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Fluid Dynamics and Hyperbolic Equations

$\mathit{\rho}\mathit{v}\mathit{v}=\mathit{\rho}{v}_{i}{v}_{j}=\mathit{\rho}\left(\begin{array}{ccc}\hfill {v}_{1}{v}_{1}\hfill & \hfill {v}_{1}{v}_{2}\hfill & \hfill {v}_{1}{v}_{3}\hfill \\ \hfill {v}_{2}{v}_{1}\hfill & \hfill {v}_{2}{v}_{2}\hfill & \hfill {v}_{2}{v}_{3}\hfill \\ \hfill {v}_{3}{v}_{1}\hfill & \hfill {v}_{3}{v}_{2}\hfill & \hfill {v}_{3}{v}_{3}\hfill \end{array}\right).$ | $(7.1)$ |

Figure 7.1: Integral of momentum flux density across the boundary surface
$\partial V$ is equal to minus the integral of rate of change of
momentum minus force density over the volume $V$. The momentum
flux density includes convective flux and stress tensor parts.

The conservation of momentum is then
$\frac{\partial}{\partial t}{\int}_{V}\mathit{\rho}\mathit{v}{d}^{3}x={\int}_{V}\mathit{F}{d}^{3}x-{\int}_{\partial V}(\mathit{\rho}\mathit{v}\mathit{v}+\mathbf{P}).\mathit{dA}.$ | $(7.2)$ |

${\int}_{V}\frac{\partial}{\partial t}(\mathit{\rho}\mathit{v})-\mathit{F}+\nabla .(\mathit{\rho}\mathit{v}\mathit{v}+\mathbf{P}){d}^{3}x=0.$ | $(7.3)$ |

$\frac{\partial}{\partial t}(\mathit{\rho}\mathit{v})-\mathit{F}+\nabla .(\mathit{\rho}\mathit{v}\mathit{v}+\mathbf{P})=0.$ | $(7.4)$ |

Figure 7.2: The transfer in the $y$-direction of $x$-momentum arises
from the rate of strain ${\mathit{dv}}_{x}/\mathit{dy}$. The rate of strain tensor is
the symmetric generalization of this form.

The traceless stress tensor $\mathit{\sigma}$
for simple fluids arises from viscosity, which relates stress to the
rate of strain tensor, see Fig. 7.2. The rate of strain tensor
is $\frac{1}{2}(\frac{\partial {v}_{i}}{\partial {x}_{j}}+\frac{\partial {v}_{j}}{\partial {x}_{i}})$. And $\mathit{\sigma}$ is proportional to
its traceless part
${\mathit{\sigma}}_{\mathit{ij}}=\mathit{\mu}[(\frac{\partial {v}_{i}}{\partial {x}_{j}}+\frac{\partial {v}_{j}}{\partial {x}_{i}})-\frac{2}{3}\nabla .\mathit{v}{\mathit{\delta}}_{\mathit{ij}}]=\mathit{\mu}{[((\nabla \mathit{v})+(\nabla \mathit{v}{)}^{T})-\frac{2}{3}(\nabla .\mathit{v})\mathbf{I}]}_{\mathit{ij}}.$ | $(7.5)$ |

$\frac{\partial}{\partial t}(\mathit{\rho}\mathit{v})+\nabla .(\mathit{\rho}\mathit{v}\mathit{v})=-\nabla .(p\mathbf{I}+\mathit{\sigma})+\mathit{F}=-\nabla p-\mathit{\mu}{\nabla}^{2}\mathit{v}-\frac{1}{3}\mathit{\mu}\nabla (\nabla .\mathit{v})+\mathit{F}.$ | $(7.6)$ |

$\frac{\partial}{\partial t}(\mathit{\rho}\mathit{v})+\nabla .(\mathit{\rho}\mathit{v}\mathit{v})=-\nabla p-\mathit{\mu}{\nabla}^{2}\mathit{v}+\mathit{F}.$ | $(7.7)$ |

