Solenoidal場はvector potentialで表

$ \bm\nabla\cdot\bm T=\bm0\iff\bm T=\bm\nabla\times\bm A\quad\text{.for }\exist\bm A

証明

$ \bm\nabla\cdot\bm\nabla\times\bm\phi={\Large\pmb\epsilon}\vdots\bm\nabla\bm\nabla\bm\phi

$ =\epsilon_{ijk}\partial_i\partial_j\bm\phi\cdot\bm e_k\bm e_k

$ = \bm 0

$ \bm\nabla\cdot\bm T=\bm 0

$ \implies\int_{\partial V}\bm T\cdot\mathrm d\bm s=\bm0\quad\text{.for }\forall V

$ \bm\nabla\times\bm\nabla\times\bm T+\bm\nabla\cdot\bm\nabla\bm T=\bm0

$ \implies\int_{\partial V}\mathrm d\bm s\times(\bm\nabla\times\bm T)+\int_{\partial V}\mathrm d\bm s\cdot\bm\nabla\bm T=\bm 0

$ \iff-{\Large\pmb\epsilon}\cdot\bm\nabla\times\bm T+\bm\nabla\bm T=\bm 0

13:37:02 ここがおかしかった

閉曲面内で湧き出しが無いことから、tensor場が任意の場で0であることは示せない

$ \iff (-2{\cal\pmb W}+{\cal\pmb I}):\bm\nabla\bm T=\bm 0

$ \iff \tilde{\cal\pmb I}:\bm\nabla\bm T=\bm 0

$ \iff \bm\nabla\bm T=\bm 0

これはおかしい結論

$ T_{ij\cdots}=\epsilon_{ikl}\partial_kA_{lj\cdots}

$ \partial_iT_{jk\cdots}=\partial_i\epsilon_{jlm}\partial_lA_{mk\cdots}

いや、正しいかも

$ \bm\nabla(\bm\nabla\times\bm A)=\partial_i\epsilon_{jlm}\partial_lA_{mk\cdots}\bm e_i\bm e_j\bm e_k\cdots

$ = \partial_i\epsilon_{012}\partial_1A_{2k\cdots}\bm e_i\bm e_0\bm e_k\cdots+

いや、やっぱ間違っている

$ \epsilon_{ijk}\partial_j\epsilon_{klm}\partial_l T_{m\cdots}=(\delta_{il}\delta_{jm}-\delta_{jl}\delta_{im})\partial_j\partial_lT_{m\cdots}=\partial_i\partial_jT_{j\cdots}-\partial_j^2 T_{i\cdots}

成分で解く

2次元の場合

$ \frac{\partial v_x}{\partial x}+\frac{\partial v_y}{\partial y}=0

$ \iff\frac{\partial}{\partial y}\left(\frac{\partial\psi}{\partial x}+v_y\right)=0\land v_x=\frac{\partial\psi}{\partial y}\quad\text{.for }\exist\psi

$ \iff\frac{\partial\psi}{\partial x}+v_y=C'(x)\land v_x=\frac{\partial\psi}{\partial y}\quad\text{.for }\exist\psi,C

$ \iff v_y=-\frac{\partial}{\partial x}(\psi+C(x))\land v_x=\frac{\partial\psi}{\partial y}\quad\text{.for }\exist\psi,C

$ \iff v_y=-\frac{\partial}{\partial x}(\psi+C(x))\land v_x=\frac{\partial}{\partial y}(\psi+C(x))\quad\text{.for }\exist\psi,C

$ \iff\begin{pmatrix}v_x\\v_y\end{pmatrix}=\begin{pmatrix}\frac{\partial\psi}{\partial y}\\-\frac{\partial\psi}{\partial x}\end{pmatrix}\quad\text{.for }\exist\psi

$ \iff\bm v=\bm\epsilon\cdot\bm\nabla\psi\quad\text{.for }\exist\psi

2次元だとscalar函数で表せるのが特徴

2024-04-14 15:43:26 ☆と同じ方法で展開する

$ \frac{\partial v_x}{\partial x}+\frac{\partial v_y}{\partial y}=0

$ \iff B_x+B_y=C_x(x)+C_y(y)\land v_x=\frac{\partial B_x}{\partial y}\land v_y=\frac{\partial B_y}{\partial x}\quad\text{.for }\exist B_x,B_y,C_x,C_y

