Past exam of the astrophysical fluid dynamics course of the University of Cambridge / 2022 / Section B / 5 / d / Solution

Let $a_j=\nu {\kappa }_j^{2}t+{\kappa }_{j}x$. Differentiating the logarithm givesFor definiteness order ${\kappa }_1<{\kappa }_2$. Then $e^{a_1}$ dominates as $x\to -\infty$, while $e^{a_2}$ dominates as $x\to +\infty$. The limiting states areWithout this ordering convention, $u(-\infty )=-2\nu \min ({\kappa }_1,{\kappa }_2)$ and $u(+\infty )=-2\nu \max ({\kappa }_1,{\kappa }_2)$. Thus the states $u_{1,2}$ correspond to $-2\nu {\kappa }_{1,2}$ after matching each label to its dominant exponential; the propagation speed does not depend on the naming of the two states.

Their exponent difference isConsequently, the Viscous Burgers travelling shock moves atSet $\Delta u=u_L-u_R=2\nu ({\kappa }_2-{\kappa }_1)>0$ and $\xi =x-Vt$. Factoring out the mean exponential gives the hyperbolic tangent profileA precise measure of the width must be stated. The natural length in the hyperbolic tangent argument isThus the layer has thickness of order $\nu /\Delta u$. A different common convention, jump divided by maximum gradient, gives $\delta =\Delta u/|u_x(0)|=8\nu /\Delta u=2\ell$; the far-tail exponential e-folding length is $\ell /2$. These differ only by their stated definitions, rather than by different solutions.

Two distinct limits require ${\kappa }_1\ne {\kappa }_2$. The source permits arbitrary real constants, but in the degenerate case ${\kappa }_1={\kappa }_2$ the solution is uniform, $u=-2\nu {\kappa }_1$, and there is no shock wave or finite shock width.