$\frac{\partial}{\partial t}(\mathit{\rho}\mathit{v})+\nabla .(\mathit{\rho}\mathit{v}\mathit{v})=\mathit{\rho}(\frac{\partial}{\partial t}\mathit{v}+\mathit{v}.\nabla \mathit{v}).$ | $(7.8)$ |

$\begin{array}{cccc}\mathit{Continuity}:\mathrm{\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}}\hfill & \hfill \frac{\partial \mathit{\rho}}{\partial t}+\frac{\partial}{\partial x}(\mathit{\rho}v)& \hfill =\hfill & 0\hfill \\ \mathit{Momentum}:\mathrm{\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}}\hfill & \hfill \frac{\partial}{\partial t}(\mathit{\rho}v)+\frac{\partial}{\partial x}(\mathit{\rho}{v}^{2})& \hfill =\hfill & -\frac{\partial}{\partial x}p\hfill \\ \mathit{State}:\mathrm{\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}}\hfill & \hfill p{\mathit{\rho}}^{-\mathit{\gamma}}& \hfill =\hfill & \mathit{const}.\hfill \end{array}$ | $(7.9)$ |

$\begin{array}{ccc}\hfill \frac{\partial \mathit{\rho}}{\partial t}& \hfill =\hfill & -\frac{\partial \mathit{\Gamma}}{\partial x}\hfill \\ \hfill \frac{\partial \mathit{\Gamma}}{\partial t}& \hfill =\hfill & -\frac{\partial}{\partial x}({\mathit{\Gamma}}^{2}/\mathit{\rho}+({p}_{0}/{\mathit{\rho}}_{0}^{\mathit{\gamma}}){\mathit{\rho}}^{\mathit{\gamma}})\hfill \end{array}.$ | $(7.10)$ |

$\frac{\partial \mathbf{u}}{\partial t}=-\frac{\partial \mathbf{f}}{\partial x}.$ | $(7.11)$ |

$\mathbf{u}=\left(\begin{array}{c}\hfill \mathit{\rho}\hfill \\ \hfill \mathit{\Gamma}\hfill \end{array}\right),\mathrm{\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}}\text{and}\mathrm{\hspace{1em}\hspace{1em}\hspace{1em}}\mathbf{f}=\left(\begin{array}{c}\hfill \mathit{\Gamma}\hfill \\ \hfill {\mathit{\Gamma}}^{2}/\mathit{\rho}+({p}_{0}/{\mathit{\rho}}_{0}^{\mathit{\gamma}}){\mathit{\rho}}^{\mathit{\gamma}}\hfill \end{array}\right)$ | $(7.12)$ |

$\frac{\partial \mathbf{u}}{\partial t}=-\frac{\partial \mathbf{f}}{\partial \mathbf{u}}\frac{\partial \mathbf{u}}{\partial x}=-\sum _{m=1,M}\frac{\partial \mathbf{f}}{\partial {u}_{m}}\frac{\partial {u}_{m}}{\partial x}=-\mathbf{J}\frac{\partial \mathbf{u}}{\partial x}.$ | $(7.13)$ |

$\mathbf{J}=\left(\begin{array}{cc}\hfill 0\hfill & \hfill 1\hfill \\ \hfill -{\mathit{\Gamma}}^{2}/{\mathit{\rho}}^{2}+\mathit{\gamma}({p}_{0}/{\mathit{\rho}}_{0}^{\mathit{\gamma}}){\mathit{\rho}}^{\mathit{\gamma}-1}\hfill & \hfill 2\mathit{\Gamma}/\mathit{\rho}\hfill \end{array}\right).$ | $(7.14)$ |

$\left|\begin{array}{cc}\hfill -\mathit{\lambda}\hfill & \hfill 1\hfill \\ \hfill -{v}^{2}+{c}_{s}^{2}\hfill & \hfill -\mathit{\lambda}+2v\hfill \end{array}\right|={\mathit{\lambda}}^{2}-2v\mathit{\lambda}+{v}^{2}-{c}_{s}^{2}=0,$ | $(7.15)$ |

$\mathit{\lambda}=v\pm {c}_{s}.$ | $(7.16)$ |

Figure 7.3: Derivatives in time ($n$) and space ($j$) implemented as
finite differences give rise to values at the half-mesh points
**x**.

