Unit V Lesson 1

Laplace Transform

Introduction and definition Laplace transform, non-standard driving functions, solving from the algebra domain.
Laplace transform examples, f(t)=1, f(t)=t.
Laplace transform examples f(t)= e^t and f(t)= sin(2t).
Laplace transform is a linear operator, an example, a book error in 7th edition, small table of transforms.
Laplace transforms existence, example of a transform of a piecewise defined function.
Laplace transforms example using linearity of transform.

is a ‘driving function’ representing some external force f(t) or an impressed voltage E(t). Piecewise continuous driving functions are not uncommon. For example, the driving function could be described graphically by either of the following:

We have no direct method of solving problems of this nature. The problems can often be solved with transform methods, and the one we will focus on is the Laplace transform.

We won’t be solving any differential equations this way until we are into Unit V Lesson 5. Before then, we need to learn how to make transforms as well as inverse transforms of various functions. The picture below tells the story of how solving equations is done. For the rest of this section and the next three as well, we will be learning how to transform and find inverse transforms of functions (the sideways arrows in the picture).

Laplace transform method can also be used to solve systems of equations.

Def: Laplace transform: If f is a function defined for , the Laplace transform of f(t), denoted L{f(t)}, is defined as L{f(t)}= provided that the improper integral converges.

Note the familiar notation of capitalizing to show an antiderivative.

Solution:

Recall an operator M is linear if M{Af(x)+Bg(x)}=AM{f(x)}+BM{g(x)}.

Note that the Laplace transform is a linear operator since integration is linear:

Take this link after trying it yourself.

 Take this link after trying it yourself.

Def: Exponential order
A function f is said to be of exponential order C if there are constants C, M>0, and T>0 such that for all t >T.

If f is a decreasing function, being of exponential order C<0 means that the graph of f lies below the graph of Me^Ct for t>T.

The proof is in the book and relies on showing that the integral representing the Laplace transform is convergent if is of exponential order C. It requires comparing the integral representing the transform and the integral of Me^Ct.

Note that L{f(t)} doesn’t exist for f(t)=1/t or f(t)=exp(t²).

Solution: With u=tⁿ and du=nt^n-1, dv=e^-st and v=-e^-st/s using integration by parts,

The first part of this can be shown to be zero with n+1 applications of L’Hopital’s rule.

Now we have to evaluate

The same process will continue, with the last integral representing L{1}=1/s from the preceding example .

L{sin(kt)}=

The integration on the first step was by parts with u=sin(kt), du=kcos(kt), dv=e^-stdt, v=-(1/s)e^-st.

The u*v part=0 since sin(0)=0. Continue on with the remaining integral:

L{sin(kt)}=

The integration on the first step was by parts with u=cos(kt), du=-ksin(kt), dv=e^-st, v=-(1/s)e^-st .

The u*v part is k/s² (If you work out the limit at infinity, it’s zero. cos(0) and e⁰ are both 1, so we have –(-1).

The problem is now at this stage:

.

This equation can be solved for the desired integral:

So we see that L{sin(kt)}=.

As an assignment, you will be asked to work out L{cos(kt)}. It is similar.

Piecewise defined functions also have Laplace transforms as the following example shows:

Solution:

Since the function is piecewise defined, we look at the three integrals:

Once we know a few rules for transforms, we can use the linearity of the Laplace transform to work out problems with ease.

Solution:

L{3t²}=3(2!)/s³ and L{sint}=1/(s²+1), so we have

L{3t²+sint }=

Now you try one:

Take this link after you’ve tried it yourself.