The Boltzmann Equation for Bose-Einstein Particles: Condensation in Finite Time

Abstract

The paper considers the problem of the Bose-Einstein condensation in finite time for isotropic distributional solutions of the spatially homogeneous Boltzmann equation for Bose-Einstein particles with the hard sphere model. We prove that if the initial datum of a solution is a function which is singular enough near the origin (the zero-point of particle energy) but still Lebesgue integrable (so that there is no condensation at the initial time), then the condensation continuously starts to occur from the initial time to every later time. The proof is based on a convex positivity of the cubic collision integral and some properties of a certain Lebesgue derivatives of distributional solutions at the origin. As applications we also study a special type of solutions and present a relation between the conservation of mass and the condensation.

Appendix

1.1 5.1 Equivalence of Distributional Solutions

Here we shall prove that the two definitions in [11] and in Definition 1.1 are equivalent. Since our basic results on existence, moment estimates, long time behavior, etc. were proven only for distributional solutions defined in [11], this clarification is necessary.

Let $W^{\sharp}:\mathbb{R}_{\ge0}^{3}\to \mathbb{R}_{\ge0}$ be defined by

(5.1)

(5.2)

where $\sigma\in{{\mathbb{S}}^{2}}$,

(5.3)

Lemma 5.1

Let W and W ^♯ be defined in (1.12)–(1.13) and (5.1)–(5.3) respectively. Then

$$ W^{\sharp}\bigl(r, r', r_*'\bigr)+W^{\sharp} \bigl(r,r_*',r'\bigr)=(2\pi)^2 W \bigl(r^2/2, r^{\prime\,2}/2,{r_*'}^2/2 \bigr)\quad\forall \bigl(r,r',r_*'\bigr) \in \mathbb{R}_{\ge0}^3. $$

(5.4)

Lemma 5.2

If $\varPsi: \mathbb{R}^{3}_{\ge0}\to \mathbb{R}$ is nonnegative measurable or satisfies some integrability conditions so that the following integrals make sense, then

(5.5)

for all v∈ℝ³∖{0}, where $\epsilon=\frac{1}{2}|\mathbf{v}|^{2}$.

Proof of Lemmas 5.1–5.2

Let

$$V\bigl(r, r', r_*'\bigr)=\frac{2\pi}{r_*'} \int _{{{\mathbb{S}}^2}}\frac{1}{|r\sigma-r'\sigma'|}\mathbf{1}_{\{ r_*'>r|\sigma\cdot\xi |\}}\mathrm{d} \sigma' ,\quad \bigl(r,r',r_*'\bigr)\in \mathbb{R}_{>0}^3 $$

with ξ defined in (5.3). We show that $V(r,r',r_{*}')$ equals to both two sides of (5.4) in $\mathbb{R}_{>0}^{3}$. First of all by coordinate rotation on the sphere we have $|r\sigma-r'\sigma'|=\sqrt{r^{2}+{r'}^{2}-2rr't}$, $|\sigma\cdot\xi|=\frac{|r-r't|}{\sqrt{r^{2}+{r'}^{2}-2rr't}}$ with t∈(−1,1) and so $W^{\sharp}(r,r_{*}',r_{*}'),V(r,r_{*}',r_{*}')$ are independent of $\sigma\in{{\mathbb{S}}^{2}}$, and

$$ V\bigl(r,r', r_*' \bigr)=\frac{(2\pi)^2}{r_*'}\int _{-1}^{1} \mathbf{1}_{\bigl\{r_*'> \frac{r|r-r't|}{\sqrt{r^2+{r'}^2-2rr't}}\bigr\}} \frac{\mathrm{d}t}{\sqrt{r^2+{r'}^2-2rr't}}. $$

(5.6)

Since the set $\{ t\in(-1,1) \mid r_{*}'=\frac{r|r-r't|}{\sqrt{r^{2}+{r'}^{2}-2rr't}}\}$ has measure zero, $W^{\sharp}(r,r',r_{*}'), V(r,r',r_{*}')$ are continuous in $\mathbb{R}_{>0}^{3}$. To prove

$$ V\bigl(r,r',r_*'\bigr)=(2\pi)^2 W \bigl(r^2/2, r^{\prime\,2}/2,{r_*'}^2/2 \bigr)\quad\forall \bigl(r,r',r_*'\bigr) \in \mathbb{R}_{>0}^3 $$

(5.7)

we observe that for any t∈(−1,1),

$$ r_*'> \frac{r|r-r't|}{\sqrt{r^2+{r'}^2-2rr't}}\quad\Longleftrightarrow \quad {r_*'}^2\bigl(r^{\prime\,2}+{r_*'}^2-r^2 \bigr)>\bigl({r_*'}^2 -r^2+rr't \bigr)^2 . $$

