긴즈버그 기준

식 $(3.26)$ $$ \Delta C = \frac{T_c^2{a_2^\prime}^2}{2a_4} $$

식 $(3.39)$ \begin{equation} \begin{split} C &= n \left[ \frac{1}{2} \left( \frac{Ta_2^\prime}{c} \right)^2 (2\pi)^-d \int d^dk^\prime (1+{k^\prime}^2)^-2 \right] \xi^{4-d} + l.s \\ &\equiv C_0\xi^{4-d} + l.s \end{split} \end{equation}

$$C_{0}\approx\left(\frac{Ta_{2}^{\prime}}{c}\right)^{2}(2\pi)^{2}$$ 로 두고 $\xi^{-1}\equiv\left(\frac{a_{2}^{\prime}}{c}\right)^{\frac{1}{2}}|T-T_{c}|^{\frac{1}{2}}$을 사용하여 $\frac{C_{0}\xi^{4-d}}{\Delta C}$을 계산해보면

\begin{equation}\notag \begin{split} \frac{C_0\xi^{4-d}}{\Delta C} &\approx \left(\frac{{Ta_2^\prime}^2}{c}\right)^2 (2\pi)^{-d} \left(\frac{a_2^\prime}{c}\right)^{\frac{d}{2}-2}|T-T_c|^{\frac{d}{2}-2}\frac{1}{\Delta C} \\ &=T^2{a_2^\prime}^{\frac{d}{2}}(2\pi)^{-d}{a_2^\prime}^{\frac{d}{2}-2}c^{-\frac{d}{2}}T_c^{\frac{d}{2}-2} \left|1-\frac{T}{T_c}\right|^{\frac{d}{2}-2}\frac{1}{\Delta C} \\ &=\frac{T^2{a_2^\prime}^{\frac{d}{2}}(2\pi)^{-d}c^{-\frac{d}{2}}T_c^{\frac{d}{2}-2}\frac{1}{\Delta C}}{\left|1-\frac{T}{T_c}\right|^{2-\frac{d}{2}}} \\ &=\left[\frac{\left\{T^2{a_2^\prime}^{\frac{d}{2}}(2\pi)^{-d}c^{-\frac{d}{2}}T_c^{\frac{d}{2}-2}\frac{1}{\Delta C}\right\}^{\frac{2}{4-d}}}{\left|1-\frac{T}{T_c}\right|}\right]^{2-\frac{d}{2}} \\ &=\left[\frac{\left\{\left(\frac{T}{T_c}\right)^{2}(2\pi)^{-d}\left(\frac{{a_2^\prime}T_c}{c}\right)^{\frac{d}{2}}\frac{1}{\Delta C}\right\}^{\frac{2}{4-d}}}{\left|1-\frac{T}{T_c}\right|}\right]^{2-\frac{d}{2}} \\ &\approx\left[\frac{\left\{(2\pi)^{-d}\left(\frac{{a_2^\prime}T_c}{c}\right)^{\frac{d}{2}}\frac{1}{\Delta C}\right\}^{\frac{2}{4-d}}}{\left|1-\frac{T}{T_c}\right|}\right]^{2-\frac{d}{2}} \end{split} \end{equation}

를 얻는다. 이제

\begin{equation}\notag \zeta_{T}\equiv\left[\frac{(2\pi\xi_{0})^{-d}}{\Delta C}\right]^{\frac{2}{4-d}},\quad\xi_{0}\equiv\left(\frac{c}{a_2^\prime T_{c}}\right)^{\frac{1}{2}} \end{equation}

로 정의하고 다시 적어보면

\begin{equation}\notag \frac{C_0\xi^{4-d}}{\Delta C}\approx\left[\frac{\zeta_T}{\left|1-\frac{T}{T_c\right|}\right]^{2-\frac{d}{2}} \end{equation} 이 된다.

이제 비동차 방정식

$$\frac{dy}{dx} + P(x)y(x) = Q(x)$$

의 특수해를 구하자. 이번에도 위와 같은 방법으로 $y(x)$를 구한다면

\begin{align*} y(x) &= y_0 - \int_0^xdx_1\left[Q(x_1)-P(x_1)y(x_1)\right] \\ &= y_0 - \int_0^xdx_1Q(x_1) + \int_0^xdx_1P(x_1)\left[y_0-\int_0^{x_1}dx_2\bigl\{Q(x_2)-P(x_2)y(x_2)\bigr\}\right] \\ &= y_0 - \int_0^xdx_1Q(x_1) + \int_0^xdx_1P(x_1)y_0 -\int_0^xdx_1P(x_1)\int_0^{x_1}dx_2Q(x_2) + \int_0^xdx_1P(x_1)\int_0^{x_1}dx_2P(x_2)\left[y_0-\int_0^{x_2}dx_3\bigl\{Q(x_3)-P(x_3)y(x_3)\bigr\}\right] + \cdots \\ \end{align*}

이고 $y_0$로 묶으면 아래와 같이 정리할 수 있다. \begin{align*} y(x) &=\left[1 + \int_0^xdx_1P(x_1) + \int_0^xdx_1P(x_1)\int_0^{x_1}dx_2P(x_2) + \int_0^xdx_1P(x_1)\int_0^{x_1}dx_2P(x_2)\int_0^{x_2}dx_3P(x_3) + \cdots\right]y_0 \\ &\quad -\left[\int_0^xdx_1Q(x_1) + \int_0^xdx_1P(x_1)\int_0^{x_1}dx_2Q(x_2) + \int_0^xdx_1P(x_1)\int_0^{x_1}dx_2P(x_2)\int_0^{x_2}dx_3Q(x_3) + \cdots\right] \\ \end{align*}

$y_0$로 묵인 항을 위에서 보인 증명을 사용하여 나타내고 아래의 항을 조금 더 정리하면 다음과 같이 특수해를 구할 수 있다.

\begin{align*} y(x) &= \sum_{n=0}^\infty\mathcal{T}\frac{1}{n!}\left[\int_0^xdx^\prime P(x^\prime)\right]^ny_0 \\ &\quad -\Biggl[\int_0^xdx_1Q(x_1) + \sum_{n=2}^\infty\Biggl\{\int_0^x\prod_{k=1}^{n-1}\left(dx_kP(x_k)\int_0^{x_k}\right)dx_nQ(x_n)\Biggr\}\Biggr] \end{align*}

동차 방정식에서 구한 해와 비동차 방정식에서 구한 특수해를 합하면 미분방정식의 일반해를 구할 수 있다. \begin{align*} y(x) &= \sum_{n=0}^\infty\mathcal{T}\frac{1}{n!}\left[(-1)^n\int_0^xdx^\prime P(x^\prime)+\int_0^xdx^\prime P(x^\prime)\right]y_0 \\ &\quad + \left[\int_0^xdx_1Q(x_1) + \sum_{n=2}^\infty\Bigl\{\int_0^x\prod_{k=1}^{n-1}\left(dx_kP(x_k)\int_0^x\right)\Bigr\}dx_nQ(x_n)\right] \\ &= \sum_0^\infty\mathcal{T}\left[\frac{1}{(2n)!}\int_0^xdx^\prime P(x^\prime)\right]y_0 - \left[\int_0^xdx_1Q(x_1) + \sum_{n=2}^\infty\Bigl\{\int_0^x\prod_{k=1}^{n-1}\left(dx_kP(x_k)\int_0^x\right)\Bigr\}dx_nQ(x_n)\right] \end{align*}