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

• Sep 22nd 2012, 02:19 AM
benb89
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!

• Sep 22nd 2012, 03:27 AM
johnsomeone
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.)
• Sep 22nd 2012, 04:15 PM
benb89
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...)
• Sep 22nd 2012, 04:16 PM
benb89
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...)