Results 1 to 4 of 4
Like Tree1Thanks
  • 1 Post By johnsomeone

Math Help - Partial derivative of (n+1) dimensional integral - Evans PDE

  1. #1
    Newbie
    Joined
    Sep 2012
    From
    Australia
    Posts
    6

    Partial derivative of (n+1) dimensional integral - Evans PDE

    Hi I'm just wondering what steps Evans takes in his PDE book on page 50.

    He defines

    u(x,t)=\int_0^t \int_{\mathbb{R}^n}\Phi(y,s)f(x-y,t-s)\,dy\,ds

    And then goes on to say:

    u_t(x,t)=\int_0^t\int_{\mathbb{R}^n}\Phi(y,s)f_t(x-y,t-s)\,dy\,ds + \int_{\mathbb{R}^n}\Phi(y,t)f(x-y,0)\,dy.

    I do not know how he got here:

    I think I can get this far:

    \frac{u(x,t+h)-u(x,t)}{h} = \frac{1}{h}\left(\int_0^t \int_{\mathbb{R}^n}\Phi(y,s)[f(x-y,t+h-s)-f(x-y,t-s)]\,dy\,ds+\int_t^{t+h}\int_{\mathbb{R}^n}\Phi(y,s)f  (x-y,t+h-s)\,dy\,ds\right)

    Taking the limit as h\rightarrow 0, I can kind of see how the left term on the RHS can go to \Phi(y,s)f_t(x-y,t-s), but even this I'm not sure of, I don't really understand, I can't really see why it shouldn't it be f_{t-s}(x-y,t-s) since isn't this the 'variables' that is having an infinitesimally small h added on to? I have no idea how the right term of the RHS of the above equation transforms as well. I think the thing that is most troubling me is that the variable t is in the integral and the integrand and what we are taking a limit of, h, is in the integral... and in the integrand once we take the 1/h back inside!

    Can someone please help?! Thanks!
    Last edited by benb89; September 22nd 2012 at 02:37 AM.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Super Member
    Joined
    Sep 2012
    From
    Washington DC USA
    Posts
    525
    Thanks
    146

    Re: Partial derivative of (n+1) dimensional integral - Evans PDE

    The general approach to \frac{d}{dt}\int_{a}^{t} \phi(t, x) dx

    Let I(t) = \int_{a}^{t} \phi(t, x) dx. Then:

    I(t+h)-I(t) = \int_{a}^{t+h} \phi(t+h, x) dx - \int_{a}^{t} \phi(t, x) dx

    = \int_{a}^{t+h} \phi(t+h, x) dx - \int_{a}^{t} \phi(t+h, x) dx + \int_{a}^{t} \phi(t+h, x) dx - \int_{a}^{t} \phi(t, x) dx

    (yes, that's the common "add and subtract the same thing" trick)

    = \int_{t}^{t+h} \phi(t+h, x) dx + \int_{a}^{t} (\phi(t+h, x) - \phi(t, x)) dx.

    Then \frac{I(t+h)-I(t)}{h} = \frac{1}{h}\int_{t}^{t+h} \phi(t+h, x) dx + \int_{a}^{t} \frac{\phi(t+h, x) - \phi(t, x)}{h} dx.

    The 2nd integral gives you the derivative. The first is the kind that shows up in the fundemental theorem of calculus. There are several ways to do it, but maybe use the mean value theorem for integrals:

    For each h, there's an x_h \in [t, t+h] (thinking of h as positive) such that \int_{t}^{t+h} \phi(t+h, x) dx = \phi(t+h, x_h)h.

    Thus as h goes to zero, x_h goes to t, and if \phi is continuous, then:

    \lim_{h \rightarrow 0} \frac{1}{h} \int_{t}^{t+h} \phi(t+h, x) dx = \lim_{h \rightarrow 0} \phi(t+h, x_h) = \phi(t, t).

    For the 2nd integral, switching limits and integrals is switching a limiting process, so in general isn't allowed. However, in most practical cases it will be permitted, I just want to warn you that this really needs a theorem to be justified (the ready availability of such theorems for the Lebesque integral is why it's preferred over the Riemann integral). Ignore that here, just for the sake of the derivation:

    \lim_{h \rightarrow 0} \int_{a}^{t} \frac{\phi(t+h, x) dx - \phi(t, x)}{h} dx = \int_{a}^{t} \left\{ \lim_{h \rightarrow 0} \frac{\phi(t+h, x) dx - \phi(t, x)}{h} \right\} dx

    = \int_{a}^{t} \frac{\partial \phi(t, x)}{\partial t} dx.

