水平集 与 二维泛函求导

数学

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由于泛函求导之后仍然是一个函数，故用变分符号$\delta$，另外$\delta$还表示$Deaviside$函数的导数，$\delta(x) = H'(x)$，应该很容易分得清

现有$F(f,f'_x,f'_y,x,y)$

泛函求导公式为

$$\begin{align*} \delta J[f] & = \iint \frac{\partial F}{\partial f}\delta f + \frac{\partial F}{\partial f'_x}\delta f'_x + \frac{\partial F}{\partial f'_y}\delta f'_y \mathrm{d}x\mathrm{d}y \\ & = \iint \frac{\partial F}{\partial f}\delta f - \frac{\mathrm{d}}{\mathrm{d}x} \left( \frac{\partial F}{\partial f'_x} \right)\delta f - \frac{\mathrm{d}}{\mathrm{d}y} \left( \frac{\partial F}{\partial f'_y} \right)\delta f \mathrm{d}x\mathrm{d}y \\ & = \iint \left( \frac{\partial F}{\partial f} - \frac{\mathrm{d}}{\mathrm{d}x} \frac{\partial F}{\partial f'_x} - \frac{\mathrm{d}}{\mathrm{d}y} \frac{\partial F}{\partial f'_y} \right)\delta f \mathrm{d}x\mathrm{d}y \\ \frac{\delta J}{\delta f} & = \frac{\partial F}{\partial f} - \frac{\mathrm{d}}{\mathrm{d}x}\frac{\partial F}{\partial f'_x} - \frac{\mathrm{d}}{\mathrm{d}y}\frac{\partial F}{\partial f'_y} \end{align*}$$

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距离函数正则化项

$$\begin{align*} |\nabla\phi| & = \sqrt{\phi'^2_x + \phi'^2_y} \\ \frac{\delta |\nabla\phi|}{\delta \phi} & = -\frac{\mathrm{d}}{\mathrm{d}x}\left( \frac{1}{2}(\phi'^2_x + \phi'^2_y)^{-\frac{1}{2}} \cdot 2 \phi'_x \right) -\frac{\mathrm{d}}{\mathrm{d}y}\left( \frac{1}{2}(\phi'^2_x + \phi'^2_y)^{-\frac{1}{2}} \cdot 2 \phi'_y \right) \\ & = -\nabla\cdot \frac{\nabla \phi}{|\nabla \phi|} \\ & = -\mathrm{div}(\frac{\nabla \phi}{|\nabla \phi|}) \end{align*}$$

$$\begin{align*} \mathcal{D}(\phi) & = \iint \frac{1}{2}(|\nabla\phi(x)-1|)^2\mathrm{d}x\mathrm{d}y \\ \frac{\delta \mathcal{D}}{\delta \phi} & = -\frac{\mathrm{d}}{\mathrm{d}x}\left((|\nabla\phi|-1)\cdot \frac{\phi'_x}{|\nabla\phi|}\right) -\frac{\mathrm{d}}{\mathrm{d}y}\left((|\nabla\phi|-1)\cdot \frac{\phi'_y}{|\nabla\phi|}\right) \\ & = -\frac{\mathrm{d}}{\mathrm{d}x}\left( \phi'_x - \frac{\phi'_x}{|\nabla\phi|} \right) - \frac{\mathrm{d}}{\mathrm{d}y}\left( \phi'_y - \frac{\phi'_y}{|\nabla\phi|} \right) \\ & = -\left\{ (\phi''_{xx} + \phi''_{yy}) - \left( \frac{\mathrm{d}}{\mathrm{d}x} \frac{\phi'_x}{|\nabla\phi|} + \frac{\mathrm{d}}{\mathrm{d}y}\frac{\phi'y}{|\nabla\phi|} \right) \right\} \\ & = - \left\{ \nabla^2\phi - \nabla\cdot \frac{\nabla\phi}{|\nabla\phi|} \right\} \\ & = - \left\{ \nabla^2\phi - \mathrm{div}(\frac{\nabla\phi}{|\nabla\phi|}) \right\} \end{align*}$$

