Difference between revisions of "Aufgaben:Problem 11"

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=Problem a)=
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==Problem==
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$$f(t)=e^{-|t|} \in L^1(\mathbb{R})$$
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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
  
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Compute the Fourier transform of \(f(t)\)
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<ol style="list-style-type:lower-latin">
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  <li>\(\partial_i u\) is constant on \(\mathbb{R}^3\), \(\forall i = 1,2,3\).</li>
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  <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>
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</ol>
  
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=Solution=
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''NOTE:'' if you choose to use the Corollary on page 3 of ”Newtonian Potential” lecture
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notes, you have to give a proof of it.
  
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$$
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==Proof==
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\begin{align}
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\hat f(x) \ &= \frac{1}{\sqrt{2\pi}} \int_{-\infty}^{\infty} \mathrm e^{-|t|}e^{-ixt}\,\mathrm dt \\
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&= \frac{1}{\sqrt{2\pi}} \int_{-\infty}^{0} \mathrm e^{t(1-ix)}\,\mathrm dt + \int_{0}^{\infty} \mathrm e^{-t(1+ix)}\,\mathrm dt \\
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&= \frac{1}{\sqrt{2\pi}} \left[ \frac{1}{1-ix}e^{t(1-ix)} \bigg \vert_{t=-\infty}^{t=0} - \frac{1}{1+ix} e^{-t(1+ix)} \bigg \vert_{t=0}^{t=\infty} \right] \\
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&= \frac{1}{\sqrt{2\pi}} \left[ \frac{2}{1+x^2} \right] \\
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&= \sqrt{\frac{2}{\pi}} \frac{1}{1+x^2}
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\end{align}
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$$
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=Problem b)=
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<div style="border:1px solid #aaa; background: #f8f8f8; padding:5px">
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Aforementioned '''corollary''' from [https://people.math.ethz.ch/~gruppe5/group5/lectures/mmp/fs15/Files/NewtonianPotential.pdf NewtonianPotential.pdf], p.3:
  
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Using the result from a), compute
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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
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$$ \int_{0}^{\infty} \frac{1}{1+x^2} \,\mathrm dx $$
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$$\left|\frac{\partial u}{\partial x_j}(a)\right| \leq \frac{C_n}{r} \sup_{y\in S_r(a)}|u(y)|$$
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and
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$$ \int_{0}^{\infty} \frac{x \sin(xt)}{1+x^2}\,\mathrm dx \ , t>0 $$
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'''Proof''' (p.3f):
  
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=Solution=
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\(0=\frac{\partial}{\partial x_j}(\Delta u) = \Delta \left(\frac{\partial}{\partial x_j} u\right)\) 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
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by using inverse fourier transform
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$$
 
$$
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\begin{align}
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|B_r(0)|\: \left|\frac{\partial u}{\partial x_j}(a)\right|
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e^{-|t|} &= \frac{1}{\sqrt{2\pi}} \int_{-\infty}^{\infty} \sqrt{\frac{2}{\pi}} \frac{1}{1+x^2} e^{ixt} \, dx \\
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\overset{(*)}{=} \left| \int_{B_r(0)} \frac{\partial u}{\partial x_j} (x+a) \: \mathrm{d}x\right|
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&= \frac{1}{\pi} \int_{-\infty}^{\infty} \frac{e^{ixt}}{1+x^2} \, dx \\
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= \left| \int_{B_r(0)} (\mathrm{div}\: V_j)(x+a) \: \mathrm{d}x\right| \\
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&= \frac{1}{\pi} \int_{0}^{\infty} \frac{e^{ixt} + e^{-ixt}}{1+x^2} \, dx \\
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= \left| \int_{S_r(0)} \langle V_j(y+a), \frac{1}{r}y \rangle \: \mathrm{d}y\right|
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&= \frac{2}{\pi} \int_{0}^{\infty} \frac{\cos(xt)}{1+x^2} \, dx
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= \left| \int_{S_r(0)} u(y+a)\frac{1}{r}y_j \: \mathrm{d}y\right|
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\end{align}  
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\leq \int_{S_r(0)} |u(y+a)| \left| \frac{1}{r}y_j \right| \mathrm{d}y \\
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\leq \int_{S_r(0)} |u(y+a)| \: \mathrm{d}y
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\leq |S_r(0)| \sup_{y\in S_r(a)}|u(y)|
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= |S_1(0)|r^{n-1} \sup_{y\in S_r(a)}|u(y)|
 
$$
 
$$
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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).
  
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thus we can set t=0
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Divide by \(|B_r(0)|=|B_1(0)|r^n\) and we're done.
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<p style="text-align:right">\(\square\)</p>
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</div>
  
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$$
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First, note that \(C_3=4\pi / \left( \frac{4}{3}\pi \right) = 3\)
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\begin{align}
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1 = \frac{2}{\pi} \int_{0}^{\infty} \frac{1}{1+x^2} \, dx \\
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\Rightarrow \int_{0}^{\infty} \frac{1}{1+x^2} \, dx = \frac{\pi}{2}
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\end{align}
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$$
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Claim:
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Let \(a\in\mathbb{R}^3\) and \(r>0\).  
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$$ \frac{d}{dt} \int_{0}^{\infty} \frac{\cos(xt)}{1+x^2} \, dx = \int_{0}^{\infty} \frac{-x \sin(xt)}{1+x^2} \, dx , t \gt 0 $$
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Proof:
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Dieser Beweis ist nicht ganz trivial. Gebt mir bescheid wenn ihr eine Idee habt. Eventuell gibt es die Möglichkeit die Fouriertransformierte der Ableitung zu verwenden und dann darauf die inverse Fouriertrafo anzuwenden. Habe aber noch ein Vorzeichenproblem, welches ich mir noch nicht erklären kann.
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use the proof for \( t>0 \)
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\(u\) is harmonic on all of \(\mathbb{R}^3\) and thus on \(\overline{B_r(a)}\), too.
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$$
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\begin{align}
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Using the corollary:
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\ \frac{d}{dt} e^{-|t|} = \frac{d}{dt} \frac{2}{\pi} \int_{0}^{\infty} \frac{\cos(xt)}{1+x^2} \, dx \\
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$$|\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)$$
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\Leftrightarrow e^{-t} = \frac{2}{\pi} \int_{0}^{\infty} \frac{x \sin(xt)}{1+x^2} \, dx \\
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Since \(u\) is harmonic on all of \(\mathbb{R}^3\), we can choose \(r\) arbitrarily large, so let \(r \rightarrow \infty\)
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\Rightarrow \int_{0}^{\infty} \frac{x \sin(xt)}{1+x^2} \, dx = \frac{\pi}{2} e^{-t} \\
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$$\Rightarrow |\partial_i u(a)| \leq 3C$$
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\end{align}
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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.
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$$
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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.
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<p style="text-align:right">\(\square\)</p>

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.

\(\square\)

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