Difference between revisions of "Aufgaben:Problem 11"

From Ferienserie MMP2
Jump to: navigation, search
(→‎Solution)
(redo add missing supremum)
 
(37 intermediate revisions by 3 users not shown)
Line 1: Line 1:
−
=Problem a)=
+
==Problem==
−
$$f(t)=e^{-|t|} \in L^1(\mathbb{R})$$
+
Let \(u\) be a harmonic function on \(\mathbb{R}^3\). Assume there exists \(C>0\), independent of \(x\), such that \(|u(x)| \leq C(1+|x|)\) on \(\mathbb{R}^3\). Show that then
  
−
Compute the Fourier transform of \(f(t)\)
+
<ol style="list-style-type:lower-latin">
 +
  <li>\(\partial_i u\) is constant on \(\mathbb{R}^3\), \(\forall i = 1,2,3\).</li>
 +
  <li>\(u\) is a linear function, i.e. \(\exists a_0, a_1, a_2, a_3 \in \mathbb{R}\) such that \(u(x) = a_0 + a_1 x_1 + a_2 x_2 + a_3 x_3.\)</li>
 +
</ol>
  
−
=Solution=
+
''NOTE:'' if you choose to use the Corollary on page 3 of ”Newtonian Potential” lecture
 +
notes, you have to give a proof of it.
  
−
$$
+
==Proof==
−
\begin{align}
+
−
\hat f(x) \ &= \frac{1}{\sqrt{2\pi}} \int_{-\infty}^{\infty} \mathrm e^{-|t|}e^{-ixt}\,\mathrm dt \\
+
−
&= \frac{1}{\sqrt{2\pi}} \int_{-\infty}^{0} \mathrm e^{t(1-ix)}\,\mathrm dt + \int_{0}^{\infty} \mathrm e^{-t(1+ix)}\,\mathrm dt \\
+
−
&= \frac{1}{\sqrt{2\pi}} \left[ \frac{1}{1-ix}e^{t(1-ix)} \bigg \vert_{t=-\infty}^{t=0} - \frac{1}{1+ix} e^{-t(1+ix)} \bigg \vert_{t=0}^{t=\infty} \right] \\
+
−
&= \frac{1}{\sqrt{2\pi}} \left[ \frac{2}{1+x^2} \right] \\
+
−
&= \sqrt{\frac{2}{\pi}} \frac{1}{1+x^2}
+
−
\end{align}
+
−
$$
+
  
−
=Problem b)=
+
<div style="border:1px solid #aaa; background: #f8f8f8; padding:5px">
 +
Aforementioned '''corollary''' from [https://people.math.ethz.ch/~gruppe5/group5/lectures/mmp/fs15/Files/NewtonianPotential.pdf NewtonianPotential.pdf], p.3:
  
−
Using the result from a), compute
+
Suppose \(u(x)\) is harmonic on the domain \(\Omega\) and the closed ball \(\overline{B_r(a)} \subset \Omega\), \(C_n=\frac{|S_1(0)|}{|B_1(0)|}\) and \(S_r(a) = \{x \in \mathbb{R}^n : |x-a|=r\}\), then
−
$$ \int_{0}^{\infty} \frac{1}{1+x^2} \,\mathrm dx $$
+
$$\left|\frac{\partial u}{\partial x_j}(a)\right| \leq \frac{C_n}{r} \sup_{y\in S_r(a)}|u(y)|$$
−
and
+
−
$$ \int_{0}^{\infty} \frac{x \sin(xt)}{1+x^2}\,\mathrm dx \ , t>0 $$
+
  
 +
'''Proof''' (p.3f):
  