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线积分项

$$\begin{align*} \mathcal{L}(\phi) & = \iint |\nabla H(\phi)|\mathrm{d}x\mathrm{d}y \qquad// H'(x)函数性质，只有在轮廓处值为1 \\ & = \iint \delta(\phi)|\nabla\phi|\mathrm{d}x\mathrm{d}y \qquad //\delta(x) = H'(x) \\ \frac{\delta \mathcal{L}}{\delta \phi} & = \delta'(\phi)|\nabla\phi| - \left\{ \frac{\mathrm{d}}{\mathrm{d}x}\left( \delta(\phi)\cdot \frac{\phi'_x}{|\nabla \phi|} \right) + \frac{\mathrm{d}}{\mathrm{d}y}\left( \delta(\phi)\cdot \frac{\phi'_y}{|\nabla \phi|} \right)\right\} \\ & = \delta'(\phi)|\nabla \phi| - \left\{ \delta'(\phi)\frac{\phi'_x}{|\nabla\phi|} + \delta(\phi)\frac{\mathrm{d}}{\mathrm{d}x}\frac{\phi'_x}{|\nabla\phi|} + \delta'(\phi)\frac{\phi'_y}{|\nabla\phi|} + \delta(\phi)\frac{\mathrm{d}}{\mathrm{d}y}\frac{\phi'_y}{|\nabla\phi|} \right\} \\ & = \delta'(\phi)|\nabla \phi| - \left\{ \delta'(\phi)\left( \frac{\phi'_x+\phi'_y}{|\nabla\phi|} \right) + \delta(\phi)\left( \frac{\mathrm{d}}{\mathrm{d}x}\frac{\phi'_x}{|\nabla\phi|} + \frac{\mathrm{d}}{\mathrm{d}y}\frac{\phi'_y}{|\nabla\phi|}\right) \right\} \\ & = \delta'(\phi)|\nabla \phi| - \delta'(\phi)|\nabla \phi| - \delta(\phi)\nabla\cdot \frac{\nabla\phi}{|\nabla\phi|} \qquad //|\nabla\phi| = \sqrt{\phi'^2_x+\phi'^2_y} \\ & = -\delta(\phi)\mathrm{div}\left( \frac{\nabla\phi}{|\nabla\phi|} \right) \end{align*}$$

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得到距离函数正则化项和线积分项的泛函求导，$10$年的那篇论文就算完事了，$08$年的那篇区域多尺度，多了一次变分出来的函数$f_1$和$f_2$，该项的能量泛函为

$$\begin{align*} \mathcal{E}(\phi, f_1, f_2) & = \iint \mathcal{E}_{(x,y)}(\phi, f_1, f_2) \mathrm{d}x\mathrm{d}y \\ & = \iint \left\{ \sum_{i=1}^2 \lambda_i \cdot \iint K(x-x',y-y')|I(x',y')-f_i(x,y)|^2M_i(\phi(x',y')) \mathrm{d}x'\mathrm{d}y' \right\}\mathrm{d}x\mathrm{d}y \\ & = \int \mathcal{E}_{(x,y)}(\phi, f_1, f_2) \mathrm{d}\mathbf{x} \\ & = \int \left\{ \sum_{i=1}^2 \lambda_i \cdot \int K(\mathbf{x-y})|I(\mathbf{y})-f_i(\mathbf{x})|^2M_i(\phi(\mathbf{y})) \mathrm{d}\mathbf{y} \right\}\mathrm{d}\mathbf{x} \end{align*}$$
原文采用向量$\mathbf{x,y}$来表示，这里为了便于理解把向量分隔开，$\mathbf{x} = (x,y);\mathbf{y} = (x',y')$
变分法求该泛函的极值，由欧拉方程得到
$$\begin{align*} f_i(\mathbf{x}) & = \frac{\int K(\mathbf{x-y})M_i(\phi(\mathbf{y}))I(\mathbf{y})\mathrm{d}\mathbf{y}}{\int K(\mathbf{x})M_i(\phi(\mathbf{y}))\mathrm{d}\mathbf{y}} \\ & = \frac{K(\mathbf{x})*[M_i(\phi(\mathbf{x}))I(\mathbf{x})]}{K(\mathbf{x})*M_i(\phi(\mathbf{x}))} \qquad //卷积表示 \end{align*}$$

