Here we consider a more simple Kolmogorov-Fokker-Planck equations \begin{equation} \partial_t u+ v\cdot D_x u -\Delta_v u =\mathrm{div}_v H+h. \end{equation} To find the fundamental solution of the equation, we consider the following Cauchy problem: \begin{equation} \left\{ \begin{alignedat}{2} \partial_t u+v\cdot D_x u-\Delta_v u&=h,\\ u(0)&=g. \end{alignedat} \right. \end{equation} By taking the Fourier transform in $(x,v)\mapsto (k,\xi)$, we have \begin{equation} \left\{ \begin{alignedat}{2} \partial_t \hat{u}-k\cdot D_{\xi} \hat{u}+|\xi|^2 \hat{u}&=\hat{h}(t,k,\xi),\\ \hat{u}(0)&=\hat{g}. \end{alignedat} \right. \end{equation} Along the characteristic $t\mapsto (t,k,\xi-tk)$, we have \begin{equation} \frac{\partial}{\partial t}\left(\hat{u}(t,k,\xi-tk \right))+|\xi-tk|^2 \hat{u}(t,k,\xi-tk)=\hat{h}(t,k,\xi-tk) \end{equation} Set \[ e(t,k,\xi)=\exp\left(-\int_0^t |\xi-\tau k|^2 d{\tau}\right).\] and note that \[ \frac{\partial}{\partial t} e(t,k,\xi)^{-1} =e(t,k,\xi)^{-1}|\xi-tk|^2. \] Multiplying $e(t,k,\xi)^{-1}$ with (4), we get \[ \frac{\partial}{\partial t}\left[e(t,k,\xi)^{-1}\hat{u}(t,k,\xi-tk) \right]=\hat{h}(t,k,\xi-tk)e(t,k,\xi)^{-1}. \] By taking integration with respect to $t$, we have \begin{equation} \begin{aligned} \hat{u}(t,k,\xi-tk)&=\hat{g}(k,\xi)e(t,k,\xi)\\ &\relphantom{=}+\int_0^t \hat{h}(s,k,\xi-sk)e(t,k,\xi)e(s,k,\xi)^{-1}ds. \end{aligned}\end{equation} Note that \[ e(t,k,\xi)e(s,k,\xi)^{-1} = \exp\left(-\int_s^t |\xi-\tau k|^2 d{\tau}\right).\] If we shift $\xi-tk$ into $\xi$, we get. \begin{equation} \begin{aligned} \hat{u}(t,k,\xi)&=\hat{g}(k,\xi+tk)e(t,k,\xi+tk)\\ &\quad +\int_0^t \hat{h}(s,k,\xi+(t-s)k)\exp\left(-\int_s^t |\xi+(t-\tau)k|^2d\tau\right)ds. \end{aligned}\end{equation} Note also that \[ e(t,k,\xi+tk)=\exp\left(-\int_0^t |\xi+(t-\tau)k|^2 d{\tau}\right)=e(t,-k,\xi)\] and \[ \exp\left(-\int_s^t |\xi+(t-\tau)k|^2d\tau \right) =\exp\left(-\int_0^{t-s} |\xi+\tau k|^2 d\tau\right)=e(t-s,-k,\xi).\] Hence we finally get \begin{equation} \hat{u}(t,k,\xi)=\hat{g}(k,\xi+tk)e(t,-k,\xi)+\int_0^t \hat{h}(s,k,\xi+(t-s)k)e(t-s,-k,\xi)ds. \end{equation} To take the inverse Fourier transform, we recall the following lemma.