−
=Solution=
+
\(0=\frac{\partial}{\partial x_j}(\Delta u) = \Delta \left(\frac{\partial}{\partial x_j} u\right)\) so that \(\frac{\partial}{\partial x_j}u\) is also harmonic on \(\Omega\). Apply the divergence theorem to the vector field \(V_j\) with components \((V_j)_k=u(x)\delta_{jk}\) to obtain
−
by using inverse fourier transform
+
 
$$
 
$$
−
\begin{align}
+
|B_r(0)|\: \left|\frac{\partial u}{\partial x_j}(a)\right|
−
e^{-|t|} &= \frac{1}{\sqrt{2\pi}} \int_{-\infty}^{\infty} \sqrt{\frac{2}{\pi}} \frac{1}{1+x^2} e^{ixt} \, dx \\
+
\overset{(*)}{=} \left| \int_{B_r(0)} \frac{\partial u}{\partial x_j} (x+a) \: \mathrm{d}x\right|
−
&= \frac{1}{\pi} \int_{-\infty}^{\infty} \frac{e^{ixt}}{1+x^2} \, dx \\
+
= \left| \int_{B_r(0)} (\mathrm{div}\: V_j)(x+a) \: \mathrm{d}x\right| \\
−
&= \frac{1}{\pi} \int_{0}^{\infty} \frac{e^{ixt} + e^{-ixt}}{1+x^2} \, dx \\
+
= \left| \int_{S_r(0)} \langle V_j(y+a), \frac{1}{r}y \rangle \: \mathrm{d}y\right|
−
&= \frac{2}{\pi} \int_{0}^{\infty} \frac{\cos(xt)}{1+x^2} \, dx
+
= \left| \int_{S_r(0)} u(y+a)\frac{1}{r}y_j \: \mathrm{d}y\right|
−
\end{align}  
+
\leq \int_{S_r(0)} |u(y+a)| \left| \frac{1}{r}y_j \right| \mathrm{d}y \\
 +
\leq \int_{S_r(0)} |u(y+a)| \: \mathrm{d}y
 +
\leq |S_r(0)| \sup_{y\in S_r(a)}|u(y)|
 +
= |S_1(0)|r^{n-1} \sup_{y\in S_r(a)}|u(y)|
 
$$
 
$$
 +
where \((*)\) follows by the ''Mean Value Property of Harmonic Functions'' (see [https://people.math.ethz.ch/~gruppe5/group5/lectures/mmp/fs15/Files/NewtonianPotential.pdf NewtonianPotential.pdf] p. 2).
  
−
thus we can set t=0 and find
+
Divide by \(|B_r(0)|=|B_1(0)|r^n\) and we're done.
 +
<p style="text-align:right">\(\square\)</p>
 +
</div>
  
−
$$
+
First, note that \(C_3=4\pi / \left( \frac{4}{3}\pi \right) = 3\)
−
\int_{0}^{\infty} \frac{1}{1+x^2} \, dx = \frac{\pi}{2}
+
−
$$
+
  
−
<hr style="border:0;border-top:thin dashed #ccc;background:none;"/>
+
Let \(a\in\mathbb{R}^3\) and \(r>0\).
  
−
Claim:
+
\(u\) is harmonic on all of \(\mathbb{R}^3\) and thus on \(\overline{B_r(a)}\), too.
−
$$ \frac{d}{dt} \int_{0}^{\infty} \frac{\cos(xt)}{1+x^2} \, dx = \int_{0}^{\infty} \frac{d}{dt} \frac{\cos(xt)}{1+x^2} \, dx $$
+
  
−
Wenn jemand einen einfacheren Weg weiss, gebt mir bescheid. Ohne partiell Integrieren klappt das Theorem für dominierende Konvergenz nicht.
+
Using the corollary:
−
<hr style="border:0;border-top:thin dashed #ccc;background:none;"/>
+
$$|\partial_i u(a)| \leq \frac{3}{r} \sup_{y\in S_r(a)}|u(y)| \leq \frac{3}{r}\sup_{y\in S_r(a)}C(1+|y|) \leq \frac{3}{r}C(1+|a|+r)=3C \left( \frac{1+|a|}{r} + 1 \right)$$
−
Proof:
+
Since \(u\) is harmonic on all of \(\mathbb{R}^3\), we can choose \(r\) arbitrarily large, so let \(r \rightarrow \infty\)
 +
$$\Rightarrow |\partial_i u(a)| \leq 3C$$
 +
We thus just showed that \(\partial_i u\) is bounded on \(\mathbb{R}^3\). As \(u\) is harmonic, so is \(\partial_i u\). Making use of Liouville's theorem (see [https://people.math.ethz.ch/~gruppe5/group5/lectures/mmp/fs15/Files/NewtonianPotential.pdf NewtonianPotential.pdf], p.4), claim a) follows directly.
  
