1.1 Main results

In [Reference Matetski, Quastel and RemenikMQR17], it is shown that for initial data $\mathfrak {h}_0$ in $\operatorname {\mathrm {\text {UC}}}$ , a space of upper semicontinuous functions with a linear growth condition (see equation (1.24)), the n-space point distribution functions from equation (1.5) are given by Fredholm determinants

(1.6) $$ \begin{align} F(t,x_1,\dotsc,x_n,r_1,\dotsc, r_n) = \det(\mathbf{I} - \mathbf{K}), \end{align} $$

where $\mathbf {K} = \mathbf {K}( t, x_1,\dotsc ,x_n,r_1,\dotsc , r_n,\mathfrak {h}_0)$ is an operator on the n-fold direct sum of $L^2(\mathbb {R}^+)$ , which is given by an explicit $n\times n$ matrix kernel $\mathbf {K}_{ij}(u_1,u_2)=\mathbf {K}_{ij}(t, x_1,\dotsc ,x_n,r_1,\dotsc , r_n,\mathfrak {h}_0,u_1,u_2)$ , $i,j=1,\dotsc ,n$ . The determinant is nonzero, so we can define $\mathbf {R}= (\mathbf {I}-\mathbf {K})^{-1}$ . Furthermore, $\mathbf {K}\mathbf {R}= \mathbf {R}\mathbf {K}=\mathbf {I}-\mathbf {R}$ , and we can let

$$ \begin{align*} Q=\mathbf{R}\mathbf{K}(0,0), \end{align*} $$

which is an $n\times n$ matrix-valued function of $t,x_1,\dotsc ,x_n,r_1,\dotsc , r_n$ and the initial height profile $\mathfrak {h}_0$ .

Let

$$ \begin{align*} \mathcal{D}_r= \partial_{r_1}+\dotsb + \partial_{r_n}, \qquad \mathcal{D}_x= \partial_{x_1}+\dotsb + \partial_{x_n}. \end{align*} $$

In Section 4.1 we will show that

(1.7) $$ \begin{align} \mathcal{D}_r\log F = \operatorname{{\mathop{\mathrm{{tr}}}}} Q. \end{align} $$

Our main result is the following:

Theorem 1.1. For any $\mathfrak {h}_0\in \operatorname {\mathrm {\text {UC}}}$ , Q and its derivative $q= \mathcal {D}_rQ$ solve the matrix Kadomtsev–Petviashvili (KP) equation

(1.8) $$ \begin{align} \partial_tq+\tfrac12\hspace{0.05em}\mathcal{D}_rq^2+\tfrac1{12}\mathcal{D}_r^3q+\tfrac14\mathcal{D}_x^2Q+\tfrac12[q,\hspace{0.05em}\mathcal{D}_xQ]=0, \end{align} $$

where $[A,B] = AB-BA$ . In particular, for the one-point marginals of the KPZ fixed point (i.e., equation (1.5) in the case $n=1$ ), $\phi =\partial _r^2 \log F$ satisfies the scalar KP-II equation

(1.9) $$ \begin{align} \partial_t \phi + \tfrac12\partial_r \phi^2 + \tfrac1{12}\partial_r^3 \phi + \tfrac14\partial_r^{-1} \partial_x^2 \phi = 0. \end{align} $$

The KP equation (1.9) was originally derived from studies of long waves in shallow water [Reference Ablowitz and SegurAS79]. It has come to be accepted as the natural $2$ -dimensional extension of the Korteweg–de Vries (KdV) equation; when $\phi $ is independent of x, corresponding in our case to flat initial data, it reduces to the KdV equation,

(1.10) $$ \begin{align} \partial_t \phi + \tfrac12\partial_r \phi^2+ \tfrac1{12}\partial_r^3 \phi=0. \end{align} $$

The KP equation is completely integrable and plays an important role in the Sato theory as the first equation in the KP hierarchy [Reference Miwa, Jimbo and DateMJD00]. The matrix KP equation (1.8) exists in the literature (see, e.g., [Reference KonopelchenkoKon82; Reference SakhnovichSak03]). None of the previous physical derivations of the KP equation seem to be related to the problem at hand, and it could well be that our evolution is through a class of functions where the equation is formally the same, because of similarities in the weakly nonlinear asymptotics, but the physics is completely different. Note that the equations are usually written with with coefficients $3$ , $1$ and $3$ replacing our $\tfrac 12$ , $\tfrac 1{12}$ and $\tfrac 14$ , which is achieved by $t\mapsto 12t$ and $\phi \mapsto \tfrac 12 \phi $ .

Theorem 1.1 follows from the form of the kernel in the determinantal formula (1.6) and the following result. In the scalar case this appeared earlier in [Reference PöppePo89].

Theorem 1.3. Let $\mathbf {K}=\mathbf {K}(t,x,r)$ be an operator on the n-fold direct sum of $L^2\left (\mathbb {R}_{\ge 0}\right )$ which is trace class uniformly in compact sets of $t>0$ , $x,r\in \mathbb {R}$ , with matrix kernel $\mathbf {K}(u,v)=\mathbf {K}(t,x,r,u,v))$ which satisfies the following:

1. (1) $\partial _r\mathbf {K}(u,v)=(\partial _{u}+\partial _{v})\mathbf {K}(u,v)$ ,

2. (2) $\partial _t\mathbf {K}(u,v)=-\tfrac 13\left (\partial _{u}^3+\partial _{v}^3\right )\mathbf {K}(u,v)$ ,

3. (3) $\partial _x\mathbf {K}(u,v)=\left (\partial _{v}^2-\partial _{u}^2\right )\mathbf {K}(u,v)$ .

Suppose in addition that $\det (\mathbf {I}-\mathbf {K})>0$ for all finite $t,x,r$ ; that $\mathbf {K}$ is real analytic in t and in each $x_i$ and $r_i$ ; and that the trace norm $\lVert \mathbf {K}\rVert _1<1$ for r in some open real interval. Then the $n\times n$ matrix $Q=(\mathbf {I}-\mathbf {K})^{-1}\mathbf {K}(0,0)$ and its derivative $q= \partial _rQ$ satisfy the matrix KP equation

(1.12) $$ \begin{align} \partial_tq+\tfrac12\hspace{0.05em}\partial_rq^2+\tfrac1{12}\partial_r^3q+\tfrac14\partial_x^2Q+\tfrac12[q,\hspace{0.05em}\partial_xQ]=0.\end{align} $$

The kernel in equation (1.6) is given explicitly (see [Reference Matetski, Quastel and RemenikMQR17, Eqn. (4.2)]) as

(1.13) $$ \begin{align} &\mathbf{K}_{ij}(t, x_1,\dotsc,x_n,r_1,\dotsc, r_n,\mathfrak{h}_0, u,v)= -e^{\left(x_j-x_i\right)\partial^2}\left(u+r_i,v+r_j\right)\mathbf{1}_{x_i<x_j}\nonumber\\ &~+ \int_{s\in\mathbb{R}^+\atop z,b\in\mathbb{R} } p_-(z,ds,db) \mathbf{S}_{t,-x_i-s} (b,u+r_i)\mathbf{S}_{t,x_j}\left(z,v+r_j\right) \nonumber\\ &~ + \int_{s\in\mathbb{R}^+\atop z,b\in\mathbb{R} } p_+(z,ds,db) \mathbf{S}_{t,-x_i} (z,u+r_i) \mathbf{S}_{t,x_j-s} \left(b, v+r_j\right)\nonumber\\ &~ - \int_{s_-,s_+\in\mathbb{R}^+\atop z,b_-,b_+\in\mathbb{R} } p_-(z,ds_-,db_-)p_+(z,ds_+,db_+) \mathbf{S}_{t,-x_i-s_-} (b_-,u+r_i) \mathbf{S}_{t,x_j-s_+} \left(b_+, v+r_j\right), \end{align} $$

where $p_\pm (z,ds,db) = p_z\left (\tau _{\mathfrak {h}_0^\pm }\in ds, B\left (\tau _{\mathfrak {h}_0^\pm }\right )\in db\right )$ are hitting measures of the hypographs of $\mathfrak {h}_0^-(x):=\mathfrak {h}_0(-x)$ and $\mathfrak {h}_0^+(x):=\mathfrak {h}_0(x)$ , $x\ge 0$ , by a Brownian motion B with diffusion coefficient $2$ , and where $\mathbf {S}_{t,x}(z_1,z_2)=\mathbf {S}_{t,x}(z_1-z_2)$ with

(1.14) $$ \begin{align} \mathbf{S}_{t,x}(z)= t^{-1/3} e^{\frac{2x^3}{3t^2}-\frac{zx}{t}}\hspace{-0.1em}\operatorname{\mathrm{\text{Ai}}}\left(-t^{-1/3}z+t^{-4/3}x^2\right). \end{align} $$

This is real analytic, and one checks directly (using $\operatorname {\mathrm {\text {Ai}}}''(z)=z\hspace {-0.1em}\operatorname {\mathrm {\text {Ai}}}(z)$ ) that

(1.15) $$ \begin{align} \partial_t\mathbf{S}_{t,x}(z) = \tfrac13 \partial_z^3\mathbf{S}_{t,x}(z),\qquad \partial_x\mathbf{S}_{t,x}(z) = \partial_z^2\mathbf{S}_{t,x}(z). \end{align} $$

Now fix $(x_1,\dots c,x_n),(r_1,\dots c,r_n)\in \mathbb {R}^n$ , introduce auxiliary variables $x,r\in \mathbb {R}$ and let

(1.16) $$ \begin{align} \widehat{\mathbf{K}}(t,x,r)=\mathbf{K}(t,x_1+x,\dots c,x_n+x,r_1+r,\dots c, r_n+r,\mathfrak{h}_0,\cdot,\cdot). \end{align} $$

The differential relations in equation (1.15) together with the fact that r enters the kernel just as a shift give Theorem 1.3(1)–(3) for $\widehat {\mathbf {K}}$ . On the other hand, it is shown in [Reference Matetski, Quastel and RemenikMQR17, Appx. A.1] that $ \Vert \widehat {\mathbf {K}}\Vert _1\longrightarrow 0$ as $r\to \infty $ and that the integrals converge absolutely and uniformly in compact balls around $t>0$ , $x\in \mathbb {R}^n$ and $r\in \mathbb {R}^n$ , and thus $\widehat {\mathbf {K}}$ is real analytic in those variables as well. Thus we have the following:

Translating the partial derivatives in r and x in equation (1.12) back to the $r_i$ s and $x_i$ s gives equation (1.8) for the kernel $\mathbf {K}$ .

There is an informal way [Reference Matetski, Quastel and RemenikMQR17] to write the kernel of the KPZ fixed point which makes Theorem 1.3(2) and (3) completely apparent (equation (1) comes simply again from the way the $r_i$ s come in the formula). For $\mathfrak {h}\in \operatorname {\mathrm {\text {UC}}}$ and $\ell _1<\ell _2$ , let

$$ \begin{align*} {\textbf{P}}_{\ell_1,\ell_2}^{\text{No hit}\,\mathfrak{h}}(u_1,u_2)\hspace{0.05em} \mathrm{d} u_2&=\mathbb{P}_{\mathbf{B}\left(\ell_1\right)=u_1}\!\left(\mathbf{B}(y)>\mathfrak{h}(y)\text{ on }[\ell_1,\ell_2],\,\mathbf{B}(\ell_2)\in\mathrm{d} u_2\right),\\ {\textbf{P}}_{\ell_1,\ell_2}^{\text{Hit}\,\mathfrak{h}}&=e^{\left(\ell_2-\ell_1\right)\partial^2}- {\textbf{P}}_{\ell_1,\ell_2}^{\text{No hit}\,\mathfrak{h}}, \end{align*} $$

where $\mathbf {B}$ is again a Brownian motion with diffusion coefficient $2$ and $\partial $ denotes the derivative operator. The Brownian scattering transform of $\mathfrak {h}$ is the formal object

(1.17) $$ \begin{align} \mathbf{K}^{\operatorname{\mathrm{\text{hypo}}}(\mathfrak{h})} =\lim_{\ell_1\to -\infty\atop \ell_2\to \infty}e^{\ell_1\partial^2}{\textbf{P}}_{\ell_1,\ell_2}^{\text{Hit}\,\mathfrak{h}}e^{-\ell_2\partial^2} =\mathbf{I}-\lim_{\ell_1\to -\infty\atop \ell_2\to \infty} e^{\ell_1\partial^2}{\textbf{P}}_{\ell_1,\ell_2}^{\text{No hit}\,\mathfrak{h}}e^{-\ell_2\partial^2}, \end{align} $$

where ${\textbf {P}}_{\ell _1,\ell _2}^{\text {Hit/No hit}\,\mathfrak {h}}$ are thought of as operators with the given integral kernels. This doesn’t make sense, since the backward heat operator is asked to act on nonanalytic functions. However, in the formula for the KPZ fixed point, $\mathbf {K}^{\operatorname {\mathrm {\text {hypo}}}(\mathfrak {h})}$ is never actually used by itself, but only after conjugation by the Airy unitary group

(1.18) $$ \begin{align} \mathbf{U}_t=e^{ -\frac13t \partial^3}, \end{align} $$

with $t\neq 0$ ; and as we explain next, in our setting the conjugated kernel is in fact well defined.

