세미나
Seminar
220404
Google Drive
원드라이브
Viscous Burgers’ equation
$$ u_{t} + u u_{x} = \mu u_{xx} $$
Let $\alpha := \mu dt / dx^{2}$.
$$ \begin{align*} & {{ u \left( t + dt, x \right) - u } \over { dt }} + u {{ u \left( t , x + dx \right) - u } \over { dx }} = \mu {{ u \left( t , x + dx \right) - 2 u + u \left( t , x - dx \right) } \over { dx^{2} }} \\ \implies & u \left( t + dt, x \right) - u + dt u {{ u \left( t , x + dx \right) - u } \over { dx }} = \alpha \left[ u \left( t , x + dx \right) - 2 u + u \left( t , x - dx \right) \right] \\ \implies & u \left( t + dt, x \right) = u - dt u {{ u \left( t , x + dx \right) - u } \over { dx }} + \alpha \left[ u \left( t , x + dx \right) - 2 u + u \left( t , x - dx \right) \right] \\ \implies & u \left( t + dt, x \right) = u \left[ 1 - dt {{ u \left( t , x + dx \right) - u } \over { dx }} \right] + \alpha \left[ u \left( t , x + dx \right) - 2 u + u \left( t , x - dx \right) \right] \\ \implies & u \left( t + dt, x \right) = u \left[ (1 - 2 \alpha ) - dt {{ u \left( t , x + dx \right) - u } \over { dx }} \right] + \alpha \left[ u \left( t , x + dx \right) + u \left( t , x - dx \right) \right] \end{align*} $$
Inviscid Burgers’ equation
$$ u_{t} + u u_{x} = 0 $$
$$ \begin{align*} & {{ u \left( t + dt, x \right) - u } \over { dt }} + u {{ u \left( t , x + dx \right) - u } \over { dx }} = 0 \\ \implies & u \left( t + dt, x \right) - u + dt u {{ u \left( t , x + dx \right) - u } \over { dx }} = 0 \\ \implies & u \left( t + dt, x \right) = u - dt u {{ u \left( t , x + dx \right) - u } \over { dx }} \\ \implies & u \left( t + dt, x \right) = u \left[ 1 - dt {{ u \left( t , x + dx \right) - u } \over { dx }} \right] \end{align*} $$
Heat equation
$$ u_{t} = \mu u_{xx} $$
Let $\alpha := \mu dt / dx^{2}$.
$$ \begin{align*} & {{ u \left( t + dt, x \right) - u } \over { dt }} = \mu {{ u \left( t , x + dx \right) - 2 u + u \left( t , x - dx \right) } \over { dx^{2} }} \\ \implies & u \left( t + dt, x \right) - u = \alpha \left[ u \left( t , x + dx \right) - 2 u + u \left( t , x - dx \right) \right] \\ \implies & u \left( t + dt, x \right) = u + \alpha \left[ u \left( t , x + dx \right) - 2 u + u \left( t , x - dx \right) \right] \\ \implies & u \left( t + dt, x \right) = (1 - 2 \alpha) u + \alpha \left[ u \left( t , x + dx \right) + u \left( t , x - dx \right) \right] \end{align*} $$
Independency of the terms
First we have to consider the updating rules as addition operattions. For Inviscid Burgers’ equation $$ \begin{align*} u \left( t + dt, x \right) =& u + f \left( u \right) \\ :=& u - u dt {{ u \left( t , x + dx \right) - u } \over { dx }} \end{align*} $$ and for Heat equation $$ \begin{align*} u \left( t + dt, x \right) =& u + g \left( u \right) \\ :=& u + \alpha \left[ u \left( t , x + dx \right) - 2 u + u \left( t , x - dx \right) \right] \end{align*} $$ then for Viscous Burgers’ equation $$ \begin{align*} & u \left( t + dt, x \right) \\ =& u \left[ (1 - 2 \alpha ) - dt {{ u \left( t , x + dx \right) - u } \over { dx }} \right] + \alpha \left[ u \left( t , x + dx \right) + u \left( t , x - dx \right) \right] \\ =& u - 2 \alpha u - u dt {{ u \left( t , x + dx \right) - u } \over { dx }} + \alpha \left[ u \left( t , x + dx \right) + u \left( t , x - dx \right) \right] \\ =& u - 2 \alpha u + f(u) + \alpha \left[ u \left( t , x + dx \right) + u \left( t , x - dx \right) \right] \\ =& u + f(u) + \alpha \left[ u \left( t , x + dx \right) - 2 u + u \left( t , x - dx \right) \right] \\ =& u + f(u) + g(u) \end{align*} $$ In conclusion, $f(u)$ and $g(u)$ must be obtained independently hence we can not apply two updating rules sequentially.
Hyperbolic?
$$ u(t + dt , x) = u \left[ (1 - 2 \alpha ) - dt {{ u \left( t , x + dx \right) - u } \over { dx }} \right] + \cdots \\ \implies 2 \alpha + dt {{ u \left( t , x + dx \right) - u } \over { dx }} < 1 \\ \implies \alpha < {{ 1 } \over { 2 }} \left[ 1 - dt {{ u \left( t , x + dx \right) - u } \over { dx }} \right] $$