−
first make some definitions:
+
Now, let \(a_i = \partial_i u = \mathrm{const.}\). We see that for \(f(x) := a_1 x_1 + a_2 x_2 + a_3 x_3\), \(\partial_i f = a_i = \partial_i u\) and since functions with identical partial derivatives can only differ by a constant, claim b) follows as well.
−
 
+
<p style="text-align:right">\(\square\)</p>
−
\(f(x,t) := \cos(xt) \) and \(g(x) := \frac{1}{1+x^2}\)
+
−
 
+
−
\( \partial^{-1}_x f(x) := \int_{0}^{x} f(x') \, dx' \)
+
−
 
+
−
use partial integration (additional term goes to \( 0 \))
+
−
$$
+
−
\int_{0}^{\infty} f(x,t)g(x) \, dx = -\int_{0}^{\infty} \partial^{-1}_x f(x,t) \partial_x g(x) \, dx \\
+
−
= \frac{2}{t} \int_{0}^{\infty} \frac{x \sin(xt)}{(1+x^2)^2} \, dx
+
−
$$
+
−
define \( G(x,t) := \sin(xt) \) and by mean value theorem for \( h \ne 0 \) we find \( \xi(x) \) between \( 0 \) and \( h \) s.t.:
+
−
$$
+
−
\frac{G(x,t+h)-G(x,t)}{h} = \partial_t G(x,t+\xi) \le |x| \\
+
−
\Rightarrow \bigg| \frac{G(x,t+h)-G(x,t)}{h} \frac{x}{(1+x^2)^2} \bigg| \le \frac{x^2}{(1+x^2)^2} \\
+
−
d_n(x,t) := \frac{G(x,t+h_n)-G(x,t)}{h_n} \frac{x}{(1+x^2)^2}
+
−
$$
+
−
with \( h_n \ne 0 \) converging to \( 0 \) we have integrable \( d_n(x) \) bounded by an integrable function
+
−
 
+
−
with dominated convergence theorem it follows that we can take the partial dervative inside the integral.
+
−
 
+
−
We do again partial integration
+
−
 
+
−
$$
+
−
-\int_{0}^{\infty} \partial_t (\partial^{-1}_x f(x,t) \partial_x g(x)) \, dx = \int_{0}^{\infty} \partial_t(f(x,t)g(x)) \, dx
+
−
$$
+
−
 
+
−
where the additional term from the partial integration goes to 0. Now we have proven the claim.
+
−
 
+
−
<hr style="border:0;border-top:thin dashed #ccc;background:none;"/>
+
−
 
+
−
use the proof for \( t>0 \)
+
−
$$
+
−
\begin{align}
+
−
\ \frac{d}{dt} e^{-|t|} = \frac{d}{dt} \frac{2}{\pi} \int_{0}^{\infty} \frac{\cos(xt)}{1+x^2} \, dx \\
+
−
\Leftrightarrow e^{-t} = \frac{2}{\pi} \int_{0}^{\infty} \frac{x \sin(xt)}{1+x^2} \, dx \\
+
−
\Rightarrow \int_{0}^{\infty} \frac{x \sin(xt)}{1+x^2} \, dx = \frac{\pi}{2} e^{-t} \\
+
−
\end{align}
+
−
$$
+

Latest revision as of 09:30, 30 July 2015

Problem

Let \(u\) be a harmonic function on \(\mathbb{R}^3\). Assume there exists \(C>0\), independent of \(x\), such that \(|u(x)| \leq C(1+|x|)\) on \(\mathbb{R}^3\). Show that then

  1. \(\partial_i u\) is constant on \(\mathbb{R}^3\), \(\forall i = 1,2,3\).
  2. \(u\) is a linear function, i.e. \(\exists a_0, a_1, a_2, a_3 \in \mathbb{R}\) such that \(u(x) = a_0 + a_1 x_1 + a_2 x_2 + a_3 x_3.\)

NOTE: if you choose to use the Corollary on page 3 of ”Newtonian Potential” lecture notes, you have to give a proof of it.