For $a\in \mathbb {R}$ , define

(1.19) $$ \begin{align} \chi_a(x)=\mathbf{1}_{x>a}\qquad\text{and}\qquad\bar\chi_a(x)=\mathbf{1}_{x\leq a}, \end{align} $$

which we also regard as multiplication operators acting on $L^2(\mathbb {R})$ . For $t>0$ the Airy semigroup acts by convolution with Airy functions – that is, $\mathbf {U}_t$ has integral kernel $\mathbf {U}_t(x,y)=t^{-1/3}\operatorname {\mathrm {\text {Ai}}}\left (t^{-1/3}(x-y)\right )$ ; its inverse $\mathbf {U}_t^{-1}=\mathbf {U}_{-t}$ equals $\mathbf {U}_t^*$ . The Airy functions are not themselves in $L^2(\mathbb {R})$ ; however, for $t>0$ and $r>-\infty $ , $ \mathbf {U}_t^{-1}\chi _{r}$ maps $L^2(\mathbb {R})$ into the domain of $e^{x \partial ^2}$ for any $x\in \mathbb {R}$ . So for $t>0$ and $r>-\infty $ , we define on $L^2([r,\infty ))$

(1.20) $$ \begin{align} \mathbf{K}^{\operatorname{\mathrm{\text{hypo}}}(\mathfrak{h})}_t= \lim_{\ell_1\to -\infty\atop \ell_2\to \infty} \mathbf{U}_t e^{\ell_1\partial^2}{\textbf{P}}_{\ell_1,\ell_2}^{\text{Hit}\,\mathfrak{h}}e^{-\ell_2\partial^2}\mathbf{U}_t^{-1}. \end{align} $$

For any $t>0$ and $r>-\infty $ , the limit on the right-hand side of equation (1.20) exists in trace class on $L^2([r,\infty ))$ [Reference Quastel and RemenikQR19; Reference Matetski, Quastel and RemenikMQR17] and defines the left-hand side as a trace class operator in this space. The limit is given by the right-hand side of equation (1.13) with $x_i=x_j=x$ and $r_i=r_j=0$ ; the convergence was proved first in [Reference Quastel and RemenikQR19]. It satisfies the semigroup property

$$ \begin{align*} \mathbf{U}_s\mathbf{K}^{\operatorname{\mathrm{\text{hypo}}}(\mathfrak{h})}_t\mathbf{U}_s^{-1}=\mathbf{K}^{\operatorname{\mathrm{\text{hypo}}}(\mathfrak{h})}_{\mathbf{t}+\mathbf{s}}, \end{align*} $$

so we can write (at least informally)

(1.21) $$ \begin{align} \mathbf{K}^{\operatorname{\mathrm{\text{hypo}}}(\mathfrak{h})}_t=\mathbf{U}_t\mathbf{K}^{\operatorname{\mathrm{\text{hypo}}}(\mathfrak{h})}\mathbf{U}_t^{-1}. \end{align} $$

Note that we avoid the problem of domains by defining the left-hand side of equation (1.20) not as a product of three operators but just as one operator with the semigroup property. In this sense the Brownian scattering operator is the germ of the semigroup. Alternatively, one can think of the Brownian scattering operator as the entire semigroup $\left (\mathbf {K}^{\operatorname {\mathrm {\text {hypo}}}(\mathfrak {h})}_t\right )_{t>0}$ from equation (1.21). The fact that equation (1.17) is formal is important, though. We will see in formula (5) that the limit of equation (1.21) as $t\searrow 0$ is not $\mathbf {K}^{\operatorname {\mathrm {\text {hypo}}}(\mathfrak {h})}$ .

From $\mathbf {K}^{\operatorname {\mathrm {\text {hypo}}}(\mathfrak {h})}$ we build an extended Brownian scattering operator acting on $L^2(\{x_1,\dotsc ,x_m\}\times \mathbb {R})$ ,

(1.22) $$ \begin{align} \mathbf{K}^{\operatorname{\mathrm{\text{hypo}}}(\mathfrak{h})}_{\text{ext}}\left(x_i,\cdot;x_j,\cdot\right)=-e^{\left(x_j-x_i\right)\partial^2}\mathbf{1}_{x_i<x_j}+e^{-x_i\partial^2}\mathbf{K}^{\operatorname{\mathrm{\text{hypo}}}(\mathfrak{h})} e^{x_j\partial^2}, \end{align} $$

with the analogous caveat that in order to make sense of it, each of the $\left (x_i,x_j\right )$ entries should be conjugated by $\mathbf {U}_t$ and surrounded by $\chi _{r_i}$ on the left and $\chi _{r_j}$ on the right. Then the kernel of the KPZ fixed point appearing on the right-hand side of equation (1.6) and given in equation (1.13) can be written as

(1.23) $$ \begin{align} \mathbf{K}_{ij}(u_1,u_2)=\mathbf{K}^{\operatorname{\mathrm{\text{hypo}}}(\mathfrak{h})}_{t,\text{ext}}\left(x_i,u_1+r_i;x_j,u_2+r_j\right)\quad\text{with}\quad \mathbf{K}^{\operatorname{\mathrm{\text{hypo}}}\left(\mathfrak{h}_0\right)}_{t,\text{ext}}=\mathbf{U}_t\mathbf{K}^{\operatorname{\mathrm{\text{hypo}}}\left(\mathfrak{h}_0\right)}_{\text{ext}}\mathbf{U}_t^{-1}, \end{align} $$

where the Airy operators act on the left and right of each entry of $\mathbf {K}^{\operatorname {\mathrm {\text {hypo}}}\left (\mathfrak {h}_0\right )}_{\text {ext}}$ . From this expression one sees that $\partial _t\mathbf {K}_{ij}=-\frac 13\left (\partial ^3\mathbf {K}_{ij}-\mathbf {K}_{ij}\partial ^3\right )$ and $\partial _{x_\ell }\mathbf {K}_{ij}=-\frac 12\left (\partial ^2\mathbf {K}_{ij}\mathbf {1}_{i=\ell }-\mathbf {K}^2_{ij}\partial ^2\mathbf {1}_{j=\ell }\right )$ , which, after integration by parts and summing over the $x_\ell $ s as before, become Theorem 1.3(2) and (3).

Remark 1.5. For compactly supported initial data, which in our context means that $\mathfrak {h}(y)=-\infty $ for y outside some compact interval, the definition of the Brownian scattering operator in equation (1.20) becomes explicit, without any limits. In fact, consider $\mathfrak {h}$ with support on some interval $[-a,a]$ and let $\ell _1<-a$ and $\ell _2>a$ . Clearly ${\textbf {P}}_{\ell _1,\ell _2}^{\text {Hit}\,\mathfrak {h}}=e^{-\left (a+\ell _1\right )\partial ^2}{\textbf {P}}_{-a,a}^{\text {Hit}\,\mathfrak {h}}e^{\left (\ell _2-a\right )\partial ^2}$ , so $\mathbf {U}_te^{\ell _1\partial ^2}{\textbf {P}}_{\ell _1,\ell _2}^{\text {Hit}\,\mathfrak {h}}e^{-\ell _2\partial ^2}\mathbf {U}_t^{-1}=\mathbf {U}_te^{-a\partial ^2}{\textbf {P}}_{-a,a}^{\text {Hit}\,\mathfrak {h}}e^{-a\partial ^2}\mathbf {U}_t^{-1}$ . Since $\mathbf {U}_te^{x\partial ^2}=\mathbf {S}_{t,x}$ and $\mathbf {U}_t^{-1}e^{x\partial ^2}=\mathbf {S}_{-t,x}$ (see equation (1.15)), this shows that

$$ \begin{align*} \mathbf{K}^{\operatorname{\mathrm{\text{hypo}}}(\mathfrak{h})}_t=\mathbf{S}_{t,-a}{\textbf{P}}_{-a,a}^{\text{Hit}\,\mathfrak{h}}\mathbf{S}_{-t,-a}. \end{align*} $$

The right-hand side coincides with the right-hand side of equation (1.13) with $i=j$ and $x_i=r_i=0$ (see [Reference Matetski, Quastel and RemenikMQR17, Sec. 4.1]).

1.2 Initial data

The natural class of initial data for our problem (the ‘ $1$ -dimensional substrate’) corresponds to functions in the spaceFootnote ²

(1.24) $$ \begin{align} &\operatorname{\mathrm{\text{UC}}}=\big\{\mathfrak{h}\!:\mathbb{R}\longrightarrow[-\infty,\infty)\!:\mathfrak{h}\text{ is upper semicontinuous,}\nonumber\\ &\qquad\qquad\qquad\quad\mathfrak{h}(x)\leq A+B\lvert x\rvert\text{ for some}\ A,B>0\ \text{and }\mathfrak{h}\not\equiv-\infty\big\}. \end{align} $$

A function is upper semicontinuous if and only if its hypograph $\operatorname {\mathrm {\text {hypo}}}(\mathfrak {h}) = \{(\mathbf {x},y): y\le \mathfrak {h}(\mathbf {x})\}$ is closed in $[-\infty ,\infty )\times \mathbb {R}$ . We endow $[-\infty ,\infty )$ with the distance $d_{\left [-\infty ,\infty \right )}(y_1,y_2) = \lvert e^{y_1} - e^{y_2}\rvert $ , and use the topology of local Hausdorff convergence, which means Hausdorff convergence of the restrictions to $-L\le x\le L$ of $\operatorname {\mathrm {\text {hypo}}}(\mathfrak {h}_n)$ to $\operatorname {\mathrm {\text {hypo}}}(\mathfrak {h})$ for each $L>0$ .

Example 1.6 Finite collection of narrow wedges

Set $a_1<a_2<\dotsb <a_k$ , $b_1,b_2,\dotsc ,b_k\in \mathbb {R}$ . Then $\mathfrak {h}=\mathfrak {d}_{\vec a}^{\vec b}$ is in $\operatorname {\mathrm {\text {UC}}}$ , with $\mathfrak {d}_{\vec a}^{\vec b}$ defined by

$$ \begin{align*}\mathfrak{d}_{\vec a}^{\vec b}(x)=b_i\quad\text{if}\ x=a_i\ \text{for some}\ i,\qquad\mathfrak{d}_{\vec a}^{\vec b}(x)=-\infty\quad\text{otherwise}.\end{align*} $$

The initial data for the one-point case (equation (1.9)) is the escarpment

(1.25) $$ \begin{align} \phi(0,x,r) = 0\quad\text{for }r\ge \mathfrak{h}_0(x),\qquad\phi(0,x,r)=-\infty\quad\text{for }r<\mathfrak{h}_0(x). \end{align} $$

These are unusual and do not fit into any well-posedness schemes known for the KP equation.Footnote ³ Although the infinity looks formal, we believe the solutions to the equations with such initial data are well posed, but we leave the proofs for future work. They also appear not to develop solitons. Since F is given by a Fredholm determinant, these initial conditions represent an entirely new class of integrable initial data for the KP equation.

The initial data for the matrix KP equation (1.8) are formally

(1.26)

where

(1.27) $$ \begin{align} {\textbf{P}}^{{\geq\mathfrak{h}_0,\leq -\mathfrak{d}_{\vec x}^{-\vec r}}}_{x_i,x_j}\!\left(r_i,r_j\right)dr_j&=\mathbb{P}_{\mathbf{B}\left(x_i\right)=r_i}\Big(\mathbf{B}(y)\geq \mathfrak{h}_0(y)~\forall\,y\in\left[x_i,x_j\right],\nonumber\\ &\mathbf{B}(x_n)\leq r_n\text{ for each }x_n\in\left(x_i,x_j\right), \mathbf{B}\left(x_j\right)\in dr_j\Big). \end{align} $$

The probability is with respect to a Brownian motion $\mathbf {B}$ with diffusivity $2$ starting at $r_i$ at time $x_i$ . This is derived in Appendix A for finite collections of narrow wedges. Unlike the scalar case, one can see immediately that the initial data as written are insufficient, because in the matrix product the $0$ and $\infty $ interact. Wrongly interpreted, it appears to produce anomalous solutions, so the initial data written would have to be augmented by at least some description of the rate of convergence to $0$ and $\infty $ in the $t\downarrow 0$ limit. We leave this also for future work.

Outline

In Section 2 we show how the famous Painlevé expressions for the Tracy–Widom distributions just arise as special examples of the KP equation, and write equations for a few other special initial data, in particular the Airy processes. In Section 3 we show that some of the explicit formulas for the $1$ -dimensional distributions of the KPZ equation (as opposed to the KPZ fixed point) also satisfy the KP equation. Section 4 contains the proof of Theorem 1.3.

