Tag Archives: half-harmonic

Corotational Harmonic map heat flow from D2 to S2

August 4, 2020 – 8:11 pm

Let {D^{2}=\left\{x=\left(x_{1}, x_{2}, 0\right) \in R^{3}:|x|^{2} \leq 1\right\}, S^{2}=\left\{x \in R^{3}:|x|^{2}=1\right\}}. Consider the initial-boundary value problem for the heat equation of harmonic maps as follows.

\displaystyle \left\{\begin{array}{l}\frac{\partial u}{\partial t}=\Delta u+|\nabla u|^{2} u \\ u(0, x)=u_{0}(x) \\ \left.u(t, \cdot)\right|_{\partial D^{2}}=\left.u_{0}\right|_{\partial D^{2}}\end{array}\right.

In [1], it is shown that if the initial map has the following symmetric form 

\displaystyle u_{0}(x)=\left(\frac{x}{|x|} \sin h_{0}(|x|), \cos h_{0}(|x|)\right)

then then the solution {u} has the form

\displaystyle u(t, x)=\left(\frac{x}{|x|} \sin h(t,|x|), \cos h(t,|x|)\right)

Let us derive the flow equation of {h} satisfies. We need to compute {\Delta u} and {|\nabla u|}. 

Remember for any function {f=f(r,\theta)} on {D^2}, one has 

\displaystyle |\nabla f|=f_r^2+\frac{1}{r^2}f_\theta^2

\displaystyle \Delta f=f_{rr}+\frac{1}{r}f_r+\frac{1}{r^2}f_{\theta\theta}

Using the above equation and {u_1=\cos\theta\sin h(r,t)}, one can derive 

\displaystyle |\nabla u_1|^2=\cos^2\theta \cos^2 h \,h_r^2+\frac{1}{r^2}\sin^2\theta \sin^2h

\displaystyle \Delta u_1=\cos\theta[\cos h \, h_{rr}-\sin h \, h_r^2+\frac{1}{r}\cos h\, h_r-\frac{1}{r^2}\sin h]

Therefore,

\displaystyle |\nabla u|^2=h_r^2+\frac{1}{r^2}\sin^2h

The flow equation {\partial_t u_1=\Delta u_1+|\nabla u|^2 u_1 } implies

\displaystyle \partial_t h=h_{rr}+\frac{1}{r}h_r-\frac{\cos h\sin h}{r^2}h

the equation of {u_2} and {u_3} implies the same equation of {h}. 

[1] Chang, K.-C., Ding, W.-Y. (2018). A Result on the Global Existence for Heat Flows of Harmonic Maps from D2 into S2, 213-223.

By Sun's World | Posted in Harmonic map, PDE, Riemannian Geo | Also tagged | Comments (0)

Non-homogeneous half-harmonic heat equation

August 2, 2020 – 2:03 pm

Consider the Possion Kernel

\displaystyle P(x,t)=\frac{\gamma_n t}{(|x|^2+t^2)^{\frac{n+1}{2}}},\quad \gamma_n=\frac{\Gamma[(n+1) / 2]}{\pi^{(n+1) / 2}}

It is know that {U(x,t)=P[u](x,t)=\int_{\mathbb{R}^n}P(x-y,t)u(y)dy} gives a (formal) solution to the Laplace equation 

\displaystyle \begin{cases}\Delta U=0\\U(x,0)=u(x)\end{cases}

Moreover the Possion Kernel satisfies {P(\cdot,t)*P(\cdot,s)=P(\cdot,t+s)}. Therefore {P_tu:=P(\cdot,t)*u(\cdot)} forms a semigroup. If {u} belongs to {L^\infty(\mathbb{R}^n)}, it is a {C_0}-semigroup. The generator of it is 

\displaystyle Au=\lim_{t\rightarrow 0^+}\frac{P_tu-u}{t}

According to Silvestre and Caffarelli, the right hand side is {-(-\Delta)^{\frac12}u}. So {A=-(-\Delta)^{\frac12}}. In semigroup notation, we write

\displaystyle e^{-(-\Delta)^{\frac12}t}=\frac{\gamma_n t}{(|x|^2+t^2)^{\frac{n+1}{2}}}.

Now suppose we have an half-harmonic heat equation 

\displaystyle u_t=-(-\Delta)^{\frac12}u,\quad u(\cdot,0)=u_0

Then its (formal) solution would be 

\displaystyle u(x,t)=e^{-(-\Delta)^{\frac12}t}u(\cdot,0)=P(\cdot,t)*u_0=U_0(x,t)

where {U_0} is the harmonic extension of {u_0} to the upper half plane {\{(x,t):t>0\}}.