We could try to
fix that by taking centered derivatives; but it turns out that may
make things worse. The scheme may become unstable. But how do we
analyse stability for this fluid? We have multiple
coupled dependent
variables. How do we deal with that?
The answer, in summary, is that we find the combinations of dependent
variables that behave in a way that is approximately uncoupled from
the other combinations of dependent variables - in other words, the
characteristic ${\mathbf{e}}_{\pm}=\left(\begin{array}{c}\hfill 1\hfill \\ \hfill v\pm {c}_{s}\hfill \end{array}\right)$ | $(7.17)$ |

$\left(\begin{array}{c}\hfill \mathit{\rho}\hfill \\ \hfill \mathit{\Gamma}\hfill \end{array}\right)={q}_{+}\left(\begin{array}{c}\hfill 1\hfill \\ \hfill v+{c}_{s}\hfill \end{array}\right)+{q}_{-}\left(\begin{array}{c}\hfill 1\hfill \\ \hfill v-{c}_{s}\hfill \end{array}\right).$ | $(7.18)$ |

$\mathbf{J}\mathbf{u}={q}_{+}\mathbf{J}{\mathbf{e}}_{+}+{q}_{-}\mathbf{J}{\mathbf{e}}_{-}={q}_{+}{\mathit{\lambda}}_{+}{\mathbf{e}}_{+}+{q}_{-}{\mathit{\lambda}}_{-}{\mathbf{e}}_{-}\mathrm{\hspace{0.5em}\hspace{0.5em}}.$ | $(7.19)$ |

$\stackrel{\u203e}{\mathbf{J}}\mathbf{q}=\left(\genfrac{}{}{0ex}{}{{q}_{+}{\mathit{\lambda}}_{+}}{{q}_{-}{\mathit{\lambda}}_{-}}\right)=\left(\begin{array}{cc}\hfill {\mathit{\lambda}}_{+}\hfill & \hfill 0\hfill \\ \hfill 0\hfill & \hfill {\mathit{\lambda}}_{-}\hfill \end{array}\right)\left(\begin{array}{c}\hfill {q}_{+}\hfill \\ \hfill {q}_{-}\hfill \end{array}\right).$ | $(7.20)$ |

$\frac{\partial {q}_{\pm}}{\partial t}=-{\mathit{\lambda}}_{\pm}\frac{\partial {q}_{\pm}}{\partial x},$ | $(7.21)$ |

Figure 7.4: Forward derivative in Time ($n$) and Centered in Space ($j$) (FTCS)
finite differences give rise to an unstable scheme for hyperbolic problems.

For stability
analysis purposes, we can suppose that
we are using the new representation (in other words $u$ stands for
each $q$ which we can consider separately in scalar equations).
But we
don't actually do the transformation to that new representation when
implementing the scheme (only when analysing its stability). The first
time through we'll do things explicitly but then take short cuts
thereafter not bringing $q$ in explicitly. We write
out the difference equation as
${\mathbf{u}}_{j}^{(n+1)}-{\mathbf{u}}_{j}^{(n)}=-\frac{\mathit{\Delta}t}{2\mathit{\Delta}x}({\mathbf{f}}_{j+1}^{(n)}-{\mathbf{f}}_{j-1}^{(n)})=-\frac{\mathit{\Delta}t}{2\mathit{\Delta}x}\mathbf{J}({\mathbf{u}}_{j+1}^{(n)}-{\mathbf{u}}_{j-1}^{(n)}),$ | $(7.22)$ |

${q}_{j}^{(n+1)}-{q}_{j}^{(n)}=-\frac{\mathit{\Delta}t}{2\mathit{\Delta}x}\mathit{\lambda}({q}_{j+1}^{(n)}-{q}_{j-1}^{(n)})$ | $(7.23)$ |