2.1 Tracy–Widom distributions

A key observation is that the GUE and GOE Tracy–Widom distributions are now seen to arise simply as special similarity solutions of the KP equation (1.9):

Example 2.1 Tracy–Widom GUE distribution

Consider $\mathfrak {h}_0=\mathfrak {d}_0$ , the narrow wedge initial condition defined as $\mathfrak {d}_0(0)=0$ and $\mathfrak {d}_0(x)=-\infty $ , $x\neq 0$ . With this choice of initial data one has $\mathfrak {h}(t,x)+x^2/t\stackrel {\text {dist}}{=}t^{1/3}\mathcal {A}\left (t^{-2/3}x\right )$ , where $\mathcal {A}$ is the Airy $_2$ process (see Section 2.3), which is stationary (in x). From this and the 1:2:3 scaling invariance of equation (1.9), it is natural to look for a self-similar solution of the form

$$ \begin{align*} \phi^{\text{nw}} (t,x,r)=t^{-2/3} \psi^{\text{nw}} \left(t^{-1/3} r + t^{-4/3}x^2\right). \end{align*} $$

This turns equation (1.9) into

(2.1) $$ \begin{align} (\psi^{\text{nw}})'''+12\psi^{\text{nw}}(\psi^{\text{nw}})'-4r(\psi^{\text{nw}})'-2\psi^{\text{nw}}=0. \end{align} $$

The transformation $\psi ^{\text {nw}}=-\mathfrak {q}^2$ takes equation (2.1) into Painlevé II:

(2.2) $$ \begin{align} \mathfrak{q}''=r\mathfrak{q}+2\mathfrak{q}^3. \end{align} $$

As $r\to -\infty $ , the solution is approximately $\phi ^{\text {nw}}(t,x,r)\sim -\left (\tfrac {r}{2t} +\tfrac {x^2}{2t^2}\right )$ , picking out the Hastings–McLeod solution $\mathfrak {q}(r)\sim -\operatorname {\mathrm {\text {Ai}}}(r)$ as $r\to \infty $ . Thus we recover

$$ \begin{align*} F(t,x,r)= F_{\text{GUE}}\left(t^{-1/3} r+t^{-4/3}x^2\right), \end{align*} $$

where $F_{\text {GUE}}$ is the GUE Tracy–Widom distribution [Reference Tracy and WidomTW94], usually written in the equivalent form

$$ \begin{align*} F_{\text{GUE}}(s) = \exp\left\{ -\int_s^\infty\!\mathrm{d} u\, (u-s)\mathfrak{q}^2(u)\right\}. \end{align*} $$

Example 2.2 Tracy–Widom GOE distribution

If $\mathfrak {h}_0(x)\equiv 0$ , the flat initial condition, there is no x dependence and equation (1.9) reduces to the KdV equation. Now we look for a self-similar solution of equation (1.10) in the form

$$ \begin{align*} \phi^{\text{fl}}(t,r) = (t/4)^{-2/3} \psi^{\text{fl}}\left((t/4)^{-1/3} r\right) \end{align*} $$

(the extra factor of $1/4$ is just to coordinate conventions with random matrix theory), obtaining the ordinary differential equation

$$ \begin{align*} \left(\psi^{\text{fl}}\right)''' +12\left(\psi^{\text{fl}}\right)'\psi^{\text{fl}}-r\left(\psi^{\text{fl}}\right)' -2\psi^{\text{fl}} =0. \end{align*} $$

Miura’s transform

$$ \begin{align*} \psi^{\text{fl}} = \tfrac12 \left( \mathfrak{q}' -\mathfrak{q}^2\right) \end{align*} $$

brings this to Painlevé II (equation (2.2)), with the same behaviour as $r\to -\infty $ . So we recover

$$ \begin{align*} F(t,x,r)= F_{\text{GOE}}\left(4^{1/3}t^{-1/3} r\right), \end{align*} $$

where $F_{\text {GOE}}$ is the GOE Tracy–Widom distribution [Reference Tracy and WidomTW96], usually written in the equivalent form

$$ \begin{align*} F_{\text{GOE}}(r) = \exp\left\{ -\frac12\int_r^\infty\!\mathrm{d} u\,\mathfrak{q}(u)\right\}F_{\text{GUE}}(r)^{1/2}. \end{align*} $$

These two examples also have the following interpretation: Let $\lambda ^{\text {max,GUE}}_N$ and $\lambda ^{\text {max,GOE}}_N$ be the largest eigenvalues of $N\times N$ matrices chosen from the Gaussian unitary and Gaussian orthogonal ensembles multiplied by $\sqrt {N}$ , so that $\lambda ^{\text {max,GUE}}_N\sim 2N + N^{1/3}\zeta _{\text {GUE}}$ and $\lambda ^{\text {max,GOE}}_N\sim 2N + N^{1/3}\zeta _{\text {GOE}}$ , with $\zeta _{\text {GUE}}$ and $\zeta _{\text {GOE}}$ the standard Tracy–Widom GUE and GOE random variables. Let

$$ \begin{align*} F_1(t,r)& = {\lim_{N\to \infty}} \mathbb{P}\left(N^{-1/3}\left( \lambda^{\text{max,GOE}}_{Nt}-2Nt\right) \le 4^{1/3}r\right)=F_{\text{GOE}}\left(4^{1/3}t^{-1/3}r\right),\\ F_2(t,x,r) &= {\lim_{N\to \infty}} \mathbb{P}\left(N^{-1/3}\left( \lambda^{\text{max,GUE}}_{Nt}-2 Nt\right) \le r+x^2/t\right)=F_{\text{GUE}}\left(t^{-1/3} r+t^{-4/3}x^2\right). \end{align*} $$

As we have seen, $\partial _r^2\log F_1$ and $\partial _r^2\log F_2$ satisfy the KP equation (1.9). In the former case, there is no dependence on x, and the KP equation reduces to the KdV equation (1.10).

2.2 Partial differential equations for other initial data

Another question is whether there are analogues of Painlevé II for other self-similar solutions. It is natural to observe $\phi $ in the frame of the inviscid solution $\tfrac 14\left (\partial _x\bar {\mathfrak {h}}\right )^2-\partial _t\bar {\mathfrak {h}}=0$ of Burgers’ equation,

$$ \begin{align*} \phi(t,x,r) := \bar\phi\left(t,x,r-\bar{\mathfrak{h}}(t,x)\right); \end{align*} $$

one obtains

(2.3) $$ \begin{align}\partial_t \bar\phi +\bar\phi\hspace{0.1em}\partial_r \bar\phi + \tfrac1{12}\partial_r^3 \bar\phi + \tfrac14\partial_r^{-1} \partial_x^2 \bar\phi -\tfrac12\partial_x\bar{\mathfrak{h}}\hspace{0.1em}\partial_x\bar\phi+ V \bar\phi= 0,\end{align} $$

with $V=-\tfrac {1}4\partial ^2_x\bar {\mathfrak {h}}$ and with initial data $\bar \phi (0,x,r) = 0$ for $r\ge 0$ and $-\infty $ for $r<0$ .

In order to get a solution for the rescaled spatial process, write

$$ \begin{align*} \phi(t,x,r) = t^{-2/3}\psi\left(t,t^{-2/3}x,t^{-1/3}\left(r- \bar{\mathfrak{h}}(t,x)\right)\right). \end{align*} $$

Then

$$ \begin{align*}t\hspace{0.05em}\partial_t\psi-\tfrac23\psi-\tfrac13r\partial_r\psi+\tfrac1{12}\partial_r^3\psi+\psi\partial_r\psi+\tfrac14\partial_r^{-1}\partial_x^2\psi -\tfrac{1}4\partial_x^2\tilde{\mathfrak{h}}\hspace{0.05em}\psi -\left(\tfrac23x +\tfrac12\partial_x\tilde{\mathfrak{h}}\right)\partial_x\psi =0,\end{align*} $$

with $\tilde {\mathfrak {h}}(t,x)=t^{-1/3}\bar {\mathfrak {h}}\left (t,t^{2/3}x\right )$ .

Example 2.3 Half-flat initial data

Consider $\mathfrak {h}_0(x)=0$ , $x\le 0$ and $\mathfrak {h}_0(x)=-\infty $ , $x>0$ . Now $\bar {\mathfrak {h}}(t,x) = -x^2/t\hspace {0.05em}\mathbf {1}_{x\geq 0}$ . There is dependence on x, though not on t. This gives rise to a partial differential equation for $\psi (x,r)$ ,

$$ \begin{align*}-\tfrac13r\partial_r\psi+\tfrac1{12}\partial_r^3\psi-\left(\tfrac16\mathbf{1}_{x\ge0}+\tfrac23\mathbf{1}_{x<0}\right)\psi+\psi\partial_r\psi+\tfrac14\partial_r^{-1}\partial_x^2\psi +\left(\tfrac13\mathbf{1}_{x\ge 0}-\tfrac23\mathbf{1}_{x<0}\right)x\partial_x\psi=0.\end{align*} $$

Remark 2.4 Lower-tail heuristics

Typically the equation is controlled on large scales by the equation with the third derivative dropped, and Burgers’ equation makes sense for such wedge-type initial data. Let $\bar \phi $ be as in equation (2.3) and

$$ \begin{align*} \check\phi = \bar\phi -\eta, \end{align*} $$

where $\eta =\tfrac {c }{t} r\mathbf {1}_{r<0}$ . Then $\bar \phi (0,x,r)=0$ and

$$ \begin{align*} \partial_t \check\phi + \tfrac12\partial_r \check\phi^2+ \tfrac1{12}\partial_r^3\check\phi+ \tfrac14\partial_r^{-1} \partial_x^2\check\phi + \partial_r\left(\eta\check\phi\right) &-\tfrac12 \partial_x \bar{\mathfrak{h}}\partial_x\check\phi-\tfrac14 \partial^2_x \bar{\mathfrak{h}}~ \check\phi \\ &+\left(\tfrac{c-1}{t} -\tfrac14 \partial^2_x \bar{\mathfrak{h}}\right) \eta - \tfrac{c }{12t} \delta'_0(r) = 0. \end{align*} $$

One hopes to set things up so that $\check \phi $ has good decay at $\pm \infty $ . Consider our two basic examples. In the flat case, $\bar {\mathfrak {h}}\equiv 0$ , and we want to take $c=1$ to make the second-to-last term drop out, which will lead to the conclusion that $\phi (t,r)\sim r/t$ , or $\log F(t,r)\sim - \tfrac 1{6t} r^3$ as $r\to -\infty $ , recovering the Tracy–Widom GOE lower tail. In the narrow wedge, case $\bar {\mathfrak {h}} = -\tfrac {x^2}{t}$ , so $\tfrac 14 \partial ^2_x \bar {\mathfrak {h}}=-\tfrac {1}{2t}$ , and we want to take $c=1/2$ leading to $\log F(t,x,r)\sim -\tfrac 1{12 t} \left (r+ \tfrac {x^2}{t}\right )^3 $ , recovering the Tracy–Widom GUE lower tail.

The conclusion is that the lower tail of the distributions can be seen directly from the ‘Burgers’ part of the KP equation, which dominates in that region.

Note that this is only intended to be a quick heuristic. If one were to try to make it rigorous, say, for the flat and narrow wedge case, one would be led through self-similar solutions back to the Painlevé equation and the standard tools there to get the left tails (see, e.g., [Reference Baik, Buckingham and DiFrancoBBD08]). Such Riemann–Hilbert methods give much finer information, for example, subleading terms. It is an interesting question whether such results can be obtained from the partial differential equation (2.3) so as to get finer information on the left tail for more general initial data.

2.3 Airy process

The Airy process $\mathcal {A}(x)$ is defined as

$$ \begin{align*} \mathcal{A}(x) := \mathfrak{h}(1,x;\mathfrak{d}_0)+x^2, \end{align*} $$

where $\mathfrak {h}(t,x;\mathfrak {d}_0)$ is the KPZ fixed point starting from a narrow wedge $\mathfrak {d}_0$ at the origin. The Airy process is stationary and the one-point distribution is the Tracy–Widom GUE distribution.

K. Johansson famously asked whether there is an equation for the multipoint distribution. Partial differential equations (PDEs) were given by [Reference Adler and van MoerbekeAM05; Reference Tracy and WidomTW03]. Starting with narrow wedge initial data, the matrix KP equation (1.8) gives us another PDE. This equation lives at the same level as the system of PDEs in [Reference Tracy and WidomTW03]; it could well be that they are equivalent, but we have not succeeded in confirming that yet, and leave it for future work.

