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| − | =Problem a)= | + | ==Problem== |
| − | $$f(t)=e^{-|t|} \in L^1(\mathbb{R})$$
| + | Let \(u\) be a harmonic function on \(\mathbb{R}^3\). Assusme 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> |
| − | =Solution=
| + | <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> |
| − | $$
| + | |
| − | \begin{align}
| + | |
| − | \hat f(x) \ &= \frac{1}{\sqrt{2\pi}} \int_{-\infty}^{\infty} \mathrm e^{-|t|}e^{-ixt}\,\mathrm dt \\
| + | |
| − | &= \frac{1}{\sqrt{2\pi}} \int_{-\infty}^{0} \mathrm e^{t(1-ix)}\,\mathrm dt + \int_{0}^{\infty} \mathrm e^{-t(1+ix)}\,\mathrm dt \\
| + | |
| − | &= \frac{1}{\sqrt{2\pi}} \left[ \frac{1}{1-ix}e^{t(1-ix)} \bigg \vert_{t=-\infty}^{t=0} - \frac{1}{1+ix} e^{-t(1+ix)} \bigg \vert_{t=0}^{t=\infty} \right] \\
| + | |
| − | &= \frac{1}{\sqrt{2\pi}} \left[ \frac{2}{1+x^2} \right] \\
| + | |
| − | &= \sqrt{\frac{2}{\pi}} \frac{1}{1+x^2}
| + | |
| − | \end{align}
| + | |
| − | $$
| + | |
| − | | + | |
| − | =Problem b)=
| + | |
| − | | + | |
| − | Using the result from a), compute
| + | |
| − | $$ \int_{0}^{\infty} \frac{1}{1+x^2} \,\mathrm dx $$
| + | |
| − | and
| + | |
| − | $$ \int_{0}^{\infty} \frac{x \sin(xt)}{1+x^2}\,\mathrm dx \ , t>0 $$
| + | |
| − | | + | |
| − | | + | |
| − | =Solution=
| + | |
| − | by using inverse fourier transform
| + | |
| − | $$
| + | |
| − | \begin{align}
| + | |
| − | e^{-|t|} &= \frac{1}{\sqrt{2\pi}} \int_{-\infty}^{\infty} \sqrt{\frac{2}{\pi}} \frac{1}{1+x^2} e^{ixt} \, dx \\
| + | |
| − | &= \frac{1}{\pi} \int_{-\infty}^{\infty} \frac{e^{ixt}}{1+x^2} \, dx \\
| + | |
| − | &= \frac{1}{\pi} \int_{0}^{\infty} \frac{e^{ixt} + e^{-ixt}}{1+x^2} \, dx \\
| + | |
| − | &= \frac{2}{\pi} \int_{0}^{\infty} \frac{\cos(xt)}{1+x^2} \, dx
| + | |
| − | \end{align}
| + | |
| − | $$
| + | |
| − | | + | |
| − | thus we can set t=0 and find
| + | |
| − | | + | |
| − | $$
| + | |
| − | \int_{0}^{\infty} \frac{1}{1+x^2} \, dx = \frac{\pi}{2}
| + | |
| − | $$
| + | |
| − | | + | |
| − | <hr style="border:0;border-top:thin dashed #ccc;background:none;"/> | + | |
| − | | + | |
| − | Claim:
| + | |
| − | $$ \frac{d}{dt} \int_{0}^{\infty} \frac{\cos(xt)}{1+x^2} \, dx = \int_{0}^{\infty} \frac{d}{dt} \frac{\cos(xt)}{1+x^2} \, dx \, , t \gt 0$$
| + | |
| − | Beachte: Wir benötigen partielle Integration, da das Integral sonst nicht Lebesgue-Integrable ist und daher das Theorem für dominierende Konvergenz versagt. (Siehe Diskussion).
| + | |
| − | <hr style="border:0;border-top:thin dashed #ccc;background:none;"/> | + | |
| − | Proof:
| + | |
| − | | + | |
| − | first make some definitions:
| + | |
| − | | + | |
| − | \(f(x,t) := \cos(xt) \) and \(g(x) := \frac{1}{1+x^2}\) | + | |
| − | | + | |
| − | \( \partial^{-1}_x f(x) := \int_{0}^{x} f(x') \, dx' \)
| + | |
| − | | + | |
| − | assume \(t \gt 0 \) and use partial integration (additional term goes to \( 0 \))
| + | |
| − | $$
| + | |
| − | \int_{0}^{\infty} f(x,t)g(x) \, dx = -\int_{0}^{\infty} \partial^{-1}_x f(x,t) \partial_x g(x) \, dx \\
| + | |
| − | = \frac{2}{t} \int_{0}^{\infty} \frac{x \sin(xt)}{(1+x^2)^2} \, dx
| + | |
| − | $$
| + | |
| − | | + | |
| − | now we try to dervate this integral w.r.t \(t\)
| + | |
| − | | + | |
| − | 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.:
| + | |
| − | $$
| + | |
| − | \left| \frac{G(x,t_0+h)-G(x,t_0)}{h} \right| = |\, G_t(x,t_0+\xi) \,| \le |x| \\ | + | |
| − | \Rightarrow \bigg| \frac{G(x,t_0+h)-G(x,t_0)}{h} \frac{x}{(1+x^2)^2} \bigg| \le \frac{x^2}{(1+x^2)^2} \\
| + | |
| − | d_n(x,t) := \frac{G(x,t+h_n)-G(x,t)}{h_n} \frac{x}{(1+x^2)^2}
