@ganeshie8

- Astrophysics

@ganeshie8

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- Astrophysics

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

- Astrophysics

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

- Astrophysics

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

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## More answers

- ganeshie8

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

- Astrophysics

Oh that should be - after the phi'

- Astrophysics

Ah yeah..you know the convolution integral right

- Astrophysics

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

- ganeshie8

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

- Astrophysics

Right that's the convolution integral

- ganeshie8

thats the only convolution i remember from ODE

- Astrophysics

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

- Astrophysics

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

- ganeshie8

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

- Astrophysics

d/dt yup

- ganeshie8

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

- ganeshie8

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

- Astrophysics

Looks good

- ganeshie8

how do you differentiate a convolution ?

- Astrophysics

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

- ganeshie8

why are we allowed to swap the derivative and integral ?

- ganeshie8

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

- Astrophysics

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

- ganeshie8

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

- ganeshie8

i don't see the connection..

- Astrophysics

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

- ganeshie8

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

- ganeshie8

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

- Astrophysics

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

- ganeshie8

we can get back to that detail later... maybe in the end of this thread :)
so do we have :
\[\phi ''(t) - \int\limits_{0}^{t} (t- \xi) \phi(\xi) d \xi = -1\]

- ganeshie8

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

- ganeshie8

\[\phi ''(t) - t*\phi = -1\]
taking laplace transform seems to be the most logical thing to do next

- Astrophysics

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

- Astrophysics

Ok it shouldn't matter yeahhh

- ganeshie8

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

- ganeshie8

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

- Astrophysics

yeah that looks a bit weird otherwise lol

- Astrophysics

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

- ganeshie8

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

- ganeshie8

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

- Astrophysics

Looks good

- ganeshie8

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

- Astrophysics

wait a sec it should be

- ganeshie8

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

- Astrophysics

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

- ganeshie8

right right

- ganeshie8

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

- Astrophysics

Yup

- ganeshie8

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

- Astrophysics

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

- ganeshie8

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

- Astrophysics

I got cost xD

- ganeshie8

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

- Astrophysics

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

- Astrophysics

Yeah should work out to be cost

- Astrophysics

But w/e you got it

- ganeshie8

yeah phi = cost looks good to me

- Astrophysics

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

- Astrophysics

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

- ganeshie8

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

- ganeshie8

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

- Astrophysics

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

- Astrophysics

Maybe I'm doing it wrong..

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