Note that the KP-II equation is written in variables t, $r=r_1+\dotsb +r_n$ and $x=x_1+\dotsb +x_n$ , the rest of the variables entering only from the boundary condition. Exploiting skew time-reversal invariance, we find an extra symmetry in the narrow wedge case. Let $\mathfrak {d}^{\vec r}_{\vec x}$ denote a multiple narrow wedge (see Example 1.6). For $\vec x\in \mathbb {R}^m$ and $z\in \mathbb {R}$ , write $\vec x+z=(x_1+z,\dotsc ,x_m+z)$ . By skew time reversibility and translation and affine invariance of the KPZ fixed point [Reference Matetski, Quastel and RemenikMQR17, Thm. 4.5],

(2.4) $$ \begin{align} F\left(t,\vec x+z,\vec r+a\right)&=\mathbb{P}\big(\mathfrak{h}(t,x_i+z;\mathfrak{d}_0)\leq r_i+a\text{ for all }i\big) =\mathbb{P}\left(\mathfrak{h}(t,\cdot;\mathfrak{d}_0)\leq-\mathfrak{d}^{-\vec r-a}_{\vec x+z}\right)\nonumber\\ &=\mathbb{P}\left(\mathfrak{h}\left(t,\cdot;\mathfrak{d}^{-\vec r-a}_{\vec x+z}\right)\leq-\mathfrak{d}_0\right) =\mathbb{P}\left(\mathfrak{h}\left(t,0;\mathfrak{d}^{-\vec r}_{\vec x+z}\right)\leq a\right)\nonumber\\ &=\mathbb{P}\left(\mathfrak{h}\left(t,-z;\mathfrak{d}^{-\vec r}_{\vec x}\right)\leq a\right). \end{align} $$

Now the right-hand side is just the one-point distribution at $-z$ with a given, fixed initial condition. So if we let $G(t,z,a)=F\left (t,\vec x+z,\vec r+a\right )$ , we see that $\partial _a^2\log G$ satisfies equation (1.9) in $(t,z,a)$ (in terms of equation (1.25), the initial data are $G(0,z,a)=-\infty $ if $z=-x_i$ for some i and $a<-r_i$ and $G(0,z,a)=0$ otherwise). But $\partial _zG(t,z,a)=\mathcal {D}_xF\left (t,\vec x+z,\vec r+a\right )$ , and similarly, $\partial _aG(t,z,a)=\mathcal {D}_rF\left (t,\vec x+z,\vec r+a\right )$ . So now setting $a=z=0$ , we deduce that $\phi =\mathcal {D}_r^2\log F$ satisfies

(2.5) $$ \begin{align} \partial_t \phi + \tfrac12 \mathcal{D}_r \phi^2 + \tfrac1{12}\mathcal{D}_r^3 \phi + \tfrac14\mathcal{D}_r^{-1} \mathcal{D}_x^2 \phi = 0. \end{align} $$

The initial data are now similarly $\phi \left (0,\vec x,\vec r\right )=-\infty $ if $x_i=0$ for some i and $r_i<0$ and $\phi \left (0,\vec x,\vec r\right )=0$ otherwise.

Note that from Theorem 1.1, we know that $\mathcal {D}_r^2\log F=\operatorname {{\mathop {\mathrm {{tr}}}}} q$ and $\partial _t\operatorname {{\mathop {\mathrm {{tr}}}}} q+\operatorname {{\mathop {\mathrm {{tr}}}}}(q\mathcal {D}_rq)+\tfrac 1{12}\mathcal {D}_r^3\operatorname {{\mathop {\mathrm {{tr}}}}} q+\tfrac 14\mathcal {D}_x^2\operatorname {{\mathop {\mathrm {{tr}}}}} Q=0$ (using $\operatorname {{\mathop {\mathrm {{tr}}}}}(AB)=\operatorname {{\mathop {\mathrm {{tr}}}}}(BA)$ ). But the foregoing argument implies then that $\operatorname {{\mathop {\mathrm {{tr}}}}} q$ itself solves the KP equation (2.5), and as a consequence, we deduce in the narrow wedge case that $\operatorname {{\mathop {\mathrm {{tr}}}}}(q\mathcal {D}_rq)=\operatorname {{\mathop {\mathrm {{tr}}}}}(q)\operatorname {{\mathop {\mathrm {{tr}}}}}(\mathcal {D}_rq)$ . This can also be proved directly using the fact that, in this case, $(\partial _u+\partial _v)\mathbf {K}^{\operatorname {\mathrm {\text {hypo}}}\left (\mathfrak {h}_0\right )}_t(u,v)$ is a rank $1$ kernel, which implies that q is a rank $1$ matrix.

An alternative derivation of equation (2.5) using the path integral formula for the KPZ fixed point can be found in Appendix B.

Next we use the 1:2:3 scaling invariance of the KPZ fixed point and stationarity of the Airy process, as in Example 2.1. Let H, $\Psi $ , $\psi $ denote the $(m+1)$ -point distribution function of the Airy process and its logarithmic derivatives,

(2.6) $$ \begin{align}H(y_1,\dotsc,y_{m},r_0,\dotsc,r_m) &= \mathbb{P}( \mathcal{A}(0)\le r_0,\mathcal{A}(y_1)\le r_1,\dotsc,\mathcal{A}(y_m)\leq r_m), \nonumber\\ \Psi&= \mathcal{D}_r \log H,\qquad \psi=\mathcal{D}_r \Psi. \end{align} $$

Since $F(t,x_0,\dotsc ,x_m,r_0,\dotsc ,r_m)=H\left (t^{-2/3}\bar x_1,\dotsc ,t^{-2/3}\bar x_m, t^{-1/3} \hat r_0, \dotsc ,t^{-1/3}\hat r_m\right )$ , with $\bar x_i=x_i-x_0$ and $\hat r_i=r_i+\tfrac {x^2_i}{t}$ , equation (2.5) leads to the following:

Theorem 2.5 Airy $_2$ -process multipoint function

The logarithmic derivatives (2.6) of the multipoint function of the Airy process satisfy the PDE

$$ \begin{align*} &\textstyle-\tfrac1{6}\psi + \tfrac12\mathcal{D}_r\psi^2 + \tfrac1{12} \mathcal{D}_r^3 \psi -\tfrac13\sum_{a=0}^mr_a\partial_{r_a}\psi-\tfrac23\sum_{a=1}^my_a\partial_{y_a}\psi\\ &\qquad\qquad\qquad\qquad\textstyle-\sum_{a=1}^m\sum_{b=0,\,b\neq a}^my_a^2\partial_{r_a}\hspace{-0.1em}\partial_{r_b}\Psi+\sum_{a,b=1,\,a\neq b}^my_ay_b\hspace{0.05em}\partial_{r_a}\hspace{-0.1em}\partial_{r_b}\Psi =0, \end{align*} $$

with boundary condition $\psi (0,\dotsc ,0,r_1,\dotsc ,r_m) =\psi ^{\text {nw}}( \min (r_1,\dotsc ,r_m))$ from equation (2.1).

In particular, for the logarithmic derivative of the Airy-process two-point function $\Psi (y,r_1,r_2)=\mathcal {D}_r\log \mathbb {P}(\mathcal {A}(0)\leq r_1,\,\mathcal {A}(y)\leq r_2)$ , we get

(2.7) $$ \begin{align} -\tfrac1{6}\psi+\tfrac12\mathcal{D}_r\psi^2+\tfrac1{12}\mathcal{D}_r^3\psi-\tfrac13(r_1\partial_{r_1}\psi+r_2\partial_{r_2}\psi)-\tfrac23y\hspace{0.05em}\partial_y\psi-y^2\partial_{r_1}\partial_{r_2}\Psi =0. \end{align} $$

In our notation, the equation derived in [Reference Adler and van MoerbekeAM05] reads

(2.8) $$ \begin{align} (r_2-r_1)\partial_{r_1}\partial_{r_2}\psi+y\hspace{0.05em}\partial_y\left(\partial_{r_1}-\partial_{r_2}\right)\psi+y^2(\partial_{r_1}-\partial_{r_2})\partial_{r_1}\partial_{r_2}\Psi +\partial_{r_1}\!\Psi\hspace{0.1em}\partial_{r_2}\hspace{-0.1em}\mathcal{D}_r\psi-\partial_{r_2}\!\Psi\hspace{0.1em}\partial_{r_1}\hspace{-0.1em}\mathcal{D}_r\psi=0. \end{align} $$

Although equations (2.7) and (2.8) have similarities, they do not appear to be equivalent. We have not succeeded in reconciling them, and we leave this for future work. Of course the Airy process has many symmetries, and it is plausible that the equations are not equivalent yet both hold.

Example 2.6 Airy $_1$ -process multipoint function

Consider now the KPZ fixed point with flat initial data, $\mathfrak {h}_0\equiv 0$ . The multipoint function satisfies

$$ \begin{align*}F\left(t,\vec x,\vec r\right)=\mathbb{P}\left(\mathcal{A}_1(0)\leq t^{-1/3}r_1,\,\mathcal{A}_1\left(t^{-2/3}\bar x_2\right)\leq t^{-1/3}r_2,\dotsc,\mathcal{A}_1\left(t^{-2/3}\bar x_m\right)\leq t^{-1/3}r_m\right) \end{align*} $$

with $\bar x_i=x_i-x_1$ , where $\mathcal {A}_1(x):=\mathfrak {h}(1,x;0)$ is the Airy $_1$ process and we have used the fact that it is stationary. We know that the left-hand side equals $\operatorname {{\mathop {\mathrm {{tr}}}}} Q$ where the $m\times m$ matrix Q and its derivative q solve equation (1.8). But the right-hand side depends only on the differences $x_i-x_1$ , so Q actually solves the (integrated) matrix KdV equation

$$ \begin{align*}\partial_tQ+\tfrac12(\mathcal{D}_rQ)^2+\tfrac1{12}\mathcal{D}_r^3Q=0.\end{align*} $$

Take now $m=2$ for simplicity. The (formal) initial data $Q(0,x_1,x_2,r_1,r_2)$ (equation (1.26)) are also invariant under $r_i,x_i\mapsto t^{-1/3}r_i,t^{-2/3}x_i$ followed by $Q\mapsto t^{-1/3}Q$ , and depend only on $\lvert x_2-x_1\rvert \,(\text {in fact, }Q_{12}(0,r_1,r_2,x_1,x_2) = \frac {\mathbf {1}_{r_1\ge 0,r_2\ge 0}}{\sqrt {4\pi \left \lvert x_2-x_1\right \rvert }}(e^{- \left (r_1+r_2\right )^2/4\left \lvert x_2-x_1\right \rvert } -e^{-\left (r_1-r_2\right )^2/4\left \lvert x_2-x_1\right \rvert } )\text { by the reflection principle} )$ . In this case we look for a matrix solution of equation (1.8) with the scaling

$$ \begin{align*}Q(t,r_1,r_2,x_1,x_2)= t^{-1/3}\bar{Q}\left(t^{-2/3}(x_2-x_1),t^{-1/3}r_1,t^{-1/3}r_2\right).\end{align*} $$

The conclusion is that

(2.9) $$ \begin{align} \mathcal{D}_r\log\mathbb{P}(\mathcal{A}_1(0)\leq r_1,\,\mathcal{A}_1(y)\leq r_2)=\operatorname{{\mathop{\mathrm{{tr}}}}}\!\left(\bar Q(y,r_1,r_2)\right), \end{align} $$

with $\bar Q(y,r_1,r_2)$ solving the matrix PDE

$$ \begin{align*}-\tfrac13\bar Q+\tfrac12\left(\mathcal{D}_r\bar Q\right)^2+\tfrac1{12}\mathcal{D}_r^3\bar Q-\tfrac13\left(r_1\partial_{r_1}\bar Q+r_2\partial_{r_2}\bar Q\right)-\tfrac23y\partial_y\bar Q=0.\end{align*} $$

It does not seem to be possible to turn this $4\times 4$ system into a closed equation for the left-hand side of equation (2.9); the fact that this works for the Airy $_2$ process is very particular to narrow wedge initial data (and follows from skew time reversibility as used in equation (2.4)).

