is the Laplace transform of

y .

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Proof.

We can evaluate the Laplace transform of

y^{'}

by using integration by parts,

\begin{aligned} L (y^{'}) (s) & = \int_{0}^{\infty} y^{'} (t) e^{- s t} d t \\ = lim_{b \to \infty} [e^{- s t} y (t) |_{0}^{b} + s \int_{0}^{b} y (t) e^{- s t} d t] \\ = lim_{b \to \infty} [e^{- s b} y (b) - y (0) + s L (y) (s)] . \end{aligned}

We claim that

lim_{b \to \infty} e^{- s b} y (b) = 0 .

Since,

y

is exponentially bounded, there exist constants

M \geq 0

and

a

such that

| y (t) | \leq M e^{a t},

for all

t

[0, \infty) .

Thus,

| e^{- s b} y (b) | \leq M e^{- (s - a) b} .

The right-hand side of this inequality as

b \to \infty

for

s > a .

Thus,

L (y^{'}) (s) = s L (y) (s) - y (0) .

The Laplace transform also behave nicely with respect to higher order derivatives.

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Theorem 6.2.2.

Let

y = y (t)

and

y^{'} (t)

be piecewise continuous, exponentially bounded functions and assume that

y^{″}

is exponentially bounded. Then for large values of

s

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L (y^{″}) (s) = s^{2} L (y) (s) - s y (0) - y^{'} (0) = s^{2} Y (s) - s y (0) - y^{'} (0),

where

Y (s)

is the Laplace transform of

y .

In general, if

y = y (t)

and all of its derivatives up to order

k - 1

are piecewise continuous, exponentially bounded functions and

y^{(k)}

is piecewise continuous, then

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\begin{matrix} L (y^{(k)}) (s) = s^{k} L (y) (s) - s^{k - 1} y (0) - \dots - s y^{(k - 2)} (0) - y^{(k - 1)} (0) \\ = s^{k} Y (s) - s^{k - 1} y (0) - s^{k - 1} y (0) - \dots - s y^{(k - 2)} (0) - y^{(k - 1)} (0) . \end{matrix}

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We can use Theorem 6.2.1 and Theorem 6.2.2 to solve initial value problems.

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Example 6.2.3.

Consider the initial value problem

\begin{aligned} y^{″} + 7 y^{'} + 10 y & = 0 \\ y (0) & = 1 \\ y^{'} (0) & = 1. \end{aligned}

By Theorem 6.2.1 and Theorem 6.2.2

\begin{matrix} L (y^{'}) (s) = s L (y) (s) - y (0) = s Y (s) - y (0) \\ L (y^{″}) (s) = s^{2} L (y) (s) - s y (0) - y^{'} (0) = s^{2} Y (s) - s y (0) - y^{'} (0) . \end{matrix}

where

Y (s)

is the Laplace transform of

y .

If we take the Laplace transform of

y^{″} + 7 y^{'} + 10 y,

we have

\begin{aligned} L (y^{″} + 7 y^{'} + 10 y) & = L (y^{″}) + 7 L (y^{'}) + 10 L (y) \\ = [s^{2} Y (s) - s y (0) - y^{'} (0)] + 7 [s Y (s) - y (0)] + 10 Y (s) \\ = [s^{2} Y (s) - s - 1] + 7 [s Y (s) - 1] + 10 Y (s) . \end{aligned}

Since

L (0),

we have the algebraic equation

[s^{2} Y (s) - s - 1] + 7 [s Y (s) - 1] + 10 Y (s) = 0.

Solving for

Y (s),

we get

Y (s) = \frac{s + 8}{s^{2} + 7 s + 10} = \frac{s + 8}{(s + 5) (s + 2)} = \frac{2}{s + 2} - \frac{1}{s + 5},

where we have used partial fractions to get the last expression.

Since the Laplace transform of

e^{a t}

1 / (s - a),

we know that

L^{- 1} (\frac{1}{s - a}) = e^{a t} .

We can now solve our initial value problem,

y (t) = 2 e^{- 2 t} - e^{- 5 t},

and our solution agrees with the one that we found at the beginning of this section.

Activity 6.2.1. Solving Linear Differential Equations with Laplace Transforms.

Use Laplace Transforms to solve each of the following initial value problems.

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(a)

x^{'} - 5 x = e^{5 t},

x (0) = 0

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(b)

x^{″} - x^{'} - 6 x = 0,

x (0) = 2,

x^{'} (0) = - 1

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(c)

x^{″} + 4 x = \sin 3 t,

x (0) = 0,

x^{'} (0) = 0

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Subsection 6.2.2 Discontinuous Functions

c \geq 0

and we define the function

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u_{c} (t) = {\begin{cases} 0 & t < c \\ 1 & t \geq c, \end{cases}

recall that the Laplace transform of

u_{c}

is given by

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L (u_{c} (t)) (s) = \int_{0}^{\infty} u_{c} (t) e^{- s t} d t = \frac{e^{- c s}}{s} .