| + | |
| − | $$
| + | |
| − | 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:
| + | |
| − | | + | |
| − | $$
| + | |
| − | \Rightarrow \lim\limits_{n \rightarrow \infty} \int_{0}^{\infty} d_n(x,t) \, dx = \int_{0}^{\infty} \partial_t \frac{x \sin(xt)}{(1+x^2)^2} \, dx \\
| + | |
| − | $$
| + | |
| − | (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)
| + | |
| − | $$
| + | |
| − | \Rightarrow \partial_t \frac{2}{t} \int_{0}^{\infty} \frac{x \sin(xt)}{(1+x^2)^2} \, dx = (\partial_t \frac{2}{t}) \int_{0}^{\infty} \frac{x \sin(xt)}{(1+x^2)^2} \, dx + \frac{2}{t} \int_{0}^{\infty} \partial_t \frac{x \sin(xt)}{(1+x^2)^2} \, dx = \int_{0}^{\infty} \bigg[(\partial_t \frac{2}{t}) \frac{x \sin(xt)}{(1+x^2)^2} + \frac{2}{t} \partial_t \frac{x \sin(xt)}{(1+x^2)^2}\bigg] \, dx = \int_{0}^{\infty} \partial_t \frac{2}{t} \frac{x \sin(xt)}{(1+x^2)^2} \, dx \\ | + | |
| − | = -\int_{0}^{\infty} \partial_t (\partial^{-1}_x f(x,t) \partial_x g(x)) \, dx
| + | |
| − | $$
| + | |
| − | | + | |
| − | | + | |
| − | Use Leibniz integral rule and then do again partial Integration (Additional term goes to \( 0 \) )
| + | |
| − | | + | |
| − | $$
| + | |
| − | -\int_{0}^{\infty} \partial_t (\partial^{-1}_x f(x,t) \partial_x g(x)) \, dx = -\int_{0}^{\infty} (\partial^{-1}_x \partial_t f(x,t)) \partial_x g(x) \, dx = \int_{0}^{\infty} \partial_t(f(x,t)g(x)) \, dx
| + | |
| − | $$
| + | |
| − | (Zum Verständnis: Die Ableitung ziehen wir nun vom ersten bis zum letzten Integral durch)
| + | |
| − | | + | |
| − | Now we have proven the claim.
| + | |
| − | | + | |
| − | use the proof for \( t>0 \)
| + | |
| − | $$
| + | |
| − | \begin{align}
| + | |
| − | \ \frac{d}{dt} e^{-|t|} = \frac{d}{dt} \frac{2}{\pi} \int_{0}^{\infty} \frac{\cos(xt)}{1+x^2} \, dx \\
| + | |
| − | \Leftrightarrow e^{-t} = \frac{2}{\pi} \int_{0}^{\infty} \frac{x \sin(xt)}{1+x^2} \, dx \\
| + | |
| − | \Rightarrow \int_{0}^{\infty} \frac{x \sin(xt)}{1+x^2} \, dx = \frac{\pi}{2} e^{-t} \\
| + | |
| − | \end{align}
| + | |
| − | $$
| + | |
| − | | + | |
| − | | + | |
| − | <hr style="border:0;border-top:thin dashed #ccc;background:none;"/> | + | |
| − | 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)
| + | |
| − | | + | |
| − | Consider $$
| + | |
| − | \begin{align}
| + | |
| − | f_n(t) = \frac{2}{\pi}\int\limits_0^n\frac{\cos{x t}}{1+x^2}\mathrm{d}x
| + | |
| − | \end{align}
| + | |
| − | $$
| + | |
| − | | + | |
| − | with \(f_n(t)\) analytic.
| + | |
| − | | + | |
| − | further $$
| + | |
| − | \begin{align}
| + | |
| − | f_n(t)\xrightarrow[n\rightarrow\infty]{glm.}\frac{2}{\pi}\int\limits_0^\infty\frac{\cos{x t}}{1+x^2}\mathrm{d}x=:f(t)
| + | |
| − | \end{align}
| + | |
| − | $$
| + | |
| − | | + | |
| − | With
| + | |
| − | $$
| + | |
| − | \begin{align}
| + | |
| − | \int\limits_0^\infty\frac{1}{1+x^2}\mathrm{d}x<\infty
| + | |
| − | \end{align}
| + | |
| − | $$
| + | |
| − | it follows
| + | |
| − | $$
| + | |
| − | \begin{align}
| + | |
| − | f'_n(t)\xrightarrow[n\rightarrow\infty]{glm.}f'(t)
| + | |
| − | \end{align}
| + | |
| − | $$
| + | |
| − | This implys:\[
| + | |
| − | \begin{align}
| + | |
| − | -\frac{2}{\pi}\int\limits_0^\infty \frac{x\sin{xt}}{1+x^2}\mathrm{d}x = \frac{\mathrm{d}}{\mathrm{d}t}\frac{2}{\pi}\int\limits_0^\infty \frac{\cos{xt}}{1+x^2}\mathrm{d}x
| + | |
| − | \end{align}
| + | |
| − | \]
| + | |
| − | | + | |
| − | And the claim is proven.
| + | |
Let \(u\) be a harmonic function on \(\mathbb{R}^3\). Assusme there exists \(C>0\), independent of \(x\), such that \(|u(x)| \leq C(1+|x|)\) on \(\mathbb{R}^3\). Show that then