${q}_{j}=q\mathrm{exp}({\mathit{ik}}_{x}j\mathit{\Delta}x),\mathrm{\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}}{f}_{j}=f\mathrm{exp}({\mathit{ik}}_{x}j\mathit{\Delta}x).$ | $(7.24)$ |

${q}_{j}^{(n+1)}=[1-\frac{\mathit{\Delta}t\mathit{\lambda}}{2\mathit{\Delta}x}({e}^{{\mathit{ik}}_{x}\mathit{\Delta}x}-{e}^{-{\mathit{ik}}_{x}\mathit{\Delta}x})]{q}_{j}^{(n)}=[1-i\frac{\mathit{\Delta}t\mathit{\lambda}}{\mathit{\Delta}x}\mathrm{sin}({k}_{x}\mathit{\Delta}x)]{q}_{j}^{(n)}.$ | $(7.25)$ |

$A=1-i\frac{\mathit{\Delta}t\mathit{\lambda}}{\mathit{\Delta}x}\mathrm{sin}({k}_{x}\mathit{\Delta}x).$ | $(7.26)$ |

Figure 7.5: Forward derivative in Time ($n$) but from the mean of the
adjacent points and Centered in Space ($j$) is the Lax Friedrichs
finite difference scheme, which is stable provided $\mathit{\Delta}t\le \mathit{\Delta}x/|\mathit{\lambda}|$.

This is then called the
Lax-Friedrichs method.
${\mathbf{u}}_{j}^{(n+1)}-({\mathbf{u}}_{j-1}^{(n)}+{\mathbf{u}}_{j+1}^{(n)})/2=-\frac{\mathit{\Delta}t}{2\mathit{\Delta}x}({\mathbf{f}}_{j+1}^{(n)}-{\mathbf{f}}_{j-1}^{(n)})=-\frac{\mathit{\Delta}t}{2\mathit{\Delta}x}\mathbf{J}({\mathbf{u}}_{j+1}^{(n)}-{\mathbf{u}}_{j-1}^{(n)}).$ | $(7.27)$ |

$A=\mathrm{cos}({k}_{x}\mathit{\Delta}x)-i\frac{\mathit{\Delta}t\mathit{\lambda}}{\mathit{\Delta}x}\mathrm{sin}({k}_{x}\mathit{\Delta}x).$ | $(7.28)$ |

$\mathit{\Delta}t\le \mathit{\Delta}x/|\mathit{\lambda}|.$ | $(7.29)$ |

${\mathbf{u}}_{j+1/2}^{(n+1/2)}=\frac{1}{2}({\mathbf{u}}_{j+1}^{(n)}+{\mathbf{u}}_{j}^{(n)})-\frac{\mathit{\Delta}t}{2\mathit{\Delta}x}({\mathbf{f}}_{j+1}^{(n)}-{\mathbf{f}}_{j}^{(n)})$ | $(7.30)$ |

${\mathbf{u}}_{j}^{(n+1)}={\mathbf{u}}_{j}^{(n)}-\frac{\mathit{\Delta}t}{\mathit{\Delta}x}({\mathbf{f}}_{j+1/2}^{(n+1/2)}-{\mathbf{f}}_{j-1/2}^{(n+1/2)}).$ | $(7.31)$ |

Figure 7.6: The Lax-Wendroff two-step scheme first (dashed lines)
generates $u$ and hence $f$ values at the half-time-step $n+\frac{1}{2}$,
by a Lax-Friedrichs advance to (**X**). Then it uses a centered time,
centered space full step advance based upon ${\mathbf{f}}^{(n+1/2)}$,
from the ${\mathbf{u}}^{(n)}$.

