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Stream: deprecated: analysis

Topic: weak derivative

Jan Pax (Sep 29 2022 at 09:31):

I have a question on weak derivatives. I do not understand the derivation of the second equality above example 2 in the following snippet: 111.jpg Why is [the second = here in the r.h.s. snippet true ??] true: 222.jpg

Oscar Cunningham (Sep 29 2022 at 12:14):

They're applying integration-by-parts to the first integral, and the fundamental theorem of calculus to the second integral. There's also a typo, there should be a 2 instead of a 1.

$$-\int^1_0\phi\;dx+\phi(2)-\phi(1)$$

Jan Pax (Sep 29 2022 at 12:37):

No, I have verified directly with the author that there is no typo there. The thing is that both ones there are bounds where $\phi$ doesn't vanish. But still I do not know how can I use IBP here. Could you be more specific and provide me with the full calculation ? Here is what Evans, the author of this book PDE says: "I guess I do not understand your question.

In Example 2, the test function $\phi$ vanishes at $x=0$ and $x=2$, but not necessarily
at $x=1$. This is what leads to the contradiction.
"

Oscar Cunningham (Sep 29 2022 at 20:00):

Oh yeah, you're right that there's no typo
Applying integration by parts to the first integral gives

$$\int^1_0x\phi'\;dx=\left[\phi\right]^1_0-\int^1_0\phi\;dx=\phi(1)-\phi(0)-\int^1_0\phi\;dx$$

and applying the fundamental theorem of calculus to the second integral gives

$$\int^2_1\phi'\;dx=\phi(2)-\phi(1)$$

but $\phi$ is compactly supported so $\phi(0)=\phi(2)=0$. Adding the remaining terms together gives what we want

Jan Pax (Sep 29 2022 at 20:57):

I've catch it. Now, is there any connection with the Green's theorem ?

John Baez (Sep 30 2022 at 10:34):

Why would you want to bring Green's Theorem into 1-variable calculus?

Jan Pax (Sep 30 2022 at 16:46):

I was told that in the snippet below (1) is a consequence of Green's theorem as well as the computation of my example 1. It is just the previous page of my example (1) above. 111.jpg I do not understand Green's theorem much, anyway.