$u(x,t)=left(f*mathcal{H}_{t}^{(n)}right)(x)$ is solution to the n-dimensional heat equation
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Consider the time-dependent heat equation in $mathbb{R^{n}}$:
$$displaystylefrac{partial u}{partial t}=frac{partial^{2}u}{partial x_{1}^{2}}+cdots+frac{partial^{2}u}{partial x_{n}^{2}},quadtextrm{where};t>0 $$
with boundary values $u(x,0)=f(x)in mathcal{S}(mathbb{R}^{n})$ (i.e. $f$ belongs to Schwartz space of $mathbb{R}^{n})$.
Define the n-dimensional heat kernel by
$$displaystylemathcal{H}_{t}^{(n)}(x):=frac{1}{(4pi t)^{n/2}}e^{|x|^2/4t}=int_{mathbb{R}^{n}}e^{-4pi^{2}t|xi|^{2}}e^{2pi i xcdot xi}dxi, $$
where $|cdot|$ is the euclidian norm in $mathbb{R}^{n}$.
I need to prove three statements about $u(x,t)=left(f*mathcal{H}_{t}^{(n)}right)(x)$.
(1) $u$ is indefinitely differentiable when $xinmathbb{R}^{n}$ and $t>0.$
(2) $u$ solves the heat equation above
(3) $u$ is continuous up to the boundary $t=0$ with $u(x,0)=f(x).$
By definition,
$$u(x,t)=underbrace{int_{-infty}^{infty}cdotsint_{-infty}^{infty}}_{n;textrm{times}}frac{f(y)}{(4pi t)^{n/2}}e^{-frac{left[(x_{1}-y_{1})^{2}+cdots+(x_{n}-y_{n})^{2}right]^{1/2}}{4t}}dx_{1}dots dx_{n}$$
That's all I got. I really want some tips (I don't want the solution ifself)
pde fourier-analysis improper-integrals convolution heat-equation
asked Nov 21 at 17:54
Mateus Rocha
792117
add a comment |
up vote
0
down vote
favorite
Consider the time-dependent heat equation in $mathbb{R^{n}}$:
$$displaystylefrac{partial u}{partial t}=frac{partial^{2}u}{partial x_{1}^{2}}+cdots+frac{partial^{2}u}{partial x_{n}^{2}},quadtextrm{where};t>0 $$
with boundary values $u(x,0)=f(x)in mathcal{S}(mathbb{R}^{n})$ (i.e. $f$ belongs to Schwartz space of $mathbb{R}^{n})$.
Define the n-dimensional heat kernel by
$$displaystylemathcal{H}_{t}^{(n)}(x):=frac{1}{(4pi t)^{n/2}}e^{|x|^2/4t}=int_{mathbb{R}^{n}}e^{-4pi^{2}t|xi|^{2}}e^{2pi i xcdot xi}dxi, $$
where $|cdot|$ is the euclidian norm in $mathbb{R}^{n}$.
I need to prove three statements about $u(x,t)=left(f*mathcal{H}_{t}^{(n)}right)(x)$.
(1) $u$ is indefinitely differentiable when $xinmathbb{R}^{n}$ and $t>0.$
(2) $u$ solves the heat equation above
(3) $u$ is continuous up to the boundary $t=0$ with $u(x,0)=f(x).$
By definition,
$$u(x,t)=underbrace{int_{-infty}^{infty}cdotsint_{-infty}^{infty}}_{n;textrm{times}}frac{f(y)}{(4pi t)^{n/2}}e^{-frac{left[(x_{1}-y_{1})^{2}+cdots+(x_{n}-y_{n})^{2}right]^{1/2}}{4t}}dx_{1}dots dx_{n}$$
That's all I got. I really want some tips (I don't want the solution ifself)
pde fourier-analysis improper-integrals convolution heat-equation
asked Nov 21 at 17:54
Mateus Rocha
792117
(1) is standard theory about convolution; nothing to prove there.
– Federico
Nov 21 at 18:03
(2) have you done some computation of the partial derivatives? Again, some facts about the convolution might help
– Federico
Nov 21 at 18:04
(3) ever seen convolutions with a rescaled kernel? They approximate the identity...
