Laplace transforms is an integral transform that assists is in solving problems in physics, engineering, and differential equations. This section will introduce Laplace transforms, and we will see how they help in solving some differential equations in the next two sections.

The formal definition of the Laplace transform is:

L( f(t) ) = F(s) = ∫ (e^(-s*t) * f(t) dt, 0, infinity)

(The "L" is usually written in curve like style)

t and s are placeholder variables. Alternatively, we see x in place of t.

L( f(x) ) = F(s) = ∫ (e^(-s*x) * f(x) dx, 0, infinity)

There are many properties of Laplace transforms, but the ones that will help us are:

(1) L( f(t) + g(t) ) = L( f(t) ) + L( g(t) )

(2) L( a*f(t) ) = a*L( f(t) )

Here is a basic table of Laplace Transforms. You can find more complete tables in math reference books and online. One such source is this table from Lamar University in Texas:

(Web page: http://tutorial.math.lamar.edu/pdf/Laplace_Table.pdf )

L( f(t) ) → F(s)

1 → 1/s

t^n → n!/(s^(n+1))

e^(a*t) → 1/(s-a)

sin(a*t) → a/(s^2 + a^2)

t*sin(a*t) → (2*a*s)/((s^2 + a^2)^2)

cos(a*t) → s/(s^2 + a^2)

t*cos(a*t) → (s^2 - a^2)/((s^2 + a^2)^2)

sinh(a*t) → a/(s^2-a^2)

cosh(a*t) → s/(s^2-a^2)

e^(a*t)*sin(b*t) → b/((s-a)^2 + b^2)

e^(a*t)*cos(b*t) → (s-a)/((s-a)^2 + b^2)

e^(a*t)*sinh(b*t) → b/((s-a)^2 - b^2)

e^(a*t)*cosh(b*t) → (s-a)/((s-a)^2 - b^2)

**Laplace Transform Examples**

1. f(t) = e^(2*t)

From e^(a*t) → 1/(s-a) , a = 2, and then:

L(e^(2*t)) = 1/(s-2)

2. f(t) = cos(3*t)

From cos(a*t) → s/(s^2 + a^2), we find that a = 3 and...

L(cos(3*t)) = s/(s^2 + 9)

3. f(x) = 2 + 3*x

Note that:

1 → 1/s

t^n → n!/(s^(n+1))

L(f(x)) = L(2 + 3*x) = L(2) + L(3*x) = 2/s + 3/s^2

4. f(x) = e^-x * sinh(2*x)

The transform to be used here is: e^(a*t)*sinh(b*t) → b/((s-a)^2 - b^2)

And here a=-1 and b = 2

L(e^-x * sinh(2*x)) = 2/((s+1)^2 - 4)

Which can also be written 2/(s^2 + 2*s - 3)

**Inverse Laplace Transforms**

L⁻¹( F(s) ) = f(t)

Often taking the inverse is more difficult. Sometimes algebraic manipulation is required to get F(s) into proper form for the inverse transformation. Practice, patience, and perhaps either a calculator with CAS capabilities or mathematical software are all useful tools for taking inverse Laplace transforms.

Let's start with a simple inverse transform.

5. 1/s^2

From t^n → n!/(s^(n+1)), we see that n+1=2, n=1, and...

L⁻¹(1/s^2) = t (or x)

6. 12/(3*s + 9)

The first thing we need to do is algebraically manipulate F(s) into a useable form. Observe that:

12/(3*s + 9) = 4/(s + 3)

Using the transformation e^(a*t) → 1/(s-a), we note that a=-3 and,

L⁻¹(12/(3*s + 9)) = L⁻¹(4/(s + 3)) = 4 * L⁻¹(1/(s + 3))

= 4*e^(-3*t)

7. (2s - 6)/(s^2 + 2*s - 3)

Note: s^2 + 2*s - 3 = (s - 1)*(s + 3)

By decomposition of fractions:

(2s - 6)/(s^2 + 2*s - 3) = -1/(s - 1) + 3/(s + 3)

Then:

L⁻¹(-1/(s - 1) + 3/(s + 3))

= L⁻¹(-1/(s - 1)) + 3 * L⁻¹(1/(s + 3))

= -e^-t + 3*e^(-3*t)

8. (s + 3)/(s^2 + 2*s + 5)

Note that: s^2 + 2*s + 1 = (s + 1)^2

By completing the square in the denominator:

s^2 + 2*s + 5 = s^2 + 2*s + 1 + 4 = (s + 1)^2 + 4

And (s + 3)/(s^2 + 2*s + 5) = (s + 3)/( (s + 1)^2 + 4 )

It looks like the forms we will use are:

e^(a*t)*sin(b*t) → b/((s-a)^2 + b^2)

e^(a*t)*cos(b*t) → (s-a)/((s-a)^2 + b^2)

We will have get the numerator ready: s + 3 = (s + 1) + 2

So we have (s + 3)/( (s + 1)^2 + 4 ) = (s+1)/((s+1)^2+4) + 3/((s+1)^2+4)

Hence:

L⁻¹( (s + 3)/(s^2 + 2*s + 5) )

= L⁻¹( (s+1)/((s+1)^2+4) + 3/((s+1)^2+4) )

= L⁻¹( (s+1)/((s+1)^2+4) ) + L⁻¹( 3/((s+1)^2+4) )

= e^(-t) * cos(2*t) + e^(-t) * sin(2*t)

Now that we had an introduction to Laplace transforms, we will use them to help solve differential equations. Until next time, Eddie!

This blog is property of Edward Shore. 2013

Do you can make a source of HP prime? thanks

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