Now consider the non-homogeneous equation 

\displaystyle u_t=-(-\Delta)^{\frac12}u+f,\quad u(\cdot,0)=0

with {f=f(x,t)} is smooth enough. By Duhamel’s principle

\displaystyle u(x,t)=\int_0^t\int_{\mathbb{R}^n}P(x-y,t-s)f(y,s)dyds

Consider a particular example in [2], that is {n=1}, {f=\frac{(x-\xi)}{(x-\xi)^2+\mu^2}=\pi P(\mu,x-\xi)}, then 

\displaystyle u(x,t)=\frac{1}{\pi}\int_0^t\int_{\mathbb{R}}\frac{t-s}{(x-y)^2+(t-s)^2}\frac{y-\xi}{(y-\xi)^2+\mu^2}dyds

\displaystyle =\frac{1}{\pi}\int_0^t\int_{\mathbb{R}}\frac{t-s}{(x-\xi-y)^2+(t-s)^2}\frac{y}{y^2+\mu^2}dyds

Using the integration formula

\displaystyle \int_{\mathbb{R}}\frac{a}{(x-y)^2+a^2}\frac{y}{y^2+\mu^2}dy=\frac{x\pi}{x^2+(a+\mu)^2}

for {a>0}, {\mu>0}. Then 

\displaystyle u(x,t)=\int_0^t \frac{(x-\xi)}{(x-\xi)^2+(t-s+\mu)^2}ds.

[1] Cabre, X., Roquejoffre, J. M. (2013). The Influence of Fractional Diffusion in Fisher-KPP Equations. Communications in Mathematical Physics, 320(3), 679-722. 

[2] Sire, Y., Wei, J., Zheng, Y. (2017). Infinite time blow-up for half-harmonic map flow from {\mathbb{R}} into {\mathbb{S}^1}, (11301374), 1-65.

By Sun's World | Posted in Analysis, Fractional, Harmonic map | Comments (0)

Simple identity on half-harmonic functions

July 31, 2020 – 7:48 pm

Suppose {s\in(0,1)}, for u in Schwartz space, define 

\displaystyle (-\Delta)^{s} u(x)=C(n, s) P . V. \int_{\mathbb{R}^{n}} \frac{u(x)-u(y)}{|x-y|^{n+2 s}} d y

It is proved in [1] that

\displaystyle (-\Delta)^{s} u=\mathscr{F}^{-1}\left(|\xi|^{2 s}(\mathscr{F} u)\right) \quad \forall \xi \in \mathbb{R}

where

\displaystyle \mathscr{F}(f)(\xi)=\hat f(\xi)=\int_{-\infty}^{\infty} f(x) e^{- i x \xi} d x.

Lemma 1 Suppose {u\in H^{1}({\mathbb{R}^1})}, then 

\displaystyle \int_{\mathbb{R}}|(-\Delta)^{\frac14}u|^2dx=\int_{\mathbb{R}}u(-\Delta)^{\frac12}u \, dx=\frac{1}{2\pi} \int_{\mathbb{R}\times \mathbb{R}} \frac{|u(x)-u(y)|^{2}}{|x-y|^{2}} d x d y

Proof: Using Plancherel identity 

\displaystyle \int_{\mathbb{R}}|(-\Delta)^{\frac14}u|^2dx=\int_{\mathbb{R}}|\sqrt{|\xi|}\hat u|^2d\xi=\int_{\mathbb{R}}|\xi| \hat u\hat ud\xi

By the Parseval’s identity 

\displaystyle \int_{\mathbb{R}}|\xi| \hat u\hat ud\xi=\int_{\mathbb{R}}(-\Delta)^{\frac12}u u dx.

Furthermore, Proposition 3.4 in [1] tells us that

\displaystyle \int_{\mathbb{R}}|\xi| \hat u^2d\xi=\frac12 C(1,\frac12) \int_{\mathbb{R}}\left(\int_{\mathbb{R}} \frac{|u(x)-u(y)|^{2}}{|x-y|^{2}} d x\right) d y

Simple calculation shows {C(1,\frac12)=1/{\pi}}. 

\Box

[1]Di Nezza, E., Palatucci, G., Valdinoci, E. (2012). Hitchhiker’s guide to the fractional Sobolev spaces. Bulletin Des Sciences Mathematiques, 136(5), 521-573.

Suppose

\displaystyle w(x)=\left(\frac{2x}{x^2+1},\frac{x^2-1}{x^2+1}\right):\mathbb{R}\to \mathbb{S}^1

then by the definition of (-\Delta)^{\frac12}, we have

\displaystyle (-\Delta )^{\frac12}w=\frac{1}{\pi}P.V.\int_{\mathbb{R}}\frac{w(x)-w(y)}{|x-y|^2}dy=\left(\frac{4x}{(1+x^2)^2},\frac{2(x^2-1)}{(1+x^2)^2}\right)

\displaystyle \frac{1}{2\pi}\int_{\mathbb{R}}\frac{|w(x)-w(y)|^2}{|x-y|^2}dy=\frac{2}{1+x^2}

It is easy to verify that

\displaystyle (-\Delta)^{\frac12}w(x)=\left(\frac{1}{2\pi}\int_{\mathbb{R}}\frac{|w(x)-w(y)|^2}{|x-y|^2}dy\right) w(x)

Therefore w is the half-harmonic map from \mathbb{R}\to \mathbb{S}^1.

If U(x)=w(x\lambda), then

\displaystyle (-\Delta )^{\frac12}U(x)=\frac{1}{\pi}P.V.\int_{\mathbb{R}}\frac{w(\lambda x)-w(\lambda y)}{|x-y|^2}dy=\lambda( (-\Delta)^{\frac12}w)(\lambda x)

\displaystyle \frac{1}{2\pi}\int_{\mathbb{R}}\frac{|U(x)-U(y)|^2}{|x-y|^2}dy=\frac{2\lambda}{1+x^2\lambda^2}

By Sun's World | Posted in Fractional, Harmonic map, PDE | Also tagged | Comments (0)