– Federico
Nov 21 at 18:05
I would suggest using vector calculus computations to simplify things..it’ll make your life easier
– DaveNine
Nov 21 at 19:49
add a comment |
up vote
0
down vote
favorite
up vote
0
down vote
favorite
Consider the time-dependent heat equation in $mathbb{R^{n}}$:
$$displaystylefrac{partial u}{partial t}=frac{partial^{2}u}{partial x_{1}^{2}}+cdots+frac{partial^{2}u}{partial x_{n}^{2}},quadtextrm{where};t>0 $$
with boundary values $u(x,0)=f(x)in mathcal{S}(mathbb{R}^{n})$ (i.e. $f$ belongs to Schwartz space of $mathbb{R}^{n})$.
Define the n-dimensional heat kernel by
$$displaystylemathcal{H}_{t}^{(n)}(x):=frac{1}{(4pi t)^{n/2}}e^{|x|^2/4t}=int_{mathbb{R}^{n}}e^{-4pi^{2}t|xi|^{2}}e^{2pi i xcdot xi}dxi, $$
where $|cdot|$ is the euclidian norm in $mathbb{R}^{n}$.
I need to prove three statements about $u(x,t)=left(f*mathcal{H}_{t}^{(n)}right)(x)$.
(1) $u$ is indefinitely differentiable when $xinmathbb{R}^{n}$ and $t>0.$
(2) $u$ solves the heat equation above
(3) $u$ is continuous up to the boundary $t=0$ with $u(x,0)=f(x).$
By definition,
$$u(x,t)=underbrace{int_{-infty}^{infty}cdotsint_{-infty}^{infty}}_{n;textrm{times}}frac{f(y)}{(4pi t)^{n/2}}e^{-frac{left[(x_{1}-y_{1})^{2}+cdots+(x_{n}-y_{n})^{2}right]^{1/2}}{4t}}dx_{1}dots dx_{n}$$
That's all I got. I really want some tips (I don't want the solution ifself)
pde fourier-analysis improper-integrals convolution heat-equation
asked Nov 21 at 17:54
Mateus Rocha
792117
Consider the time-dependent heat equation in $mathbb{R^{n}}$:
$$displaystylefrac{partial u}{partial t}=frac{partial^{2}u}{partial x_{1}^{2}}+cdots+frac{partial^{2}u}{partial x_{n}^{2}},quadtextrm{where};t>0 $$
with boundary values $u(x,0)=f(x)in mathcal{S}(mathbb{R}^{n})$ (i.e. $f$ belongs to Schwartz space of $mathbb{R}^{n})$.
Define the n-dimensional heat kernel by
$$displaystylemathcal{H}_{t}^{(n)}(x):=frac{1}{(4pi t)^{n/2}}e^{|x|^2/4t}=int_{mathbb{R}^{n}}e^{-4pi^{2}t|xi|^{2}}e^{2pi i xcdot xi}dxi, $$
where $|cdot|$ is the euclidian norm in $mathbb{R}^{n}$.
I need to prove three statements about $u(x,t)=left(f*mathcal{H}_{t}^{(n)}right)(x)$.
(1) $u$ is indefinitely differentiable when $xinmathbb{R}^{n}$ and $t>0.$
(2) $u$ solves the heat equation above
(3) $u$ is continuous up to the boundary $t=0$ with $u(x,0)=f(x).$
By definition,
$$u(x,t)=underbrace{int_{-infty}^{infty}cdotsint_{-infty}^{infty}}_{n;textrm{times}}frac{f(y)}{(4pi t)^{n/2}}e^{-frac{left[(x_{1}-y_{1})^{2}+cdots+(x_{n}-y_{n})^{2}right]^{1/2}}{4t}}dx_{1}dots dx_{n}$$
That's all I got. I really want some tips (I don't want the solution ifself)
pde fourier-analysis improper-integrals convolution heat-equation
pde fourier-analysis improper-integrals convolution heat-equation
asked Nov 21 at 17:54
Mateus Rocha
792117
asked Nov 21 at 17:54
Mateus Rocha
792117
asked Nov 21 at 17:54
Mateus Rocha
792117
asked Nov 21 at 17:54
Mateus Rocha
792117
asked Nov 21 at 17:54
Mateus Rocha
792117
792117
(1) is standard theory about convolution; nothing to prove there.
