Question

How can the Laplace transformation L{f(ct)} (where c is a constant) be derived using the definition of a Laplace transformation? (As opposed to finding it by using a table of known Laplace transforms?)

Answer 1

Calculus 3?

Answer 2

DE

Answer 3

whats the definition of a laplace transform?

Answer 4

\[L\{f(ct)\}=\int^{\infty}_{0}f(ct)~e^{-st}~dt\]

Answer 5

do you know how to work a by parts ... using a table?

Answer 6

yes I do I have that the answer is (1/c)F(s/c) I just have to show how to get there and that's what I'm confused on

Answer 7

if we work the by parts, im sure we will get to that

Answer 8

okay so if i start the integral should i do integration by parts?

Answer 9

u= f(ct) ; du=c*f'(ct) ; dv=e^-st dt ; v=(-1/s)e^-st

Answer 10

\[ \Large \begin{array}c & & \int dt\\(+/-) & d/dt & e^{-st}\\+ & f(ct) & -\frac{1}{s}e^{-st}\\- & c f'(ct) & \frac{1}{s^2}e^{-st}\\+ & c^2 f''(ct) & -\frac{1}{s^3}e^{-st}\\- & c^3 f'''c(t) & \frac{1}{s^4}e^{-st}\\\end{array} \]

Answer 11

c got a little out of the function at the end lol

Answer 12

integration by parts gives us a pattern that gets established right?

Answer 13

yeah i see that pattern, i'm just not sure how the answer in the book got (1/c) on the outside of the inverse laplace transform of (s/c) ?

Answer 14

what you're saying totally makes sense cause the power of c is increasing so i'm not really sure?

Answer 15

\[\int_{0}^{\infty}f(ct)e^{-st}~dt=-\underbrace{\sum_{n=0}^{\infty}c^nf^{(n)}(c\infty)\frac{e^{-s\infty}}{s^{n+1}}}_{=0}+\sum_{n=0}^{\infty}c^nf^{(n)}(c0)\frac{e^{0}}{s^{n+1}}\]

Answer 16

hmm, the other option is to derivative e^(-st) and integrate f(ct) but that might be complicated ....

Answer 17

okay so the second summation looks very similar to the t^n la place transformation correct?

Answer 18

yes

Answer 19

now, are you sure you typed the correct results?

Answer 20

i'm trying to use the equation drawer but its not working let me see if i can get it to work just to double check

Answer 21

equation button is not that well equiped. knowing latex allows you to type it in directly to this reply box and avoids the hunt and peck

Answer 22

\[inverse laplace = f(ct) ; F(s) [laplace] = \frac{ 1 }{ c }F(\frac{ s }{ c }) , c > 0\]

Answer 23

.. yeah, thats what i thought you typed :) now working it according to the definition

Answer 24

okay so we're on the right track :)

Answer 25

so for my answer i would just write the integral above equals the summation and that should satisfy the proof?

Answer 26

int u dv = uv - int v du

let u = e^(-st), du = -s e^(-st) dt
let v = 1/c F(ct) + K ; dv = f(ct)

sorry, really really had to go ... do something

Answer 27

haha that okay thanks a lot for your help!

Answer 28

ohhhhh okay now i see the 1/c can't believe i didn't see that before

Answer 29

\[\int_{0}^{\infty}f(ct)e^{-st}~dt=\left(\frac 1c F(ct)+K\right)e^{-st}|^{\infty}_{0}+\frac sc\int_{0}^{\infty} \left(F(ct)+K\right) e^{-st}~dt\]
trying to code it really lags my computer

Answer 30

i think K is immaterial here since its a definant integral

Answer 31

I didn't even think of making the exponential function u ,okay perfect I got it!

Answer 32

got any time for another laplace problem? ;)

Answer 33

you got it? good, becuase between trying to code and keep up with mathing it ... i tend to get lost inbetween lol

Answer 34

maybe, whats teh problem?

Answer 35

y'' + y = [t, <= t < 1 ] & [0, 1 <= t < infinity] y(0)=0 y'(0)=0 ?