We can use this information to solve initial value problems with discontinuous functions.

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Example 6.2.4.

Consider the initial value problem

\begin{aligned} y^{'} + y & = u_{3} (t) \\ y (0) & = 1. \end{aligned}

If we take the Laplace transform of both sides of

y^{'} + y = u_{3} (t),

we obtain

s Y (s) - y (0) + Y (s) = \frac{e^{- 3 s}}{s} .

Using the fact that

y (0) = 1

and solving for

Y (s),

we get

Y (s) = \frac{1}{s + 1} + \frac{e^{- 3 s}}{s (s + 1)} .

Therefore,

y (t) = L^{- 1} (\frac{1}{s + 1}) + L^{- 1} (\frac{e^{- 3 s}}{s (s + 1)}) .

The inverse Laplace transform of the first term is

L^{- 1} (\frac{1}{s + 1}) = e^{- t} .

To compute the inverse Laplace transform of the second term, recall from Example 6.1.7 that if

L (f) = F (s),

then

L (u_{a} (t) f (t - a)) = e^{- a s} F (s) .

Using partial fractions to obtain

\frac{1}{s (s + 1)} = \frac{1}{s} - \frac{1}{s - 1} .

Hence,

L^{- 1} (\frac{e^{- 3 s}}{s (s + 1)}) = L^{- 1} (\frac{e^{- 3 s}}{s}) - L^{- 1} (\frac{e^{- 3 s}}{s + 1}) = u_{3} (t) - L^{- 1} (\frac{e^{- 3 s}}{s + 1}) .

g (t) = u_{3} (t) e^{- (t - 3)},

then the Laplace transform of

g (t)

L (g) = e^{- 3 s} L (e^{- t}) = \frac{e^{- 3 s}}{s + 1} .

Thus,

y (t) = e^{- t} + u_{3} (t) (1 - e^{- (t - 3)}) .

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Subsection 6.2.3 Forced Harmonic Oscillators

Example 6.2.5.

Consider the forced harmonic oscillator

\begin{matrix} y^{″} + 4 y = 3 \cos t \\ y (0) = y^{'} (0) = 0. \end{matrix}

Taking the Laplace transform of both sides of the equation

y^{″} + 4 y = 3 \cos t,

we obtain

L (y^{″}) + 4 L (y) = 3 L (\cos t)

s^{2} Y (s) - s y (0) - y^{'} (0) + 4 Y (s) = \frac{3 s}{s^{2} + 1},

where

L (y) (s) = Y (s) .

Substituting the initial conditions and solving for

Y,

we have

Y (s) = \frac{3 s}{(s^{2} + 4) (s^{2} + 1)} = \frac{- s}{s^{2} + 4} + \frac{s}{s^{2} + 1},

where the last expression was obtained using partial fractions. Taking the inverse Laplace transform, we have our solution

y (t) = L^{- 1} (\frac{- s}{s^{2} + 4}) + L^{- 1} (\frac{s}{s^{2} + 1}) = - \cos 2 t + \cos t .

Example 6.2.6.

Now let us consider a harmonic oscillator with discontinuous forcing,

\begin{matrix} y^{″} + 2 y^{'} + 5 y = h (t) \\ y (0) = y^{'} (0) = 0, \end{matrix}

where

h (t)

is given by

h (t) = {\begin{cases} 5 & t < 7 \\ 0 & t \geq 7. \end{cases}

That is,

h (t) = 5 (1 - u_{7} (t)) .

We may consider this to be a mass-spring system sliding on a table, where the mass is one unit, the spring constant is 5, and the damping coefficient is 2. When

t < 7

the table is tilted so that gravity provides a force of 5 units when stretching the spring. At time

t = 7,

the table is suddenly returned to the level position.

Taking the Laplace transform of both sides of

y^{″} + 2 y^{'} + 5 y = h (t),

we obtain

[s^{2} Y (s) - s y (0) - y^{'} (0)] + 2 [s Y (s) - y (0)] + 5 Y (s) = L (h),

where

L (y) (s) = Y (s) .

Substituting the initial conditions and evaluating the Laplace transform on the right, we have

(s^{2} + 2 s + 5) Y (s) = 5 (\frac{1}{s} - \frac{e^{- 7 s}}{s}) .

Solving for

Y (s),

we have

Y (s) = \frac{5}{s (s^{2} + 2 s + 5)} - \frac{5 e^{- 7 s}}{s (s^{2} + 2 s + 5)}

and

y = L^{- 1} (\frac{5}{s (s^{2} + 2 s + 5)} - \frac{5 e^{- 7 s}}{s (s^{2} + 2 s + 5)}) .

