Can I get a little math help in here? - GoFuckYourself.com

Malicious Biz · 05-19-2006, 06:34 PM

Xkdjdskjsdkjsfkjsfkjsf

In class recently, my professor gave a proof that any constant coefficient partial differential equation has a weak solution on a bounded subset of $R^n$. That is, if
$|\alpha| = \alpha_1 + \alpha_2 + ... + \alpha_n$
and
$L = \sum_{|\alpha|<k}\frac{\partial^{\alpha_1}}{\parti al x_1}\frac{\partial^{\alpha_2}}{\partial x_2}...\frac{\partial^{\alpha_n}}{\partial x_n}$
then for any bounded $\Omega \subset R^d$ and $f \in L^2(\Omega)$, there exists some $u \in L^2(\Omega)$ that is a solution to the PDE $Lu = f$ in the weak sense. In fact, there is a bounded operator $K$ such that $LKf = f$ in the weak sense for any $f$.

Now, I'm going to outline the proof and point out what I don't understand and why, and hopefully someone can help me.

$\textbf{Lemma: }$ $\exists c \in R$ such that $\varphi \in C^\infty_0(\Omega) \Rightarrow \|\varphi\|_{L^2(\Omega)} \le c\|L^*\varphi\|_{L^2(\Omega)}$.

That is, for any infinitely differentiable function of compact support, the norm of the image of the function under the adjoint of $L$ is bounded below by a constant multiple of the norm of the original function, and this constant depends only on the space $\Omega$.

Thing I Don't Understand #1: Why is this the most difficult part of the theorem? (We'll get into more detail with this later when I return to prove the lemma.)

Now, let's assume the lemma for now. It seems that part of the purpose of this lemma is to demonstrate that $L^*$ is one-to-one, and therefore $L$ is onto. However, I have no idea why it is important that we know this. That's Thing I Don't Understand #2.

So we consider the pre-Hilbert space $H_0$ which is the space of all functions that are $C^\infty_0(\Omega)$, and let $H$ be the closure of that space. Take the norm to be $\|f\|_0 = \|L^*f\|_{L^2(\Omega)}$ (notice that $L^*: H \rightarrow L^2(\Omega)$.

So we consider the linear functional $l: \varphi \rightarrow (\varphi, f)$ for some fixed $f$ and $\varphi \in H_0$ (the usual $L^2$ inner product, I believe). Then

$|(\varphi, f)| \le \|\phi\|_{L^2(\Omega)}\|f\|_{L^2(\Omega)} \le c\|L^*\varphi\|_{L^2(\Omega)}\|f\|_{L^2(\Omega)} = c\|\varphi\|_0\|f\|_{L^2(\Omega)}$.

I don't get the point of this, though I can see why it is true (Cauchy-Schwartz and the Lemma). I also don't get what follows, though I do know the Riesz representation theorem. The claim is that there exists some $U \in H$ such that $l(\varphi) = (L^*\varphi, L^*U)$.

Then we pick $u = L^*U$, and thus $(\varphi, f) = (L^*\varphi, u)$, so $u$ is the weak solution we wanted.

Now, we also need to prove that lemma. Observe that if $(u, L^*\varphi) = (f, \varphi)$ for all $\varphi \in C^\infty_0(\Omega)$, then we can use the Plancherel identity to see that $(\widehat{u}, \widehat{L^*\varphi}) = (\widehat{f}, \widehat{\varphi})$. But derivatives get turned into polynomials, so what we'd like to see is that $\|\widehat{\varphi}\| \le c\|P\widehat{\varphi}\|$ where $P$ is the polynomial corresponding to $L^*$. (Stein calls it the characteristic polynomial; is this really the characteristic polynomial of that linear operator, by the standard lin alg definition?)

Anyway, we now have two sub-lemmas.

Lemma 1a: suppose $\Omega = (-M, M)$ (a subset of the real line) and $g \in L^2(R)$ is supported on $[-M, M]$. Then
$\widehat{g}(\xi) = \int_{-M}^M g(x) e^{-2\pi i\xi x}dx$
can be extended to an analytic function on the complex plane, and furthermore
$\int |\widehat{g}(\xi + i\eta)|^2d\xi \le e^{4\pi M|\eta|}\|g\|^2_{L^2}$.

