題目

問題 7. 應用 9.1.4 小節描述的下降法，但針對二維的拉普拉斯方程，并從三維的 Coulomb 勢出發
KaTeX parse error: Invalid delimiter: '{"type":"ordgroup","mode":"math","loc":{"lexer":{"input":" U_3(x,y,z) = -\\frac{1}{4\\pi}\\big{(}x^2 + y^2 + z^2\\big{)}^{-\\frac{1}{2}}, ","settings":{"displayMode":true,"leqno":false,"fleqn":false,"throwOnError":true,"errorColor":"#cc0000","macros":{},"colorIsTextColor":false,"strict":"warn","maxSize":null,"maxExpand":1000,"allowedProtocols":["http","https","mailto","_relative"]},"tokenRegex":{},"catcodes":{"%":14}},"start":33,"end":36},"body":[{"type":"atom","mode":"math","family":"open","loc":{"lexer":{"input":" U_3(x,y,z) = -\\frac{1}{4\\pi}\\big{(}x^2 + y^2 + z^2\\big{)}^{-\\frac{1}{2}}, ","settings":{"displayMode":true,"leqno":false,"fleqn":false,"throwOnError":true,"errorColor":"#cc0000","macros":{},"colorIsTextColor":false,"strict":"warn","maxSize":null,"maxExpand":1000,"allowedProtocols":["http","https","mailto","_relative"]},"tokenRegex":{},"catcodes":{"%":14}},"start":34,"end":35},"text":"("}]}' after '\big' at position 34: …ac{1}{4\pi}\big{?(?}?x^2 + y^2 + z^2…
(4)
推導出二維的對數勢
KaTeX parse error: Invalid delimiter: '{"type":"ordgroup","mode":"math","loc":{"lexer":{"input":" U_2(x,y) = \\frac{1}{2\\pi}\\log\\big{(}x^2 + y^2\\big{)}^{\\frac{1}{2}}. ","settings":{"displayMode":true,"leqno":false,"fleqn":false,"throwOnError":true,"errorColor":"#cc0000","macros":{},"colorIsTextColor":false,"strict":"warn","maxSize":null,"maxExpand":1000,"allowedProtocols":["http","https","mailto","_relative"]},"tokenRegex":{},"catcodes":{"%":14}},"start":34,"end":37},"body":[{"type":"atom","mode":"math","family":"open","loc":{"lexer":{"input":" U_2(x,y) = \\frac{1}{2\\pi}\\log\\big{(}x^2 + y^2\\big{)}^{\\frac{1}{2}}. ","settings":{"displayMode":true,"leqno":false,"fleqn":false,"throwOnError":true,"errorColor":"#cc0000","macros":{},"colorIsTextColor":false,"strict":"warn","maxSize":null,"maxExpand":1000,"allowedProtocols":["http","https","mailto","_relative"]},"tokenRegex":{},"catcodes":{"%":14}},"start":35,"end":36},"text":"("}]}' after '\big' at position 35: …}{2\pi}\log\big{?(?}?x^2 + y^2\big{)…
(5)
提示： 你需要計算發散的積分 $${\int }_0^{\infty }{U}_3(x,y,z)dz$$。作為替代，考慮 $${\int }_0^N{U}_3(x,y,z)dz$$，減去一個常數（例如 $${\int }_0^N{U}_3(1,0,z)dz$$），然后取極限 $$N\to \infty$$。

注意： 在給定的 $$U_2(x,y,z)$$ 中，變量 $$z$$ 是多余的，因為二維勢只依賴于 $$x$$ 和 $$y$$，因此已修正為 $$U_2(x,y)$$。

