Past exam of the mathematics course of the University of Cambridge / 2013 / iii / Paper 6 / 5

For every test function $\varphi \in C_c^{\infty }({\mathbb{R}}_t\times {\mathbb{R}}_x^3\times {\mathbb{R}}_v^3)$, the distributional weak solution satisfiesThere is no initial-time boundary term because the test functions are on the whole time line. In divergence notation the flux in the seven coordinates is $(f,vf,0)$, whose divergence is ${\partial }_{t}f+v\cdot {\nabla }_{x}f$. For a general $N$-dimensional divergence the introductory flux must be vector-valued; the scalar codomain printed for $A$ is dimensionally inconsistent when $N>1$. The transport identity above has the correct flux dimensions.

The shear $(t,x,v)\mapsto (t,x+tv,v)$ is a smooth diffeomorphism with determinant one and sends compact sets to compact sets. Thus $F,G$ are locally integrable functions. On a compact time interval $I$ and compact label set $K\subset {\mathbb{R}}^6$, the integral ${\int }_K{\int }_I|G|dtdxdv$ is finite. The Fubini theorem, followed by a countable exhaustion of time intervals and label sets, shows that $G(\cdot ,x,v)\in L_{\mathrm{loc}}^1(\mathbb{R})$ for almost every $(x,v)$.

To transform the weak equation, choose $\varphi (t,x,v)=\rho (x-tv,v)\psi (t)$, where $\rho \in C_c^{\infty }({\mathbb{R}}^6)$ and $\psi \in C_c^{\infty }(\mathbb{R})$. Including a velocity cutoff in $\rho$ ensures the required compact support. The transport derivative of this test function is $\rho (x-tv,v){\psi }^{\prime }(t)$. Change variables to the characteristic labels to obtainThe fundamental test-function identity makes the bracket zero almost everywhere. Choose a countable dense family of time test functions on each bounded interval and extend by continuity; there is then a single negligible label set outside which ${\partial }_{t}F=G$ as one-dimensional distributions.

For each remaining label, subtract the primitive $H(t)={\int }_0^{t}G(s)ds$. The distributional derivative of $F-H$ is zero, so it is a constant almost everywhere. Choose the resulting absolutely continuous representative along free characteristics, $\tilde{F}(t)=C+H(t)$. The fundamental theorem of calculus givesThe representative qualification is essential: an arbitrary locally integrable representative can be modified on one time slice without changing the distributional equation, and would not satisfy the identity for every pair of times.

Interpret $L$ as the velocity operator appearing in the equation and in the stated quadratic-form hypothesis. Let $E(t)=\int \int |f(t,x,v){|}^{2}dxdv$. Smoothness turns the distributional equation into the usual equation. Multiply it by $f$ and integrate; for complex functions use the real part of the inner product. Spatial integration by parts gives $\int v\cdot {\nabla }_x(f^2)=0$, while the coercivity hypothesis at each $x$ givesThus $e^{2\delta t}E(t)$ is nonincreasing. The coercive energy estimate for kinetic transport isNo division by $E$ is required, so the zero solution causes no exception. Compact spatial support, or sufficient decay to eliminate the boundary flux, is enough; compact support in time is unnecessary.

Strict positivity throughout phase space is incompatible with compact support. The following calculation uses positive smooth rapidly decaying solutions with all displayed integrals justified; nonnegative cases use the corresponding entropy limits where justified.

Write $B=|{\mathbb{S}}^2|$ and use integration over $(v,v_{\ast },\sigma )$. The two symmetrized weak collision identities areandThey follow from particle interchange and the angular exchange for elastic collisions. They are the weak identities needed here; no invalid fixed-angle Jacobian is used.

The collision invariants $1$, $v$ and $|v{|}^2$ obey ${\varphi }^{\prime }+{\varphi }_{\ast }^{\prime }=\varphi +{\varphi }_{\ast }$. Momentum follows from $v^{\prime }+v_{\ast }^{\prime }=v+v_{\ast }$, and energy from $|v^{\prime }{|}^2+|v_{\ast }^{\prime }{|}^2=|v{|}^2+|v_{\ast }{|}^2$. HenceFor the Boltzmann H functional $H=\int \int f\log fdxdv$, the spatial transport contribution is a boundary divergence and the term $\int Q$ is zero. Set $a=f^{\prime }{f}_{\ast }^{\prime }$, $b=f{f}_{\ast }>0$. The second weak identity with $\varphi =\log f$ gives the Boltzmann H theoremIndeed this integrand with its negative sign is $b(1-X)\log X$ for $X=a/b>0$, which is nonpositive. The kinetic H functional decreases; the physical entropy with opposite sign increases.

A local Maxwellian can be written as $\mathcal{M}=\exp (a+b\cdot v+c|v{|}^2)$, with $c<0$ and coefficients depending on $(t,x)$. Its logarithm is a linear combination of collision invariants, so ${\mathcal{M}}^{\prime }{\mathcal{M}}_{\ast }^{\prime }=\mathcal{M}{\mathcal{M}}_{\ast }$ and $Q(\mathcal{M},\mathcal{M})=0$ at each spatial point.

Comparing powers of $v$ in $({\partial }_t+v\cdot {\nabla }_x)\log \mathcal{M}=0$ givesHere $\mathrm{sym}\nabla b$ is the symmetric part of a matrix formed by the coefficients of the quadratic polynomial. For constants $A,\alpha ,\beta >0$, choose $a=\log A-\alpha |x{|}^2$, $b=2\alpha tx$ and $c=-\beta -\alpha t^2$. All four coefficient equations hold. The resulting expanding Gaussian solution of the Boltzmann equation isDirect verification is particularly simple: $x-tv$ and $v$ are constant along free characteristic curves, so the transport derivative is zero. At fixed $(t,x)$ its logarithm is the displayed Maxwellian quadratic, so the collision term is also zero. Its spatial dependence is nonconstant, it is positive and rapidly decaying, and its Gaussian velocity integral is finite because $\beta +\alpha t^2>0$. It is not a compactly supported example, nor does this final part require one.