We consider Walsh’s conformal map from the complement of a compact set <InlineEquation ID="IEq1"> <EquationSource Format="TEX">$E = \cup _{j=1}^\ell E_j$</EquationSource> <EquationSource Format="MATHML"><math> <mrow> <mi>E</mi> <mo>=</mo> <msubsup> <mo>∪</mo> <mrow> <mi>j</mi> <mo>=</mo> <mn>1</mn> </mrow> <mi>ℓ</mi> </msubsup> <msub> <mi>E</mi> <mi>j</mi> </msub> </mrow> </math></EquationSource> </InlineEquation> with <InlineEquation ID="IEq2"> <EquationSource Format="TEX">$\ell $</EquationSource> <EquationSource Format="MATHML"><math> <mi>ℓ</mi> </math></EquationSource> </InlineEquation> components onto a lemniscatic domain <InlineEquation ID="IEq3"> <EquationSource Format="TEX">$\widehat{\mathbb {C}} \setminus L$</EquationSource> <EquationSource Format="MATHML"><math> <mrow> <mover accent="true"> <mi mathvariant="double-struck">C</mi> <mo stretchy="true">^</mo> </mover> <mo lspace="0.15em" rspace="0.15em" stretchy="false">\</mo> <mi>L</mi> </mrow> </math></EquationSource> </InlineEquation>, where L has the form <InlineEquation ID="IEq4"> <EquationSource Format="TEX">$L = \{ w \in \mathbb {C}: \prod _{j=1}^\ell |w - a_j|^{m_j} \le {{\,\textrm{cap}\,}}(E) \}$</EquationSource> <EquationSource Format="MATHML"><math> <mrow> <mi>L</mi> <mo>=</mo> <mo stretchy="false">{</mo> <mi>w</mi> <mo>∈</mo> <mi mathvariant="double-struck">C</mi> <mo>:</mo> <msubsup> <mo>∏</mo> <mrow> <mi>j</mi> <mo>=</mo> <mn>1</mn> </mrow> <mi>ℓ</mi> </msubsup> <mo stretchy="false">|</mo> <mi>w</mi> <mo>-</mo> <msub> <mi>a</mi> <mi>j</mi> </msub> <msup> <mo stretchy="false">|</mo> <msub> <mi>m</mi> <mi>j</mi> </msub> </msup> <mo>≤</mo> <mrow> <mspace width="0.166667em" /> <mtext>cap</mtext> <mspace width="0.166667em" /> </mrow> <mrow> <mo stretchy="false">(</mo> <mi>E</mi> <mo stretchy="false">)</mo> </mrow> <mo stretchy="false">}</mo> </mrow> </math></EquationSource> </InlineEquation>. We prove that the exponents <InlineEquation ID="IEq5"> <EquationSource Format="TEX">$m_j$</EquationSource> <EquationSource Format="MATHML"><math> <msub> <mi>m</mi> <mi>j</mi> </msub> </math></EquationSource> </InlineEquation> appearing in L satisfy <InlineEquation ID="IEq6"> <EquationSource Format="TEX">$m_j = \mu _E(E_j)$</EquationSource> <EquationSource Format="MATHML"><math> <mrow> <msub> <mi>m</mi> <mi>j</mi> </msub> <mo>=</mo> <msub> <mi>μ</mi> <mi>E</mi> </msub> <mrow> <mo stretchy="false">(</mo> <msub> <mi>E</mi> <mi>j</mi> </msub> <mo stretchy="false">)</mo> </mrow> </mrow> </math></EquationSource> </InlineEquation>, where <InlineEquation ID="IEq7"> <EquationSource Format="TEX">$\mu _E$</EquationSource> <EquationSource Format="MATHML"><math> <msub> <mi>μ</mi> <mi>E</mi> </msub> </math></EquationSource> </InlineEquation> is the equilibrium measure of E. When E is the union of <InlineEquation ID="IEq8"> <EquationSource Format="TEX">$\ell $</EquationSource> <EquationSource Format="MATHML"><math> <mi>ℓ</mi> </math></EquationSource> </InlineEquation> real intervals, we derive a fast algorithm for computing the centers <InlineEquation ID="IEq9"> <EquationSource Format="TEX">$a_1, \ldots , a_\ell $</EquationSource> <EquationSource Format="MATHML"><math> <mrow> <msub> <mi>a</mi> <mn>1</mn> </msub> <mo>,</mo> <mo>…</mo> <mo>,</mo> <msub> <mi>a</mi> <mi>ℓ</mi> </msub> </mrow> </math></EquationSource> </InlineEquation>. For <InlineEquation ID="IEq10"> <EquationSource Format="TEX">$\ell = 2$</EquationSource> <EquationSource Format="MATHML"><math> <mrow> <mi>ℓ</mi> <mo>=</mo> <mn>2</mn> </mrow> </math></EquationSource> </InlineEquation>, the formulas for <InlineEquation ID="IEq11"> <EquationSource Format="TEX">$m_1, m_2$</EquationSource> <EquationSource Format="MATHML"><math> <mrow> <msub> <mi>m</mi> <mn>1</mn> </msub> <mo>,</mo> <msub> <mi>m</mi> <mn>2</mn> </msub> </mrow> </math></EquationSource> </InlineEquation> and <InlineEquation ID="IEq12"> <EquationSource Format="TEX">$a_1, a_2$</EquationSource> <EquationSource Format="MATHML"><math> <mrow> <msub> <mi>a</mi> <mn>1</mn> </msub> <mo>,</mo> <msub> <mi>a</mi> <mn>2</mn> </msub> </mrow> </math></EquationSource> </InlineEquation> are explicit. Moreover, we obtain the conformal map numerically. Our approach relies on the real and complex Green’s functions of <InlineEquation ID="IEq13"> <EquationSource Format="TEX">$\widehat{\mathbb {C}} \setminus E$</EquationSource> <EquationSource Format="MATHML"><math> <mrow> <mover accent="true"> <mi mathvariant="double-struck">C</mi> <mo stretchy="true">^</mo> </mover> <mo lspace="0.15em" rspace="0.15em" stretchy="false">\</mo> <mi>E</mi> </mrow> </math></EquationSource> </InlineEquation> and <InlineEquation ID="IEq14"> <EquationSource Format="TEX">$\widehat{\mathbb {C}} \setminus L$</EquationSource> <EquationSource Format="MATHML"><math> <mrow> <mover accent="true"> <mi mathvariant="double-struck">C</mi> <mo stretchy="true">^</mo> </mover> <mo lspace="0.15em" rspace="0.15em" stretchy="false">\</mo> <mi>L</mi> </mrow> </math></EquationSource> </InlineEquation>.

