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Differential equations
Course: Differential equations > Unit 3
Lesson 1: Laplace transformLaplace transform 2
Laplace transform of e^at. Created by Sal Khan.
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How do you compare complex and real number? I mean he writes s>a...(8 votes)- For those who are wondering this happens at4:37.
Well firstly, 'a' can also be complex, and secondly, in maths you cannot compare complex numbers, either with real numbers or with other complex numbers.
So what he did is a little more in depth that what he wrote down, and I'm sure that most people don't realise this, as he hasn't defined 's' as a complex number yet.
Let's say that 's = a + bi', and 'a = c + di', then that part on the inside of the integral becomes 'exp((a-s)t) = exp((c + di - a - bi)t)'. Grouping the real and imaginary terms we get 'exp((c-a)t+i(d-b)t)'. We can then split the exponent as such: 'exp((c-a)t)exp(i(d-b)t)'.
Looking at this, as t tends to infinity, we know that the complex exponential will just go round in circles (from Euler's identity, 'exp(i*theta)'). But if we look at the real exponential, we can see that it will either go to infinity or to zero, depending on the value of '(c-a)'. So something that goes around in circles times zero will be zero, and something that goes around in circles times something that tends to infinity will also tend to infinity.
So to make a slight correction (which is only necessary if you want to be strictly mathematical), by writing 's>a', he actually means 'Re(s)>Re(a)', which is entirely possible to compare.
I hope this explanation is clear.(46 votes)
At2:10, why is e^(-st) * e^(at) combine to e^(a-s)t instead of e^(a-s) ?? Wouldn't the -t and t combine to cancel??(4 votes)- For multiplying terms which have the same base, (x^a) * (x^b) = x^(a + b). So,
e^(-st) * e^(at) = e^(-st + at). Take out the common factor, "t" from (-st + at) =>
t(a-s), which gives us e^(a-s)t(11 votes)
What about the case when a = s? We get 0*t, when t tends to infinity, do we consider it an indeterminate form in this case?(8 votes)- in doing these transformations it is assumed s>a by definition otherwise the function diverges.(1 vote)
What is the difference between ordinary differential equations and partial differential equations?(2 votes)- Ordinary differential equations involve quantities that vary with respect to a single variable.
Partial differential equations involve quantities that vary with respect to more than one variable. For example the quantity of interest could be thermal energy across a metal surface and the variables are time and space.
Note: When I say quantity I am reffering to the output of a function
Thumbs Up!(5 votes)
How would you go about the Laplace transform of e^(t+2) ?(1 vote)
Notice that e^(t+2) could be written as (e^2)*e^t, since e^2 is a constant, you can use the linearity of the Laplace transform, and find (e^2)*L(e^t).
This would equal
(e^2)*1/(s-1).(5 votes)
- Is 's' here something like the 'plus C' in indefinite integrals? I mean it can be nearly any constant, isn't is?(2 votes)
Not exactly, it was my understanding that the 's' that appears in laplace transforms is actually another variable in a different dimension or with different units of measure (i.e. t -> time domain, while s -> frequency domain).(3 votes)
- What if s=a? would this mean the limit/Laplace Transform DNE because you would have -1/0?(2 votes)
yes, correct. The Transform still does not work.(1 vote)
- why does he change the expression to be a limit of a as before?(2 votes)
It is often a formality to explicitly show why the next steps happen. It is like a justification. And sometimes, with complex systems, it is necessary to show a limit, else the function may behave unexpectedly! :O However, often times we can see what the limit does without putting it down and we don't bother writing it at all! :)(1 vote)
- At4:30I understand what he is doing, but why does he evaluate to infinity? In the first video, he solves for the limit at a constant. Why is this no longer required?(2 votes)
- He directly evaluates to infinity because the limit isn't absolutely necessary to preform the substitution of infinity in the Laplace transform equation.