Using partial fractions, we can rewrite the first term as

\frac{5}{s (s^{2} + 2 s + 5)} = \frac{1}{s} - \frac{s + 2}{s^{2} + 2 s + 5} .

The inverse Laplace transform of

1 / s

is 1. To find the inverse Laplace transform of the second term, we complete the square of the denominator,

\begin{aligned} \frac{s + 2}{s^{2} + 2 s + 5} & = \frac{s + 2}{(s + 1)^{2} + 4} \\ = \frac{s + 1}{(s + 1)^{2} + 4} + \frac{1}{(s + 1)^{2} + 4} \\ = \frac{s + 1}{(s + 1)^{2} + 4} + \frac{1}{2} \frac{2}{(s + 1)^{2} + 4} . \end{aligned}

Consequently,

L^{- 1} (\frac{s + 2}{s^{2} + 2 s + 5}) = e^{- t} \cos 2 t + \frac{1}{2} e^{- t} \sin 2 t = e^{- t} (\cos 2 t + \frac{1}{2} \sin 2 t) .

and

L^{- 1} (\frac{5}{s (s^{2} + 2 s + 5)}) = L^{- 1} (\frac{1}{s} - \frac{s + 2}{s^{2} + 2 s + 5}) = 1 - e^{- t} (\cos 2 t + \frac{1}{2} \sin 2 t) .

We can compute the inverse Laplace transform of

\frac{5 e^{- 7 s}}{s (s^{2} + 2 s + 5)}

using the Heaviside function

u_{7} (t)

and the inverse Laplace transform that we just calculated to obtain

L^{- 1} (\frac{5 e^{- 7 s}}{s (s^{2} + 2 s + 5)}) = u_{7} (t) (1 - e^{- (t - 7)} (\cos 2 (t - 7) + \frac{1}{2} \sin 2 (t - 7))) .

Therefore, the solution to our original initial value problem is

y (t) = 1 - e^{- t} (\cos 2 t + \frac{1}{2} \sin 2 t) - u_{7} (t) (1 - e^{- (t - 7)} (\cos 2 (t - 7) + \frac{1}{2} \sin 2 (t - 7))) .

Activity 6.2.2. Solving Differential Equations with Laplace Transforms with Discontinuous Forcing Functions.

Use Laplace Transforms to solve each of the following initial value problems.

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(a)

x^{'} + x = u_{3} (t),

x (0) = 2

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(b)

x^{″} + x = h (t),

x (0) = 0,

x^{'} (0) = 0

where

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h (t) = {\begin{cases} 0 & t < 1 \\ 2 & t \geq 1. \end{cases}

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Subsection 6.2.4 Important Lessons

Using the Laplace transform including how the transform behaves, we can solve initial value problems such as

$\begin{aligned} a y^{″} (t) + b y^{'} (t) + c y (t) & = f (t) \\ y (0) & = y_{0} \\ y^{'} (0) & = y_{0}^{'} \end{aligned}$

even when $f (t)$ is discontinuous.
Let $y = y (t)$ be a piecewise continuous, exponentially bounded function and assume that $y^{'}$ is also exponentially bounded. Then for large values of $s$

$L (y^{'}) (s) = s L (y) (s) - y (0) = s Y (s) - y (0),$

where $Y (s)$ is the Laplace transform of $y .$
Let $y = y (t)$ and $y^{'} (t)$ be piecewise continuous, exponentially bounded functions and assume that $y^{″}$ is exponentially bounded. Then for large values of $s$

$L (y^{″}) (s) = s^{2} L (y) (s) - s y (0) - y^{'} (0) = s^{2} Y (s) - s y (0) - y^{'} (0),$

where $Y (s)$ is the Laplace transform of $y .$ In general, if $y = y (t)$ and all of its derivatives up to order $k - 1$ are piecewise continuous, exponentially bounded functions and $y^{(k)}$ is piecewise continuous, then

$\begin{matrix} L (y^{(k)}) (s) = s^{k} L (y) (s) - s^{k - 1} y (0) - \dots - s y^{(k - 2)} (0) - y^{(k - 1)} (0) \\ = s^{k} Y (s) - s^{k - 1} y (0) - s^{k - 1} y (0) - \dots - s y^{(k - 2)} (0) - y^{(k - 1)} (0) . \end{matrix}$

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Reading Questions 6.2.5 Reading Questions

1.

Describe in your own words how the Laplace transform can be used to solve an initial value problem.

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2.

Explain what an exponentially bounded function is. Give an example.

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Exercises 6.2.6 Exercises

Solving Initial Value Problems.

Solve the initial problems in Exercise Group 6.2.6.1–8 using the Laplace transform.