(5.8)

If ${r'}^{2}+{r_{*}'}^{2}-r^{2}\le0$, then (5.8) and (5.6) imply $V(r,r',r_{*}')=0$ and by definition of W we also have $W(r^{2}/2, r^{\prime\,2}/2,{r_{*}'}^{2}/2)=0$. Suppose ${r'}^{2}+{r_{*}'}^{2}-r^{2}>0$ and let $r_{*}=\sqrt{r^{\prime\,2}+{r_{*}'}^{2}-r^{2} }$. By changing variable $s=\sqrt{r^{2}+{r'}^{2}-2rr't}$ we compute

where the third equality follows from the identity $(r'+r)(r'-r)= (r_{*}+r_{*}')(r_{*}-r_{*}')$. This proves (5.7). Next we prove that

$$ V\bigl(r,r',r_*'\bigr)=W^{\sharp} \bigl(r,r',r_*'\bigr)+W^{\sharp } \bigl(r,r_*',r'\bigr)\quad \forall \bigl(r,r',r_*'\bigr) \in\mathbb{R}_{>0}^3. $$

(5.9)

To do this, we have to go back to derivations in [10, 11] there we used two formulas: The first one is the Carleman’s representation:

$$ \int_{{{{\mathbb{R}}^3}\times{\mathbb{S}}^2}}\big|(\mathbf{v}-\mathbf{v}_*)\cdot\omega \big|\varPhi\bigl({\mathbf{v}'},\mathbf{v}'_*\bigr)\mathrm{d}\omega\mathrm{d} \mathbf{v}_*= 2\int_{{\mathbb{{R}}^3}} \frac{\mathrm{d}\mathbf{x}}{|\mathbf{x}|}\int _{\mathbb{R}^{2} (\mathbf{x})} \varPhi(\mathbf{v}-\mathbf{x},\mathbf{v}-\mathbf{y})\mathrm{d}^{\bot}\mathbf{y} $$

(5.10)

where ℝ²(x)={y∈ℝ³|y⊥x} for x≠0, and d^⊥ y is the Lebesgue measure on ℝ²(x). The second one is

$$ \int_{{{\mathbb{R}}^3}}\mathrm{d} \mathbf{x} \int_{\mathbb{R}^2(\mathbf{x})} \varPhi(\mathbf{x},\mathbf{y}) \mathrm{d}^{\bot}\mathbf{y} =\int_{{\mathbb{{R}}^3}} \mathrm{d} \mathbf{y}\int_{\mathbb{R}^2(\mathbf{y})}\frac{|\mathbf{x}|}{|\mathbf{y}|}\varPhi(\mathbf{x}, \mathbf{y})\mathrm{d}^{\bot}\mathbf{x}. $$

(5.11)

Here Φ is nonnegative and continuous in (ℝ³∖{0})×(ℝ³∖{0}). Now fix any r>0, let $\mathbf{v}=r\sigma,\sigma\in{{\mathbb{S}}^{2}}$, and take any $0\le\psi\in C(\mathbb{R}_{\ge0}^{2})$. Using (5.10), (5.11), and the identity

$$\frac{1}{|\mathbf{x}|}=\frac{|\mathbf{v}-\mathbf{y}|}{|\mathbf{x}||\mathbf{v}-\mathbf{y}|+ |\mathbf{y}||\mathbf{v}-\mathbf{x}|} +\frac{|\mathbf{y}|}{|\mathbf{x}|}\cdot \frac{|\mathbf{v}-\mathbf{x}|}{|\mathbf{x}||\mathbf{v}-\mathbf{y}|+|\mathbf{y}||\mathbf{v}-\mathbf{x}|} ,\quad\mathbf{x}\in\mathbb{R}^3\setminus\{0\},\ \mathbf{y}\in\mathbb{R}^2(x) $$

we compute (see [11] for details)

(5.12)

Similarly using (5.10) and recalling X=|rσ−r′σ′| we also have

(5.13)

Since $W^{\sharp}(r,r',r_{*}')+W^{\sharp}(r,r_{*}',r')$ and $V(r,r',r_{*}')$ are both continuous in $(r',r_{*}')\in\mathbb{R}_{>0}^{2}$, the equality (5.9) follows from comparing (5.12) and (5.13).

Thus (5.4) holds for all $(r,r',r_{*}')\in\mathbb{R}_{>0}^{3}$. On the other hand, by definition of W and W ^♯, it is easily checked that (5.4) holds also for all $(r,r',r_{*}')\in\mathbb{R}_{\ge0}^{3}\setminus \mathbb{R}_{>0}^{3}$.