– Federico
Nov 21 at 18:03
(2) have you done some computation of the partial derivatives? Again, some facts about the convolution might help
– Federico
Nov 21 at 18:04
(3) ever seen convolutions with a rescaled kernel? They approximate the identity...
– Federico
Nov 21 at 18:05
I would suggest using vector calculus computations to simplify things..it’ll make your life easier
– DaveNine
Nov 21 at 19:49
add a comment |
(1) is standard theory about convolution; nothing to prove there.
– Federico
Nov 21 at 18:03
(2) have you done some computation of the partial derivatives? Again, some facts about the convolution might help
– Federico
Nov 21 at 18:04
(3) ever seen convolutions with a rescaled kernel? They approximate the identity...
– Federico
Nov 21 at 18:05
I would suggest using vector calculus computations to simplify things..it’ll make your life easier
– DaveNine
Nov 21 at 19:49
(1) is standard theory about convolution; nothing to prove there.
– Federico
Nov 21 at 18:03
(1) is standard theory about convolution; nothing to prove there.
– Federico
Nov 21 at 18:03
(2) have you done some computation of the partial derivatives? Again, some facts about the convolution might help
– Federico
Nov 21 at 18:04
(2) have you done some computation of the partial derivatives? Again, some facts about the convolution might help
– Federico
Nov 21 at 18:04
(3) ever seen convolutions with a rescaled kernel? They approximate the identity...
– Federico
Nov 21 at 18:05
(3) ever seen convolutions with a rescaled kernel? They approximate the identity...
– Federico
Nov 21 at 18:05
I would suggest using vector calculus computations to simplify things..it’ll make your life easier
– DaveNine
Nov 21 at 19:49
I would suggest using vector calculus computations to simplify things..it’ll make your life easier
– DaveNine
Nov 21 at 19:49
add a comment |
1 Answer
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This is an easy consequence of the fact that $f'*g = f*g'$ for any $f$ and $g$, as well as the fact that the heat kernel is $C^{infty}$ for $t>0$.
This can be shown with the above fact also, namely we have that $$(partial_t-Delta)u = f*((partial_t-Deltamathcal){H}) = f*0=0$$ since the heat kernel is a solution to the homogenous heat equation.
This is pretty tough to prove and the difficulty varies depending on space $f$ lies in. I will try to sketch a proof. You are trying to show that $lim_{tto0}u(x,t) = f(x)$. The way this typically goes is you divide up the region into an $varepsilon$-ball centered at $x$ and its compliment then show that as $tto 0$, the integral of the outer part goes to $0$ and the interior approaches $f(x)$. For the outer part, use the fact that the integral converges (i.e. the heat kernel decreases rapidly) and the fact that $f$ is rapidly decreasing (Schwartz). For the inner part, you can use some the continuity of $f$ to relate the value of $f$ in this region to $f(x)$. You will use the fact that the heat kernel integrates to 1 at this point somewhere.
answered Nov 21 at 22:55
whpowell96
5315
add a comment |
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1 Answer
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This is an easy consequence of the fact that $f'*g = f*g'$ for any $f$ and $g$, as well as the fact that the heat kernel is $C^{infty}$ for $t>0$.
This can be shown with the above fact also, namely we have that $$(partial_t-Delta)u = f*((partial_t-Deltamathcal){H}) = f*0=0$$ since the heat kernel is a solution to the homogenous heat equation.
This is pretty tough to prove and the difficulty varies depending on space $f$ lies in. I will try to sketch a proof. You are trying to show that $lim_{tto0}u(x,t) = f(x)$. The way this typically goes is you divide up the region into an $varepsilon$-ball centered at $x$ and its compliment then show that as $tto 0$, the integral of the outer part goes to $0$ and the interior approaches $f(x)$. For the outer part, use the fact that the integral converges (i.e. the heat kernel decreases rapidly) and the fact that $f$ is rapidly decreasing (Schwartz). For the inner part, you can use some the continuity of $f$ to relate the value of $f$ in this region to $f(x)$. You will use the fact that the heat kernel integrates to 1 at this point somewhere.
answered Nov 21 at 22:55
whpowell96
5315
add a comment |
up vote
1
down vote
accepted
This is an easy consequence of the fact that $f'*g = f*g'$ for any $f$ and $g$, as well as the fact that the heat kernel is $C^{infty}$ for $t>0$.