The first step is like a Lax-Friedrichs half-step to the half-way
positions. Then the fluxes are evaluated again, at the half-step
times and positions using the ${\mathbf{u}}^{(n+1/2)}$ values, to find the
${\mathbf{f}}^{(n+1/2)}$. Those are used in the second step to advance all
the way from $n$ to $n+1$ in a properly centered manner.
The amplification factor for the combined step can be shown to be
$A=1-i\frac{\mathit{\Delta}t\mathit{\lambda}}{\mathit{\Delta}x}\mathrm{sin}({k}_{x}\mathit{\Delta}x)+{\left(\frac{\mathit{\Delta}t\mathit{\lambda}}{\mathit{\Delta}x}\right)}^{2}[\mathrm{cos}({k}_{x}\mathit{\Delta}x)-1]$ | $(7.32)$ |

Start with the formula for the first time half-step: eq. (7.30). For stability analysis (but not in implementing an actual numerical scheme), approximate the Jacobian matrix locally as uniform, and substitute $\mathbf{f}=\mathbf{J}\mathbf{u}$ at all the required mesh positions, deriving

$\begin{array}{cc}\hfill {\mathbf{u}}_{j+1/2}^{(n+1/2)}& =\frac{1}{2}({\mathbf{u}}_{j+1}^{(n)}+{\mathbf{u}}_{j}^{(n)})-\frac{\mathit{\Delta}t}{2\mathit{\Delta}x}\mathbf{J}({\mathbf{u}}_{j+1}^{(n)}-{\mathbf{u}}_{j}^{(n)})\hfill \\ \hfill & =\frac{1}{2}[(\mathbf{I}-\frac{\mathit{\Delta}t}{\mathit{\Delta}x}\mathbf{J}){\mathbf{u}}_{j+1}^{(n)}+(\mathbf{I}+\frac{\mathit{\Delta}t}{\mathit{\Delta}x}\mathbf{J}){\mathbf{u}}_{j}^{(n)}].\hfill \end{array}$ | $(7.33)$ |

${\mathbf{u}}_{j}^{(n+1)}={\mathbf{u}}_{j}^{(n)}-\frac{\mathit{\Delta}t}{\mathit{\Delta}x}\mathbf{J}({\mathbf{u}}_{j+1/2}^{(n+1/2)}-{\mathbf{u}}_{j-1/2}^{(n+1/2)}).$ | $(7.34)$ |

$\begin{array}{ccc}\multicolumn{1}{c}{{\mathbf{u}}_{j}^{(n+1)}-{\mathbf{u}}_{j}^{(n)}}& =\hfill & -\frac{\mathit{\Delta}t}{2\mathit{\Delta}x}\mathbf{J}[(\mathbf{I}-\frac{\mathit{\Delta}t}{\mathit{\Delta}x}\mathbf{J}){\mathbf{u}}_{j+1}^{(n)}+(\mathbf{I}+\frac{\mathit{\Delta}t}{\mathit{\Delta}x}\mathbf{J}){\mathbf{u}}_{j}^{(n)}\hfill \\ \multicolumn{1}{c}{}& \hfill & \mathrm{\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}}\mathrm{\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}}-(\mathbf{I}-\frac{\mathit{\Delta}t}{\mathit{\Delta}x}\mathbf{J}){\mathbf{u}}_{j}^{(n)}-(\mathbf{I}+\frac{\mathit{\Delta}t}{\mathit{\Delta}x}\mathbf{J}){\mathbf{u}}_{j-1}^{(n)}]\hfill & \hfill \hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}(7.35)\\ \multicolumn{1}{c}{}& =\hfill & -\frac{\mathit{\Delta}t}{2\mathit{\Delta}x}\mathbf{J}[({\mathbf{u}}_{j+1}^{(n)}-{\mathbf{u}}_{j-1}^{(n)})-\frac{\mathit{\Delta}t}{\mathit{\Delta}x}\mathbf{J}({\mathbf{u}}_{j+1}^{(n)}-2{\mathbf{u}}_{j}^{(n)}+{\mathbf{u}}_{j-1}^{(n)})].\hfill \end{array}$ |

Now we consider an eigenmode of $\mathbf{J}$, so we can substitute the eigenvalue $\mathit{\lambda}$ for $\mathbf{J}$, everywhere in the above expression. And we consider a spatial Fourier mode, for which ${\mathbf{u}}_{j}\propto {e}^{{\mathit{ik}}_{x}j\mathit{\Delta}x}$. The equation can then be written

