前回計算した $1$ 階の導関数は

\begin{align} \partial_pg(y)=\dfrac{k(1)}{r}\int_0^{2\pi}\langle p,e(\theta)\rangle f(re(\theta)+y)\,d\theta-\int_0^1\dfrac{X\partial k(X)}{r}\bigg\{\int_0^{2\pi}\langle p,e(\theta)\rangle f(rXe(\theta)+y)\,d\theta\bigg\}\,dX \end{align}

でした。今回はもう一回微分した $\partial_q\partial_pg$ を計算します。第 $1$ 項の計算は簡単ですね。

\begin{align} &\dfrac{k(1)}{r}\int_0^{2\pi}\langle p,e(\theta)\rangle \{\langle q,e(\theta)\rangle\partial_{e(\theta)}f(re(\theta)+y)+\langle q,e(\theta+\pi/2)\rangle\partial_{e(\theta+\pi/2)}f(re(\theta)+y)\}\,d\theta\\ &=\dfrac{k(1)}{r}\int_0^{2\pi}\langle p,e(\theta)\rangle \bigg\{\langle q,e(\theta)\rangle\partial_rf(re(\theta)+y)+\dfrac{\langle q,e(\theta+\pi/2)\rangle}{r}\partial_{\theta}f(re(\theta)+y)\bigg\}\,d\theta\\ &=\dfrac{k(1)}{r}\int_0^{2\pi}\langle p,e(\theta)\rangle\langle q,e(\theta)\rangle\partial_rf(re(\theta)+y)\,d\theta\\ &\quad -\dfrac{k(1)}{r^2}\int_0^{2\pi}\{\langle p,e(\theta+\pi/2)\rangle\langle q,e(\theta+\pi/2)\rangle-\langle p,e(\theta)\rangle\langle q,e(\theta)\rangle\}f(re(\theta)+y)\,d\theta \end{align}

一方、第 $2$ 項は

\begin{align} &-\int_0^1\dfrac{X\partial k(X)}{r^2}\bigg\{\int_0^{2\pi}\langle p,e(\theta)\rangle\langle q,e(\theta)\rangle\partial_Xf(rXe(\theta)+y)\bigg\}\,dX\\ &-\int_0^1\dfrac{\partial k(X)}{r^2}\bigg\{\int_0^{2\pi}\langle p,e(\theta)\rangle\langle q,e(\theta+\pi/2)\rangle\partial_{\theta}f(rXe(\theta)+y)\bigg\}\,dX\\ &=-\dfrac{1}{r^2}\int_0^{2\pi}\langle p,e(\theta)\rangle\langle q,e(\theta)\rangle\bigg\{\partial k(1)f(re(\theta)+y)-\int_0^1\{\partial k(X)+X\partial^2k(X)\}f(rXe(\theta)+y)\,dX\bigg\}\,d\theta\\ &+\int_0^1\dfrac{\partial k(X)}{r^2}\bigg\{\int_0^{2\pi}\{\langle p,e(\theta+\pi/2)\rangle\langle q,e(\theta+\pi/2)\rangle-\langle p,e(\theta)\rangle\langle q,e(\theta)\rangle\}f(rXe(\theta)+y)\,d\theta\bigg\}\,dX \end{align}

