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Team:HokkaidoU Japan/Projects/Modelling
From 2014.igem.org
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\[ \frac{d}{dt} \left( \matrix{\delta X\cr \delta Y\cr \delta Z\cr} \right)=\left( \matrix{-k_{bind}-b_X & -k_{bind} & k_{unbind} \cr -k_{bind} & -k_{bind}-b_Y & k_{unbind} \cr k_{bind} & k_{bind} & -k_{unbind}-b_Z} \right) \left( \matrix{\delta X\cr \delta Y\cr \delta Z\cr} \right) \] | \[ \frac{d}{dt} \left( \matrix{\delta X\cr \delta Y\cr \delta Z\cr} \right)=\left( \matrix{-k_{bind}-b_X & -k_{bind} & k_{unbind} \cr -k_{bind} & -k_{bind}-b_Y & k_{unbind} \cr k_{bind} & k_{bind} & -k_{unbind}-b_Z} \right) \left( \matrix{\delta X\cr \delta Y\cr \delta Z\cr} \right) \] | ||
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| + | この行列をAとおく。det(A-\lambda I)=0を解いて\lambda<0を示し、解が安定であることを証明する。 | ||
| + | 次に具体的に解を求めて云々。 | ||
</html> | </html> | ||
Latest revision as of 15:48, 27 September 2014
\[ \zeta(s) = \sum_{n=1}^\infty\frac{1}{n^s} \] \[ \sigma_1 = \left( \matrix{ 0 & 1 \cr 1 & 0 } \right), \sigma_2 = \left( \matrix{ 0 & -i \cr i & 0 } \right), \sigma_3 = \left( \matrix{ 1 & 0 \cr 0 & -1} \right) は,以下の式を満たす. \left[ \frac{\sigma_i}{2}~,~\frac{\sigma_j}{2} \right] = i\varepsilon_{ijk} \frac{\sigma_k}{2} \] \begin{cases} \dot{X}=a_X-k_{bind}XY+k_{unbind}Z-b_XX & \\ \dot{Y}=a_Y-k_{bind}XY+k_{unbind}Z-b_YY & \\ \dot{Z}=k_{bind}XY-k_{unbind}Z-b_ZZ & \end{cases} \[ \frac{d}{dt} \left( \matrix{\delta X\cr \delta Y\cr \delta Z\cr} \right)=\left( \matrix{-k_{bind}-b_X & -k_{bind} & k_{unbind} \cr -k_{bind} & -k_{bind}-b_Y & k_{unbind} \cr k_{bind} & k_{bind} & -k_{unbind}-b_Z} \right) \left( \matrix{\delta X\cr \delta Y\cr \delta Z\cr} \right) \] この行列をAとおく。det(A-\lambda I)=0を解いて\lambda<0を示し、解が安定であることを証明する。 次に具体的に解を求めて云々。