3.1 Narrow wedge solution of the KPZ equation

Let ${\textrm {h}}_{\text {nw}}$ be the narrow wedge solution of equation (1.1) with $\lambda =\nu =\tfrac 14$ and $\sigma =1$ . In other words, ${\textrm {h}}_{\text {nw}}=\log Z$ , where Z is the fundamental solution of the stochastic heat equation with multiplicative noise:

(3.1) $$ \begin{align} \partial_t Z = \tfrac14 \partial^2_xZ + \xi Z,\qquad Z(0,x)=\delta_0(x). \end{align} $$

The KPZ generating function is

(3.2) $$ \begin{align} G_{\text{nw}}(t,x,r) = \mathbb{E}\left[\exp\!\left\{-e^{{\textrm{h}}_{\text{nw}}\left(t,x\right)+\frac{t}{12}-r}\right\}\right]. \end{align} $$

The distribution of ${\textrm {h}}_{\text {nw}}(t,x)$ was computed in 2010 in [Reference Amir, Corwin and QuastelACQ11; Reference Sasamoto and SpohnSS10; Reference DotsenkoDot10; Reference Calabrese, Doussal and RossoCDR10], with the resultFootnote ⁴ that $G_{\text {nw}}(t,x,r)=\det (\mathbf {I} - \mathbf {K})_{L^2(\mathbb {R}^+)}$ with

$$ \begin{align*}\mathbf{K}(u,v)=\int_{-\infty}^\infty dy\,t^{-2/3}\frac{1}{1+e^{y}}\operatorname{\mathrm{\text{Ai}}}\left(t^{-1/3}(u+r-y)+t^{-4/3}x^2\right)\operatorname{\mathrm{\text{Ai}}}\left(t^{-1/3}(v+r-y)+t^{-4/3}x^2\right).\end{align*} $$

If we conjugate the operator by multiplying the kernel by $e^{\left (v-u\right )x/t}$ , the value of the Fredholm determinant doesn’t change – that is, we also have

$$ \begin{align*}G_{\text{nw}}(t,x,r)=\det\left(\mathbf{I} - \widetilde{\mathbf{K}}\right)_{L^2(\mathbb{R}^+)}\quad\text{with}\quad\widetilde{\mathbf{K}}\left(u,v\right)=e^{(v-u)x/t}\mathbf{K}(u,v).\end{align*} $$

One checks directly that $\widetilde {\mathbf {K}}$ satisfies the differential relations Theorem 1.3(1)–(3). This can also be seen by writing $\widetilde {\mathbf {K}}(u,v) = \mathbf {U}_te^{-x\partial ^2}\hspace {0.05em}\mathbf {M}\hspace {0.05em} e^{x\partial ^2}\mathbf {U}_t^{-1}(u+r,v+r)$ , with $\mathbf {M}$ the multiplication operator $\mathbf {M} f(u) = (1+e^{u})^{-1}f(u)$ and $\mathbf {U}_t$ the Airy unitary operator defined in equation (1.18). We then have the following:

The initial condition is $\lim _{t\searrow 0}\phi _{\text {nw}}\left (t,x,r-\tfrac {x^2}{t} -\log \sqrt {\pi t} \right ) = -e^{-r}$ . This suggests defining the x-independent shifted variable $\hat \phi _{\text {nw}}(t,r )= \phi _{\text {nw}}\left (t,x,r-\tfrac {x^2}{t} -\log \sqrt {\pi t} \right )$ , which now satisfies the cylindrical KdV equation

$$ \begin{align*} \partial_t \hat \phi_{\text{nw}} +\tfrac1{2t}\partial_r \hat \phi_{\text{nw}}+ \hat \phi_{\text{nw}}\partial_r \hat \phi_{\text{nw}}+ \tfrac1{12}\partial_r^3 \hat \phi_{\text{nw}} +\tfrac1{2t} \hat \phi_{\text{nw}}= 0,\qquad \hat \phi_{\text{nw}}(0,r) =-e^{-r}. \end{align*} $$

3.2 KPZ equation with spiked/half-Brownian initial data

Consider now the solution ${\textrm {h}}_{b}$ of equation (1.1) with $\lambda =\nu = \tfrac 14$ , $\sigma =1$ and m-spiked initial data, where $b=(b_1,\dotsc ,b_m)\in \mathbb {R}^m$ are the spike parameters. When $m=1$ , this corresponds to half-Brownian initial data (more precisely, at the level of the stochastic heat equation one sets $Z(0,x)=e^{B(x)+b_1x}\mathbf {1}_{x\geq 0}$ , where $B(x)$ is a Brownian motion with diffusivity 2); for the definition in the general case $m\geq 1$ , we refer to [Reference Borodin, Corwin and FerrariBCF14, Defn. 1.9]. Define $G_b$ as in equation (3.2) (with ${\textrm {h}}_b$ in place of ${\textrm {h}}_{\text {nw}}$ ). Then from [Reference Borodin, Corwin and FerrariBCF14, Thm. 1.10] we get that $G_b(t,0,r)=\det (\mathbf {I} - \mathbf {K}_0)_{L^2\left [0,\infty \right )}$ (see footnote 4 again), with

$$ \begin{align*} \mathbf{K}_0(u,v) =\frac{1}{(2\pi{\textrm{i}})^{2}}\int_{\mathcal{C}_t} d\eta\int_{\mathcal{C}_t'} d\xi\,\frac{t^{-1/3}\pi}{\sin\left(t^{-1/3}\pi(\xi-\eta)\right)}\frac{e^{\xi^3/3-\left(u+t^{-1/3}r\right)\xi}}{e^{\eta^3/3-\left(v+t^{-1/3}r\right)\eta}}\prod_{k=1}^m\frac{\Gamma\left(t^{-1/3}\eta-b_k\right)}{\Gamma\left(t^{-1/3}\xi-b_k\right)}, \end{align*} $$

where $\mathcal {C}_t$ goes from $-\frac 14t^{1/3}-{\textrm {i}}\infty $ to $-\tfrac 14t^{1/3}+{\textrm {i}}\infty $ crossing the real axis to the right of $t^{1/3}b_1,\dotsc ,t^{1/3}b_m$ and $\mathcal {C}_t'=\mathcal {C}_t+\frac 12t^{1/3}$ . We scale $(\eta ,\xi )\mapsto \left (t^{1/3}\eta ,t^{1/3}\xi \right )$ and $(u,v)\mapsto \left (t^{-1/3}u,t^{-1/3}v\right )$ (in the Fredholm determinant) so that $\mathbf {K}_0(u,v)$ is now given as

$$ \begin{align*}\int_{\mathcal{C}_1} d\eta\int_{\mathcal{C}_1'} d\xi\,\frac{\pi}{\sin(\pi(\xi-\eta))}\frac{e^{t\xi^3/3-(u+r)xi}}{e^{t\eta^3/3-(v+r)\eta}}\prod_{k=1}^m\!\frac{\Gamma(\eta-b_k)}{\Gamma(\xi-b_k)}.\end{align*} $$

Since ${\textrm {h}}_{b+x/t}(t,x)+x^2/t$ is stationary in x (see [Reference Borodin, Corwin and FerrariBCF14, Rem. 1.14]), we may write $G_{b}(t,x,r)=\det (\mathbf {I}-\mathbf {K})_{L^2\left [0,\infty \right )}$ , with

$$ \begin{align*} \mathbf{K}(u,v)=\frac{1}{(2\pi{\textrm{i}})^{2}}\int_{\widetilde{\mathcal{C}}_1} d\xi\int_{\widetilde{\mathcal{C}}_1'} d\eta\,\frac{\pi}{\sin(\pi(\xi-\eta))}\frac{e^{t\xi^3/3-\left(u+r+x^2/t\right)\xi}}{e^{t\eta^3/3-\left(v+r+x^2/t\right)\eta}}\prod_{k=1}^m\frac{\Gamma(\eta-b_k-x/t)}{\Gamma(\xi-b_k-x/t)}. \end{align*} $$

Here the contour $\widetilde {\mathcal {C}}_1$ has to cross the real axis to the right of $b_i+x/t$ for all i. Conjugating the kernel by $e^{ux/t}$ as in the previous case (that is, multiplying the kernel by $e^{(u-v)x/t}$ ) – and changing variables $\eta \longmapsto \eta +x/t$ , $\xi \longmapsto \xi +x/t$ , we get

$$ \begin{align*} G_{b}(t,x,r)=\det\left(\mathbf{I}-\widetilde{\mathbf{K}}\right)_{L^2\left[0,\infty\right)}, \end{align*} $$

with

$$ \begin{align*}\widetilde{\mathbf{K}}(u,v)=\frac{1}{(2\pi{\textrm{i}})^{2}}\int_{\mathcal{C}_1} d\xi\int_{\mathcal{C}_1'} d\eta\,\frac{\pi}{\sin(\pi(\xi-\eta))}\frac{e^{t\xi^3/3+x\xi^2-(u+r)\xi}}{e^{t\eta^3/3+x\eta^2-(v+r)\eta}}\prod_{k=1}^m\frac{\Gamma(\eta-b_k)}{\Gamma(\xi-b_k)}.\end{align*} $$

As in the previous case, $\tilde {\mathbf {K}}$ satisfies the necessary differential relations, so we have the following:

3.3 KPZ equation with two-sided Brownian initial data

In [Reference Le DoussalLD20] (which appeared about a month after the first version of the present article), Le Doussal suggests that a certain modified generating function for the solution ${\textrm {h}}_{w_\pm }$ of equation (1.1) with the same scaling as in the of regoing and initial data of the form ${\textrm {h}}_{w_\pm }(0,x)=\mathbf {B}(x)+w_-x\mathbf {1}_{x<0}+w_+x\mathbf {1}_{x\geq 0}$ with $\mathbf {B}$ a double-sided Brownian motion with $\mathbf {B}(0)=0$ and $w_->w_+$ should also satisfy the KP equation. This is actually true, as we explain next. Define

$$ \begin{align*}\widetilde G_{w_\pm}(t,x,r)=\Gamma(w_--w_+)^{-1}\mathbb{E}\left[2\hspace{0.05em} e^{\frac12\left(w_--w_+\right)\left({\textrm{h}}_{w_\pm}\left(t,x\right)+\frac{t}{12}-r\right)}K_{w_+-w_-}\left(2e^{\frac12\left({\textrm{h}}_{w_\pm}\left(t,x\right)+\frac{t}{12}-r\right)}\right)\right],\end{align*} $$

where $K_\nu $ is the modified Bessel function of order $\nu $ . This modified generating function $\widetilde G_{w_\pm }$ can alternatively be expressed as the analog of equation (3.2) where the KPZ height function ${\textrm {h}}_{w_\pm }$ is replaced by a randomly shifted height function ${\textrm {h}}_{w_\pm }(t,x)+\Upsilon $ :

$$ \begin{align*}\widetilde G_{w_\pm}(t,x,r)=\mathbb{E}\left[\exp\!\left\{-e^{{\textrm{h}}_{w_\pm}\left(t,x\right)+\Upsilon+\frac{t}{12}-r}\right\}\right],\end{align*} $$

with $\Upsilon $ an independent log-gamma random variable with parameter $w_--w_+$ ; that is, $e^{-\Upsilon }$ has density $\Gamma (w_--w_+)^{-1}x^{w_--w_+-1}e^{-x}$ . Explicit formulas for the distribution of this shifted height function were obtained in [Reference Imamura and SasamotoIS12; Reference Imamura and SasamotoIS13] using the nonrigorous replica method. A similar formula, which is the one we will use, was obtained rigorously later on [Reference Borodin, Corwin, Ferrari and VetőBor+15]; the equality between the two expressions for $\widetilde G_{w_\pm }$ is essentially [Reference Borodin, Corwin, Ferrari and VetőBor+15, Corollary 2.6] (see also [Reference Borodin, Corwin, Ferrari and VetőBor+15, Remark 2.10]).

From [Reference Borodin, Corwin, Ferrari and VetőBor+15, Thm. 2.9] (and footnote 4 again), we have $\widetilde G_{w_\pm }(t,x,r)=\det (\mathbf {I}-\mathbf {K})_{L^2\left [0,\infty \right )}$ , with

$$ \begin{align*} &\mathbf{K}(u,v)=\frac{1}{(2\pi{\textrm{i}})^{2}}\int_{\mathcal{C}_t} d\eta\int_{\tilde{\mathcal{C}}_t} d\xi\,\frac{t^{-1/3}\pi}{\sin\left(t^{-1/3}\pi(\xi-\eta)\right)}\frac{e^{\xi^3/3-\left(u+t^{-1/3}\left(r+\frac{x^2}{t}\right)\right)\xi}}{e^{\eta^3/3-\left(v+t^{-1/3}\left(r+\frac{x^2}{t}\right)\right)\eta}}\\ &\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\times\frac{\Gamma\left(w_-+\frac{x}{t}-t^{-1/3}\xi\right)}{\Gamma\left(t^{-1/3}\xi-w_+-\frac{x}{t}\right)}\frac{\Gamma\left(t^{-1/3}\eta-w_+-\frac{x}{t}\right)}{\Gamma\left(w_-+\frac{x}{t}-t^{-1/3}\eta\right)}, \end{align*} $$

where $\mathcal {C}_t$ goes from $-\frac 14t^{1/3}-{\textrm {i}}\infty $ to $-\frac 14t^{1/3}+{\textrm {i}}\infty $ crossing the real axis between $t^{1/3}w_+$ and $t^{1/3}w_-$ , and $\tilde {\mathcal {C}}_t$ goes from $\frac 14t^{1/3}-{\textrm {i}}\infty $ to $\frac 14t^{1/3}+{\textrm {i}}\infty $ staying to the right of $\mathcal {C}_t$ and also crossing the real axis between $t^{1/3}w_+$ and $t^{1/3}w_-$ . Changing variables as in the last example leads to

$$ \begin{align*} \widetilde G_{w_\pm}(t,x,r)=\det\left(\mathbf{I}-\widetilde{\mathbf{K}}\right)_{L^2\left[0,\infty\right)}, \end{align*} $$

with

$$ \begin{align*}\widetilde{\mathbf{K}}(u,v)=\frac{1}{(2\pi{\textrm{i}})^{2}}\int_{\mathcal{C}_1} d\eta\int_{\tilde{\mathcal{C}}_1} d\xi\,\frac{\pi}{\sin(\pi(\xi-\eta))}\frac{e^{t\xi^3/3+x\xi^2-(u+r)\xi}}{e^{t\eta^3/3+x\eta^2-(v+r)\eta}}\frac{\Gamma(w_--\xi)}{\Gamma(\xi-w_+)}\frac{\Gamma(\eta-w_+)}{\Gamma(w_--\eta)}.\end{align*} $$

As in the previous examples, $\widetilde {\mathbf {K}}$ satisfies the hypotheses of Theorem 1.3, so we have the following:

Our next two examples are related to the setting of Theorem 3.2. Here we abandon the setting of the KPZ equation and go back to random matrix distributions and the KPZ fixed point.