${\mathbf{u}}_{j}^{(n+1)}-{\mathbf{u}}_{j}^{(n)}=-\frac{\mathit{\Delta}t}{2\mathit{\Delta}x}\mathit{\lambda}[({e}^{{\mathit{ik}}_{x}\mathit{\Delta}x}-{e}^{-{\mathit{ik}}_{x}\mathit{\Delta}x})+\frac{\mathit{\Delta}t}{\mathit{\Delta}x}\mathit{\lambda}({e}^{{\mathit{ik}}_{x}\mathit{\Delta}x}-2+{e}^{-{\mathit{ik}}_{x}\mathit{\Delta}x})]{\mathbf{u}}_{j}^{(n)},$ | $(7.36)$ |

${\mathbf{u}}_{j}^{(n+1)}=\{1-\frac{\mathit{\Delta}t\mathit{\lambda}}{\mathit{\Delta}x}i\mathrm{sin}({k}_{x}\mathit{\Delta}x)+{\left(\frac{\mathit{\Delta}t\mathit{\lambda}}{\mathit{\Delta}x}\right)}^{2}[\mathrm{cos}({k}_{x}\mathit{\Delta}x)-1]\}{\mathbf{u}}_{j}^{(n)}.$ | $(7.37)$ |

$\begin{array}{ccc}\multicolumn{1}{c}{|A{|}^{2}}& =\hfill & {\{1+{\left(\frac{\mathit{\Delta}t\mathit{\lambda}}{\mathit{\Delta}x}\right)}^{2}[\mathrm{cos}({k}_{x}\mathit{\Delta}x)-1]\}}^{2}+{\{\frac{\mathit{\Delta}t\mathit{\lambda}}{\mathit{\Delta}x}\mathrm{sin}({k}_{x}\mathit{\Delta}x)\}}^{2}\hfill \\ \multicolumn{1}{c}{}& =\hfill & 1+{\left(\frac{\mathit{\Delta}t\mathit{\lambda}}{\mathit{\Delta}x}\right)}^{2}[2\mathrm{cos}({k}_{x}\mathit{\Delta}x)-2+{\mathrm{sin}}^{2}({k}_{x}\mathit{\Delta}x)]\hfill \\ \multicolumn{1}{c}{}& \hfill & \mathrm{\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}}+{\left(\frac{\mathit{\Delta}t\mathit{\lambda}}{\mathit{\Delta}x}\right)}^{4}[\mathrm{cos}({k}_{x}\mathit{\Delta}x)-1{]}^{2}\hfill & \hfill \hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}(7.38)\\ \multicolumn{1}{c}{}& =\hfill & 1+[-{\left(\frac{\mathit{\Delta}t\mathit{\lambda}}{\mathit{\Delta}x}\right)}^{2}+{\left(\frac{\mathit{\Delta}t\mathit{\lambda}}{\mathit{\Delta}x}\right)}^{4}][\mathrm{cos}({k}_{x}\mathit{\Delta}x)-1{]}^{2}.\hfill \end{array}$ |

Thus $|A{|}^{2}\le 1$ provided $\frac{\mathit{\Delta}t|\mathit{\lambda}|}{\mathit{\Delta}x}\le 1$, which is the stability criterion.

We use the density $\mathit{\rho}$ and the flux density $\mathit{\Gamma}$ as the elements of the state vector $\mathbf{u}$. In 3-dimensions, a vector quantity like $\mathit{\Gamma}$ has three components, ${\mathit{\Gamma}}_{\mathit{\alpha}},\hspace{0.5em}\mathit{\alpha}=1,2,3$. So the state vector has a total of four.

