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 and find
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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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\int_{0}^{\infty} \frac{1}{1+x^2} \, dx = \frac{\pi}{2}
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$$
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<hr style="border:0;border-top:thin dashed #ccc;background:none;"/>
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Let \(a\in\mathbb{R}^3\) and \(r>0\).
  
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Claim:
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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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$$ \frac{d}{dt} \int_{0}^{\infty} \frac{\cos(xt)}{1+x^2} \, dx = \int_{0}^{\infty} \frac{d}{dt} \frac{\cos(xt)}{1+x^2} \, dx \, , t \gt 0$$
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Beachte: Wir benötigen partielle Integration, da das Integral sonst nicht Lebesgue-Integrable ist und daher das Theorem für dominierende Konvergenz versagt. (Siehe Diskussion).
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<hr style="border:0;border-top:thin dashed #ccc;background:none;"/>
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Proof:
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first make some definitions:
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Using the corollary:
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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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\(f(x,t) := \cos(xt) \) and \(g(x) := \frac{1}{1+x^2}\)
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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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$$\Rightarrow |\partial_i u(a)| \leq 3C$$  
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\( \partial^{-1}_x f(x) := \int_{0}^{x} f(x') \, dx' \)
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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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assume \(t \gt 0 \) and use partial integration (additional term goes to \( 0 \))
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$$
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\int_{0}^{\infty} f(x,t)g(x) \, dx = -\int_{0}^{\infty} \partial^{-1}_x f(x,t) \partial_x g(x) \, dx \\
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= \frac{2}{t} \int_{0}^{\infty} \frac{x \sin(xt)}{(1+x^2)^2} \, dx
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$$
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now we try to dervate this integral w.r.t \(t\)
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define \( G(x,t) := \sin(xt) \), fix \(t_0 \gt 0 \) and by mean value theorem for \( h \ne 0 \) we find \( \xi(x) \) between \( 0 \) and \( h \) s.t.:
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$$
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\left| \frac{G(x,t_0+h)-G(x,t_0)}{h} \right| = |\, G_t(x,t_0+\xi) \,| \le |x| \\
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\Rightarrow \bigg| \frac{G(x,t_0+h)-G(x,t_0)}{h} \frac{x}{(1+x^2)^2} \bigg| \le \frac{x^2}{(1+x^2)^2} \\
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d_n(x,t) := \frac{G(x,t+h_n)-G(x,t)}{h_n} \frac{x}{(1+x^2)^2}
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$$
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with arbitrary sequence \( h_n \ne 0 \) converging to \( 0 \) we have integrable \( d_n(x) \) dominated by an integrable function. Thus we use dominated convergence theorem:
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$$
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\Rightarrow \lim\limits_{n \rightarrow \infty} \int_{0}^{\infty} d_n(x,t) \, dx = \int_{0}^{\infty} \partial_t \frac{x \sin(xt)}{(1+x^2)^2} \, dx \\
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$$
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(Ich habe hier noch die Produktregel ausgeführt, würde aber nur den ersten und letzten Term aufschreiben und einfach erwähnen wie das folgt. Ist ja ziemlich offensichtlich)
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$$
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\Rightarrow \partial_t \frac{2}{t} \int_{0}^{\infty} \frac{x \sin(xt)}{(1+x^2)^2} \, dx = (\partial_t \frac{2}{t}) \int_{0}^{\infty} \frac{x \sin(xt)}{(1+x^2)^2} \, dx + \frac{2}{t} \int_{0}^{\infty} \partial_t \frac{x \sin(xt)}{(1+x^2)^2} \, dx = \int_{0}^{\infty} \bigg[(\partial_t \frac{2}{t}) \frac{x \sin(xt)}{(1+x^2)^2} + \frac{2}{t} \partial_t \frac{x \sin(xt)}{(1+x^2)^2}\bigg] \, dx = \int_{0}^{\infty} \partial_t \frac{2}{t} \frac{x \sin(xt)}{(1+x^2)^2} \, dx \\
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= -\int_{0}^{\infty} \partial_t (\partial^{-1}_x f(x,t) \partial_x g(x)) \, dx
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$$
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Use Leibniz integral rule and then do again partial Integration (Additional term goes to \( 0 \) )
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$$
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-\int_{0}^{\infty} \partial_t (\partial^{-1}_x f(x,t) \partial_x g(x)) \, dx = -\int_{0}^{\infty} (\partial^{-1}_x \partial_t  f(x,t)) \partial_x g(x) \, dx =  \int_{0}^{\infty} \partial_t(f(x,t)g(x)) \, dx
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$$
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Now we have proven the claim.
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use the proof for \( t>0 \)
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$$
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\begin{align}
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\ \frac{d}{dt} e^{-|t|} = \frac{d}{dt} \frac{2}{\pi} \int_{0}^{\infty} \frac{\cos(xt)}{1+x^2} \, dx \\
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\Leftrightarrow e^{-t} = \frac{2}{\pi} \int_{0}^{\infty} \frac{x \sin(xt)}{1+x^2} \, dx \\
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\Rightarrow \int_{0}^{\infty} \frac{x \sin(xt)}{1+x^2} \, dx = \frac{\pi}{2} e^{-t} \\
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\end{align}
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$$
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<hr style="border:0;border-top:thin dashed #ccc;background:none;"/>
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Alternativer Beweis (Dies ist nur eine Skizze. Ich bin mir da etwas unsicher(Siehe Diskussion für mehr Infos). Ich halte mich bisher an den obigen Beweis, der vom Aufwand her auch noch ok ist. Ev. kommt jemand anders hier noch auf eine einfachere Lösung, dann ergänze ich das noch)
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Consider $$
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\begin{align}
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f_n(t) = \frac{2}{\pi}\int\limits_0^n\frac{\cos{x t}}{1+x^2}\mathrm{d}x
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\end{align}
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$$
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with \(f_n(t)\) analytic.
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further $$
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\begin{align}
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f_n(t)\xrightarrow[n\rightarrow\infty]{glm.}\frac{2}{\pi}\int\limits_0^\infty\frac{\cos{x t}}{1+x^2}\mathrm{d}x=:f(t)
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\end{align}
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$$
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With
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$$
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\begin{align}
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\int\limits_0^\infty\frac{1}{1+x^2}\mathrm{d}x<\infty
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\end{align}
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$$
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it follows
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$$
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\begin{align}
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f'_n(t)\xrightarrow[n\rightarrow\infty]{glm.}f'(t)
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\end{align}
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$$
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This implys:\[
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\begin{align}
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-\frac{2}{\pi}\int\limits_0^\infty \frac{x\sin{xt}}{1+x^2}\mathrm{d}x = \frac{\mathrm{d}}{\mathrm{d}t}\frac{2}{\pi}\int\limits_0^\infty \frac{\cos{xt}}{1+x^2}\mathrm{d}x
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\end{align}
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\]
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And the claim is proven.
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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.
 +
<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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