把该$f_1,f_2$带入上式，这时$f_1,f_2$为定值，下面的求导不应该拆解开，设

$$\begin{align*} \mathcal{E}^i_\mathbf{x}(\phi) & = \int K(\mathbf{x-y})|I(\mathbf{y})-f_i(\mathbf{x})|^2M_i(\phi(\mathbf{y})) \mathrm{d}\mathbf{y} \\ \mathcal{E}^i(\phi) & = \int \mathcal{E}^i_\mathbf{x}(\phi)\mathrm{d}\mathbf{x} \end{align*}$$

下面的说法是错误的，只是根据结果臆想的方法，具体这个公式究竟是怎么出来的还不知道
对其求变分

$$\begin{align*} \frac{\delta \mathcal{E}^i_\mathbf{x}}{\delta \phi}[\mathbf{y}] & = \frac{\delta \mathcal{E}^i_\mathbf{x}}{\delta M_i}\frac{\delta M_i}{\delta \phi}[\mathbf{y}] \quad //这是个关于\mathbf{y}的函数 \\ & = K(\mathbf{x-y})|I(\mathbf{y})-f_i(\mathbf{x})|^2 \cdot M'_i(\phi) \end{align*}$$
那么对应的$\mathcal{E}^i$的变分对应于关于$\mathbf{x}$的函数，需要把$\mathbf{y}$积分掉
$$\begin{align*} \frac{\delta \mathcal{E}^i}{\delta \phi}[\mathbf{x}] & = \int \frac{\delta \mathcal{E}^i_\mathbf{x}}{\delta \phi}[\mathbf{y}] \mathrm{d}\mathbf{y} \\ & = \int \frac{\delta\mathcal{E}^i_\mathbf{x}}{\delta M_i}\frac{\delta M_i}{\delta\phi}[\mathbf{y}] \mathrm{d}\mathbf{y} \\ & = M_i'(\phi) \int K(\mathbf{x-y})|I(\mathbf{x})-f_i(\mathbf{y})|^2\mathrm{d}\mathbf{y} \end{align*}$$

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整理一下：

$$\begin{align*} & \phi(x) \\ & M(\phi) \\ \frac{\delta M}{\delta \phi} & = M'(\phi) \\ \delta M & = M'(\phi) \cdot \delta \phi \\ & g(M,y,x) \\ G(x) & = \int g(M,y,x)\mathrm{d}y \\ \delta G(x) & = \int \frac{\partial g}{\partial M}\delta M \mathrm{d}y \\ & = \int \frac{\partial g}{\partial M}M'(\phi)\delta \phi\mathrm{d}y \\ & f(G,\phi,x) \\ F & = \int f(G,\phi,x)\mathrm{d}x \\ \delta F & = \int \frac{\partial f}{\partial G}\delta G + \frac{\partial f}{\partial \phi}\delta \phi \mathrm{d}x \\ & = \int \frac{\partial f}{\partial G}\frac{\delta G}{\delta \phi}\delta \phi + \frac{\partial f}{\partial \phi}\delta \phi \mathrm{d}x \\ & = \int \frac{\partial f}{\partial G}\left\{ \int \frac{\partial g}{\partial M}M'(\phi) \delta \phi \mathrm{d}y \right\} + \frac{\partial f}{\partial \phi}\delta \phi \mathrm{d}x \\ \frac{\delta F}{\delta \phi} & = \frac{\partial f}{\partial G} \end{align*}$$

接下来怎么搞。。。