Green function of kinetic Kolmogorov-Fokker-Planck equations
September 26, 2022
Here we consider a more simple Kolmogorov-Fokker-Planck equations
∂tu+v⋅Dxu−Δvu=divvH+h.
To find the fundamental solution of the equation, we consider the following Cauchy problem:
{∂tu+v⋅Dxu−Δvuu(0)=h,=g.
By taking the Fourier transform in (x,v)↦(k,ξ), we have
{∂tu^−k⋅Dξu^+∣ξ∣2u^u^(0)=h^(t,k,ξ),=g^.
Along the characteristic t↦(t,k,ξ−tk), we have
∂t∂(u^(t,k,ξ−tk))+∣ξ−tk∣2u^(t,k,ξ−tk)=h^(t,k,ξ−tk)
Set
e(t,k,ξ)=exp(−∫0t∣ξ−τk∣2dτ).
and note that
∂t∂e(t,k,ξ)−1=e(t,k,ξ)−1∣ξ−tk∣2.
Multiplying e(t,k,ξ)−1 with (4), we get
∂t∂[e(t,k,ξ)−1u^(t,k,ξ−tk)]=h^(t,k,ξ−tk)e(t,k,ξ)−1.
By taking integration with respect to t, we have
u^(t,k,ξ−tk)=g^(k,ξ)e(t,k,ξ)\relphantom=+∫0th^(s,k,ξ−sk)e(t,k,ξ)e(s,k,ξ)−1ds.
Note that
e(t,k,ξ)e(s,k,ξ)−1=exp(−∫st∣ξ−τk∣2dτ).
If we shift ξ−tk into ξ, we get.
u^(t,k,ξ)=g^(k,ξ+tk)e(t,k,ξ+tk)+∫0th^(s,k,ξ+(t−s)k)exp(−∫st∣ξ+(t−τ)k∣2dτ)ds.
Note also that
e(t,k,ξ+tk)=exp(−∫0t∣ξ+(t−τ)k∣2dτ)=e(t,−k,ξ)
and
exp(−∫st∣ξ+(t−τ)k∣2dτ)=exp(−∫0t−s∣ξ+τk∣2dτ)=e(t−s,−k,ξ).
Hence we finally get
u^(t,k,ξ)=g^(k,ξ+tk)e(t,−k,ξ)+∫0th^(s,k,ξ+(t−s)k)e(t−s,−k,ξ)ds.
To take the inverse Fourier transform, we recall the following lemma.
Lemma. If t,β>0, then
(2π)d/21∫Rde−β∣ξ∣2teix⋅ξdξ=(2βt)d/21exp(−4βt∣x∣2).
A change of variable shows that
(2π)d1∫R2de(t,−k,ξ)eik⋅xeiξ⋅vdkdξ=(2π)dtd1∫R2dexp(−∣ξ∣2t−121∣k∣2t)eix⋅(k/t)eiv⋅(ξ−k/2)dkdξ=td1((2π)d/21∫Rdexp(−∣ξ∣2t)eiξ⋅vdξ)((2π)d/21∫Rdexp(−121∣k∣2t)ei(x/t−v/2)⋅kdk)=td1((2t)d/21exp(−4t∣v∣2))((t/6)d/21exp(−t/3∣x/t−v/2∣2))=t2d3d/2exp(−4t∣v∣2−t33∣∣x−2tv∣∣2)=t2d3d/2exp(−41∣∣t1/2v∣∣2−3∣∣t3/2x−2t1/2v∣∣2)=t2d3d/2G(t3/2x,t1/2v),
where
G(x,v)=exp(−41∣v∣2−3∣∣x−2v∣∣2).
To proceed further, we define the modified convolution ∗t by
h∗tj(x,v)=∫R2dh(y,w)j(x−y−tw,v−w)dwdy.
Observe that if j~(x,v)=j(x+tv,v), then h∗tj(x,v)=h∗j~(x−tv,v). Hence it follows from Young's convolution inequality that
∥h∗tj∥Lr(R2d)≤∥h∥Lp(R2d)∥j∥Lq(R2d)
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
(h∗tj)(k,ξ)=(2π)dh^(k,ξ+tk)j^(k,ξ).
Proof. Recall that the Fourier transform of the convolution
(f∗g)(x)=∫Rdf(x−y)g(y)dy
is
(f∗g)(ξ)=(2π)d/2f^(ξ)g^(ξ).
By taking Fourier transform, we have
h∗tj(k,ξ)=(2π)d1∫R2d(h∗j~)(x−tv,v)e−ik⋅xe−iξ⋅vdxdv=(2π)d1∫Rde−iξ⋅ve−ik⋅(tv)[∫Rd(h∗j~)(x,v)e−ik⋅xdx]dv=(2π)d1∫R2de−i(ξ+tk)⋅ve−ik⋅x(h∗j~)(x,v)dxdv=(2π)dh^(k,ξ+tk)j~^(k,ξ+tk).
Since j~(x,v)=j(x+tv,v), it follows that
=(2π)dh^(k,ξ+tk)j^(k,ξ).
From this property and (7), we have
u(t,x,v)=(G∗tg)(x,v)+∫0t∫R2dh(τ)∗t−τG(t−τ,⋅,⋅)dτ,
where
G(t,x,v)=(4π2t43)d/2G(t3/2x,t1/2v).
We list some properties of the fundamental solution.
Proposition. The function G and G have the following properties:
The function G is C∞ and decays polynomially at infinity. Moreover, G and all its derivatives are integrable in R2d.
For every t>0, ∫R2dG(t,x,v)dvdx=1.
Both functions are nonnegative: G≥0 and G≥0.
For any p≥1, we have
∥G(t,⋅,⋅)∥Lp(R2d)∥DvG(t,⋅,⋅)∥Lp(R2d)≈t−d(1+1/2)(1−1/p)∥G∥Lp(R2d)≈t−d(1+1/2)(1−1/p)−1/2∥DvG∥Lp(R2d).
Proof. (1) -- (3) are immediate from the definition.
To show (4), a change of variable shows that
∥G(t,⋅,⋅)∥Lp(R2d)=(4π2t43)d/2(∫R2d∣∣G(t3/2x,t1/2v)∣∣pdxdv)1/p=(4π23)d/2t−2d(1−1/p)∥G∥Lp(R2d).
To show the second estimate, we note that
DvG(t,x,v)=(4π2t43)d/2DvG(t3/2x,t1/2v)t1/21.
A change of variable shows that
∥DvG(t,⋅,⋅)∥Lp(R2d)=(4π23)d/2t−2d(1−1/p)−1/2∥DvG∥Lp(R2d).
This completes the proof.