Lemma. If $t,\beta>0$, then \[ \frac{1}{(2\pi)^{d/2}} \int_{\mathbb{R}^d} e^{-\beta|\xi|^2 t}e^{ix\cdot \xi} d{\xi}=\frac{1}{(2\beta t)^{d/2}}\exp\left(-\frac{|x|^2}{4\beta t}\right).\]
A change of variable shows that \begin{align} &\frac{1}{(2\pi)^{d}}\int_{\mathbb{R}^{2d}} e(t,-k,\xi)e^{ik\cdot x} e^{i\xi \cdot v} d{k}d\xi\\ &=\frac{1}{(2\pi)^{d}t^d} \int_{\mathbb{R}^{2d}} \exp\left(-|\xi|^2 t-\frac{1}{12}|k|^2 t\right)e^{ix\cdot (k/t)} e^{iv\cdot (\xi-k/2)} d{k}d\xi \nonumber\\ &=\frac{1}{t^d} \left(\frac{1}{(2\pi)^{d/2}}\int_{\mathbb{R}^d} \exp(-|\xi|^2 t)e^{i\xi \cdot v}d{\xi} \right)\left(\frac{1}{(2\pi)^{d/2}} \int_{\mathbb{R}^d} \exp\left(-\frac{1}{12}|k|^2 t\right)e^{i(x/t-v/2)\cdot k}d{k} \right) \nonumber\\ &=\frac{1}{t^d} \left(\frac{1}{(2t)^{d/2}} \exp\left(-\frac{|v|^2}{4t}\right)\right)\left(\frac{1}{(t/6)^{d/2}} \exp\left(-\frac{|x/t-v/2|^2}{t/3} \right) \right) \nonumber\\ &=\frac{3^{d/2}}{t^{2d}} \exp\left(-\frac{|v|^2}{4t}-\frac{3\left|x-\frac{t}{2}v\right|^2}{t^3}\right) \nonumber\\ &=\frac{3^{d/2}}{t^{2d}} \exp\left(-\frac{1}{4} \left|\frac{v}{t^{1/2}}\right|^2-{3\left|\frac{x}{t^{3/2}}-\frac{v}{2t^{1/2}}\right|^2}\right) \nonumber\\ &=\frac{3^{d/2}}{t^{2d}}\mathcal{G}\left(\frac{x}{t^{3/2}},\frac{v}{t^{1/2}}\right), \nonumber \end{align} where \[ \mathcal{G}(x,v)=\exp\left(-\frac{1}{4}|v|^2-3\left|x-\frac{v}{2}\right|^2\right).\] To proceed further, we define the modified convolution $*_t$ by \[ h*_t j(x,v)=\int_{\mathbb{R}^{2d}} h(y,w)j(x-y-tw,v-w)dwdy.\] Observe that if $\tilde{j}(x,v)=j(x+tv,v)$, then $h*_t j(x,v)=h*\tilde{j}(x-tv,v)$. Hence it follows from Young's convolution inequality that \begin{equation} \Vert{h*_t j}\Vert_{L^{r}(\mathbb{R}^{2d})}\leq \Vert{h}\Vert_{L^{p}(\mathbb{R}^{2d})}\Vert{j}\Vert_{L^{q}(\mathbb{R}^{2d})} \end{equation} independently of $t$, where $1+1/r=1/p+1/q$. If we take a Fourier transform to the convolution, then we have the following:
Proposition. For $t>0$, we have \[ \widehat{(h*_t j)}(k,\xi)=(2\pi)^d\hat{h}(k,\xi+tk)\hat{j}(k,\xi). \]
Proof. Recall that the Fourier transform of the convolution \[ (f*g)(x)=\int_{\mathbb{R}^d} f(x-y)g(y)d{y} \] is \[ \widehat{(f*g)}(\xi)=(2\pi)^{d/2}\hat{f}(\xi)\hat{g}(\xi).