    Putting it together:

    Then I'(t) = \lim_{h \rightarrow 0} \frac{I(t+h)-I(t)}{h}

    = \lim_{h \rightarrow 0} \frac{1}{h}\int_{t}^{t+h} \phi(t+h, x) dx + \lim_{h \rightarrow 0} \int_{a}^{t} \frac{\phi(t+h, x) dx - \phi(t, x)}{h} dx.

    Therefore \frac{d}{dt}\int_{a}^{t} \phi(t, x) dx = \phi(t, t) + \int_{a}^{t} \frac{\partial \phi(t, x)}{\partial t} dx.

    It's sloppy, but that's the gist of what's going on.

    In your case, that becomes (note that x is considered a constant for these purposes, so, for clarity, I won't even include it in the function notation):

    u(t) = \int_{0}^{t} \phi(t, s) ds, where \phi(t, s) = \int_{\mathbb{R}^n} \Phi(y,s)f(x-y, t-s) dy.

    Thus u_t(t) = \phi(t, t) + \int_{0}^{t} \frac{\partial \phi(t, s)}{\partial t} ds.

    The first term is \phi(t, t) = \int_{\mathbb{R}^n} \Phi(y,(t))f(x-y, t-(t)) dy = \int_{\mathbb{R}^n} \Phi(y,t)f(x-y, 0) dy.

    The 2nd term gives the \int_0^t \left\{ \int_{\mathbb{R}^n} \Phi(y,s)f_t(x-y, t-s) dy \right\} ds.

    Thus u_t(x, t) = \int_{\mathbb{R}^n} \Phi(y,t)f(x-y, 0) dy + \int_0^t \int_{\mathbb{R}^n} \Phi(y,s)f_t(x-y, t-s) dy ds.

    (Again, that's not going to always hold - you'd need to add some conditions to know in which cases it holds. It's a derivation, not a proof.)
    Last edited by johnsomeone; September 22nd 2012 at 06:25 AM.
    Thanks from benb89
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Newbie
    Joined
    Sep 2012
    From
    Australia
    Posts
    6

    Re: Partial derivative of (n+1) dimensional integral - Evans PDE

    You, Sir,

    are quite possibly my favourite person on this planet right now. That is wonderful. Thankyou so much. Are you a math PhD? I hope you're not an undergrad - or you have really made me feel like an IDIOT at the moment. A happy and relieved idiot.

    I am doing a course in measure theory at the moment. I have seen first hand how simple the Lebesgue integral makes things. It's a wonderful construction.

    Again, I am very very appreciative. I have another question about this section in Evans as well. If you are a PDE expert - you may be the one to help me .

    Thanks from Australia!

    (My parents are visiting Washington DC soon I should tell them to go find you and thank you personally... but that would be weird...)
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Newbie
    Joined
    Sep 2012
    From
    Australia
    Posts
    6

    Re: Partial derivative of (n+1) dimensional integral - Evans PDE

    You, Sir,

    are quite possibly my favourite person on this planet right now. That is wonderful. Thankyou so much. Are you a math PhD? I hope you're not an undergrad - or you have really made me feel like an IDIOT at the moment. A happy and relieved idiot.

    I am doing a course in measure theory at the moment. I have seen first hand how simple the Lebesgue integral makes things. It's a wonderful construction.

    Again, I am very very appreciative. I have another question about this section in Evans as well. If you are a PDE expert - you may be the one to help me .

    Thanks from Australia!

    (My parents are visiting Washington DC soon I should tell them to go find you and thank you personally... but that would be weird...)
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Replies: 2
    Last Post: July 16th 2011, 11:57 AM
  2. Looking for help with PDE (Evans, Brezis), will pay for your time
    Posted in the Differential Equations Forum
    Replies: 1
    Last Post: September 4th 2009, 08:13 AM
  3. Partial differential equation-wave equation - dimensional analysis
    Posted in the Differential Equations Forum
    Replies: 3
    Last Post: August 28th 2009, 11:39 AM
  4. 2-dimensional integral
    Posted in the Calculus Forum
    Replies: 7
    Last Post: January 21st 2009, 07:10 AM
  5. 3 Dimensional Volume Integral
    Posted in the Calculus Forum
    Replies: 2
    Last Post: April 14th 2007, 07:34 PM

Search Tags


/mathhelpforum @mathhelpforum