Question

@ganeshie8

Answer 1

Ooh it shows college...anyways wanna do a fun convolution integral problem

Answer 2

\[\phi'(t) = 1/2 \int\limits_{0}^{t} (t- \xi)^2 \phi(\xi) = -t , \phi(0)=1\]

Answer 3

So solve the integro differential using laplace and then I think the funner part is by differentiating the integro diff equations a sufficient amount of times convert it into a ivp and then verify the solution as it should be the same as the first one

Answer 4

I have no clue what these are, could you teach me some pretending that il pick up fast :)

Answer 5

Oh that should be - after the phi'

Answer 6

Ah yeah..you know the convolution integral right

Answer 7

I don't know too much about all the symbols haha but I thought this problem was fun, but this is called the integro - differential equations where both derivatives and integrals of a known function appear

Answer 8

\[f*g = \int\limits_{0}^{t} f(u)g(t-u)\,du\]

?

Answer 9

Right that's the convolution integral

Answer 10

thats the only convolution i remember from ODE

Answer 11

\[\phi'(t) - 1/2 \int\limits\limits_{0}^{t} (t- \xi)^2 \phi(\xi) = -t , \phi(0)=1\] fixed

Answer 12

Yeah haha the theory in DE's is pretty difficult I think

Answer 13

you want to differentiate both sides with respect to \(t\) is it ?

Answer 14

d/dt yup

Answer 15

that might give us a nice form for the right hand side to use convolution

Answer 16

\[t^2*\phi=\int\limits\limits_{0}^{t} (t- \xi)^2 \phi(\xi) d\xi \]

?

Answer 17

Looks good

Answer 18

how do you differentiate a convolution ?

Answer 19

Same way as everything else \[\phi ''(t) - 1/2 \int\limits_{0}^{t} 2(t- \xi) \phi(\xi) d \xi = -1\] taking the derivative respect to t

Answer 20

why are we allowed to swap the derivative and integral ?

Answer 21

FTC tells me \[\dfrac{d}{d\color{red}{t}} \int\limits_0^{\color{red}{t}} f(\xi)\,d\xi = f(t)\]

Answer 22

Well we're really using the relation between laplace transform and convolution

Answer 23

\[F(s)G(s) = \mathcal{L(f*g)}\]

that one ?

Answer 24

i don't see the connection..

Answer 25

It's because the integral is the convolution between f(t)=t^2 and phi(t)

Answer 26

right, that still doesn't explain how we can exchange derivative and integral

Answer 27

maybe lets just proceed assuming that swap is legitimate...

Answer 28

Haha, wait no I want to know, I was just doing the math not thinking about that

Answer 29

we can get back to that detail later... maybe in the end of this thread :)

so do we have :
\[\phi ''(t) - \int\limits_{0}^{t} (t- \xi) \phi(\xi) d \xi = -1\]

Answer 30

next you want to take laplace transform both sides is it ?

Answer 31

\[\phi ''(t) - t*\phi = -1\]

taking laplace transform seems to be the most logical thing to do next

Answer 32

Oh ok I see you're going with taking the derivative first then laplace, I used laplace right away so I had \[\phi'(t) - 1/2 (t^2 \phi(t)) = t\] because of the ivp

Answer 33

Ok it shouldn't matter yeahhh

Answer 34

i don't want to take derivative first, i don't believe we can exchange derivative and integral...

Answer 35

lets go with ur equation may be :
\[\phi'(t) - 1/2 (t^2* \phi(t)) = -t\]

Answer 36

yeah that looks a bit weird otherwise lol

Answer 37

I see it's part of this https://en.wikipedia.org/wiki/Volterra_integral_equation ok, keep going

Answer 38

\[s\Phi(s) - 1-1/2 \mathcal{L}(t^2)\mathcal{L}(\phi(t)) = -\mathcal{L}(t)\]

Answer 39

\[s\Phi(s) - 1- \dfrac{\mathcal{L}({\phi}(t))}{s^3} = -\dfrac{1}{s^2}\]

Answer 40

Looks good

Answer 41

i can isolate \(\Phi(s)\)
but what next ?

Answer 42

wait a sec it should be

Answer 43

what to do with that \(\mathcal{L}(\phi(t))\) ?

Answer 44

\[s \Phi(s) - 1 - \frac{ \Phi(s) }{ s^3 }=-1/s^2\] right

Answer 45

right right

Answer 46

we can solve \(\phi(t)\) then

Answer 47

Yup

Answer 48

\(\Phi(s) = \dfrac{1}{s^2+1}\)

Answer 49

I got \[\frac{ s }{ s^2+1 }\]

Answer 50

\(\phi(t) = \sin t\)

Answer 51

I got cost xD

Answer 52

i may be wrong, im doing all my calculations in my unreliable head..

Answer 53

\[\Phi(s) (s-1/s^3) = -\frac{ 1 }{ s^2 }+1\]

Answer 54

Yeah should work out to be cost

Answer 55

But w/e you got it

Answer 56

yeah phi = cost looks good to me

Answer 57

I actually spent the least amount of time on convolution but it seems pretty useful

Answer 58

Oh so we can do \[\frac{ d }{ dt } (\int\limits_{0}^{t} (t- \xi)\phi(\xi)d \xi )\] and apply the fundamental theorem of calc and product rule

Answer 59

to my knowledge \(\frac{ d }{ dt } (\int\limits_{0}^{t} (t- \xi)\phi(\xi)d \xi )\) is \(0\)

Answer 60

i may be off... could you elaborate a bit..

Answer 61

\[\frac{ d }{ dt } (\int\limits\limits_{0}^{t} (t- \xi)\phi(\xi)d \xi ) = \frac{ d }{ dt }(\int\limits_{0}^{t}t \phi(\xi)d \xi))-\frac{ d }{ dt }(\int\limits_{0}^{t}\xi \phi(\xi) d \xi))\]

Answer 62

Maybe I'm doing it wrong..