Lemma 1b: If $F$ is holomorphic on the complex plane, then for any polynomial $Q(z) = z^m + a_{m-1}z^{m-1} + ... + a_1z + a_0$, we have

$|F(0)|^2 \le \frac{1}{2\pi}\int_0^{2\pi}|Q(e^{i\theta})F(e^{i\t heta})|^2d\theta$.

If we take $Q = P$, $F(z) = \widehat{g}(\xi + i\eta)$, $\eta = \sin(\phi)$, shift $\xi$ by $\cos(\phi)$, and finally apply Lemma 1b to the result of Lemma 1a, then we have the result of the original Lemma 1.

Now, there was a lot of information and very little proof near the end there. I'm not even certain everything is right as I don't have any intuition for functions on the complex numbers, so I can't check this very well. Lemma 1a seems reasonable to me, but I have no idea about the reason for 1b, except that the professor was saying some vague thing about the polynomial not causing any harm when $|z| = 1$ and this is just like the much more basic averaging result about holomorphic functions.

So, I'm not looking for a full explanation of this theorem, because it is pretty long and hardcore, but if there's any particular part of my confusion that you think you might be able to address, that would be great. Thanks!

GatorB · 05-19-2006, 06:35 PM

the answer is 7

Evil Doer · 05-19-2006, 06:40 PM

why are people just posting shit from google?

http://www.artofproblemsolving.com/F...st-498561.html

BigCashCrew · 05-19-2006, 06:40 PM

smoke some meth and figure it out

Malicious Biz · 05-19-2006, 06:41 PM

Quote:

Originally Posted by BigCashCrew

smoke some meth and figure it out

RayBonga · 05-19-2006, 06:42 PM

You are so smart!!!!!!!!!!!!!!!!!!!!!!!

I love you!!!!!!!!!!!!!!!!!!!!!!!

Malicious Biz · 05-19-2006, 06:43 PM

Quote:

Originally Posted by RayBonga

You are so smart!!!!!!!!!!!!!!!!!!!!!!!

I love you!!!!!!!!!!!!!!!!!!!!!!!

I LOVE YOU TOO, LETS GET MARRIED

BoyAlley · 05-19-2006, 06:45 PM

Quote:

Originally Posted by Evil Doer

why are people just posting shit from google?

http://www.artofproblemsolving.com/F...st-498561.html

Cuz they think it make dem looks smart!

Evil Doer · 05-19-2006, 06:46 PM

Quote:

Originally Posted by BoyAlley

Cuz they think it make dem looks smart!

too bad it's not helping

GatorB · 05-19-2006, 06:48 PM

Did you ask for MATH help or METH help?

Malicious Biz · 05-19-2006, 06:56 PM

Quote:

Originally Posted by GatorB

Did you ask for MATH help or METH help?

THE MATH HELP I CAN HANDLE MY METH JUST FINE THANKS FOR ASKING ANYWAY OK

2HousePlague · 05-19-2006, 06:59 PM

The pre-Hilbert Space is awesome --

2hp

biftek · 05-19-2006, 07:10 PM

as has already been shown that this is just a straight cut and paste
but why the fuck come to a fucking adult webmaster forum and ask for fucking maths fucking help

but i suppose it was a creative way to boost the spam count

OzMan · 05-19-2006, 07:13 PM

Apply the Gram-Schmidt Orthogonalization Procedure to the sequence and er Bob's yer uncle!

MaddCaz · 05-19-2006, 07:18 PM

im so fonfused...

MaddCaz · 05-19-2006, 07:19 PM

x+y=z squared

Linkster · 05-19-2006, 07:22 PM

Actually the answer is pretty clear - there is a theory of complex analysis in more than one complex dimension where the analytic properties such as power series expansion still remain true whereas most of the geometric properties of functions in one complex dimension such as youve demonstrated above are no longer true - need to change your theory a little to get to the correct approach

05-19-2006, 06:40 PM	#3
Evil Doer Confirmed User Join Date: Dec 2004 Location: ICQ: 251-911-362 Posts: 915	why are people just posting shit from google? http://www.artofproblemsolving.com/F...st-498561.html __________________ see sig above mine

05-19-2006, 06:59 PM	#12
2HousePlague CURATOR Join Date: Jul 2004 Location: the attic Posts: 14,572	The pre-Hilbert Space is awesome -- 2hp __________________ tada!