This can be shown with the above fact also, namely we have that $$(partial_t-Delta)u = f*((partial_t-Deltamathcal){H}) = f*0=0$$ since the heat kernel is a solution to the homogenous heat equation.
This is pretty tough to prove and the difficulty varies depending on space $f$ lies in. I will try to sketch a proof. You are trying to show that $lim_{tto0}u(x,t) = f(x)$. The way this typically goes is you divide up the region into an $varepsilon$-ball centered at $x$ and its compliment then show that as $tto 0$, the integral of the outer part goes to $0$ and the interior approaches $f(x)$. For the outer part, use the fact that the integral converges (i.e. the heat kernel decreases rapidly) and the fact that $f$ is rapidly decreasing (Schwartz). For the inner part, you can use some the continuity of $f$ to relate the value of $f$ in this region to $f(x)$. You will use the fact that the heat kernel integrates to 1 at this point somewhere.
answered Nov 21 at 22:55
whpowell96
5315
add a comment |
up vote
1
down vote
accepted
up vote
1
down vote
accepted
This is an easy consequence of the fact that $f'*g = f*g'$ for any $f$ and $g$, as well as the fact that the heat kernel is $C^{infty}$ for $t>0$.
This can be shown with the above fact also, namely we have that $$(partial_t-Delta)u = f*((partial_t-Deltamathcal){H}) = f*0=0$$ since the heat kernel is a solution to the homogenous heat equation.
This is pretty tough to prove and the difficulty varies depending on space $f$ lies in. I will try to sketch a proof. You are trying to show that $lim_{tto0}u(x,t) = f(x)$. The way this typically goes is you divide up the region into an $varepsilon$-ball centered at $x$ and its compliment then show that as $tto 0$, the integral of the outer part goes to $0$ and the interior approaches $f(x)$. For the outer part, use the fact that the integral converges (i.e. the heat kernel decreases rapidly) and the fact that $f$ is rapidly decreasing (Schwartz). For the inner part, you can use some the continuity of $f$ to relate the value of $f$ in this region to $f(x)$. You will use the fact that the heat kernel integrates to 1 at this point somewhere.
answered Nov 21 at 22:55
whpowell96
5315
This is an easy consequence of the fact that $f'*g = f*g'$ for any $f$ and $g$, as well as the fact that the heat kernel is $C^{infty}$ for $t>0$.
This can be shown with the above fact also, namely we have that $$(partial_t-Delta)u = f*((partial_t-Deltamathcal){H}) = f*0=0$$ since the heat kernel is a solution to the homogenous heat equation.
This is pretty tough to prove and the difficulty varies depending on space $f$ lies in. I will try to sketch a proof. You are trying to show that $lim_{tto0}u(x,t) = f(x)$. The way this typically goes is you divide up the region into an $varepsilon$-ball centered at $x$ and its compliment then show that as $tto 0$, the integral of the outer part goes to $0$ and the interior approaches $f(x)$. For the outer part, use the fact that the integral converges (i.e. the heat kernel decreases rapidly) and the fact that $f$ is rapidly decreasing (Schwartz). For the inner part, you can use some the continuity of $f$ to relate the value of $f$ in this region to $f(x)$. You will use the fact that the heat kernel integrates to 1 at this point somewhere.
answered Nov 21 at 22:55
whpowell96
5315
answered Nov 21 at 22:55
whpowell96
5315
answered Nov 21 at 22:55
whpowell96
5315
answered Nov 21 at 22:55
whpowell96
5315
5315
add a comment |
add a comment |
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(1) is standard theory about convolution; nothing to prove there.
– Federico
Nov 21 at 18:03
(2) have you done some computation of the partial derivatives? Again, some facts about the convolution might help
– Federico
Nov 21 at 18:04
(3) ever seen convolutions with a rescaled kernel? They approximate the identity...
– Federico
Nov 21 at 18:05
I would suggest using vector calculus computations to simplify things..it’ll make your life easier
– DaveNine
Nov 21 at 19:49