Difference between revisions of "Aufgaben:Problem 11"
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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> | <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> | </ol> | ||
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| + | ''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 Sketch== | ==Proof Sketch== | ||
Proof in 2 dimensions: http://www.math.columbia.edu/~savin/c12d.pdf | Proof in 2 dimensions: http://www.math.columbia.edu/~savin/c12d.pdf | ||
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| + | ---- | ||
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| + | Unrelated attempt: | ||
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| + | <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 (proof on p.3f): | ||
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| + | If \(u(x)\) harmonic on 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 | ||
| + | $$\left|\frac{\partial u}{\partial x_j}(a)\right| \leq \frac{C_n}{r} \sup_{y\in S_r(a)}|u(y)|$$ | ||
| + | </div> | ||
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| + | First, note that \(C_3=\frac{4\pi}{\frac{4}{3}\pi}=3\) | ||
| + | |||
| + | 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: | ||
| + | $$|\partial_i u(a)| \leq \frac{3}{r} \sup_{y\in S_r(a)}|u(y)| \leq \frac{3}{r}C(1+|y|) \leq \frac{3}{r}C(1+|a|+r)=3C(\frac{1+|a|}{r} + 1)$$ | ||
| + | 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. Not quite sure how b) doesn't result directly from a) without further ado. | ||
Revision as of 12:14, 9 June 2015
Problem
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
- \(\partial_i u\) is constant on \(\mathbb{R}^3\), \(\forall i = 1,2,3\).
- \(u\) 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 Sketch
Proof in 2 dimensions: http://www.math.columbia.edu/~savin/c12d.pdf
Unrelated attempt:
Aforementioned corollary from NewtonianPotential.pdf, p.3 (proof on p.3f):
If \(u(x)\) harmonic on 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 $$\left|\frac{\partial u}{\partial x_j}(a)\right| \leq \frac{C_n}{r} \sup_{y\in S_r(a)}|u(y)|$$
First, note that \(C_3=\frac{4\pi}{\frac{4}{3}\pi}=3\)
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: $$|\partial_i u(a)| \leq \frac{3}{r} \sup_{y\in S_r(a)}|u(y)| \leq \frac{3}{r}C(1+|y|) \leq \frac{3}{r}C(1+|a|+r)=3C(\frac{1+|a|}{r} + 1)$$ 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. Not quite sure how b) doesn't result directly from a) without further ado.