$ \iff\exist B_x,B_y,C_x,C_y\begin{dcases}(B_y-C_y(y))+(B_x-C_x(x))=0\\v_x=\frac{\partial}{\partial y}(B_x-C_x(x))\\v_y=\frac{\partial}{\partial x}(B_y-C_y(y))\end{dcases}

$ \iff\exist B_x,B_y,C_x,C_y\begin{dcases}(B_y-C_y(y))+(B_x-C_x(x))=0\\\begin{pmatrix}v_x\\v_y\end{pmatrix}=\begin{pmatrix}\frac{\partial}{\partial y}\\-\frac{\partial}{\partial x}\end{pmatrix}(B_x-C_x(x))\end{dcases}

$ \iff\bm v=\bm\epsilon\cdot\bm\nabla A\quad\text{.for }\exist A

3次元

$ \frac{\partial v_x}{\partial x}+\frac{\partial v_y}{\partial y}+\frac{\partial v_z}{\partial z}=0

$ \iff\frac{\partial}{\partial z}\left(\frac{\partial\psi}{\partial x}+\frac{\partial\xi}{\partial y}+v_z\right)=0\land v_x=\frac{\partial\psi}{\partial z}\land v_y=\frac{\partial\xi}{\partial z}\quad\text{.for }\exist\psi,\xi

$ \iff\frac{\partial\psi}{\partial x}+\frac{\partial\xi}{\partial y}+v_z=c(x,y)\land v_x=\frac{\partial\psi}{\partial z}\land v_y=\frac{\partial\xi}{\partial z}\quad\text{.for }\exist\psi,\xi,c

$ \iff\frac{\partial}{\partial y}\left(\frac{\partial\chi}{\partial x}+\xi+\gamma+C(x,y)\right)=0\land v_x=\frac{\partial^2\chi}{\partial z\partial y}\land v_y=\frac{\partial\xi}{\partial z}\land v_z=\frac{\partial\gamma}{\partial y}\quad\text{.for }\exist\chi,\xi,\gamma,C

$ \iff\frac{\partial\chi}{\partial x}+\xi+\gamma+C(x,y)+D(z,x)=0\land v_x=\frac{\partial^2\chi}{\partial z\partial y}\land v_y=\frac{\partial\xi}{\partial z}\land v_z=\frac{\partial\gamma}{\partial y}\quad\text{.for }\exist\chi,\xi,\gamma,C,D

最初から3成分とも2階偏微分させたほうが楽そう

$ \iff\frac{\partial}{\partial x}\left(\frac{\partial^2\psi}{\partial y\partial z}+\frac{\partial^2\xi}{\partial y\partial z}+\frac{\partial^2\chi}{\partial y\partial z}\right)=0\land v_x=\frac{\partial^2\psi}{\partial y\partial z}\land v_y=\frac{\partial^2\xi}{\partial z\partial x}\land v_z=\frac{\partial^2\chi}{\partial x\partial y}\quad\text{.for }\exist\psi,\xi,\chi

$ \iff\frac{\partial}{\partial y}\left(\frac{\partial\psi}{\partial z}+\frac{\partial\xi}{\partial z}+\frac{\partial\chi}{\partial z}\right)=C(y,z)\land v_x=\frac{\partial^2\psi}{\partial y\partial z}\land v_y=\frac{\partial^2\xi}{\partial z\partial x}\land v_z=\frac{\partial^2\chi}{\partial x\partial y}\quad\text{.for }\exist\psi,\xi,\chi,C

$ \iff\frac{\partial}{\partial z}\left(\psi+\xi+\chi\right)=C(y,z)+D(z,x)\land v_x=\frac{\partial^2\psi}{\partial y\partial z}\land v_y=\frac{\partial^2\xi}{\partial z\partial x}\land v_z=\frac{\partial^2\chi}{\partial x\partial y}\quad\text{.for }\exist\psi,\xi,\chi,C,D

$ \iff\psi+\xi+\chi=C(y,z)+D(z,x)+E(x,y)\land v_x=\frac{\partial^2\psi}{\partial y\partial z}\land v_y=\frac{\partial^2\xi}{\partial z\partial x}\land v_z=\frac{\partial^2\chi}{\partial x\partial y}\quad\text{.for }\exist\psi,\xi,\chi,C,D,E

$ \iff\exist\psi,\xi,\chi,C,D,E\begin{dcases}\psi+\xi+\chi=C(y,z)+D(z,x)+E(x,y)\\ v_x=\frac{\partial^2}{\partial y\partial z}(C-\xi-\chi)\\ v_y=\frac{\partial^2}{\partial z\partial x}(D-\chi-\psi)\\ v_z=\frac{\partial^2}{\partial x\partial y}(E-\psi-\xi)\end{dcases}

