Astrophysics
  • Astrophysics
@ganeshie8
Mathematics
  • Stacey Warren - Expert brainly.com
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schrodinger
  • schrodinger
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Astrophysics
  • Astrophysics
Ooh it shows college...anyways wanna do a fun convolution integral problem
Astrophysics
  • Astrophysics
\[\phi'(t) = 1/2 \int\limits_{0}^{t} (t- \xi)^2 \phi(\xi) = -t , \phi(0)=1\]
Astrophysics
  • 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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ganeshie8
  • ganeshie8
I have no clue what these are, could you teach me some pretending that il pick up fast :)
Astrophysics
  • Astrophysics
Oh that should be - after the phi'
Astrophysics
  • Astrophysics
Ah yeah..you know the convolution integral right
Astrophysics
  • 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
  • ganeshie8
\[f*g = \int\limits_{0}^{t} f(u)g(t-u)\,du\] ?
Astrophysics
  • Astrophysics
Right that's the convolution integral
ganeshie8
  • ganeshie8
thats the only convolution i remember from ODE
Astrophysics
  • Astrophysics
\[\phi'(t) - 1/2 \int\limits\limits_{0}^{t} (t- \xi)^2 \phi(\xi) = -t , \phi(0)=1\] fixed
Astrophysics
  • Astrophysics
Yeah haha the theory in DE's is pretty difficult I think
ganeshie8
  • ganeshie8
you want to differentiate both sides with respect to \(t\) is it ?
Astrophysics
  • Astrophysics
d/dt yup
ganeshie8
  • ganeshie8
that might give us a nice form for the right hand side to use convolution
ganeshie8
  • ganeshie8
\[t^2*\phi=\int\limits\limits_{0}^{t} (t- \xi)^2 \phi(\xi) d\xi \] ?
Astrophysics
  • Astrophysics
Looks good
ganeshie8
  • ganeshie8
how do you differentiate a convolution ?
Astrophysics
  • 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
  • ganeshie8
why are we allowed to swap the derivative and integral ?
ganeshie8
  • ganeshie8
FTC tells me \[\dfrac{d}{d\color{red}{t}} \int\limits_0^{\color{red}{t}} f(\xi)\,d\xi = f(t)\]
Astrophysics
  • Astrophysics
Well we're really using the relation between laplace transform and convolution
ganeshie8
  • ganeshie8
\[F(s)G(s) = \mathcal{L(f*g)}\] that one ?
ganeshie8
  • ganeshie8
i don't see the connection..
Astrophysics
  • Astrophysics
It's because the integral is the convolution between f(t)=t^2 and phi(t)
ganeshie8
  • ganeshie8
right, that still doesn't explain how we can exchange derivative and integral
ganeshie8
  • ganeshie8
maybe lets just proceed assuming that swap is legitimate...
Astrophysics
  • Astrophysics
Haha, wait no I want to know, I was just doing the math not thinking about that
ganeshie8
  • 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
  • ganeshie8
next you want to take laplace transform both sides is it ?
ganeshie8
  • ganeshie8
\[\phi ''(t) - t*\phi = -1\] taking laplace transform seems to be the most logical thing to do next
Astrophysics
  • 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
  • Astrophysics
Ok it shouldn't matter yeahhh
ganeshie8
  • ganeshie8
i don't want to take derivative first, i don't believe we can exchange derivative and integral...
ganeshie8
  • ganeshie8
lets go with ur equation may be : \[\phi'(t) - 1/2 (t^2* \phi(t)) = -t\]
Astrophysics
  • Astrophysics
yeah that looks a bit weird otherwise lol
Astrophysics
  • Astrophysics
I see it's part of this https://en.wikipedia.org/wiki/Volterra_integral_equation ok, keep going
ganeshie8
  • ganeshie8
\[s\Phi(s) - 1-1/2 \mathcal{L}(t^2)\mathcal{L}(\phi(t)) = -\mathcal{L}(t)\]
ganeshie8
  • ganeshie8
\[s\Phi(s) - 1- \dfrac{\mathcal{L}({\phi}(t))}{s^3} = -\dfrac{1}{s^2}\]
Astrophysics
  • Astrophysics
Looks good
ganeshie8
  • ganeshie8
i can isolate \(\Phi(s)\) but what next ?
Astrophysics
  • Astrophysics
wait a sec it should be
ganeshie8
  • ganeshie8
what to do with that \(\mathcal{L}(\phi(t))\) ?
Astrophysics
  • Astrophysics
\[s \Phi(s) - 1 - \frac{ \Phi(s) }{ s^3 }=-1/s^2\] right
ganeshie8
  • ganeshie8
right right
ganeshie8
  • ganeshie8
we can solve \(\phi(t)\) then
Astrophysics
  • Astrophysics
Yup
ganeshie8
  • ganeshie8
\(\Phi(s) = \dfrac{1}{s^2+1}\)
Astrophysics
  • Astrophysics
I got \[\frac{ s }{ s^2+1 }\]
ganeshie8
  • ganeshie8
\(\phi(t) = \sin t\)
Astrophysics
  • Astrophysics
I got cost xD
ganeshie8
  • ganeshie8
i may be wrong, im doing all my calculations in my unreliable head..
Astrophysics
  • Astrophysics
\[\Phi(s) (s-1/s^3) = -\frac{ 1 }{ s^2 }+1\]
Astrophysics
  • Astrophysics
Yeah should work out to be cost
Astrophysics
  • Astrophysics
But w/e you got it
ganeshie8
  • ganeshie8
yeah phi = cost looks good to me
Astrophysics
  • Astrophysics
I actually spent the least amount of time on convolution but it seems pretty useful
Astrophysics
  • 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
  • ganeshie8
to my knowledge \(\frac{ d }{ dt } (\int\limits_{0}^{t} (t- \xi)\phi(\xi)d \xi )\) is \(0\)
ganeshie8
  • ganeshie8
i may be off... could you elaborate a bit..
Astrophysics
  • 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
  • Astrophysics
Maybe I'm doing it wrong..

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