About the optimal constant in the parabolic Harnack inequality
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For the heat equation $partial_t u = Delta u$ there exists the following version of the harnack inequality. Given $0<t_1<t_2$ and $x_1,x_2 in mathbb{R}^d$ it holds that
$$u(t_1,x_1) le u(t_2,x_2) left( frac{t_2}{t_1}right) ^{frac{d}{2}} exp left( frac{|x_2-x_1|_2^2}{4(t_2-t_1)} right). $$
For many authors it is a well known fact that this is sharp. Some authors argue that for the heat kernel
$$u(t,x) = left( frac{1}{4pi t}right) ^{frac{d}{2}} exp left( frac{-|x|_2^2}{4t} right)$$
above inequality is an equality. When trying to verify that i get stuck at
$$ exp left( frac{|x_2-x_1|_2^2}{4(t_2-t_1)} - frac{|x_2|_2^2}{4t_2} +frac{|x_1|_2^2}{4t_1} right) overset{!}{=} 1$$ which clearly doesn't hold. So my question is: Why is above Harnack inequality sharp?
heat-equation
asked Nov 10 at 8:22
user33
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For the heat equation $partial_t u = Delta u$ there exists the following version of the harnack inequality. Given $0<t_1<t_2$ and $x_1,x_2 in mathbb{R}^d$ it holds that
$$u(t_1,x_1) le u(t_2,x_2) left( frac{t_2}{t_1}right) ^{frac{d}{2}} exp left( frac{|x_2-x_1|_2^2}{4(t_2-t_1)} right). $$
For many authors it is a well known fact that this is sharp. Some authors argue that for the heat kernel
$$u(t,x) = left( frac{1}{4pi t}right) ^{frac{d}{2}} exp left( frac{-|x|_2^2}{4t} right)$$
above inequality is an equality. When trying to verify that i get stuck at
$$ exp left( frac{|x_2-x_1|_2^2}{4(t_2-t_1)} - frac{|x_2|_2^2}{4t_2} +frac{|x_1|_2^2}{4t_1} right) overset{!}{=} 1$$ which clearly doesn't hold. So my question is: Why is above Harnack inequality sharp?
heat-equation
asked Nov 10 at 8:22
user33
12
add a comment |
up vote
0
down vote
favorite
up vote
0
down vote
favorite
For the heat equation $partial_t u = Delta u$ there exists the following version of the harnack inequality. Given $0<t_1<t_2$ and $x_1,x_2 in mathbb{R}^d$ it holds that
$$u(t_1,x_1) le u(t_2,x_2) left( frac{t_2}{t_1}right) ^{frac{d}{2}} exp left( frac{|x_2-x_1|_2^2}{4(t_2-t_1)} right). $$
For many authors it is a well known fact that this is sharp. Some authors argue that for the heat kernel
$$u(t,x) = left( frac{1}{4pi t}right) ^{frac{d}{2}} exp left( frac{-|x|_2^2}{4t} right)$$
above inequality is an equality. When trying to verify that i get stuck at
$$ exp left( frac{|x_2-x_1|_2^2}{4(t_2-t_1)} - frac{|x_2|_2^2}{4t_2} +frac{|x_1|_2^2}{4t_1} right) overset{!}{=} 1$$ which clearly doesn't hold. So my question is: Why is above Harnack inequality sharp?
heat-equation
asked Nov 10 at 8:22
user33
12
For the heat equation $partial_t u = Delta u$ there exists the following version of the harnack inequality. Given $0<t_1<t_2$ and $x_1,x_2 in mathbb{R}^d$ it holds that
$$u(t_1,x_1) le u(t_2,x_2) left( frac{t_2}{t_1}right) ^{frac{d}{2}} exp left( frac{|x_2-x_1|_2^2}{4(t_2-t_1)} right). $$
For many authors it is a well known fact that this is sharp. Some authors argue that for the heat kernel
$$u(t,x) = left( frac{1}{4pi t}right) ^{frac{d}{2}} exp left( frac{-|x|_2^2}{4t} right)$$
above inequality is an equality. When trying to verify that i get stuck at
$$ exp left( frac{|x_2-x_1|_2^2}{4(t_2-t_1)} - frac{|x_2|_2^2}{4t_2} +frac{|x_1|_2^2}{4t_1} right) overset{!}{=} 1$$ which clearly doesn't hold. So my question is: Why is above Harnack inequality sharp?
heat-equation
heat-equation
asked Nov 10 at 8:22
user33
12
asked Nov 10 at 8:22
user33
12
edited Nov 22 at 18:06
asked Nov 10 at 8:22
user33
12
asked Nov 10 at 8:22
user33
12
asked Nov 10 at 8:22
user33
12
12
add a comment |
add a comment |
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