从三维Coulomb势到二维对数势的下降法推导

题目

问题 7. 应用 9.1.4 小节描述的下降法，但针对二维的拉普拉斯方程，并从三维的 Coulomb 势出发
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(4)
推导出二维的对数势
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(5)
提示： 你需要计算发散的积分 $${\int }_0^{\infty }{U}_3(x,y,z)dz$$。作为替代，考虑 $${\int }_0^N{U}_3(x,y,z)dz$$，减去一个常数（例如 $${\int }_0^N{U}_3(1,0,z)dz$$），然后取极限 $$N\to \infty$$。

注意： 在给定的 $$U_2(x,y,z)$$ 中，变量 $$z$$ 是多余的，因为二维势只依赖于 $$x$$ 和 $$y$$，因此已修正为 $$U_2(x,y)$$。

解答问题

下降法的核心思想是通过对额外维度（此处为 $$z$$ 轴）积分，从高维（三维）拉普拉斯方程的基本解（Coulomb 势）推导出低维（二维）的基本解（对数势）。给定三维 Coulomb 势：
$$U_3(x,y,z)=-\frac{1}{4\pi }(x^2+y^2+z^2{)}^{-1/2}.$$
目标是推导二维对数势：
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…

直接积分 $${\int }_{-\infty }^{\infty }{U}_3(x,y,z)dz$$ 是发散的，因此需按提示进行正则化：考虑有限范围 $$[0,N]$$ 的积分，减去一个参考常数（例如在点 $$(1,0)$$ 处的积分），然后取 $$N\to \infty$$。然而，由于 $$U_3$$ 是偶函数（关于 $$z$$ 对称），为获得正确的幅值，需考虑全对称积分范围 $$[-N,N]$$（这相当于将提示中的半范围积分乘以 2）。定义：
$$I_N(x,y)={\int }_{-N}^N{U}_3(x,y,z)dz.$$
令 $$r^2=x^2+y^2$$，则：
$$I_N(x,y)={\int }_{-N}^N-\frac{1}{4\pi }(r^2+z^2{)}^{-1/2}dz.$$
被积函数是偶函数，因此：
$$I_N(x,y)=-\frac{1}{4\pi }\cdot 2{\int }_0^N(r^2+z^2{)}^{-1/2}dz=-\frac{1}{2\pi }{\int }_0^N(r^2+z^2{)}^{-1/2}dz.$$
计算积分：
$${\int }_0^N(r^2+z^2{)}^{-1/2}dz={\left[\log \left(z+\sqrt{z^2+r^2}\right)\right]}_0^N=\log \left(N+\sqrt{N^2+r^2}\right)-\log r.$$
所以：
$$I_N(x,y)=-\frac{1}{2\pi }\left[\log \left(N+\sqrt{N^2+r^2}\right)-\log r\right].$$
在参考点 $$(x,y)=(1,0)$$（即 $$r=1$$）：
$$I_N(1,0)=-\frac{1}{2\pi }\left[\log \left(N+\sqrt{N^2+1}\right)-\log 1\right]=-\frac{1}{2\pi }\log \left(N+\sqrt{N^2+1}\right),$$
因为 $$\log 1=0$$。

定义正则化后的势：
$$V_N(x,y)=I_N(x,y)-I_N(1,0).$$
代入表达式：
$$V_N(x,y)=-\frac{1}{2\pi }\left[\log \left(N+\sqrt{N^2+r^2}\right)-\log r\right]+\frac{1}{2\pi }\log \left(N+\sqrt{N^2+1}\right).$$
简化：
$$V_N(x,y)=-\frac{1}{2\pi }\left[\log \left(N+\sqrt{N^2+r^2}\right)-\log r-\log \left(N+\sqrt{N^2+1}\right)\right]=-\frac{1}{2\pi }\left[\log \left(\frac{N+\sqrt{N^2+r^2}}{N+\sqrt{N^2+1}}\right)-\log r\right].$$

取极限 $$N\to \infty$$。分析比值：
$$\frac{N+\sqrt{N^2+r^2}}{N+\sqrt{N^2+1}}.$$
当 $$N\to \infty$$，使用渐近展开 $$\sqrt{N^2+a^2}=N\sqrt{1+(a/N{)}^2}=N\left(1+\frac{a^2}{2N^2}+O(N^{-4})\right)$$，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：
$$\frac{N+\sqrt{N^2+r^2}}{N+\sqrt{N^2+1}}=\frac{2N+\frac{r^2}{2N}+O(N^{-3})}{2N+\frac{1}{2N}+O(N^{-3})}=\frac{1+\frac{r^2}{4N^2}+O(N^{-4})}{1+\frac{1}{4N^2}+O(N^{-4})}=1+\frac{r^2-1}{4N^2}+O(N^{-4}).$$
取对数：
$$\log \left(\frac{N+\sqrt{N^2+r^2}}{N+\sqrt{N^2+1}}\right)=\log \left(1+\frac{r^2-1}{4N^2}+O(N^{-4})\right)=\frac{r^2-1}{4N^2}+O(N^{-4})\to 0\text{as}N\to \infty .$$
所以：
$$\underset{N\to \infty }{\lim }{V}_N(x,y)=-\frac{1}{2\pi }\left[0-\log r\right]=\frac{1}{2\pi }\log r.$$
这正是二维对数势：
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总结

通过下降法，从三维 Coulomb 势 $$U_3(x,y,z)=-\frac{1}{4\pi }(x^2+y^2+z^2{)}^{-1/2}$$ 出发，考虑对称积分范围 $$[-N,N]$$，正则化后取极限 $$N\to \infty$$，成功推导出二维对数势 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…。正则化步骤通过减去参考点 $$(1,0)$$ 处的积分消除了发散项，确保了极限收敛。