ええ……ここからどう展開するんだ？

あ、思いついた！

$ \iff\exist\psi,\xi,\chi,C,D,E\begin{dcases}\psi+\xi+\chi=C(y,z)+D(z,x)+E(x,y)\\ v_x=\frac{\partial^2}{\partial y\partial z}(C-\xi+D-\chi)\\ v_y=\frac{\partial^2}{\partial z\partial x}(D-\chi+E-\psi)\\ v_z=\frac{\partial^2}{\partial x\partial y}(E-\psi+C-\xi)\end{dcases}

$ \because\frac{\partial^2D}{\partial y\partial z}=\frac{\partial^2E}{\partial z\partial x}=\frac{\partial^2C}{\partial x\partial y}=0

$ \iff\exist\psi,\xi,\chi,C,D,E\begin{dcases}\psi+\xi+\chi=C(y,z)+D(z,x)+E(x,y)\\v_x=\frac{\partial}{\partial y}\frac{\partial}{\partial z}(D-\chi)+\frac{\partial}{\partial z}\frac{\partial}{\partial y}(C-\xi)\\v_y=\frac{\partial}{\partial z}\frac{\partial}{\partial x}(E-\psi)+\frac{\partial}{\partial x}\frac{\partial}{\partial z}(D-\chi)\\v_z=\frac{\partial}{\partial x}\frac{\partial}{\partial y}(C-\xi)+\frac{\partial}{\partial y}\frac{\partial}{\partial x}(E-\psi)\end{dcases}

いや、これだと反対称にならないか……

展開し直し

$ \iff\exist B_x,B_y,B_z,C_x,C_y,C_z\begin{dcases}(B_x-C_x(x,y))+(B_y-C_y(y,z))+(B_z-C_z(z,x))=0\\ v_x=\frac{\partial^2}{\partial y\partial z}(B_x-C_x)\\ v_y=\frac{\partial^2}{\partial z\partial x}(B_y-C_y)\\ v_z=\frac{\partial^2}{\partial x\partial y}(B_z-C_z)\end{dcases}

$ \iff\exist B_x,B_y,B_z,C_x,C_y,C_z\begin{dcases}v_x=\frac{\partial}{\partial y}\frac{\partial}{\partial z}(-(B_y-C_y))-\frac{\partial}{\partial z}\frac{\partial}{\partial y}(B_z-C_z)\\ v_y=\frac{\partial0}{\partial z}-\frac{\partial}{\partial x}\frac{\partial}{\partial z}(-(B_y-C_y))\\ v_z=\frac{\partial}{\partial x}\frac{\partial}{\partial y}(B_z-C_z)-\frac{\partial0}{\partial y}\end{dcases}

$ \iff\exist A_y,A_z\begin{pmatrix}v_x\\v_y\\v_z\end{pmatrix}=\begin{pmatrix}\frac{\partial A_z}{\partial y}-\frac{\partial A_y}{\partial z}\\\frac{\partial0}{\partial z}-\frac{\partial A_z}{\partial x}\\\frac{\partial A_y}{\partial x}-\frac{\partial 0}{\partial y}\end{pmatrix}

$ \implies\exist\bm A;\bm v=\bm\nabla\times\bm A

なんとか求まった

2024-04-15 01:14:29 3変数のまま解き進められないか？

$ \iff\exist A_x,A_y,A_z\begin{dcases}A_x+A_y+A_z=0\\ v_x=\frac{\partial^2A_x}{\partial y\partial z}\\ v_y=\frac{\partial^2A_y}{\partial z\partial x}\\ v_z=\frac{\partial^2A_z}{\partial x\partial y}\end{dcases}

$ \iff\exist A_x,A_y,A_z\begin{dcases}A_x+A_y+A_z=0\\ v_x=\frac{\partial^2(-A_y)}{\partial y\partial z}-\frac{\partial^2A_z}{\partial z\partial y}\\ v_y=-\frac{\partial^2A_z}{\partial z\partial x}-\frac{\partial^2A_x}{\partial x\partial z}\\ v_z=-\frac{\partial^2A_x}{\partial x\partial y}-\frac{\partial^2A_y}{\partial y\partial x}\end{dcases}

あー、そもそも偏微分もつじつま合わせできなくなるのか