3.4 The Baik–Ben Arous–Péché distribution for spiked random matrices

Consider $G_b$ as in the case of spiked initial data. It is known (see [Reference Borodin, Corwin and FerrariBCF14, Cor. 1.15] for the case $t=1,x=0$ ; the general case follows in the same way or by scaling and shift invariance) that

$$ \begin{align*}\lim_{\varepsilon\to0}G_{\varepsilon^{1/2}b}\left(\varepsilon^{-3/2}t,\varepsilon^{-1}x,\varepsilon^{-1/2}r\right)=F_{\text{BBP},t^{1/3}b-t^{-2/3}x}\left(t^{-1/3}r+t^{-4/3}x^2\right),\end{align*} $$

where $F_{\text {BBP},b}$ is the Baik–Ben Arous–Péché (BBP) distribution arising from spiked (unitarily invariant) random matrices [Reference Baik, Ben Arous and PéchéBBP05]. On the other hand, by the scaling invariance of the KP-II equation (see Remark 1.2), for each fixed $\varepsilon>0$ , $\phi ^\varepsilon _b(t,x,r):=\partial _r^2\log G_{\varepsilon ^{1/2}b}\left (\varepsilon ^{-3/2}t,\varepsilon ^{-1}x,\varepsilon ^{-1/2}r\right )$ satisfies the KP-II equation as well. As a consequence, one expects the following to hold:

Theorem 3.4. Define

$$ \begin{align*}\tilde F_{\text{BBP},b}(t,x,r)=F_{\text{BBP},t^{1/3}b-t^{-2/3}x}\left(t^{-1/3}r+t^{-4/3}x^2\right).\end{align*} $$

Then $\phi _{\text {BBP},b}:=\partial _r^2\log \tilde F_{\text {BBP},b}$ satisfies the KP-II equation (1.9).

This is indeed the case, which can be checked in a similar way as before using Theorem 1.3 and the explicit Fredholm determinant formula for $F_{\text {BBP},b}$ (see, e.g., [Reference Borodin, Corwin and FerrariBCF14, Eqn. (1.4)]).

3.5 KPZ fixed point with half-Brownian initial data

Consider the KPZ fixed point $\mathfrak {h}$ with half-Brownian initial data $\mathfrak {h}_0(x)=\mathbf {B}(-x)$ for $x\leq 0$ and $\mathfrak {h}_0(x)=-\infty $ for $x>0$ , where $\mathbf {B}$ is a Brownian motion started at $0$ with diffusivity $2$ . In the setting of [Reference Matetski, Quastel and RemenikMQR17], this corresponds to starting TASEP with a product measure with density $\frac 12$ on the negative integers and no particles on the positive integers, and known results [Reference Corwin, Ferrari and PéchéCFP10] in this case give $\mathbb {P}(\mathfrak {h}(1,x)\leq r)=F_{\text {BBP},x}\left (r+x^2\right )$ . By the scaling invariance of the KPZ fixed point, in this case for $F_{\text {half-BM}}(t,x,r)=\mathbb {P}_{\mathfrak {h}_0}(\mathfrak {h}(t,x)\leq r)$ this gives

$$ \begin{align*}F_{\text{half-BM}}(t,r,x)=F_{\text{BBP},t^{-2/3}x}\left(t^{-1/3}r+t^{-4/3}x^2\right).\end{align*} $$

Comparing with Theorem 3.4 we deduce the following:

Similar statements can be written for more general spiked initial conditions in this setting as well as for multipoint distributions (now in terms of the matrix KP equation as in Theorem 1.1).

4.1 The logarithmic derivative

Let

$$ \begin{align*}\Phi(t,r,x)=\partial_r\log\det(\mathbf{I}-\mathbf{K})_{L^2\left(\mathbb{R}^+\right)\oplus\dotsm\oplus L^2\left(\mathbb{R}^+\right)},\end{align*} $$

where $\mathbf {K}$ is a matrix kernel satisfying the assumptions of Theorem 1.3. From now on we omit the subscript on the Fredholm determinant and traces.

Given an operator $\mathbf {A}$ acting on the n-fold direct sum of $L^2(\mathbb {R}^+)$ with matrix kernel $(\mathbf {A}_{ab}(u,v))_{a,b=1}^n$ , we will write $\mathrm {d}_1\mathbf {A}$ and $\mathrm {d}_2\mathbf {A}$ for the operators with kernels, given by

$$ \begin{align*}\mathrm{d}_1\mathbf{A}(u,v)=\big(\partial_u\mathbf{K}_{ab}(u,v)\big)_{a,b=1,\dotsc,n}\quad\text{and}\quad\mathrm{d}_2\mathbf{A}(u,v)=\big(\partial_v\mathbf{K}_{ab}(u,v)\big)_{a,b=1,\dotsc,n}.\end{align*} $$

Note that this is just a notational device; $\mathrm {d}_i$ is not meant to denote an operator. By Theorem 1.3(1),

$$ \begin{align*} \Phi(t,r,x)&=\partial_{r}\hspace{-0.1em}\log(\det(\mathbf{I}-\mathbf{K}))=-\operatorname{{\mathop{\mathrm{{tr}}}}}\left((\mathbf{I}-\mathbf{K})^{-1}\partial_{r}\mathbf{K}\right)\\ &=-\operatorname{{\mathop{\mathrm{{tr}}}}}\left((\mathbf{I}-\mathbf{K})^{-1}(\mathrm{d}_1+\mathrm{d}_2)\mathbf{K}\right)=-\operatorname{{\mathop{\mathrm{{tr}}}}}\left((\mathbf{I}-\mathbf{K})^{-1}\mathrm{d}_1\mathbf{K}+\mathrm{d}_2\left((\mathbf{I}-\mathbf{K})^{-1}\mathbf{K}\right)\right)\\ &=-\operatorname{{\mathop{\mathrm{{tr}}}}}\left(\mathrm{d}_1\mathbf{K}(\mathbf{I}-\mathbf{K})^{-1}+\mathrm{d}_2\left((\mathbf{I}-\mathbf{K})^{-1}\mathbf{K}\right)\right)=-\operatorname{{\mathop{\mathrm{{tr}}}}}\left((\mathrm{d}_1+\mathrm{d}_2)\left((\mathbf{I}-\mathbf{K})^{-1}\mathbf{K}\right)\right)\\ &=-\sum_a\int_0^\infty d\xi\,\partial_\xi\left((\mathbf{I}-\mathbf{K})^{-1}\mathbf{K}\right)_{aa}(\xi,\xi)=\sum_a\left((\mathbf{I}-\mathbf{K})^{-1}\mathbf{K}\right)_{aa}(0,0), \end{align*} $$

where we used the cyclicity of the trace. Introducing the notation

$$ \begin{align*}[\mathbf{A}]=\big(\mathbf{A}_{a,b}(0,0)\big)_{a,b=1,\dotsc,n},\end{align*} $$

this tells us that $\Phi $ can be expressed as an n-dimensional trace,

$$ \begin{align*}\Phi(t,r,x)=\operatorname{{\mathop{\mathrm{{tr}}}}}[\mathbf{R}\mathbf{K}]\qquad\text{with}\quad\mathbf{R}=(\mathbf{I}-\mathbf{K})^{-1}.\end{align*} $$

Note here that $\mathbf {I}-\mathbf {K}$ is invertible because the determinant is nonzero. This proves equation (1.7) (after identifying $\partial _r$ with $\mathcal {D}_r$ as in Section 1.1) with, in the present notation,

$$ \begin{align*} Q=[\mathbf{R}\mathbf{K}]. \end{align*} $$

4.2 Formulas for the partial derivatives

The goal is now is to use Theorem 1.3(1)–(3) to show that algebraically, Q satisfies the matrix KP equation (1.8).

Write $\mathbf {K}'=\partial _r\mathbf {K}$ . Using the general formula $\partial _a(\mathbf {I}-\mathbf {A}(a))^{-1}=(\mathbf {I}-\mathbf {A}(a))^{-1}\partial _a\mathbf {A}(a)(\mathbf {I}-\mathbf {A}(a))^{-1}$ for an operator $\mathbf {A}(a)$ depending smoothly on a parameter a, together with the identity

$$ \begin{align*}\mathbf{K}\mathbf{R}=\mathbf{R}\mathbf{K}=\mathbf{R}-\mathbf{I},\end{align*} $$

which we will use repeatedly, we have

$$ \begin{align*} \partial_rQ&=[\partial_r(\mathbf{R}\mathbf{K})]=[\mathbf{R}\mathbf{K}'\mathbf{R}\mathbf{K}+\mathbf{R}\mathbf{K}']=[\mathbf{R}\mathbf{K}'\mathbf{R}]\\ \text{and similarly}\\ \partial_r^2Q&=2[\mathbf{R}\mathbf{K}'\mathbf{R}\mathbf{K}'\mathbf{R}]+[\mathbf{R}\mathbf{K}''\mathbf{R}],\\ \partial_r^3Q&=6[\mathbf{R}\mathbf{K}'\mathbf{R}\mathbf{K}'\mathbf{R}\mathbf{K}'\mathbf{R}]+3[\mathbf{R}\mathbf{K}''\mathbf{R}\mathbf{K}'\mathbf{R}]+3[\mathbf{R}\mathbf{K}'\mathbf{R}\mathbf{K}''\mathbf{R}]+[\mathbf{R}\mathbf{K}'''\mathbf{R}],\\ \partial_tQ&=[\mathbf{R}\partial_t\mathbf{K}\mathbf{R}]. \end{align*} $$

Next we want to compute $(\partial _rQ)^2$ . Note that, in general,

$$ \begin{align*} ([\mathbf{A}][\mathbf{B}])_{a,b}&=-\sum_{c}\int_0^{\infty}d\eta\,\partial_\eta(\mathbf{A}_{ac}(0,\eta)\mathbf{B}_{cb}(\eta,0))\\ &=-\sum_{c}\int_0^{\infty}d\eta\,\big(\partial_\eta\mathbf{A}_{ac}(0,\eta)\mathbf{B}_{cb}(\eta,0)+\mathbf{A}_{ac}(0,\eta)\partial_\eta\mathbf{B}_{cb}(\eta,0)\big)\\ &=-(\mathrm{d}_2\mathbf{A}\mathbf{B})_{ab}(0,0)-(\mathbf{A}\mathrm{d}_1\mathbf{B})_{ab}(0,0), \end{align*} $$

so that the following integration-by-parts formula holds:

(4.1) $$ \begin{align} [\mathbf{A}][\mathbf{B}]=-[\mathbf{A}\mathrm{d}_1\mathbf{B}+\mathrm{d}_2\mathbf{A}\mathbf{B}]. \end{align} $$

We will use this in the formula

$$ \begin{align*} (\partial_rQ)^2&=([\mathbf{R}\mathbf{K}'\mathbf{R}\mathbf{K}]+[\mathbf{R}\mathbf{K}'])^2 \\ &=[\mathbf{R}\mathbf{K}'\mathbf{R}\mathbf{K}]^2+[\mathbf{R}\mathbf{K}'\mathbf{R}\mathbf{K}][\mathbf{R}\mathbf{K}']+[\mathbf{R}\mathbf{K}'][\mathbf{R}\mathbf{K}'\mathbf{R}\mathbf{K}]+[\mathbf{R}\mathbf{K}']^2. \end{align*} $$

Using $\mathrm {d}_2(\mathbf {K}_a\mathbf {K}_b)=\mathbf {K}_a\mathrm {d}_2\mathbf {K}_b$ , the first term equals

$$ \begin{align*} -[\mathbf{R}\mathbf{K}'\mathbf{R}(\mathrm{d}_2\mathbf{K}\mathbf{R}&+\mathbf{K}\mathrm{d}_1\mathbf{R})\mathbf{K}'\mathbf{R}\mathbf{K}]\\ &=-[\mathbf{R}\mathbf{K}'\mathbf{R}\mathbf{K}'\mathbf{R}\mathbf{K}'\mathbf{R}\mathbf{K}]+[\mathbf{R}\mathbf{K}'\mathbf{R}(\mathrm{d}_1\mathbf{K}\mathbf{R}-\mathbf{K}\mathrm{d}_1\mathbf{R})\mathbf{K}'\mathbf{R}\mathbf{K}]. \end{align*} $$