Now let Ψ be the function in Lemma 5.2. Applying (5.13) and (5.7) with change of variables $r'=\sqrt{2\epsilon'}, r_{*}'=\sqrt{2\epsilon_{*}'}$ and denoting $\epsilon=\frac{1}{2}r^{2}=\frac{1}{2}|\mathbf{v}|^{2}$ with v∈ℝ³∖{0} we have

This completes the proof. □

Let $J_{B}[\varphi](\rho,\rho_{*}), K_{B}[\varphi](r,r',r_{*}')$ be defined as in pp. 1605–1609 of [11] for the hard sphere model B(v−v _∗,ω)=λ|(v−v _∗)⋅ω|(λ>0), i.e.

$$ J_B[\varphi](\rho,\rho_*)=\frac{\lambda}{(4\pi)^2} \int _{{{{\mathbb{S}}^2}\times{{\mathbb{S}}^2}\times{\mathbb{{S}}^2}}} \bigl[\varphi\bigl(\big|\mathbf{v}'\big|^2 \bigr)- \varphi\bigl(|\mathbf{v}|^2\bigr) \bigr]\big|(\mathbf{v}-\mathbf{v}_*)\cdot \omega\big|\mathrm{d}\omega\mathrm{d}\sigma_*\mathrm{d}\sigma $$

(5.14)

with v=ρσ,v _∗=ρ _∗ σ _∗;

$$ K_B[\varphi]\bigl(r,r',r_*'\bigr)= \frac{4\lambda}{(4\pi )^2}W^{\sharp}\bigl(r,r',r_*' \bigr) \Delta\varphi\bigl(r^2, r^{\prime\,2},{r_*'}^2 \bigr) $$

(5.15)

where Δφ(x,y,z) are defined in (1.11), i.e.

$$\Delta\varphi(x,y,z)=\varphi(x)+\varphi(x_*)-\varphi(y)-\varphi (z), \quad x_* = (y+z-x)_{+}. $$

Let $F\in\mathcal{B}_{1}^{+}(\mathbb{R}_{\ge0}), \bar{F}\in\mathcal{B}_{2}^{+}(\mathbb{R}_{\ge0})$ be defined through the following relation (each one determines the other according to Riesz’s representation theorem):

$$ 4\pi\sqrt{2}\int_{\mathbb{R}_{\ge0}}\psi(\epsilon)\mathrm{d}F(\epsilon)= \int_{\mathbb{R}_{\ge0}}\psi\bigl(r^2/2\bigr)\mathrm{d}\bar {F}(r) \quad\forall \psi\in C_{-1}(\mathbb{R}_{\ge0}) $$

(5.16)

where C ₋₁(ℝ_≥0)={ψ∈C(ℝ_≥0)|sup _ϵ≥0(1+ϵ)⁻¹|ψ(ϵ)|<∞}. In the special case that F and $\bar{F}$ are given by $\mathrm{d}F(\epsilon)= f(\epsilon)\sqrt{\epsilon} \mathrm{d}\epsilon$, $\mathrm{d}\bar{F}(r)=4\pi f(r^{2}/2) r^{2}\mathrm{d} r$, the above relation is just the change of variable: $4\pi\sqrt{2}\psi(\epsilon)\mathrm{d}F(\epsilon)=\psi(r^{2}/2)\mathrm{d}\bar{F}(r)$.

Lemma 5.3

Given any $\varphi\in C_{b}^{2}(\mathbb{R}_{\ge0})$. Let B(v−v _∗,ω)=λ|(v−v _∗)⋅ω| (λ>0), $\mathcal{J}[\varphi],\mathcal{K}[\varphi]$ and J _B [φ],K _B [φ] be defined by (1.9), (1.10) and (5.14), (5.15) respectively. Then for all $\epsilon, \epsilon',\epsilon_{*}'\in \mathbb{R}_{\ge0}$

$$ 4\pi\sqrt{2}\lambda\mathcal{J}[\varphi]\bigl(\epsilon', \epsilon_*'\bigr)=J_B\bigl[\varphi(\cdot/2)\bigr]\bigl( \sqrt {2\epsilon '},\sqrt{2\epsilon_*'}\, \bigr)+J_B\bigl[\varphi(\cdot/2)\bigr]\bigl(\sqrt{2 \epsilon_*'}, \sqrt{2\epsilon'}\,\bigr) , $$

(5.17)

(5.18)

