Understand Laplace Transformations and Delta Functions in Calculus

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In summary, the lecturer applied a laplace transformation to: \frac{d^2x}{dt^2} - 5\frac{dx}{dt} + 6x = 3H(t) - 3H(t - 6) However, I do not understand what happened or what the result was. Can someone please explain this to me? The Dirac delta function is just a special case of the Heaviside function. It makes sense in the context of generalized functions. Finally, I take the inverse transform of the laplace transformation to get the final result.
  • #1
Zurtex
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My Calculus lecturer is pretty bad, so I just wanted to check over some stuff. First of all what use are Laplace transformations? Secondly he applied a laplace transformation to:

[tex]\frac{d^2x}{dt^2} - 5\frac{dx}{dt} + 6x = 3H(t) - 3H(t - 6)[/tex]

Where H(t) is the heavy side step function. However I really don't understand what went on there, could some one please explain how you at least start this?

Finally he defined the dirac delta function as being, 0 everywhere, except at a and that:

[tex]\int_{-\infty}^{\infty} \delta (x) dx= 1[/tex]

Does that make any sense at all?
 
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  • #2
Well the idea of using the Laplace transform on typical integrodifferential equations is to effectively turn a scary calculus problem into an algebra problem...

The second order DE you have is pretty easy to transform... I tend to use tables of Laplace transforms...

Simply transform the second order derivative appropriately... I hope that you happen to know the initial conditions of the situation being described...

Then apply it to the first order derivative, and finally to the third term.

For the 3H(t-6), you will need to recall that a laplace transform here will result in a shift denoted by the 6.

I'll tex out an appropriate transform for you in a bit.
 
  • #3
As for the delta function, it makes perfect sense in the context of generalised functions. In particular, it is the generalised derivative of the Heaviside function. Generalised functions are tremendously important in finding approximate solutions to PDEs. Mathworld has quite a nice exposition on generalised functions if you're interested http://mathworld.wolfram.com/GeneralizedFunction.html
 
  • #4
Zurtex said:
My Calculus lecturer is pretty bad, so I just wanted to check over some stuff. First of all what use are Laplace transformations? Secondly he applied a laplace transformation to:

[tex]\frac{d^2x}{dt^2} - 5\frac{dx}{dt} + 6x = 3H(t) - 3H(t - 6)[/tex]

Where H(t) is the heavy side step function.

We can do this one. First things first:

1. It's called the Heaviside function which is just an impulse right:

[tex]H(x)=\left\{\begin{array}{cc}0,&\mbox{ if }x\leq 0\\1, &\mbox{ if }x>0\end{array}\right[/tex]

In the case above, 3H(t)-3H(t-6) is an impulse of size 3 from t=0 to t=6, then it drops to 0. See this?

Now,

[tex]\mathcal{L}\{H(t-a)\}=\frac{e^{-as}}{s}[/tex]

Right?

2. In regards to LaPlace Transforms, need to consider initial conditions. Let's just say:

y(0)=a
y'(0)=b


3. Can you take the LaPlace transform of the following? It's easy right? Just look in your ODE book and they'll have formulas for the 1'st and 2'nd derivatives.

[tex]\frac{d^2x}{dt^2} - 5\frac{dx}{dt} + 6x [/tex]

I'll do the first one for you:

Let:

[tex]\mathcal{L}\{x(t)\}=f(s)[/tex]

Then:

[tex]\mathcal{L}\{x''\}=s^2f-sa-b[/tex]

Right?

4. Now combine all of this and get a messy expression for f(s) in terms of the variable s from all the expressions above. Can you do this?

Then we'll take the inverse transform and we're done.
 
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  • #5
Oh yea, same dif for the Dirac delta function:

[tex]\frac{d^2x}{dt^2} - 5\frac{dx}{dt} + 6x = \delta(t-a)[/tex]

However, in this case instead of subjecting the system to a constant impulse of a certain duration, we strike it instantaneously like hitting a pendulum with a hammer. The LaPlace transform of the delta function is also a simple expression. That integral you gave is just an instrument used to describe it.
 
  • #6
:rolleyes: I don't actually have an "ODE book". The intial conditions where [itex]x(0) = 0 \quad \dot{x}(0)=2[/itex]. Thanks for trying to explain but I still have no idea what is going on, I'll ask someone to take me through an example.
 
  • #7
Lonewolf said:
As for the delta function, it makes perfect sense in the context of generalised functions. In particular, it is the generalised derivative of the Heaviside function. Generalised functions are tremendously important in finding approximate solutions to PDEs. Mathworld has quite a nice exposition on generalised functions if you're interested http://mathworld.wolfram.com/GeneralizedFunction.html
As it is presented to most students, it is completely meaningless.
Textbook authors, or at the very least, lecturers, should be much clearer on this issue.
In particular, that we want to use a convenient formalism, the justification of which is too difficult to get into.
The meaninglessness of the standard way of using the delta function is not something which should be sought to be swept under the carpet (as often happens), it should be openly admitted, and references to texts/sites on generalized functions should be made.
 