Walsh’s Conformal Map onto Lemniscatic Domains for Several Intervals

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2025年1月21日

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Klaus Schiefermayr,
Olivier Sète

摘要

We consider Walsh’s conformal map from the complement of a compact set $E = \cup _{j=1}^\ell E_j$ $E = \cup_{j = 1}^{ℓ} E_{j}$ with $\ell $ $ℓ$ components onto a lemniscatic domain $\widehat{\mathbb {C}} \setminus L$ $\hat{C} \ L$ , where L has the form $L = \{ w \in \mathbb {C}: \prod _{j=1}^\ell |w - a_j|^{m_j} \le {{\,\textrm{cap}\,}}(E) \}$ $L = {w \in C : \prod_{j = 1}^{ℓ} | w - a_{j} |^{m_{j}} \leq cap (E)}$ . We prove that the exponents $m_j$ $m_{j}$ appearing in L satisfy $m_j = \mu _E(E_j)$ $m_{j} = μ_{E} (E_{j})$ , where $\mu _E$ $μ_{E}$ is the equilibrium measure of E. When E is the union of $\ell $ $ℓ$ real intervals, we derive a fast algorithm for computing the centers $a_1, \ldots , a_\ell $ $a_{1}, \dots, a_{ℓ}$ . For $\ell = 2$ $ℓ = 2$ , the formulas for $m_1, m_2$ $m_{1}, m_{2}$ and $a_1, a_2$ $a_{1}, a_{2}$ are explicit. Moreover, we obtain the conformal map numerically. Our approach relies on the real and complex Green’s functions of $\widehat{\mathbb {C}} \setminus E$ $\hat{C} \ E$ and $\widehat{\mathbb {C}} \setminus L$ $\hat{C} \ L$ .