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1.

y^{″} - 2 y^{'} - 3 y = 3 e^{2 t},

y (0) = 1,

y^{'} (0) = 0

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2.

y^{″} - y^{'} - 2 y = 4 x^{2},

y (0) = - 1,

y^{'} (0) = 1

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3.

\frac{d^{2} x}{d x^{2}} - 6 \frac{d x}{d t} + 25 x = 64 e^{- t},

x (0) = 1,

x^{'} (0) = - 2

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4.

y^{″} + 16 y = 2 \sin 2 t,

y (0) = 1,

y^{'} (0) = 0

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5.

y^{″} + 16 y = 2 \sin 4 t,

y (0) = 1,

y^{'} (0) = 0

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6.

y^{″} + 2 y^{'} + y = 2 e^{- t},

y (0) = - 1,

y^{'} (0) = 3

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7.

y^{″} + 6 y^{'} + 8 y = \cos 3 t,

y (0) = - 2,

y^{'} (0) = 1

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8.

u^{″} + ω_{0}^{2} y = \cos ω t,

ω^{2} \neq ω_{0}^{2},

u (0) = 1,

u^{'} (0) = - 1

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9.

Find the solution of the initial value problem

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\begin{matrix} 2 y^{″} + y^{'} + 2 y = g (t) \\ y (0) = y^{'} (0) = 0 \end{matrix}

where

g (t)

is defined by

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g (t) = u_{5} (t) - u_{20} (t) = {\begin{cases} 1, & 5 \leq t < 20, \\ 0, & 0 \leq t < 5 and t \geq 20. \end{cases}

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10.

Let

y = y (t)

and

y^{'} (t)

be piecewise continuous, exponentially bounded functions and assume that

y^{″}

is exponentially bounded.

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Prove that

$L (y^{″}) (s) = s^{2} L (y) (s) - s y (0) - y^{'} (0) = s^{2} Y (s) - s y (0) - y^{'} (0)$

for large values of $s,$ where $Y (s)$ is the Laplace transform of $y .$
If $y = y (t)$ and all of its derivatives up to order $k - 1$ are piecewise continuous, exponentially bounded functions and $y^{(k)}$ is piecewise continuous, prove that

$\begin{matrix} L (y^{(k)}) (s) = s^{k} L (y) (s) - s^{k - 1} y (0) - \dots - s y^{(k - 2)} (0) - y^{(k - 1)} (0) \\ = s^{k} Y (s) - s^{k - 1} y (0) - s^{k - 1} y (0) - \dots - s y^{(k - 2)} (0) - y^{(k - 1)} (0) . \end{matrix}$

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11.

Prove Theorem 6.2.2.

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12.

Let

f (t)

be a peicewise continuous function for

t \geq 0 .

In addition, suppose that

f (t)

satisfies the condition

| f (t) | \leq M e^{c t}

with

t \geq 0 .

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Prove that

$L (\int_{0}^{t} f (τ) d τ) = \frac{1}{s} L (f (t) = \frac{F (s)}{s} .$
Show that

$L^{- 1} (\frac{F (s)}{s}) = \int_{0}^{t} f (τ) d τ$

follows from (a).

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13. Solving Linear Systems Using Laplace Transforms.

Consider the linear system

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\begin{aligned} u^{'} + u - v & = 0 \\ v^{'} - u + v & = 2 \\ u (0) = 1, & v (0) = 2. \end{aligned}

If we let

L (u (s)) = U (s)

and

L (v (s)) = V (s)

and take the Laplace transform of the two differential equations, we obtain

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\begin{aligned} [s U (s) - 1] + U (s) - V (s) & = 0 \\ [s V (s) - 2] - U (s) + V (s) & = \frac{2}{s} . \end{aligned}

Verify that if we solve the system

$\begin{aligned} [s U (s) - 1] + U (s) - V (s) & = 0 \\ [s V (s) - 2] - U (s) + V (s) & = \frac{2}{s} \end{aligned}$

for $U (s)$ and $V (s),$ we obtain

$U (s) = \frac{s + 1}{s^{2}} and V (s) = \frac{2 s + 1}{s^{2}} .$
Find $u (t)$ and $v (t) .$
Solve the linear system

$\begin{aligned} u^{'} + 4 u - 6 v & = 0 \\ v^{'} + 3 u - 5 v & = 0 \\ u (0) = 3, & v (0) = 2 \end{aligned}$

using Laplace transforms.
Solve the linear system

$\begin{aligned} w^{'} - y & = 0 \\ w + y^{'} + z & = 1 \\ w - y + z^{'} & = 2 \sin t \\ w (0) = 1, & y (0) = 1, z (0) = 1 \end{aligned}$

using Laplace transforms.

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