$\mathbf{u}=\left(\begin{array}{c}\hfill \mathit{\rho}\hfill \\ \hfill \mathit{\Gamma}\hfill \end{array}\right)=\left(\begin{array}{c}\hfill \mathit{\rho}\hfill \\ \hfill {\mathit{\Gamma}}_{1}\hfill \\ \hfill {\mathit{\Gamma}}_{2}\hfill \\ \hfill {\mathit{\Gamma}}_{3}\hfill \end{array}\right)$ | $(7.39)$ |

$\frac{\partial \mathit{\rho}}{\partial t}=-\nabla .(\mathit{\rho}\mathit{v})=-\nabla .\mathit{\Gamma},$ | $(7.40)$ |

$\frac{\partial \mathit{\Gamma}}{\partial t}=-\nabla .(\mathit{\rho}\mathit{v}\mathit{v})-\nabla p=-\nabla .(\mathit{\Gamma}\mathit{\Gamma}/\mathit{\rho}+\mathbf{I}p).$ | $(7.41)$ |

$\frac{\partial \mathbf{u}}{\partial t}=-\nabla .\mathbf{f},$ | $(7.42)$ |

$\mathbf{f}=\left(\begin{array}{c}\hfill \mathit{\Gamma}\hfill \\ \hfill \mathit{\Gamma}{\mathit{\Gamma}}_{1}/\mathit{\rho}+p{\hat{\mathit{x}}}_{1}\hfill \\ \hfill \mathit{\Gamma}{\mathit{\Gamma}}_{2}/\mathit{\rho}+p{\hat{\mathit{x}}}_{2}\hfill \\ \hfill \mathit{\Gamma}{\mathit{\Gamma}}_{3}/\mathit{\rho}+p{\hat{\mathit{x}}}_{3}\hfill \end{array}\right)=\left(\begin{array}{c}\hfill \mathit{\Gamma}\hfill \\ \hfill \mathit{\Gamma}\mathit{\Gamma}/\mathit{\rho}+p\mathbf{I}\hfill \end{array}\right).$ | $(7.43)$ |

$\begin{array}{ccc}\multicolumn{1}{c}{\nabla .\mathbf{f}}& =\hfill & \frac{{\hat{\mathit{x}}}_{1}}{2\mathit{\Delta}{x}_{1}}.({\mathbf{f}}_{(i+1)\mathit{jk}}^{(n)}-{\mathbf{f}}_{(i-1)\mathit{jk}}^{(n)})+\frac{{\hat{\mathit{x}}}_{2}}{2\mathit{\Delta}{x}_{2}}.({\mathbf{f}}_{i(j+1)k}^{(n)}-{\mathbf{f}}_{i(j-1)k}^{(n)})\hfill \\ \multicolumn{1}{c}{}& \hfill & \mathrm{\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}}\mathrm{\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}}+\frac{{\hat{\mathit{x}}}_{3}}{2\mathit{\Delta}{x}_{3}}.({\mathbf{f}}_{\mathit{ij}(k+1)}^{(n)}-{\mathbf{f}}_{\mathit{ij}(k-1)}^{(n)}).\hfill & \hfill \hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}(7.44)\end{array}$ |

We are told that the eigenvalue of the state system is $\mathit{\lambda}$ and the eigenmode is longitudinal${}^{44}$. Therefore, for a plane wave proportional to $\mathrm{exp}(i\mathit{k}.\mathit{x})$ that is an eigenmode of the state, each state-component of $\mathbf{f}$ is oriented in the spatial-direction $\mathit{k}$. Write the unit vector $\hat{\mathit{k}}=({\hat{k}}_{1},{\hat{k}}_{3},{\hat{k}}_{3})$, and $\mathit{k}=k\hat{\mathit{k}}$. Then for this plane wave we can replace each ${\hat{\mathit{x}}}_{\mathit{\alpha}}.\mathbf{f}$ with $\mathit{\lambda}{\hat{k}}_{\mathit{\alpha}}\mathbf{u}$ to obtain

$\nabla .\mathbf{f}=\frac{1}{2}[\frac{\mathit{\lambda}{\hat{k}}_{1}}{\mathit{\Delta}{x}_{1}}({\mathbf{u}}_{(i+1)\mathit{jk}}^{(n)}-{\mathbf{u}}_{(i-1)\mathit{jk}}^{(n)})-\frac{\mathit{\lambda}{\hat{k}}_{2}}{\mathit{\Delta}{x}_{2}}({\mathbf{u}}_{i(j+1)k}^{(n)}-{\mathbf{u}}_{i(j-1)k}^{(n)})-\frac{\mathit{\lambda}{\hat{k}}_{3}}{\mathit{\Delta}{x}_{3}}({\mathbf{u}}_{\mathit{ij}(k+1)}^{(n)}-{\mathbf{u}}_{\mathit{ij}(k-1)}^{(n)})].$ | $(7.45)$ |