以上より $2$ 階の導関数は

\begin{align} \partial_q\partial_pg&=\dfrac{k(1)}{r}\int_0^{2\pi}\langle p,e(\theta)\rangle\langle q,e(\theta)\rangle\partial_rf(re(\theta)+y)\,d\theta\\ &\quad -\dfrac{k(1)}{r^2}\int_0^{2\pi}\{\langle p,e(\theta+\pi/2)\rangle\langle q,e(\theta+\pi/2)\rangle-\langle p,e(\theta)\rangle\langle q,e(\theta)\rangle\}f(re(\theta)+y)\,d\theta\\ &\quad -\dfrac{1}{r^2}\int_0^{2\pi}\langle p,e(\theta)\rangle\langle q,e(\theta)\rangle\bigg\{\partial k(1)f(re(\theta)+y)-\int_0^1\{\partial k(X)+X\partial^2k(X)\}f(rXe(\theta)+y)\,dX\bigg\}\,d\theta\\ &+\int_0^1\dfrac{\partial k(X)}{r^2}\bigg\{\int_0^{2\pi}\{\langle p,e(\theta+\pi/2)\rangle\langle q,e(\theta+\pi/2)\rangle-\langle p,e(\theta)\rangle\langle q,e(\theta)\rangle\}f(rXe(\theta)+y)\,d\theta\bigg\}\,dX \end{align}

かなり複雑な式になってしまいましたが、独立な場合すなわち $(p,q)=(e_1,e_1),(p,q)=(e_2,e_2),(p,q)=(e_1,e_2)$ の場合を具体的に書いてみます。

\begin{align} \partial^2_1g&=\dfrac{k(1)}{r}\int_0^{2\pi}\cos^2\theta\partial_rf(re(\theta)+y)\,d\theta\\ &\quad -\dfrac{k(1)}{r^2}\int_0^{2\pi}(\sin^2\theta-\cos^2\theta)f(re(\theta)+y)\,d\theta\\ &\quad -\dfrac{1}{r^2}\int_0^{2\pi}\cos^2\theta\bigg\{\partial k(1)f(re(\theta)+y)-\int_0^1\{\partial k(X)+X\partial^2k(X)\}f(rXe(\theta)+y)\,dX\bigg\}\,d\theta\\ &+\int_0^1\dfrac{\partial k(X)}{r^2}\bigg\{\int_0^{2\pi}(\sin^2\theta-\cos^2\theta)f(rXe(\theta)+y)\,d\theta\bigg\}\,dX \end{align} \begin{align} \partial^2_2g&=\dfrac{k(1)}{r}\int_0^{2\pi}\sin^2\theta\partial_rf(re(\theta)+y)\,d\theta\\ &\quad -\dfrac{k(1)}{r^2}\int_0^{2\pi}(\cos^2\theta-\sin^2\theta)f(re(\theta)+y)\,d\theta\\ &\quad -\dfrac{1}{r^2}\int_0^{2\pi}\sin^2\theta\bigg\{\partial k(1)f(re(\theta)+y)-\int_0^1\{\partial k(X)+X\partial^2k(X)\}f(rXe(\theta)+y)\,dX\bigg\}\,d\theta\\ &+\int_0^1\dfrac{\partial k(X)}{r^2}\bigg\{\int_0^{2\pi}(\cos^2\theta-\sin^2\theta)f(rXe(\theta)+y)\,d\theta\bigg\}\,dX \end{align} \begin{align} \partial_1\partial_2g&=\dfrac{k(1)}{r}\int_0^{2\pi}\cos\theta\sin\theta\partial_rf(re(\theta)+y)\,d\theta\\ &\quad +\dfrac{2k(1)}{r^2}\int_0^{2\pi}\sin\theta\cos\theta f(re(\theta)+y)\,d\theta\\ &\quad -\dfrac{1}{r^2}\int_0^{2\pi}\cos\theta\sin\theta\bigg\{\partial k(1)f(re(\theta)+y)-\int_0^1\{\partial k(X)+X\partial^2k(X)\}f(rXe(\theta)+y)\,dX\bigg\}\,d\theta\\ &-2\int_0^1\dfrac{\partial k(X)}{r^2}\bigg\{\int_0^{2\pi}\sin\theta\cos\theta f(rXe(\theta)+y)\,d\theta\bigg\}\,dX \end{align}

今回はここまで。計算練習のような回になってしまいましたが、次回はデコンボリューションの方針である $\psi$ を計算することに整合する導関数を考えましょう。(written by J.Nishigata)

研究11- 畳込みの 2 階導関数