  • #8
arildno said:
As it is presented to most students, it is completely meaningless.
Textbook authors, or at the very least, lecturers, should be much clearer on this issue.
In particular, that we want to use a convenient formalism, the justification of which is too difficult to get into.
The meaninglessness of the standard way of using the delta function is not something which should be sought to be swept under the carpet (as often happens), it should be openly admitted, and references to texts/sites on generalized functions should be made.

I agree completely. Generalized functions are too often neglected, and more often than not, a university will offer no course on generalized functions in the undergraduate mathematics curriculum. This is a shame, considering that they are so useful. The integral as presented is little more than a notational convinience.
 
  • #9
I find that even if generalized functions do not get the treatment they deserve in undergraduate studies, there is at least one aspect of "normal" maths which should have been commented on better in physics courses/undergraduate math courses:
And that has to do with switching the order of limiting operations.
All too often, this is just blithely done without comment; I think students would have understood the delta function formalism better if their lecturers and textbook authors had made some explicit comments and examples here (in particular, by showing examples where such switching fails/ is meaningless, as is the case with the delta "function").
Basically, it would show the students that the maths they've got has some limits; there are certain things we would like to do, but where we're lacking the proper tools to do so.
The first explicit meeting with functionals is perhaps when the student learn calculus of variations; prior to that stage, he at least deserves to understand why there are still a lot of interesting maths to be learnt..:wink:

So, even if a rigourous development of distribution theory is out of the question, there is still a lot which could be improved upon, IMO.
 
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  • #10
Zurtex said:
:rolleyes: I don't actually have an "ODE book". The intial conditions where [itex]x(0) = 0 \quad \dot{x}(0)=2[/itex]. Thanks for trying to explain but I still have no idea what is going on, I'll ask someone to take me through an example.

Alright, how about a pendulum, just sitting there, doing nothing, hanging at rest, clock is ticking. At 1 second I strike it sideways with a hammer. Off it goes . . .

It's motion is described (approximately) by:

[tex]y^{''}+3y^{'}+2y=\delta(t-1)[/tex]

Note the delta function, it's a quick jolt (at t=1) to a "damped vibrating system" initially at rest. Therefore:

y(0)=0
y'(0)=0

We'll take the Laplace Transform of both sides of the ODE to figure out it's motion, that is, y (in radians) as a function of time.

So . . .

[tex]\mathcal{L}\{y''+3y'+2y=\delta(t-1)\}[/tex]

It's very easy to take the transform of each member of the equation. With:

[tex]\mathcal{L}\{y(t)\}=f(s)[/tex]

We get:

[tex]s^2f+3sf+2f=e^s[/tex]

Solving for f(s) we have:

[tex]f(s)=(\frac{1}{s+1}-\frac{1}{s+2})e^s[/tex]

Taking the inverse transform:

[tex]\mathcal{L}^{-1}\{f(s)=(\frac{1}{s+1}-\frac{1}{s+2})e^s\}[/tex]

We get:

[tex]y(t)=(e^{1-t}-e^{2(1-t)})H(t-1)[/tex]

Yep, that's Heaviside.

Look at the attached plot. Doesn't it look like a pendulum, originally at rest (in water, whatever), and then suddenly struck?

Laplace transforms offer a convenient method to solve problems like these which are not approachable by other methods. Now I know the engineering people in here are going to think, "you got that one right out of Kreyszig didn't you". So . . .
 

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  • #11
Oh wow cool thanks :smile:

Makes much sense now :smile:
 
  • #12
Zurtex said:
Oh wow cool thanks :smile:

Makes much sense now :smile:

I think the original equation is not reflective of a real damped system. Should have a positive damping effect like:

[tex]\frac{d^2x}{dt^2} + 5\frac{dx}{dt} + 6x = 3H(t) - 3H(t - 6)[/tex]

You can do this one using Laplace Transforms. A bit messy but doable. You'll think these are a piece of cake once you do a few. The attached is what I get when you plug in x(0)=0, x'(0)=2. Note the pulse during the six second interval.
 

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  • #13
Thanks Zurtex. Now I know what this one meant:

[tex] m\ddot{x} + b\dot{x}^2 + mglSin(x) = C\delta (x - x_m_a_x) [/tex]

Funny how things work out . . .
 
  • #14
saltydog said:
Thanks Zurtex. Now I know what this one meant:

[tex] m\ddot{x} + b\dot{x}^2 + mglSin(x) = C\delta (x - x_m_a_x) [/tex]

Funny how things work out . . .
haha, well I help people on this forum because it helps me better understand the maths and even realize things about the topic that I didn't when explaining it :smile: Or at least it gives me good practise.
 

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