Similarly, the fourth term equals

$$ \begin{align*}-[\mathbf{R}\mathbf{K}''\mathbf{R}\mathbf{K}']+[\mathbf{R}(\mathrm{d}_1\mathbf{K}'\mathbf{R}-\mathbf{K}'\mathrm{d}_1\mathbf{R})\mathbf{K}']\end{align*} $$

and the two middle ones equal

$$ \begin{align*} &-[\mathbf{R}\mathbf{K}'\mathbf{R}\mathbf{K}'\mathbf{R}\mathbf{K}']-[\mathbf{R}\mathbf{K}''\mathbf{R}\mathbf{K}'\mathbf{R}\mathbf{K}]\\ &\qquad\qquad+[\mathbf{R}\mathbf{K}'\mathbf{R}(\mathrm{d}_1\mathbf{K}\mathbf{R}-\mathbf{K}\mathrm{d}_1\mathbf{R})\mathbf{K}']+[\mathbf{R}(\mathrm{d}_1\mathbf{K}'\mathbf{R}-\mathbf{K}'\mathrm{d}_1\mathbf{R})\mathbf{K}'\mathbf{R}\mathbf{K}]. \end{align*} $$

Using this together with Theorem 1.3(1) and (2), which in the current notation read $\partial _r\mathbf {K}=(\mathrm {d}_1+\mathrm {d}_2)\mathbf {K}$ and $\partial _t\mathbf {K}=-\frac 13\left (\mathrm {d}_1^3+\mathrm {d}_2^3\right )\mathbf {K}$ , we get

$$ \begin{align*} \begin{array}{llr} &12\partial_tQ(t,r,x)+\partial_r^3Q(t,r,x)+6(\partial_rQ(t,r,x))^2&\\ &=-\,4\left[\mathbf{R}\left(\mathrm{d}_1^3+\mathrm{d}_2^3\right)\mathbf{K}\mathbf{R}\right]+6[\mathbf{R}\mathbf{K}'\mathbf{R}\mathbf{K}'\mathbf{R}\mathbf{K}'\mathbf{R}]+3[\mathbf{R}\mathbf{K}''\mathbf{R}\mathbf{K}'\mathbf{R}] &\quad{-(1)+(2)+(3)}\\ &\qquad+\,3[\mathbf{R}\mathbf{K}'\mathbf{R}\mathbf{K}''\mathbf{R}]+\left[\mathbf{R}\left(\mathrm{d}_1^3+\mathrm{d}_2^3+3\mathrm{d}_1^2\mathrm{d}_2+3\mathrm{d}_1\mathrm{d}_2^2\right)\mathbf{K}\mathbf{R}\right]\quad &{+(4)+(5)}\\ &\quad-\,6[\mathbf{R}\mathbf{K}''\mathbf{R}\mathbf{K}']+6[\mathbf{R}(\mathrm{d}_1\mathbf{K}'\mathbf{R}-\mathbf{K}'\mathrm{d}_1\mathbf{R})\mathbf{K}']&\quad{-(6)+(7)}\\ &\quad-\,6[\mathbf{R}\mathbf{K}'\mathbf{R}\mathbf{K}'\mathbf{R}\mathbf{K}']-6[\mathbf{R}\mathbf{K}''\mathbf{R}\mathbf{K}'\mathbf{R}\mathbf{K}]&\quad{-(8)-(9)}\\ &\quad+\,6[\mathbf{R}\mathbf{K}'\mathbf{R}(\mathrm{d}_1\mathbf{K}\mathbf{R}-\mathbf{K}\mathrm{d}_1\mathbf{R})\mathbf{K}']&\quad{+(10)}\\ &\quad+\,6[\mathbf{R}(\mathrm{d}_1\mathbf{K}'\mathbf{R}-\mathbf{K}'\mathrm{d}_1\mathbf{R})\mathbf{K}'\mathbf{R}\mathbf{K}]&\quad{+(11)}\\ &\quad-\,6[\mathbf{R}\mathbf{K}'\mathbf{R}\mathbf{K}'\mathbf{R}\mathbf{K}'\mathbf{R}\mathbf{K}]+6[\mathbf{R}\mathbf{K}'\mathbf{R}(\mathrm{d}_1\mathbf{K}\mathbf{R}-\mathbf{K}\mathrm{d}_1\mathbf{R})\mathbf{K}'\mathbf{R}\mathbf{K}{}].&\quad{-(12)+(13)} \end{array} \end{align*} $$

The element we have marked (12) equals $6[\mathbf {R}\mathbf {K}'\mathbf {R}\mathbf {K}'\mathbf {R}\mathbf {K}'\mathbf {R}]-6[\mathbf {R}\mathbf {K}'\mathbf {R}\mathbf {K}'\mathbf {R}\mathbf {K}']$ , so

$$ \begin{align*} S_{\text{I}}&:=-\text{(1)}+\text{(2)}+\text{(5)}-\text{(8)}-\text{(12)}=-3\left[\mathbf{R}\left(\mathrm{d}_1^3+\mathrm{d}_2^3-\mathrm{d}_1^2\mathrm{d}_2-\mathrm{d}_1\mathrm{d}_2^2\right)\mathbf{K}\mathbf{R}\right]\\ &=-3\left[\mathbf{R}(\mathrm{d}_1-\mathrm{d}_2)^2(\mathrm{d}_1+\mathrm{d}_2)\mathbf{K}\mathbf{R}\right]=-3\left[\mathbf{R}(\mathrm{d}_1-\mathrm{d}_2)^2\mathbf{K}'\mathbf{R}\right]. \end{align*} $$

Similarly,

$$ \begin{align*} S_{\text{II}}:=+\text{(3)}+\text{(4)}-\text{(6)}-\text{(9)}=-3[\mathbf{R}\mathbf{K}''\mathbf{R}\mathbf{K}'\mathbf{R}]+3[\mathbf{R}\mathbf{K}'\mathbf{R}\mathbf{K}''\mathbf{R}] \end{align*} $$

and

$$ \begin{align*} S_{\text{III}}&:=+\text{(7)}+\text{(10)}+\text{(11)}+\text{(13)}\\ &=+6[\mathbf{R}(\mathrm{d}_1\mathbf{K}'\mathbf{R}-\mathbf{K}'\mathrm{d}_1\mathbf{R})\mathbf{K}'\mathbf{R}]+6[\mathbf{R}\mathbf{K}'\mathbf{R}(\mathrm{d}_1\mathbf{K}\mathbf{R}-\mathbf{K}\mathrm{d}_1\mathbf{R})\mathbf{K}'\mathbf{R}]. \end{align*} $$

So

$$ \begin{align*} \begin{aligned} &4\partial_tQ(t,r,x)+\tfrac13\partial_r^3Q(t,r,x)+2(\partial_rQ(t,r,x))^2=\tfrac13(S_{\text{I}}+S_{\text{II}}+S_{\text{III}})\\ &\quad=-[\mathbf{R}\mathbf{K}''{\mathbf{R}}\mathbf{K}'{\mathbf{R}}]+[\mathbf{R}\mathbf{K}'{\mathbf{R}}\mathbf{K}''{\mathbf{R}}]-\left[\mathbf{R}\left(\mathrm{d}_1^3+\mathrm{d}_2^3-\mathrm{d}_1^2\mathrm{d}_2-\mathrm{d}_1\mathrm{d}_2^2\right)\mathbf{K}\mathbf{R}\right]\\ &\qquad+2[\mathbf{R}(\mathrm{d}_1\mathbf{K}'{\mathbf{R}}-\mathbf{K}'{\mathrm{d}}_1\mathbf{R})\mathbf{K}'{\mathbf{R}}]+2[\mathbf{R}\mathbf{K}'{\mathbf{R}}(\mathrm{d}_1\mathbf{K}\mathbf{R}-\mathbf{K}\mathrm{d}_1\mathbf{R})\mathbf{K}'{\mathbf{R}}]\\ &\quad=-[\mathbf{R}(\mathrm{d}_2-\mathrm{d}_1)\mathbf{K}'{\mathbf{R}}\mathbf{K}'{\mathbf{R}}]+[\mathbf{R}\mathbf{K}'{\mathbf{R}}\mathbf{K}''{\mathbf{R}}]-\left[\mathbf{R}\left(\mathrm{d}_1^3+\mathrm{d}_2^3-\mathrm{d}_1^2\mathrm{d}_2-\mathrm{d}_1\mathrm{d}_2^2\right)\mathbf{K}\mathbf{R}\right]\\ &\qquad-2[\mathbf{R}\mathbf{K}'{\mathbf{R}}\mathrm{d}_1\mathbf{R}\mathbf{K}'{\mathbf{R}}]+2[\mathbf{R}\mathbf{K}'{\mathbf{R}}\mathrm{d}_1\mathbf{K}\mathbf{R}\mathbf{K}'{\mathbf{R}}]. \end{aligned} \end{align*} $$

Now using $-[\mathbf {R}\mathbf {K}'\mathbf {R}\mathrm {d}_1\mathbf {R}\mathbf {K}'\mathbf {R}]+[\mathbf {R}\mathbf {K}'\mathbf {R}\mathrm {d}_1\mathbf {K}\mathbf {R}\mathbf {K}'\mathbf {R}]=-[\mathbf {R}\mathbf {K}'\mathbf {R}\mathrm {d}_1(\mathbf {I}-\mathbf {K})\mathbf {R}\mathbf {K}'\mathbf {R}]$ , which equals $-[\mathbf {R}\mathbf {K}'\mathbf {R}\mathrm {d}_1\mathbf {K}'\mathbf {R}]$ , yields

(4.2) $$ \begin{align} &4\partial_tQ(t,r,x)+\tfrac13\partial_r^3Q(t,r,x)+2(\partial_rQ(t,r,x))^2\nonumber\\ &=-[\mathbf{R}(\mathrm{d}_2-\mathrm{d}_1)\mathbf{K}'\mathbf{R}\mathbf{K}'\mathbf{R}]-[\mathbf{R}\mathbf{K}'\mathbf{R}(\mathrm{d}_1-\mathrm{d}_2)\mathbf{K}'\mathbf{R}]-\left[\mathbf{R}(\mathrm{d}_1-\mathrm{d}_2)^2\mathbf{K}'\mathbf{R}\right]. \end{align} $$

Next we add the derivatives in the $x_i$ variables. As for $\partial _r^2Q$ , we have

(4.3) $$ \begin{align} \partial_x^2Q=2[\mathbf{R}\partial_x\mathbf{K}\mathbf{R}\partial_x\mathbf{K}\mathbf{R}]+\left[\mathbf{R}\partial_x^2\mathbf{K}\mathbf{R}\right]. \end{align} $$

On the other hand, if we apply $\partial _r$ to equation (4.2) and use Theorem 1.3(3) to write $(\mathrm {d}_2-\mathrm {d}_1)\mathbf {K}'=\partial _x\mathbf {K}$ , we get

$$ \begin{align*} \begin{split} &-[\mathbf{R}\mathbf{K}'\mathbf{R}\partial_x\mathbf{K}\mathbf{R}\mathbf{K}'\mathbf{R}]-[\mathbf{R}\partial_x\mathbf{K}'\mathbf{R}\mathbf{K}'\mathbf{R}]-2[\mathbf{R}\partial_x\mathbf{K}\mathbf{R}\mathbf{K}'\mathbf{R}\mathbf{K}'\mathbf{R}]-[\mathbf{R}\partial_x\mathbf{K}\mathbf{R}\mathbf{K}''\mathbf{R}]\hskip-9pt\\ &\quad+2[\mathbf{R}\mathbf{K}'\mathbf{R}\mathbf{K}'\mathbf{R}\partial_x\mathbf{K}\mathbf{R}]+[\mathbf{R}\mathbf{K}''\mathbf{R}\partial_x\mathbf{K}\mathbf{R}]+[\mathbf{R}\mathbf{K}'\mathbf{R}\partial_x\mathbf{K}'\mathbf{R}]+[\mathbf{R}\mathbf{K}'\mathbf{R}\partial_x\mathbf{K}\mathbf{R}\mathbf{K}'\mathbf{R}]\hskip-9pt\\ &\quad+[\mathbf{R}\mathbf{K}'\mathbf{R}(\mathrm{d}_1-\mathrm{d}_2)\partial_x\mathbf{K}\mathbf{R}]+[\mathbf{R}(\mathrm{d}_1-\mathrm{d}_2)\partial_x\mathbf{K}'\mathbf{R}]+[\mathbf{R}(\mathrm{d}_1-\mathrm{d}_2)\partial_x\mathbf{K}\mathbf{R}\mathbf{K}'\mathbf{R}]. \end{split} \end{align*} $$