解答問題

下降法的核心思想是通過對額外維度（此處為 $$z$$ 軸）積分，從高維（三維）拉普拉斯方程的基本解（Coulomb 勢）推導出低維（二維）的基本解（對數勢）。給定三維 Coulomb 勢：
$$U_3(x,y,z)=?\frac{1}{4\pi }(x^2+y^2+z^2{)}^{?1/2}.$$
目標是推導二維對數勢：
KaTeX parse error: Invalid delimiter: '{"type":"ordgroup","mode":"math","loc":{"lexer":{"input":" U_2(x,y) = \\frac{1}{2\\pi} \\log \\big{(} (x^2 + y^2)^{1/2} \\big{)} = \\frac{1}{2\\pi} \\log r, \\quad \\text{其中} \\quad r = \\sqrt{x^2 + y^2}. ","settings":{"displayMode":true,"leqno":false,"fleqn":false,"throwOnError":true,"errorColor":"#cc0000","macros":{},"colorIsTextColor":false,"strict":"warn","maxSize":null,"maxExpand":1000,"allowedProtocols":["http","https","mailto","_relative"]},"tokenRegex":{},"catcodes":{"%":14}},"start":36,"end":39},"body":[{"type":"atom","mode":"math","family":"open","loc":{"lexer":{"input":" U_2(x,y) = \\frac{1}{2\\pi} \\log \\big{(} (x^2 + y^2)^{1/2} \\big{)} = \\frac{1}{2\\pi} \\log r, \\quad \\text{其中} \\quad r = \\sqrt{x^2 + y^2}. ","settings":{"displayMode":true,"leqno":false,"fleqn":false,"throwOnError":true,"errorColor":"#cc0000","macros":{},"colorIsTextColor":false,"strict":"warn","maxSize":null,"maxExpand":1000,"allowedProtocols":["http","https","mailto","_relative"]},"tokenRegex":{},"catcodes":{"%":14}},"start":37,"end":38},"text":"("}]}' after '\big' at position 37: …2\pi} \log \big{?(?}? (x^2 + y^2)^{1…

直接積分 $${\int }_{?\infty }^{\infty }{U}_3(x,y,z)dz$$ 是發散的，因此需按提示進行正則化：考慮有限范圍 $$[0,N]$$ 的積分，減去一個參考常數（例如在點 $$(1,0)$$ 處的積分），然后取 $$N\to \infty$$。然而，由于 $$U_3$$ 是偶函數（關于 $$z$$ 對稱），為獲得正確的幅值，需考慮全對稱積分范圍 $$[?N,N]$$（這相當于將提示中的半范圍積分乘以 2）。定義：
$$I_N(x,y)={\int }_{?N}^N{U}_3(x,y,z)dz.$$
令 $$r^2=x^2+y^2$$，則：
$$I_N(x,y)={\int }_{?N}^N?\frac{1}{4\pi }(r^2+z^2{)}^{?1/2}dz.$$
被積函數是偶函數，因此：
$$I_N(x,y)=?\frac{1}{4\pi }?2{\int }_0^N(r^2+z^2{)}^{?1/2}dz=?\frac{1}{2\pi }{\int }_0^N(r^2+z^2{)}^{?1/2}dz.$$
計算積分：
$${\int }_0^N(r^2+z^2{)}^{?1/2}dz={\left[\log \left(z+\sqrt{z^2+r^2}\right)\right]}_0^N=\log \left(N+\sqrt{N^2+r^2}\right)?\log r.$$
所以：
$$I_N(x,y)=?\frac{1}{2\pi }\left[\log \left(N+\sqrt{N^2+r^2}\right)?\log r\right].$$
在參考點 $$(x,y)=(1,0)$$（即 $$r=1$$）：
$$I_N(1,0)=?\frac{1}{2\pi }\left[\log \left(N+\sqrt{N^2+1}\right)?\log 1\right]=?\frac{1}{2\pi }\log \left(N+\sqrt{N^2+1}\right),$$
因為 $$\log 1=0$$。

定義正則化后的勢：
$$V_N(x,y)=I_N(x,y)?I_N(1,0).$$
代入表達式：
$$V_N(x,y)=?\frac{1}{2\pi }\left[\log \left(N+\sqrt{N^2+r^2}\right)?\log r\right]+\frac{1}{2\pi }\log \left(N+\sqrt{N^2+1}\right).$$
簡化：
$$V_N(x,y)=?\frac{1}{2\pi }\left[\log \left(N+\sqrt{N^2+r^2}\right)?\log r?\log \left(N+\sqrt{N^2+1}\right)\right]=?\frac{1}{2\pi }\left[\log \left(\frac{N+\sqrt{N^2+r^2}}{N+\sqrt{N^2+1}}\right)?\log r\right].$$