${\mathbf{u}}_{\mathit{ijk}}^{(n+1)}-{\mathbf{u}}_{s}^{(n)}=-\mathit{\Delta}t\nabla .\mathbf{f}=-\sum _{\mathit{\alpha}}\frac{i\mathit{\Delta}t\mathit{\lambda}{\hat{k}}_{\mathit{\alpha}}}{\mathit{\Delta}{x}_{\mathit{\alpha}}}\mathrm{sin}(k{\hat{k}}_{\mathit{\alpha}}\mathit{\Delta}{x}_{\mathit{\alpha}}){\mathbf{u}}_{\mathit{ijk}},$ | $(7.46)$ |

$A=\sum _{\mathit{\alpha}}[\frac{1}{3}\mathrm{cos}(k{\hat{k}}_{\mathit{\alpha}}\mathit{\Delta}{x}_{\mathit{\alpha}})-\frac{i\mathit{\Delta}t\mathit{\lambda}{\hat{k}}_{\mathit{\alpha}}}{\mathit{\Delta}{x}_{\mathit{\alpha}}}\mathrm{sin}(k{\hat{k}}_{\mathit{\alpha}}\mathit{\Delta}{x}_{\mathit{\alpha}})].$ | $(7.47)$ |

$\mathit{\Delta}t|\mathit{\lambda}|\sum _{\mathit{\alpha}}\frac{{\hat{k}}_{\mathit{\alpha}}^{2}}{{\hat{k}}_{\mathit{\alpha}}\mathit{\Delta}{x}_{\mathit{\alpha}}}=\frac{\mathit{\Delta}t|\mathit{\lambda}|}{\mathit{\Delta}}\le 1.$ | $(7.48)$ |

1. Prove equation 7.29, the amplification factor for the Lax Friedrichs scheme.

2. Consider an isothermal gas in one dimension. It obeys the equations

$\begin{array}{cccc}\mathit{Continuity}:\mathrm{\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}}\hfill & \hfill \frac{\partial \mathit{\rho}}{\partial t}+\frac{\partial}{\partial x}(\mathit{\rho}v)& \hfill =\hfill & 0\hfill \\ \mathit{Momentum}:\mathrm{\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}}\hfill & \hfill \frac{\partial}{\partial t}(\mathit{\rho}v)+\frac{\partial}{\partial x}(\mathit{\rho}{v}^{2})& \hfill =\hfill & -\frac{\partial}{\partial x}p\hfill \\ \mathit{State}:\mathrm{\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}}\hfill & \hfill p& \hfill =\hfill & \mathit{\rho}(\mathit{kT}/m)\hfill \end{array}$ |

with $\mathit{kT}/m$ simply a constant equal to the ratio of the temperature in energy units to the gas molecule mass $m$. (a) Convert this into the form of a state and flux vector equation

$\frac{\partial \mathbf{u}}{\partial t}=-\frac{\partial \mathbf{f}}{\partial x}.$ |

where

$\mathbf{u}=\left(\begin{array}{c}\hfill \mathit{\rho}\hfill \\ \hfill \mathit{\Gamma}\hfill \end{array}\right)$ |

is the state vector ($\mathit{\Gamma}=\mathit{\rho}v$) and you should give the flux vector $\mathbf{f}$. (b) Calculate the Jacobian matrix $\mathbf{J}=\partial \mathbf{f}/\partial \mathbf{u}$. (c) Find its eigenvalues.

3. Find the finite difference form and CFL stability when the linearized eigenmode is longitudinal with eigenvalue $\mathit{\lambda}$, for the Lax-Wendroff scheme in two space-dimensions.

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