Answer 36

is that a heaviside? pieced function?

Answer 37

yes piecewise defined

Answer 38

i have some of it worked out i'm going to try to upload a picture of it ?

Answer 39

y'' + y = t, 0 < t < 1

y'' + y = 0, 1 < t < inf

Answer 40

L{y''} + L{y} = L{t}

L{y''} + L{y} = L{0}

or are you sposed to do definition and not table?

Answer 41

on the first one it is 0 <= t < 1
1 <= t < infinity

Answer 42

i can use the table

Answer 43

your work is looking fine, but the thing is, the interval is not from 0 to inf sooo, you might have to split it

\[\int_{0}^{inf}f(t)e^{-st}dt=\int_{0}^{1}f(t)e^{-st}dt+\int_{1}^{inf}f(t)e^{-st}dt\]

Answer 44

yeah you're right. so my answer of y = t - sin(t) would only solve the first part of the piece wise function ? or is that part wrong ?

Answer 45

it would solve it IF there wasnt a change in function, instead of e^(inf) in the normal operations of things, we stop at e^(-s) which doesnt zero out

Answer 46

alright i'll start from the beginning

Answer 47

use the definiton and youll see how it alters the setup

Answer 48

i'm hopeless lol

Answer 49

L{y''(t)}, [0,1]

int[0,1] y''(t) e^(-st) dt

u = e^(-st) v = y'(t) dt
du = -s e^(-st) dv = y''(t) dt


int[0,1] y''(t) e^(-st) dt = [y'(1) e^(-s)- y'(0)] +s int[0,1] y' e^(-st) dt

L{y''(t)}, [0,1] = y'(1) e^(-s) +s L{y''(t)}, [0,1]

does this make sense?

Answer 50

y' on the end, got a little '''' happy lol

Answer 51

okay about to upload my work

Answer 52

this thing needs emojis or something lol

Answer 53

really close just a little off

Answer 54

sorry i was having a hard time using the integral way because my teacher never taught it that way

Answer 55

im trying to paper it too ... let review, mainly for my sake :)

\[\left.L\{y'(t)\}\right|_{0}^{a}=\int_{0}^{a}y'(t)e^{-st}dt=\\
~~~~~~~~\left[y(a)e^{-sa}-y(0)e^{-s(0)}\right]+s\int_{0}^{a}y(t)e^{-st}dt\]
\[\left.L\{y'(t)\}\right|_{0}^{a}=y(a)e^{-sa}-y(0)+sL\{y(t)\}|_{0}^{a} \]

Answer 56

as a approaches infinity, this is the usual laplace that we are familiar with yes?

Answer 57

yes

Answer 58

becasue we have the function split up along intervals, then this is a process that we can do.

Answer 59

a to infnity just gives us

y(inf) e^{-inf}-y(a) e^{-sa} right?

Answer 60

right

Answer 61

not sure what im getting at here, i just know what the laplace is altered when we dont do a complete interval from 0 to 1 :)

Answer 62

hmm me either but i definitely see the direction you're going in

Answer 63

i have to go :( but i think if i spend some more time on it i can get it, thank you very much for all your help!

Answer 64

ill continue to work at it just becasue i like the question :) have fun

Answer 65

okay thank you!!

Answer 66

@amistre64 A more straightforward approach to establish the transform of \(f(ct)\) might be to substitution \(u=ct\), then you get
\[\begin{align*}\mathcal{L}\{f(ct)\}&=\int_0^\infty f(ct)~e^{-st}\,dt\\\\
&=\frac{1}{c}\int_0^\infty f(u)~e^{-su/c}\,du\\\\
&=\frac{1}{c}F\left(\frac{s}{c}\right)\end{align*}\]
where the last line is recognizing that the \(s\) argument in the integral is replaced with \(\dfrac{s}{c}\).

Answer 67

yeah, i was a bit stymied at how we got the s/c in there :)

the change of variable makes it more sensible to me know.