05-19-2006, 07:18 PM	#15
MaddCaz Confirmed User Join Date: Mar 2006 Location: Illinois Posts: 9,483	im so fonfused... __________________ BigCocks.com - MatureWomen.com - Tranny.com - DrunkGirls.com - TeenGirls.com - MonsterCock.com and many more... Click here to see them all!

05-19-2006, 07:19 PM	#16
MaddCaz Confirmed User Join Date: Mar 2006 Location: Illinois Posts: 9,483	x+y=z squared __________________ BigCocks.com - MatureWomen.com - Tranny.com - DrunkGirls.com - TeenGirls.com - MonsterCock.com and many more... Click here to see them all!

05-19-2006, 07:22 PM	#17
Linkster Confirmed User Join Date: Feb 2003 Location: DeltaHell Posts: 3,216	Actually the answer is pretty clear - there is a theory of complex analysis in more than one complex dimension where the analytic properties such as power series expansion still remain true whereas most of the geometric properties of functions in one complex dimension such as youve demonstrated above are no longer true - need to change your theory a little to get to the correct approach __________________ Porn Links Crak for your bank account - Get Addicted FREE LIVE CAM SHOWS

05-19-2006, 06:34 PM	#1
Malicious Biz So Fucking Banned Join Date: Mar 2005 Location: Put Shoe On Head Plz Posts: 4,575	Can I get a little math help in here? Xkdjdskjsdkjsfkjsfkjsf In class recently, my professor gave a proof that any constant coefficient partial differential equation has a weak solution on a bounded subset of $R^n$. That is, if $\|\alpha\| = \alpha_1 + \alpha_2 + ... + \alpha_n$ and $L = \sum_{\|\alpha\|<k}\frac{\partial^{\alpha_1}}{\parti al x_1}\frac{\partial^{\alpha_2}}{\partial x_2}...\frac{\partial^{\alpha_n}}{\partial x_n}$ then for any bounded $\Omega \subset R^d$ and $f \in L^2(\Omega)$, there exists some $u \in L^2(\Omega)$ that is a solution to the PDE $Lu = f$ in the weak sense. In fact, there is a bounded operator $K$ such that $LKf = f$ in the weak sense for any $f$. Now, I'm going to outline the proof and point out what I don't understand and why, and hopefully someone can help me. $\textbf{Lemma: }$ $\exists c \in R$ such that $\varphi \in C^\infty_0(\Omega) \Rightarrow \\|\varphi\\|_{L^2(\Omega)} \le c\\|L^\varphi\\|_{L^2(\Omega)}$. That is, for any infinitely differentiable function of compact support, the norm of the image of the function under the adjoint of $L$ is bounded below by a constant multiple of the norm of the original function, and this constant depends only on the space $\Omega$. Thing I Don't Understand #1: Why is this the most difficult part of the theorem? (We'll get into more detail with this later when I return to prove the lemma.) Now, let's assume the lemma for now. It seems that part of the purpose of this lemma is to demonstrate that $L^$ is one-to-one, and therefore $L$ is onto. However, I have no idea why it is important that we know this. That's Thing I Don't Understand #2. So we consider the pre-Hilbert space $H_0$ which is the space of all functions that are $C^\infty_0(\Omega)$, and let $H$ be the closure of that space. Take the norm to be $\\|f\\|_0 = \\|L^f\\|_{L^2(\Omega)}$ (notice that $L^: H \rightarrow L^2(\Omega)$. So we consider the linear functional $l: \varphi \rightarrow (\varphi, f)$ for some fixed $f$ and $\varphi \in H_0$ (the usual $L^2$ inner product, I believe). Then $\|(\varphi, f)\| \le \\|\phi\\|_{L^2(\Omega)}\\|f\\|_{L^2(\Omega)} \le c\\|L^\varphi\\|_{L^2(\Omega)}\\|f\\|_{L^2(\Omega)} = c\\|\varphi\\|_0\\|f\\|_{L^2(\Omega)}$. I don't get the point of this, though I can see why it is true (Cauchy-Schwartz and the Lemma). I also don't get what follows, though I do know the Riesz representation theorem. The claim is that there exists some $U \in H$ such that $l(\varphi) = (L^\varphi, L^U)$. Then we pick $u = L^U$, and thus $(\varphi, f) = (L^\varphi, u)$, so $u$ is the weak solution we wanted. Now, we also need to prove that lemma. Observe that if $(u, L^\varphi) = (f, \varphi)$ for all $\varphi \in C^\infty_0(\Omega)$, then we can use the Plancherel identity to see that $(\widehat{u}, \widehat{L^\varphi}) = (\widehat{f}, \widehat{\varphi})$. But derivatives get turned into polynomials, so what we'd like to see is that $\\|\widehat{\varphi}\\| \le c\\|P\widehat{\varphi}\\|$ where $P$ is the polynomial corresponding to $L^$. (Stein calls it the characteristic polynomial; is this really the characteristic polynomial of that linear operator, by the standard lin alg definition?) Anyway, we now have two sub-lemmas. Lemma 1a: suppose $\Omega = (-M, M)$ (a subset of the real line) and $g \in L^2(R)$ is supported on $[-M, M]$. Then $\widehat{g}(\xi) = \int_{-M}^M g(x) e^{-2\pi i\xi x}dx$ can be extended to an analytic function on the complex plane, and furthermore $\int \|\widehat{g}(\xi + i\eta)\|^2d\xi \le e^{4\pi M\|\eta\|}\\|g\\|^2_{L^2}$. Lemma 1b: If $F$ is holomorphic on the complex plane, then for any polynomial $Q(z) = z^m + a_{m-1}z^{m-1} + ... + a_1z + a_0$, we have $\|F(0)\|^2 \le \frac{1}{2\pi}\int_0^{2\pi}\|Q(e^{i\theta})F(e^{i\t heta})\|^2d\theta$. If we take $Q = P$, $F(z) = \widehat{g}(\xi + i\eta)$, $\eta = \sin(\phi)$, shift $\xi$ by $\cos(\phi)$, and finally apply Lemma 1b to the result of Lemma 1a, then we have the result of the original Lemma 1. Now, there was a lot of information and very little proof near the end there. I'm not even certain everything is right as I don't have any intuition for functions on the complex numbers, so I can't check this very well. Lemma 1a seems reasonable to me, but I have no idea about the reason for 1b, except that the professor was saying some vague thing about the polynomial not causing any harm when $\|z\| = 1$ and this is just like the much more basic averaging result about holomorphic functions. So, I'm not looking for a full explanation of this theorem, because it is pretty long and hardcore, but if there's any particular part of my confusion that you think you might be able to address, that would be great. Thanks!