Consequently if $F\in\mathcal{B}_{1}^{+}(\mathbb{R}_{\ge0}), \bar{F}\in\mathcal{B}_{2}^{+}(\mathbb{R}_{\ge0})$ satisfy (5.16), then

$$ (4\pi)^3\sqrt{2}\lambda\int_{\mathbb{R}_{\ge0}^2}\mathcal{J}[ \varphi]\mathrm{d}^2F=\int_{\mathbb{R}_{\ge0}^2}J_B \bigl[\varphi(\cdot /2)\bigr]\mathrm{d}^2\bar{F}, $$

(5.19)

$$ (4\pi)^3\sqrt{2}\lambda\int_{\mathbb{R}_{\ge 0}^3}\mathcal{K}[ \varphi ]\mathrm{d}^3F=\int_{\mathbb{R}_{\ge0}^3}K_B \bigl[\varphi(\cdot/2)\bigr]\mathrm{d}^3{\bar{F}}. $$

(5.20)

Here as before $(\mathrm{d}^{2}\bar{F})(\rho, \rho)=\mathrm{d}\bar{F}(\rho)\mathrm{d}\bar{F}(\rho_{*}), (\mathrm{d}^{3}\bar{F})(r,r',r_{*}')=\mathrm{d}\bar {F}(r')\mathrm{d}\bar{F}(r_{*}')\mathrm{d}\bar{F}(r)$.

Proof

By symmetry $\Delta\varphi(r^{2},{r'}^{2},{r_{*}'}^{2} )=\Delta \varphi(r^{2},{r_{*}'}^{2},{r'}^{2})$ and (5.4) we have

This proves (5.18). To prove (5.17), we first use (5.14), (1.11) to get

(5.21)

with v=ρσ,v _∗=ρ _∗ σ _∗. Here we used $|\mathbf{v}'|^{2}+|\mathbf{v}_{*}'|^{2}=|\mathbf{v}|^{2}+|\mathbf{v}_{*}|^{2}$ so that

$$\varphi\bigl(\big|\mathbf{v}'\big|^2/2\bigr)+\varphi\bigl(\big|\mathbf{v}_*'\big|^2/2\bigr) -\varphi\bigl(|\mathbf{v}|^2/2 \bigr)-\varphi\bigl(|\mathbf{v}_*|^2/2\bigr) =\Delta\varphi\bigl(\big|\mathbf{v}'\big|^2/2, |\mathbf{v}|^2/2,|\mathbf{v}_*|^2/2\bigr). $$

Let $\psi\in C_{c}(\mathbb{R}_{\ge0}^{2})$. By change of variables $\epsilon'=\frac{1}{2}\rho^{2}$, $\epsilon_{*}'=\frac{1}{2}\rho_{*}^{2}$ and using the fact that $(\mathbf{v},\mathbf{v}_{*})\to(\mathbf{v}',\mathbf{v}_{*}')$ (for fixed ω) is an orthogonal transform (see (1.2), (1.3)) and finally using (5.5) we compute from (5.21),

Since $J_{B}[\varphi](\rho,\rho_{*}), \mathcal{J}[\varphi](\epsilon ',\epsilon _{*}')$ are both continuous on $\mathbb{R}_{\ge0}^{2}$, it follows that (5.17) holds true.

Next from the estimates in page 1610 of [11] and (2.2), (2.3) above we see that the integrals in (5.19)–(5.20) are absolutely convergent. Therefore applying Fubini theorem and (5.16)–(5.18) we obtain

 □

Definition 5.1

[11]

Let $\{ \bar{F}_{t}\}_{t\ge0}\subset\mathcal{B}_{2}^{+}(\mathbb{R}_{\ge0})$. We say that $\{\bar{F}_{t}\}_{t\ge0}$, or simply $\bar{F}_{t}$, is an isotropic distributional solution of Eq. (1.1) for the hard sphere model (1.5) with the initial datum ${\bar{F}}_{0}$ if $\bar{F}_{t}$ satisfies the following (i)–(iii):

• (i) $\sup_{t\in[0,T]}\int_{\mathbb{R}_{\ge0}}(1+r^{2})\mathrm{d}\bar{F}_{t}(r)<\infty$ ∀0<T<∞,

• (ii) the function $t\mapsto\int_{\mathbb{R}_{\ge0}}\varphi (r^{2})\mathrm{d}\bar{F}_{t}(r)$ belongs to C ¹([0,∞)) for all $\varphi\in C^{2}_{b}(\mathbb{R}_{\ge0})$, and

• (iii)
  $$\frac{\mathrm{d}}{\mathrm{d}t}\int_{\mathbb{R}_{\ge0}}\varphi\bigl(r^2 \bigr)\mathrm{d}\bar{F}_t(r)=\int_{\mathbb{R}_{\ge0}^2}J_B[ \varphi]\mathrm{d}^2\bar {F}_t+\int_{\mathbb{R}_{\ge0}^3}K_B[ \varphi]\mathrm{d}^3\bar {F}_t,\quad t\in[0,\infty) $$

  for all $\varphi\in C^{2}_{b}(\mathbb{R}_{\ge0})$.