Proof

Aforementioned corollary from NewtonianPotential.pdf, p.3:

Suppose \(u(x)\) is harmonic on the domain \(\Omega\) and the closed ball \(\overline{B_r(a)} \subset \Omega\), \(C_n=\frac{|S_1(0)|}{|B_1(0)|}\) and \(S_r(a) = \{x \in \mathbb{R}^n : |x-a|=r\}\), then $$\left|\frac{\partial u}{\partial x_j}(a)\right| \leq \frac{C_n}{r} \sup_{y\in S_r(a)}|u(y)|$$

Proof (p.3f):

\(0=\frac{\partial}{\partial x_j}(\Delta u) = \Delta \left(\frac{\partial}{\partial x_j} u\right)\) so that \(\frac{\partial}{\partial x_j}u\) is also harmonic on \(\Omega\). Apply the divergence theorem to the vector field \(V_j\) with components \((V_j)_k=u(x)\delta_{jk}\) to obtain $$ |B_r(0)|\: \left|\frac{\partial u}{\partial x_j}(a)\right| \overset{(*)}{=} \left| \int_{B_r(0)} \frac{\partial u}{\partial x_j} (x+a) \: \mathrm{d}x\right| = \left| \int_{B_r(0)} (\mathrm{div}\: V_j)(x+a) \: \mathrm{d}x\right| \\ = \left| \int_{S_r(0)} \langle V_j(y+a), \frac{1}{r}y \rangle \: \mathrm{d}y\right| = \left| \int_{S_r(0)} u(y+a)\frac{1}{r}y_j \: \mathrm{d}y\right| \leq \int_{S_r(0)} |u(y+a)| \left| \frac{1}{r}y_j \right| \mathrm{d}y \\ \leq \int_{S_r(0)} |u(y+a)| \: \mathrm{d}y \leq |S_r(0)| \sup_{y\in S_r(a)}|u(y)| = |S_1(0)|r^{n-1} \sup_{y\in S_r(a)}|u(y)| $$ where \((*)\) follows by the Mean Value Property of Harmonic Functions (see NewtonianPotential.pdf p. 2).

Divide by \(|B_r(0)|=|B_1(0)|r^n\) and we're done.

\(\square\)

First, note that \(C_3=4\pi / \left( \frac{4}{3}\pi \right) = 3\)

Let \(a\in\mathbb{R}^3\) and \(r>0\).

\(u\) is harmonic on all of \(\mathbb{R}^3\) and thus on \(\overline{B_r(a)}\), too.

Using the corollary: $$|\partial_i u(a)| \leq \frac{3}{r} \sup_{y\in S_r(a)}|u(y)| \leq \frac{3}{r}\sup_{y\in S_r(a)}C(1+|y|) \leq \frac{3}{r}C(1+|a|+r)=3C \left( \frac{1+|a|}{r} + 1 \right)$$ Since \(u\) is harmonic on all of \(\mathbb{R}^3\), we can choose \(r\) arbitrarily large, so let \(r \rightarrow \infty\) $$\Rightarrow |\partial_i u(a)| \leq 3C$$ We thus just showed that \(\partial_i u\) is bounded on \(\mathbb{R}^3\). As \(u\) is harmonic, so is \(\partial_i u\). Making use of Liouville's theorem (see NewtonianPotential.pdf, p.4), claim a) follows directly.

Now, let \(a_i = \partial_i u = \mathrm{const.}\). We see that for \(f(x) := a_1 x_1 + a_2 x_2 + a_3 x_3\), \(\partial_i f = a_i = \partial_i u\) and since functions with identical partial derivatives can only differ by a constant, claim b) follows as well.

\(\square\)

Retrieved from "http://ferienserie.ch/index.php?title=Aufgaben:Problem_11&oldid=1566"