Mittag-Leffler family for an integro-differential equation

In a few months ago I post a note about a family of an integro-partial differential equations (2), there the main objective was to expose, without details, a first work to treat questions such as existence and symmetries of solutions in certain critical space, namely, homogeneous Morrey Spaces. On that work the Mittag-Leffler family below was central on the estimates,

\displaystyle \widehat{u}(t,\xi)=\mathbb{E}_{\alpha}(-t^{\alpha}\vert\xi\vert^2)\widehat{\varphi}(\xi) \ \ \ \ \ (1)

which is an integral form of the integro-partial differential equation

\displaystyle u_t(t,x)=\int_{0}^{t}r_{\alpha-1}(t-s) \Delta_{x}u(s,x)ds,\; u(0,x)=\varphi(x) \ \ \ \ \ (2)

where {1\leq\alpha<2}, {\Delta_{x}=\Sigma_{j=1}^{n}(\partial/\partial x_j)^2}, {u(t,x)=(u_{1}(t,x),\cdots,u_{n}(t,x))} for {t\in [0,\infty)} and {x\in\mathbb{R}^{n}} and {r_{\alpha}(s)=s^{\alpha-1}/\Gamma(\alpha)}, here {\Gamma(\alpha)} stands for Gamma function. In this post we will prove (1). Before, we need some properties of the Laplace transform. Given a “well” real-valued function {u_j(\cdot,x)} on {[0,\infty]}, we define the Laplace transform as

\displaystyle \mathcal{L}(u_j(\cdot,x))(s)=\int_0^{\infty}e^{-st}u_j(t,x)dt. \ \ \ \ \ (3)

Given {\lambda\in (0,\infty)}, let  {\partial_tu, u,v:[0,\lambda]\times\mathbb{R}^{n}\rightarrow\mathbb{R}^n} be certain  {\mathbb{R}^n}-valued functions for each {t\in(0,\lambda)} fixed, then

\displaystyle \mathcal{L}(\partial_tu)(s)=-u(0,x)+s\mathcal{L}(u)(s) \ \ \ \ \ (4)


\displaystyle \mathcal{L}(u\ast^{t}v)(s)=\mathcal{L}(u)(s)\mathcal{L}(v)(s) \ \ \ \ \ (5)

where {\ast^{t}} stands for convolution operator on variable {t}, that is, {u\ast^{t}v(t)=\int_0^t u(t-s,x)v(s,x)ds}. Hence, applying the Laplace transform in (2), we obtain

\displaystyle s\mathcal{L}(u)(s)-\varphi(x)=\mathcal{L}(r_{\alpha-1})(s)\mathcal{L}(\Delta_{x} u)(s). \ \ \ \ \ (6)

In view of

\displaystyle \begin{array}{rcl} \mathcal{L}(t^{d})(s)=\int_0^{\infty}e^{-ts}t^{d}dt=\frac{\Gamma(d+1)}{s^{d+1}}, \;\; (d>0) \end{array}


\displaystyle \begin{array}{rcl} \mathcal{L}(\frac{\partial^2}{\partial x_i^2}u)(s)=\frac{\partial^2}{\partial x_i^2}\mathcal{L}(u)(s) \end{array}

we have that the equation (6) can be written as

\displaystyle s\mathcal{L}(u)(s)-\varphi(x)=s^{1-\alpha}\Delta_{x}\mathcal{L}(u)(s). \ \ \ \ \ (7)

For {j=1,\cdots n}, let {u_j\in \mathcal{S}(\mathbb{R}^{n})} (Schwartz space) and let {\widehat{u_j}} be the Fourier transform of {u_j} given by

\displaystyle \widehat{u_j}(t,\xi)=\int_{\mathbb{R}^{n}}e^{-ix\cdot \xi}u_j(t,x)dx. \ \ \ \ \ (8)

If {u_j\in C^{\infty}(\mathbb{R}^{n})} with compact support, easily seems that

\displaystyle \widehat{\frac{\partial^{2} u_j}{\partial x_i^{2}}}(t,\xi)=-\xi_j^{2}\,\widehat{u_j}(t,\xi)\Rightarrow \widehat{\Delta_{x} u}(t,\xi)=-\vert \xi\vert^2\,\widehat{u}(t,\xi). \ \ \ \ \ (9)

As {\widehat{\mathcal{L}(u)(s)}(\xi)=\mathcal{L}(\widehat{u}(\cdot,\xi))(s)} it follows from (7) and (9) that

\displaystyle s\mathcal{L}(\widehat{u}(\cdot,\xi))(s)=\widehat{\varphi}(\xi)-s^{1-\alpha}\vert \xi\vert^2 \mathcal{L}(\widehat{u}(\cdot,\xi))(s)\nonumber \ \ \ \ \ (10)

which yields

\displaystyle \mathcal{L}(\widehat{u}(\cdot,\xi))(s)=\frac{\widehat{\varphi}(\xi)}{s+s^{1-\alpha}\vert\xi\vert^2}=\frac{s^{\alpha-1}}{s^{\alpha}+\vert\xi\vert^2} \widehat{\varphi}(\xi). \ \ \ \ \ (11)

This motivates to define Mittag-Leffler function as a complex integral on certain curve. Indeed, firstly notice that the complex valued function {f(z)=\frac{e^zz^{\alpha-1}}{z^{\alpha}+\vert\xi\vert^2}} has singularities in