Proposition 5.1

Let $F_{t}\in\mathcal{B}^{+}_{1}(\mathbb{R}_{\ge0})$ and $\bar{F}_{t}\in\mathcal{B}^{+}_{2}(\mathbb{R}_{\ge0})$ satisfy the relation (5.16) for every t∈[0,∞). Then F _t is a distributional solution of Eq. (1.6) in terms of Definition 1.1, if and only if $\bar{F}_{t}$ is an isotropic distributional solution of Eq. (1.1) in terms of Definition 5.1.

Proof

Recall $B(\mathbf{v}-\mathbf{v}_{*},\omega)=\frac{1}{(4\pi )^{2}}|\langle\mathbf{v}-\mathbf{v}_{*},\omega\rangle|$, i.e. $\lambda=\frac {1}{(4\pi)^{2}}$. By assumption and (5.19)–(5.20) we have

for all t∈[0,∞) and all $\varphi\in C^{2}_{b}(\mathbb{R}_{\ge 0})$. These lead to the conclusion of the proposition. □

1.2 5.2 Integral Identities

Proposition 5.2

Let $\varphi\in C_{b}^{1,1}(\mathbb{R}_{\ge0})$ be a convex function satisfying lim _x→∞ φ(x)=0. Then

$$ \int_{0<y<z}\frac{1}{yz}\bigl[\varphi(z+y)+\varphi (z-y)-2\varphi(z)\bigr] \mathrm{d}y\mathrm{d}z=\frac{\pi^2}{12}\varphi(0) $$

(5.22)

$$ \int_{0<x<y<z}\frac{\sqrt{x}}{(xyz)^{7/6}}\bigl[\varphi (z+y-x)+\varphi (z+x-y)-2\varphi(z)\bigr] \mathrm{d}x\mathrm{d}y\mathrm{d}z=B_1 \varphi(0) $$

(5.23)

(5.24)

where 0<B ₁,B ₂<∞ are universal constants:

Proof

First of all the three integrals are well-defined since their integrands are nonnegative due to the convexity of φ. For instance, for the third integral, according to (2.10) and renaming x,y,z as z,x,y we have, for all 0<x<y<z<x+y,

(5.25)

By assumption lim _x→∞ φ(x)=0 and the convexity of φ we have $0\le-\varphi'(x)\le2\frac{\varphi(x/2)-\varphi(x)}{x}$ for all x>0 and so lim _x→∞ xφ′(x)=0 hence $\int_{0}^{\infty }x\varphi''(x)\mathrm{d}x=\varphi(0)$.

Let I ₀,I ₁,I ₂ denote the left hand sides of (5.22), (5.23), (5.24) respectively. Using (2.8), (2.9) with x=0 and changing variable y→zy we compute

and, for 0<y, s,t<1,

$$\int_{0}^{\infty} z \varphi'' \bigl(z\bigl(1+(s-t)y\bigr)\bigr)\mathrm{d}z=\frac{1}{[1+(s-t)y]^2} \int _{0}^{\infty} z \varphi''(z) \mathrm{d}z=\frac{1}{[1+(s-t)y]^2}\varphi(0). $$

Thus

$$I_0= \varphi(0)\int_{0}^{1}\!\!\int_{0}^1\!\!\int_{0}^1 \frac{y\mathrm{d}y\mathrm{d}s\mathrm{d}t}{[1+(s-t)y]^2}=-\varphi(0)\int_{0}^1 \frac{1}{y}\log\bigl( 1-y^2\bigr)\mathrm{d}y= \frac {\pi^2}{12}\varphi(0) . $$

Similarly using (2.8), (2.9) and changing variable (x,y)→(zx,zy) we have

and using (5.25) and changing variable (x,y)→(zx,zy)

 □

Remark 5.1

(1) Rough estimate gives 0<B ₁<3, 0<B ₂<5.

(2) From the proof of (5.22)–(5.24) one sees that (5.22)–(5.24) still hold true if the convexity assumption on φ is replaced by lim _x→∞ xφ′(x)=0 and $\int_{0}^{\infty }x|\varphi ''(x)|\mathrm{d}x<\infty$.