Note that the first and eighth terms cancel. We want to add $\partial _x^2Q(t,r,x)$ . Since $(\mathrm {d}_1-\mathrm {d}_2)\partial _x\mathbf {K}'=\partial _x\left (\mathrm {d}_1^2-\mathrm {d}_2^2\right )\mathbf {K}=-\partial _x^2\mathbf {K}$ , the next-to-last term in the last expression cancels the second bracket on the right-hand side of equation (4.3). Using additionally $\partial _x\mathbf {K}'+(\mathrm {d}_1-\mathrm {d}_2)\partial _x\mathbf {K}=2\mathrm {d}_1\partial _x\mathbf {K}$ and $-\partial _x\mathbf {K}'+(\mathrm {d}_1-\mathrm {d}_2)\partial _x\mathbf {K}=-2\mathrm {d}_2\partial _x\mathbf {K}$ and writing

$$ \begin{align*} q=\partial_rQ, \end{align*} $$

we deduce that

(4.4) $$ \begin{align} &4\partial_tq+\tfrac13\partial_r^3q+2(q\hspace{0.05em}\partial_rq+\partial_rq\hspace{0.1em} q)+\partial_x^2Q(t,r,x)\nonumber\\ &\quad=2\hspace{0.1em}\Big(\hspace{-0.1em}\!-[\mathbf{R}\mathrm{d}_2\partial_x\mathbf{K}\mathbf{R}\mathbf{K}'\mathbf{R}]-[\mathbf{R}\partial_x\mathbf{K}\mathbf{R}\mathbf{K}'\mathbf{R}\mathbf{K}'\mathbf{R}]-\tfrac12[\mathbf{R}\partial_x\mathbf{K}\mathbf{R}\mathbf{K}''\mathbf{R}]\nonumber\\ &\qquad+[\mathbf{R}\mathbf{K}'\mathbf{R}\mathbf{K}'\mathbf{R}\partial_x\mathbf{K}\mathbf{R}]+\tfrac12[\mathbf{R}\mathbf{K}''\mathbf{R}\partial_x\mathbf{K}\mathbf{R}]+[\mathbf{R}\mathbf{K}'\mathbf{R}\mathrm{d}_1\partial_x\mathbf{K}\mathbf{R}]\nonumber\\ &\qquad+[\mathbf{R}\partial_x\mathbf{K}\mathbf{R}\partial_x\mathbf{K}\mathbf{R}]\Big). \end{align} $$

We claim that the right-hand side equals two times

(4.5) $$ \begin{align} &-\big( [\mathbf{R}\partial_x\mathbf{K}\mathbf{R}\mathrm{d}_1\mathbf{K}\mathbf{R}\mathbf{K}'\mathbf{R}] + [\mathbf{R}\partial_x\mathrm{d}_2\mathbf{K}\mathbf{R}\mathbf{K}'\mathbf{R}]\big)\nonumber\\ &+ \big( [\mathbf{R}\mathbf{K}'\mathbf{R}\mathrm{d}_2\mathbf{K}\mathbf{R}\partial_x\mathbf{K}\mathbf{R}] + [\mathbf{R}\mathbf{K}'\mathbf{R}\mathrm{d}_1\partial_x\mathbf{K}\mathbf{R}]\big)\nonumber\\ &-\big( [\mathbf{R}\partial_x\mathbf{K}\mathbf{R}\mathrm{d}_2\mathbf{K}\mathbf{R}\mathbf{K}'\mathbf{R}] + [\mathbf{R}\partial_x\mathbf{K}\mathbf{R} \mathrm{d}_1\mathbf{K}'\mathbf{R}]\big)\nonumber\\ &+ \big( [\mathbf{R}\mathbf{K}'\mathbf{R}\mathrm{d}_1\mathbf{K}\mathbf{R}\partial_x\mathbf{K}\mathbf{R}] + [\mathbf{R}\mathrm{d}_2\mathbf{K}'\mathbf{R}\partial_x\mathbf{K}\mathbf{R}]\big). \end{align} $$

To see this, express the right-hand side of equation (4.4) as $2(r_1+r_2+\dotsb +r_7)$ , express formula (4.5) as $q_1+q_2+\dotsb +q_8$ and note first that $r_1=q_2$ , $r_6=q_4$ , $r_2=q_1+q_5$ and $r_4=q_3+q_7$ . On the other hand, we have $r_3=-\frac 12[\mathbf {R}\partial _x\mathbf {K}\mathbf {R}(\mathrm {d}_1+\mathrm {d}_2)\mathbf {K}'\mathbf {R}]=q_6+\frac 12[\mathbf {R}\partial _x\mathbf {K}\mathbf {R}(\mathrm {d}_1-\mathrm {d}_2)\mathbf {K}'\mathbf {R}]=q_6-\frac 12r_7$ and similarly $r_5=\frac 12[\mathbf {R}(\mathrm {d}_1+\mathrm {d}_2)\mathbf {K}'\mathbf {R}\partial _x\mathbf {K}\mathbf {R}]=q_8+\frac 12[\mathbf {R}(\mathrm {d}_1-\mathrm {d}_2)\mathbf {K}'\mathbf {R}\partial _x\mathbf {K}\mathbf {R}]=q_8-\frac 12r_7$ . This gives $r_3+r_5+r_7=q_6+q_8$ , and finishes proving the claim.

Integrating by parts (i.e., using equation (4.1)) within each parenthesis in formula (4.5), we get

$$ \begin{align*} 2\partial_tq+\tfrac16\partial_r^3q&+(q\hspace{0.05em}\partial_rq+\partial_rq\hspace{0.1em} q)+\tfrac12\partial_x^2Q(t,r,x)\nonumber\\ =&-[\mathbf{R}\partial_x\mathbf{K}(\mathbf{R}\mathrm{d}_1\mathbf{K}\mathbf{R}-\mathrm{d}_1\mathbf{R}) \mathbf{K}'\mathbf{R}] + [\mathbf{R}\partial_x\mathbf{K}][\mathbf{R}\mathbf{K}'\mathbf{R}]\nonumber\\ &\quad + [\mathbf{R}\mathbf{K}'(\mathbf{R}\mathrm{d}_2\mathbf{K}\mathbf{R}-\mathrm{d}_2\mathbf{R})\partial_x\mathbf{K}\mathbf{R}] - [\mathbf{R}\mathbf{K}'\mathbf{R}][\partial_x\mathbf{K}\mathbf{R}]\nonumber\\ &\quad - [\mathbf{R}\partial_x\mathbf{K}(\mathbf{R}\mathrm{d}_2\mathbf{K}\mathbf{R}-\mathrm{d}_2\mathbf{R})\mathbf{K}'\mathbf{R}] + [\mathbf{R}\partial_x\mathbf{K}\mathbf{R}][\mathbf{K}'\mathbf{R}] \nonumber\\ &\quad + [\mathbf{R}\mathbf{K}'(\mathbf{R}\mathrm{d}_1\mathbf{K}\mathbf{R}-\mathrm{d}_1\mathbf{R})\partial_x\mathbf{K}\mathbf{R}] - [\mathbf{R}\mathbf{K}'][\mathbf{R}\partial_x\mathbf{K}\mathbf{R}]. \end{align*} $$

Write this as $s_1+\dotsb +s_8$ . Notice that $s_1+s_5$ yields a term involving

$$ \begin{align*}( \mathbf{R}\mathrm{d}_1\mathbf{K} \mathbf{R}-\mathrm{d}_1\mathbf{R})+ (\mathbf{R}\mathrm{d}_2\mathbf{K}\mathbf{R}-\mathrm{d}_2\mathbf{R}) = \mathbf{R}(\mathbf{K}\mathrm{d}_1\mathbf{K}+\mathrm{d}_2\mathbf{K}\mathbf{K})\mathbf{R}-(\mathrm{d}_1+\mathrm{d}_2)\mathbf{I}, \end{align*} $$

where we have used $\mathbf {R}\mathbf {K}=\mathbf {K}\mathbf {R}=\mathbf {R}-\mathbf {I}$ again, and thus integrating by parts one more time we get

$$ \begin{align*} s_1+ &s_5=-[\mathbf{R}\partial_x\mathbf{K}(\mathbf{R}(\mathbf{K}\mathrm{d}_1\mathbf{K}+\mathrm{d}_2\mathbf{K}\mathbf{K})\mathbf{R}-(\mathrm{d}_1+\mathrm{d}_2)\mathbf{I})\mathbf{K}'\mathbf{R}]\\ &=[\mathbf{R}\partial_x\mathbf{K}\mathbf{R}\mathbf{K}][\mathbf{K}\mathbf{R}\mathbf{K}'\mathbf{R}]+[\mathbf{R}\partial_x\mathbf{K}((\mathrm{d}_1+\mathrm{d}_2)\mathbf{I})\mathbf{K}'\mathbf{R}]\\ &=([\mathbf{R}\partial_x\mathbf{K}\mathbf{R}]-[\mathbf{R}\partial_x\mathbf{K}])([\mathbf{R}\mathbf{K}'\mathbf{R}]-[\mathbf{K}'\mathbf{R}])+[\mathbf{R}\partial_x\mathbf{K}((\mathrm{d}_1+\mathrm{d}_2)\mathbf{I})\mathbf{K}'\mathbf{R}]\\ &=[\mathbf{R}\partial_x\mathbf{K}\mathbf{R}][\mathbf{R}\mathbf{K}'\mathbf{R}]-s_2-s_6+[\mathbf{R}\partial_x\mathbf{K}][\mathbf{K}'\mathbf{R}]+[\mathbf{R}\partial_x\mathbf{K}((\mathrm{d}_1+\mathrm{d}_2)\mathbf{I})\mathbf{K}'\mathbf{R}]. \end{align*} $$

In a similar fashion, we get

$$ \begin{align*}s_3+s_7=-[\mathbf{R}\mathbf{K}'\mathbf{R}][\mathbf{R}\partial_x\mathbf{K}\mathbf{R}]-s_4-s_8-[\mathbf{R}\mathbf{K}'][\partial_x\mathbf{K}\mathbf{R}]-[\mathbf{R}\mathbf{K}'((\mathrm{d}_1+\mathrm{d}_2)\mathbf{I})\partial_x\mathbf{K}\mathbf{R}].\end{align*} $$

Therefore,

$$ \begin{align*} 2\partial_tq&+\tfrac16\partial_r^3q+(q\hspace{0.05em}\partial_rq+\partial_rq\hspace{0.1em} q)+\tfrac12\partial_x^2Q(t,r,x)\\ &\quad=[\mathbf{R}\partial_x\mathbf{K}][\mathbf{K}'\mathbf{R}]+[\mathbf{R}\partial_x\mathbf{K}((\mathrm{d}_1+\mathrm{d}_2)\mathbf{I})\mathbf{K}'\mathbf{R}]-[\mathbf{R}\mathbf{K}'][\partial_x\mathbf{K}\mathbf{R}]\\ &\qquad-[\mathbf{R}\mathbf{K}'((\mathrm{d}_1+\mathrm{d}_2)\mathbf{I})\partial_x\mathbf{K}\mathbf{R}]+[\mathbf{R}\partial_x\mathbf{K}\mathbf{R}][\mathbf{R}\mathbf{K}'\mathbf{R}]-[\mathbf{R}\mathbf{K}'\mathbf{R}][\mathbf{R}\partial_x\mathbf{K}\mathbf{R}]. \end{align*} $$

In order to complete the proof we note that if $\mathbf {A}$ and $\mathbf {B}$ are nice kernels, then integrating by parts we get

$$ \begin{align*} [\mathbf{A}((\mathrm{d}_1+\mathrm{d}_2)\mathbf{I})\mathbf{B}]=[\mathbf{A}\mathrm{d}_1\mathbf{B}]+[\mathbf{A}\mathrm{d}_2\mathbf{I}\hspace{0.05em}\mathbf{B}]=-[\mathbf{A}][\mathbf{B}], \end{align*} $$

which immediately yields

$$ \begin{align*} 2\partial_tq+\tfrac16\partial_r^3q+(q\hspace{0.05em}\partial_rq+\partial_rq\hspace{0.1em} q)+\tfrac12\partial_x^2Q(t,r,x)=[\mathbf{R}\partial_x\mathbf{K}\mathbf{R}][\mathbf{R}\mathbf{K}'\mathbf{R}]-[\mathbf{R}\mathbf{K}'\mathbf{R}][\mathbf{R}\partial_x\mathbf{K}\mathbf{R}].\hskip-8pt \end{align*} $$

The right-hand side equals $\partial _xQ\hspace {0.1em} q-q\hspace {0.05em}\partial _xQ$ , so the equation becomes equation (1.8), as needed.

4.3 Proof of Theorem 1.3

For r inside the interval where the trace norm of $\mathbf {K}$ is strictly less than $1$ , everything in the previous section is well defined, because that condition ensures that $(\mathbf {I}-\mathbf {K})^{-1}-\mathbf {I}$ is trace class, and therefore the series for the brackets are convergent. So the algebraic computations hold pointwise for such r.

Since the kernel $\mathbf {K}(u,v)$ is real analytic in $t>0$ , x and r, the Fredholm determinant $\det (\mathbf {I}-\mathbf {K}(t,x,r))$ is as well, since it is given by its Fredholm series, each of whose terms is real analytic and which converges uniformly because $\mathbf {K}$ is trace class uniformly in compact sets of $t,x,r$ . Since the determinant never vanishes, $\partial ^2_r \log \det (\mathbf {I}-\mathbf {K})$ is also real analytic, as are all the terms in the KP equation (either the scalar or matrix version). Therefore the left-hand side of the KP equation (1.8) (or equation (1.9)) is a real analytic function. We have proved that this function vanishes for r in an open interval, and therefore it vanishes everywhere.