\displaystyle a_{\alpha}(\xi)=|\xi|^{\frac{2}{\alpha}}e^{\frac{i\pi}{\alpha}} ,\;\;\;b_{\alpha}(\xi)=|\xi|^{\frac{2}{\alpha}}e^{-\frac{i\pi}{\alpha}},\text{ for }\xi\in\mathbb{R}^{n}. \ \ \ \ \ (12)

Let {\gamma} be the standard Hankel’s curve in {\mathbb{C}}, positive oriented, such that {a_{\alpha}(\xi)\in Int(\gamma)} and {b_{\alpha}(\xi)\in Int(\gamma)}, that is, let {\gamma=r_1+r_2+C_r} be a parametrized curve given by {r_1(t)=te^{i\theta}}, {r_2(t)=te^{-i\theta}} for {t\in (r,\infty)} and {C_r(t)=r e^{it}} for {t\in (-\theta,\theta)}. As we want to chose {\gamma} such that {a_{\alpha}(\xi)\in Int(\gamma)} and {b_{\alpha}(\xi)\in Int(\gamma)}, we shall suppose that {r^{\alpha}>\vert \xi\vert^2>\epsilon^{\alpha}>0} and one defines Mittag-Leffler function as

\displaystyle \begin{array}{rcl} \mathbb{E}_{\alpha,\beta}(-\vert\xi\vert^2)=\frac{1}{2\pi i}\int_{\gamma}\frac{e^zz^{\alpha-\beta}}{z^{\alpha}+\vert\xi\vert^2}dz,\; (\alpha>0,\beta>0). \end{array}

Using residue theorem, a characterization very important of this definition was obtained in [Fujita] (see also [Hirata-Miao][Fujita2]).

Proposition 1 If {1<\alpha<2} and {\beta=1}, we have

\displaystyle \begin{array}{rcl} L^{1}(\mathbb{R}^{n})\ni\mathbb{E}_{\alpha}(-|\xi|^{2})=\frac{1}{\alpha}(\exp(a_{\alpha}(\xi))+\exp(b_{\alpha}(\xi)))+l_{\alpha}(\xi) \end{array}


\displaystyle l_{\alpha}(\xi)= \begin{cases} \frac{\sin(\alpha\pi)}{\pi}\int_{0}^{\infty}\frac{|\xi|^{2}s^{\alpha-1}e^{-s} }{s^{2\alpha}+2|\xi|^{2}s^{\alpha}\cos(\alpha\pi)+|\xi|^{4}}ds & \text{ if }\xi\neq0\\ 1-\frac{2}{\alpha}, & \text{ if }\xi=0. \end{cases} \ \ \ \ \ (13)

Some corollaries can be obtained by Lemma above. Indeed, taking {t=\vert\xi\vert^{\frac{2}{\alpha}}s^{\frac{1}{\alpha}}} one has

\displaystyle \begin{array}{rcl} \vert \mathbb{E}_{\alpha}(-|\xi|^{2})\vert&\leq& \frac{2}{\alpha}+\vert l_{\alpha}(\xi)\vert\\ &\leq&\frac{2}{\alpha}+\frac{\sin(\alpha\pi)}{\pi}\int_{0}^{\infty}\frac{e^{-\vert\xi\vert^{\frac{2}{\alpha}}s^{\frac{1}{\alpha}}}}{s^{2}+2s\cos(\alpha\pi)+1}ds\\ &\leq&\frac{2}{\alpha}+\frac{\sin(\alpha\pi)}{\pi}\int_{0}^{\infty}\frac{1}{s^{2}+2s\cos(\alpha\pi)+1}ds\\ &=&\frac{2}{\alpha} +(1-\frac{2}{\alpha})=1, \end{array}

more general (see de Almeida, Ferrreira, L.F.C).

Proposition 2 Let {1\leq\alpha<2} and {0\leq\delta<2.} There is {C>0} such that

\displaystyle \left\vert \frac{\partial^{k}}{\partial\xi^{k}}\left[ \left\vert \xi\right\vert ^{\delta}\mathbb{E}_{\alpha}(-|\xi|^{2})\right] \right\vert \leq C\left\vert \xi\right\vert ^{-\left\vert k\right\vert },\text{ } \ \ \ \ \ (14)

for all {k\in(\mathbb{N}\cup\{0\})^{n}} with {\left\vert k\right\vert \leq\lbrack n/2]+1} and for all {\xi\neq0.}

Using an exercise,

\displaystyle \int_{0}^{\infty}e^{-t}\mathbb{E}_{\alpha}(at^{\alpha})dt=\frac{1}{1-a},\;\; (r^{\alpha}>\vert a\vert>0)\text{ and } (1< \alpha< 2),\nonumber \ \ \ \ \ (15)

we have

\displaystyle \mathcal{L}(\mathbb{E}_{\alpha}(-t^{\alpha}\vert\xi\vert^2))(s)=\frac{s^{\alpha-1}}{s^{\alpha}+\vert\xi\vert^2}. \ \ \ \ \ (16)

Taking in (11) the inverse Laplace transform and using (16), we obtain {\widehat{u}(t,\xi)=\mathbb{E}_{\alpha}(-t^{\alpha}\vert\xi\vert^2)\widehat{\varphi}(\xi).}


Leave a Reply

Fill in your details below or click an icon to log in: Logo

You are commenting using your account. Log Out / Change )

Twitter picture

You are commenting using your Twitter account. Log Out / Change )

Facebook photo

You are commenting using your Facebook account. Log Out / Change )

Google+ photo

You are commenting using your Google+ account. Log Out / Change )

Connecting to %s