He uses a constant in the first video to make it easier to understand.(1 vote)
- can any one tell me how -1/s-a = 1/s+a what is the math behind this ?(2 votes)
- Multiply -1 on the numerator and denominator(1 vote)
4:37Video transcript
Let's keep doing some Laplace
transforms. For one, it's good to see where a lot of those
Laplace transform tables you'll see later on actually
come from, and it just makes you comfortable with
the mathematics. Which is really just kind of
your second semester calculus mathematics, but it makes you
comfortable with the whole notion of what we're doing. So first of all, let me just
rewrite the definition of the Laplace transform. So it's the L from Laverne
& Shirley. So the Laplace transform of some
function of t is equal to the improper integral from 0 to
infinity of e to the minus st times our function. Times our function of t, and
that's with respect to dt. So let's do another
Laplace transform. Let's say that we want to take
the Laplace transform-- and now our function f of t, let's
say it is e to the at. Laplace transform
of e to the at. Well we just substituted it into
this definition of the Laplace transform. And this is all going to be
really good integration practice for us. Especially integration
by parts. Almost every Laplace transform
problem turns into an integration by parts problem. Which, as we learned long ago,
integration by parts is just the reverse product rule. So anyway. This is equal to the integral
from 0 to infinity. e to the minus st times
e to the at, right? That's our f of t. dt. Well this is equal to just
adding the exponents because we have the same base. The integral from 0
to infinity of e to the a minus stdt. And what's the antiderivative
of this? Well that's equal to what? With respect to C. So it's equal to-- a minus s,
that's just going to be a constant, right? So we can just leave it
out on the outside. 1/a minus s times e to the a
minus st. And we're going to evaluate that from t is equal to
infinity or the limit as t approaches infinity to
t is equal to 0. And I could have put this inside
the brackets, but it's just a constant term, right? None of them have t's in them,
so I can just pull them out. And so this is equal to 1/a
minus s times-- now we essentially have to evaluate
t at infinity. So what is the limit
at infinity? Well we have two cases
here, right? If this exponent-- if this a
minus s is a positive number, if a minus s is greater than
0, what's going to happen? Well as we approach infinity,
e to the infinity just gets bigger and bigger and
bigger, right? Because it's e to an infinitely
positive exponent. So we don't get an answer. And when you do improper
integrals, when you take the limit to infinity and it doesn't
come to a finite number, the limit doesn't
approach anything, that means that k the improper
integral diverges. And so there is no limit. And to some degree, we can say
that the Laplace transform is not defined with a minus s is
greater than 0 or when a is greater than s. Now what happens if a minus
s is less than 0? Well then this is going
to be some negative number here, right? And then if we take e to an
infinitely negative number, well then that does approach
something. That approaches 0. And we saw that in the
previous video. And I hope you understand
what I'm saying, right? e to an infinity negative number
approaches 0, while e to an infinitely positive
number is just infinity. So that doesn't really
converge on anything. So anyway. If I assumed that a minus s is
less than 0, or a is less than s, and this is the assumption I
will make, just so that this improper integral actually
converges to something. So if a minus s is less than
0, and this is a negative number, e to the a minus s
times-- well t, where t approaches infinity will be 0. Minus this integral
evaluated at 0. So when you value this
at 0, what happens? T equals 0. This whole thing becomes
e to the 0 is 1. And we are left with what? Minus 1/a minus s. And that's just the same
thing as 1/(s-a). So we have our next entry in our
Laplace transform table. And that is the Laplace
transform. The Laplace transform of e to
the at is equal to 1/(s-a) as long as we make the
assumption that s is greater than a. This is true when s is greater
than a, or a is less than s. You could view it either way. So that's our second entry in
our Laplace transform table. Fascinating. And actually, let's relate this
to our previous entry in our Laplace transform
table, right? What was our first entry in our
Laplace transform table? It was Laplace transform of
1 is equal to 1/s, right? Well isn't 1 just the same
thing as e to the 0? So we could have said that this
is the Laplace-- I know I'm running out of space, but
I'll do it here in purple. We could have said Laplace
transform of 1 is the same thing as e to the 0
times t, right? And that equals 1/s. And luckily it's good to see
that that is consistent. And actually, remember, we even
made the condition when s is greater than 0, right? We assumed that s is greater
than 0 this example. Here again, you say s
is greater than 0. This is completely consistent
with this one, right? Because if a is equal to 0, then
the Laplace transform of e to the 0 is just
1/s minus 0. That's just 1/s. And we have to assume that
s is greater than zero. So really these are kind of the
same entry in our Laplace transform table. But it's always nice in
mathematics when we see that two results we got in trying
to do slightly different problems actually are,
in some ways, connected or the same result. Anyway I'll see you in the
next video and we'll keep trying to build our table of
Laplace transforms. And maybe three or four videos from now
I'll actually show you how these transforms are extremely
useful in solving all sorts of differential equations. See you soon.