\] By taking Fourier transform, we have \begin{align} \widehat{h*_t j}(k,\xi)&=\frac{1}{(2\pi)^d}\int_{\mathbb{R}^{2d}} (h*\tilde{j})(x-tv,v)e^{-ik\cdot x} e^{-i\xi \cdot v} d{x}dv\\ &=\frac{1}{(2\pi)^d} \int_{\mathbb{R}^d} e^{-i\xi \cdot v} e^{-ik\cdot (tv)} \left[\int_{\mathbb{R}^d} (h*\tilde{j})(x,v)e^{-ik\cdot x}d{x} \right] dv \\ &=\frac{1}{(2\pi)^d}\int_{\mathbb{R}^{2d}} e^{-i(\xi+tk)\cdot v} e^{-ik\cdot x} (h*\tilde{j})(x,v)d{x}dv\\ &=(2\pi)^{d} \hat{h}(k,\xi+tk)\hat{\tilde{j}}(k,\xi+tk). \end{align} Since $\tilde{j}(x,v)=j(x+tv,v)$, it follows that \begin{align} &=(2\pi)^d \hat{h}(k,\xi+tk)\hat{j}(k,\xi). \end{align} From this property and (7), we have \[ u(t,x,v)=(G*_t g)(x,v)+\int_0^t \int_{\mathbb{R}^{2d}} h(\tau)*_{t-\tau}G(t-\tau,\cdot,\cdot)d\tau, \] where \[ G(t,x,v)=\left(\frac{3}{4\pi^2 t^4} \right)^{d/2}\mathcal{G}\left(\frac{x}{t^{3/2}},\frac{v}{t^{1/2}} \right). \] We list some properties of the fundamental solution.
Proposition. The function $G$ and $\mathcal{G}$ have the following properties:
  • The function $\mathcal{G}$ is $C^\infty$ and decays polynomially at infinity. Moreover, $\mathcal{G}$ and all its derivatives are integrable in $\mathbb{R}^{2d}$.
  • For every $t>0$, $\int_{\mathbb{R}^{2d}} G(t,x,v)d{v}dx=1$.
  • Both functions are nonnegative: $G\geq 0$ and $\mathcal{G}\geq 0$.
  • For any $p\geq 1$, we have \begin{equation} \begin{aligned} \Vert{G(t,\cdot,\cdot)}\Vert_{L^{p}(\mathbb{R}^{2d})}&\approx t^{-d(1+1/2)(1-1/p)} \Vert{\mathcal{G}}\Vert_{L^{p}(\mathbb{R}^{2d})}\\ \Vert{D_v G(t,\cdot,\cdot)}\Vert_{L^{p}(\mathbb{R}^{2d})}&\approx t^{-d(1+1/2)(1-1/p)-1/2} \Vert{D_v\mathcal{G}}\Vert_{L^{p}(\mathbb{R}^{2d})}. \end{aligned}\end{equation}
Proof. (1) -- (3) are immediate from the definition. To show (4), a change of variable shows that \begin{equation} \begin{aligned} \Vert G(t,\cdot,\cdot)\Vert_{L^{p}(\mathbb{R}^{2d})}&=\left(\frac{3}{4\pi^2 t^4}\right)^{d/2} \left(\int_{\mathbb{R}^{2d}} \left|\mathcal{G}\left(\frac{x}{t^{3/2}},\frac{v}{t^{1/2}}\right)\right|^p d{x}dv\right)^{1/p}\\ &= \left(\frac{3}{4\pi^2}\right)^{d/2} t^{-2d(1-1/p)} \Vert{\mathcal{G}}\Vert_{L^{p}(\mathbb{R}^{2d})}. \end{aligned}\end{equation} To show the second estimate, we note that \[ D_v G(t,x,v) = \left(\frac{3}{4\pi^2 t^4}\right)^{d/2} D_v \mathcal{G}\left(\frac{x}{t^{3/2}},\frac{v}{t^{1/2}}\right) \frac{1}{t^{1/2}}.\] A change of variable shows that \[ \Vert{D_v G(t,\cdot,\cdot)}\Vert_{L^{p}(\mathbb{R}^{2d})} = \left(\frac{3}{4\pi^2}\right)^{d/2}t^{-2d(1-1/p)-1/2}\Vert{D_v \mathcal{G}}\Vert_{L^{p}(\mathbb{R}^{2d})}. \] This completes the proof.