Difference between revisions of "Aufgaben:Problem 11"
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notes, you have to give a proof of it. | notes, you have to give a proof of it. | ||
| − | ==Proof | + | ==Proof== |
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<div style="border:1px solid #aaa; background: #f8f8f8; padding:5px"> | <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 | + | Aforementioned '''corollary''' from [https://people.math.ethz.ch/~gruppe5/group5/lectures/mmp/fs15/Files/NewtonianPotential.pdf 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 | |
$$\left|\frac{\partial u}{\partial x_j}(a)\right| \leq \frac{C_n}{r} \sup_{y\in S_r(a)}|u(y)|$$ | $$\left|\frac{\partial u}{\partial x_j}(a)\right| \leq \frac{C_n}{r} \sup_{y\in S_r(a)}|u(y)|$$ | ||
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| + | '''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 | ||
| + | $$ | ||
| + | |B_r(0)|\: \left|\left(\frac{\partial}{\partial x_j} u\right)\right| | ||
| + | = \left| \int_{B_r(0)} \left(\frac{\partial}{\partial x_j} u\right)(x+a) \: \mathrm{d}x\right| | ||
| + | = \left| \int_{B_r(0)} (\mathrm{div}\: V_j)(x+a) \: \mathrm{d}x\right| \\ | ||
| + | = \left| \int_{S_r(0)} \langle V_j(y+a), \frac{1}{r}y \rangle \: \mathrm{d}y\right| | ||
| + | = \left| \int_{S_r(0)} u(y+a)\frac{1}{r}y_j \: \mathrm{d}y\right| | ||
| + | \leq \int_{S_r(0)} |u(y+a)| \left| \frac{1}{r}y_j \right| \mathrm{d}y \\ | ||
| + | \leq \int_{S_r(0)} |u(y+a)| \: \mathrm{d}y | ||
| + | \leq |S_r(0)| \sup_{y\in S_r(a)}|u(y)| | ||
| + | = |S_1(0)|r^{n-1} \sup_{y\in S_r(a)}|u(y)| | ||
| + | $$ | ||
| + | Divide by \(|B_r(0)|=|B_1(0)|r^n\) and we're done. | ||
| + | <p style="text-align:right">\(\square\)</p> | ||
</div> | </div> | ||
| − | First, note that \(C_3= | + | 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: | + | Let \(a\in\mathbb{R}^3\) and \(r>0\). |
| − | $$|\partial_i u(a)| \leq \frac{3}{r} \sup_{y\in S_r(a)}|u(y)| \leq \frac{3}{r}C(1+|y|) \leq \frac{3}{r}C(1+|a|+r)=3C(\frac{1+|a|}{r} + 1)$$ | + | |
| + | \(u\) is harmonic on all of \(\mathbb{R}^3\) and thus on \(\overline{B_r(a)}\), too. | ||
| + | |||
| + | Using the corollary: | ||
| + | $$|\partial_i u(a)| \leq \frac{3}{r} \sup_{y\in S_r(a)}|u(y)| \leq \frac{3}{r}C(1+|y|) \leq \frac{3}{r}C(1+|a|+r)=3C \left( \frac{1+|a|}{r} + 1 \right)$$ | ||
Since \(u\) is harmonic on all of \(\mathbb{R}^3\), we can choose \(r\) arbitrarily large, so let \(r \rightarrow \infty\) | Since \(u\) is harmonic on all of \(\mathbb{R}^3\), we can choose \(r\) arbitrarily large, so let \(r \rightarrow \infty\) | ||
$$\Rightarrow |\partial_i u(a)| \leq 3C$$ | $$\Rightarrow |\partial_i u(a)| \leq 3C$$ | ||
| − | 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. | + | 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. |
| + | |||
| + | 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> | ||
Revision as of 20:22, 15 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
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 $$ |B_r(0)|\: \left|\left(\frac{\partial}{\partial x_j} u\right)\right| = \left| \int_{B_r(0)} \left(\frac{\partial}{\partial x_j} u\right)(x+a) \: \mathrm{d}x\right| = \left| \int_{B_r(0)} (\mathrm{div}\: V_j)(x+a) \: \mathrm{d}x\right| \\ = \left| \int_{S_r(0)} \langle V_j(y+a), \frac{1}{r}y \rangle \: \mathrm{d}y\right| = \left| \int_{S_r(0)} u(y+a)\frac{1}{r}y_j \: \mathrm{d}y\right| \leq \int_{S_r(0)} |u(y+a)| \left| \frac{1}{r}y_j \right| \mathrm{d}y \\ \leq \int_{S_r(0)} |u(y+a)| \: \mathrm{d}y \leq |S_r(0)| \sup_{y\in S_r(a)}|u(y)| = |S_1(0)|r^{n-1} \sup_{y\in S_r(a)}|u(y)| $$ 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: $$|\partial_i u(a)| \leq \frac{3}{r} \sup_{y\in S_r(a)}|u(y)| \leq \frac{3}{r}C(1+|y|) \leq \frac{3}{r}C(1+|a|+r)=3C \left( \frac{1+|a|}{r} + 1 \right)$$ 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.
\(\square\)
