Zeldovich–Taylor flow

From the second equation described above, it follows that when ${\displaystyle v=0}$, ${\displaystyle \xi =c}$. More precisely, as ${\displaystyle v\rightarrow 0}$, that equation can be approximated as[5]

${\displaystyle (\ln v)'=2c^{2}/[\xi (\xi ^{2}-c^{2})].}$

As ${\displaystyle v\rightarrow 0}$, ${\displaystyle \ln v\rightarrow -\infty }$ and ${\displaystyle (\ln v)'\rightarrow \infty }$ if ${\displaystyle \xi }$ decreases as ${\displaystyle v\rightarrow 0}$. The left hand side of the above equation can become positive infinity only if ${\displaystyle \xi \rightarrow c}$. Thus, when ${\displaystyle \xi }$ decreases to the value ${\displaystyle \xi =c_{0}}$, the gas comes to rest (Here ${\displaystyle c_{0}}$ is the sound speed corresponding to ${\displaystyle v=0}$). Thus, the rarefaction motion occurs for ${\displaystyle c_{0}<\xi \leq D}$ and there is no fluid motion for ${\displaystyle 0\leq \xi \leq c_{0}}$.

Rewrite the second equation as

${\displaystyle v{\frac {d\xi }{dv}}={\frac {1}{2}}\xi \left[{\frac {(\xi -v)^{2}}{c^{2}}}-1\right].}$

In the neighborhood of the weak discontinuity, the quantities to the first order (such as ${\displaystyle v,\,\xi -c_{0},\,c-c_{0}}$) reduces the above equation to

${\displaystyle v{\frac {d}{dv}}(\xi -c_{0})=(\xi -c_{0})-(v+c-c_{0}).}$

At this point, it is worth mentioning that in general, disturbances in gases are propagated with respect to the gas at the local sound speed. In other words, in the fixed frame, the disturbances are propagated at the speed ${\displaystyle v+c}$ (the other possibility is ${\displaystyle v-c}$ although it is of no interest here). If the gas is at rest ${\displaystyle v=0}$, then the disturbance speed is ${\displaystyle c_{0}}$. This is just a normal sound wave propagation. If however ${\displaystyle v}$ is non-zero but a small quantity, then one find the correction for the disturbance propagation speed as ${\displaystyle v+c=c_{0}+\alpha _{0}v}$ obtained using a Taylor series expansion, where ${\displaystyle \alpha _{0}}$ is a necessarily a positive constant (for ideal gas, ${\displaystyle \alpha _{0}=(\gamma +1)/2}$, where ${\displaystyle \gamma }$ is the specific heat ratio). This means that the above equation can be written as

${\displaystyle v{\frac {d}{dv}}(\xi -c_{0})-(\xi -c_{0})=\alpha _{0}v}$

whose solution is

${\displaystyle \xi -c_{0}=\alpha _{0}v\ln(A/v)}$

where ${\displaystyle A}$ is a constant. This determines ${\displaystyle v(\xi )}$ implicitly in the neighborhood of the week discontinuity where ${\displaystyle v}$ is small. This equation shows that at ${\displaystyle \xi =c_{0}}$, ${\displaystyle v=0}$, ${\displaystyle dv/d\xi =0}$, but all higher-order derivatives are discontinuous. In the above equation, subtract ${\displaystyle v+c-c_{0}}$ from the left-hand side and ${\displaystyle \alpha _{0}v}$ from the right-hand side to obtain

${\displaystyle \xi -v-c=(\xi -c_{0})-(v+c-c_{0})=\alpha _{0}v[\ln(A/v)-1]}$

which implies that ${\displaystyle \xi -v>c}$ if ${\displaystyle v}$ is a small quantity. It can be shown that the relation ${\displaystyle \xi -v>c}$ not only holds for small ${\displaystyle v}$, but throughout the rarefaction wave.

First let us show that the relation ${\displaystyle \xi -v>c}$ is not only valid near the weak discontinuity, but throughout the region. If this inequality is not maintained, then there must be a point where ${\displaystyle \xi -v=c,\,v\neq 0}$ between the weak discontinuity and the detonation front. The second governing equation implies that at this point ${\displaystyle v'}$ must be infinite or, ${\displaystyle d\xi /dv=0}$. Let us obtain ${\displaystyle d^{2}\xi /dv^{2}}$ by taking the second derivative of the governing equation. In the resulting equation, impose the condition ${\displaystyle \xi -v=c,\,v\neq 0,\,d\xi /dv=0}$ to obtain ${\displaystyle d^{2}\xi /dv^{2}=-\alpha _{0}\xi /c_{0}v\neq 0}$. This implies that ${\displaystyle \xi (v)}$ reaches a maximum at this point which in turn implies that ${\displaystyle v(\xi )}$ cannot exist for ${\displaystyle \xi }$ greater than the maximum point considered since otherwise ${\displaystyle v(\xi )}$ would be multi-valued. The maximum point at most can be corresponded to the outer boundary (detonation front). This means that ${\displaystyle \xi -v-c}$ can vanish only on the boundary and it is already shown that ${\displaystyle \xi -v-c}$ is positive near the weak discontinuity, ${\displaystyle \xi -v-c}$ is positive everywhere in the region except the boundaries where it can vanish.

Note that near the detonation front, we must satisfy the condition ${\displaystyle \xi -v=c,\,v\neq 0}$. The value evaluated at ${\displaystyle \xi =D}$ for the function ${\displaystyle \xi -v}$, i.e., ${\displaystyle D-v(D)}$ is nothing but the velocity of the detonation front with respect to the gas velocity behind it. For a detonation front, the condition ${\displaystyle D-v(D)\leq c(D)}$ must always be met, with the equality sign representing Chapman–Jouguet detonations and the inequalities representing over-driven detonations. The analysis describing the point ${\displaystyle \xi -v=c,\,v\neq 0}$ must correspond to the detonation front.