取極限 $$N\to \infty$$。分析比值：
$$\frac{N+\sqrt{N^2+r^2}}{N+\sqrt{N^2+1}}.$$
當 $$N\to \infty$$，使用漸近展開 $$\sqrt{N^2+a^2}=N\sqrt{1+(a/N{)}^2}=N\left(1+\frac{a^2}{2N^2}+O(N^{?4})\right)$$，所以：
$$N+\sqrt{N^2+r^2}=N+N\left(1+\frac{r^2}{2N^2}+O(N^{?4})\right)=2N+\frac{r^2}{2N}+O(N^{?3}),$$
$$N+\sqrt{N^2+1}=2N+\frac{1}{2N}+O(N^{?3}).$$
因此：
$$\frac{N+\sqrt{N^2+r^2}}{N+\sqrt{N^2+1}}=\frac{2N+\frac{r^2}{2N}+O(N^{?3})}{2N+\frac{1}{2N}+O(N^{?3})}=\frac{1+\frac{r^2}{4N^2}+O(N^{?4})}{1+\frac{1}{4N^2}+O(N^{?4})}=1+\frac{r^2?1}{4N^2}+O(N^{?4}).$$
取對數：
$$\log \left(\frac{N+\sqrt{N^2+r^2}}{N+\sqrt{N^2+1}}\right)=\log \left(1+\frac{r^2?1}{4N^2}+O(N^{?4})\right)=\frac{r^2?1}{4N^2}+O(N^{?4})\to 0\text{as}N\to \infty .$$
所以：
$$\underset{N\to \infty }{\lim }{V}_N(x,y)=?\frac{1}{2\pi }\left[0?\log r\right]=\frac{1}{2\pi }\log r.$$
這正是二維對數勢：
KaTeX parse error: Invalid delimiter: '{"type":"ordgroup","mode":"math","loc":{"lexer":{"input":" U_2(x,y) = \\frac{1}{2\\pi} \\log r = \\frac{1}{2\\pi} \\log \\big{(} (x^2 + y^2)^{1/2} \\big{)}. ","settings":{"displayMode":true,"leqno":false,"fleqn":false,"throwOnError":true,"errorColor":"#cc0000","macros":{},"colorIsTextColor":false,"strict":"warn","maxSize":null,"maxExpand":1000,"allowedProtocols":["http","https","mailto","_relative"]},"tokenRegex":{},"catcodes":{"%":14}},"start":60,"end":63},"body":[{"type":"atom","mode":"math","family":"open","loc":{"lexer":{"input":" U_2(x,y) = \\frac{1}{2\\pi} \\log r = \\frac{1}{2\\pi} \\log \\big{(} (x^2 + y^2)^{1/2} \\big{)}. ","settings":{"displayMode":true,"leqno":false,"fleqn":false,"throwOnError":true,"errorColor":"#cc0000","macros":{},"colorIsTextColor":false,"strict":"warn","maxSize":null,"maxExpand":1000,"allowedProtocols":["http","https","mailto","_relative"]},"tokenRegex":{},"catcodes":{"%":14}},"start":61,"end":62},"text":"("}]}' after '\big' at position 61: …2\pi} \log \big{?(?}? (x^2 + y^2)^{1…

總結

通過下降法，從三維 Coulomb 勢 $$U_3(x,y,z)=?\frac{1}{4\pi }(x^2+y^2+z^2{)}^{?1/2}$$ 出發，考慮對稱積分范圍 $$[?N,N]$$，正則化后取極限 $$N\to \infty$$，成功推導出二維對數勢 KaTeX parse error: Invalid delimiter: '{"type":"ordgroup","mode":"math","loc":{"lexer":{"input":" U_2(x,y) = \\frac{1}{2\\pi} \\log \\big{(} (x^2 + y^2)^{1/2} \\big{)} ","settings":{"displayMode":true,"leqno":false,"fleqn":false,"throwOnError":true,"errorColor":"#cc0000","macros":{},"colorIsTextColor":false,"strict":"warn","maxSize":null,"maxExpand":1000,"allowedProtocols":["http","https","mailto","_relative"]},"tokenRegex":{},"catcodes":{"%":14}},"start":36,"end":39},"body":[{"type":"atom","mode":"math","family":"open","loc":{"lexer":{"input":" U_2(x,y) = \\frac{1}{2\\pi} \\log \\big{(} (x^2 + y^2)^{1/2} \\big{)} ","settings":{"displayMode":true,"leqno":false,"fleqn":false,"throwOnError":true,"errorColor":"#cc0000","macros":{},"colorIsTextColor":false,"strict":"warn","maxSize":null,"maxExpand":1000,"allowedProtocols":["http","https","mailto","_relative"]},"tokenRegex":{},"catcodes":{"%":14}},"start":37,"end":38},"text":"("}]}' after '\big' at position 37: …2\pi} \log \big{?(?}? (x^2 + y^2)^{1…。正則化步驟通過減去參考點 $$(1,0)$$ 處的積分消除了發散項，確保了極限收斂。