05-19-2006, 06:35 PM	#2
GatorB The Demon & 12clicks Industry Role: Join Date: Oct 2001 Location: SallyRand is a FAGGOT Posts: 18,208	the answer is 7

05-19-2006, 06:40 PM	#4
BigCashCrew Registered User Join Date: Aug 2005 Posts: 3,570	smoke some meth and figure it out

05-19-2006, 06:42 PM	#6
RayBonga too cool for highschool Join Date: Nov 2005 Location: East side, West side, Worldwide! Posts: 12,164	You are so smart!!!!!!!!!!!!!!!!!!!!!!! I love you!!!!!!!!!!!!!!!!!!!!!!!

05-19-2006, 06:48 PM	#10
GatorB The Demon & 12clicks Industry Role: Join Date: Oct 2001 Location: SallyRand is a FAGGOT Posts: 18,208	Did you ask for MATH help or METH help?

05-19-2006, 07:10 PM	#13
biftek So Fucking Banned Join Date: Jan 2005 Location: Victoria, Australia Posts: 1,030	as has already been shown that this is just a straight cut and paste but why the fuck come to a fucking adult webmaster forum and ask for fucking maths fucking help but i suppose it was a creative way to boost the spam count

05-19-2006, 07:13 PM	#14
OzMan Confirmed User Join Date: Sep 2003 Location: Los Begas Posts: 9,162	Apply the Gram-Schmidt Orthogonalization Procedure to the sequence and er Bob's yer uncle!