Heat kernel derivation from heat equation

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While reading this paper, I couldn't understand how they derived heat kernel $H_t$ from the heat equation,

$$ fracdeltau_tdeltat = - mathcalLu_t $$

When I take the integration, I can derive the equation of heat $u(t)$ at every node but I am unable to get the heat kernel. I would be very thankful if someone could give an overview explanation or some pointers as to how they got from equation (1) to (3).

linear-algebra heat-equation laplacian

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asked Jul 30 at 6:04

learner

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up vote
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favorite

While reading this paper, I couldn't understand how they derived heat kernel $H_t$ from the heat equation,

$$ fracdeltau_tdeltat = - mathcalLu_t $$

When I take the integration, I can derive the equation of heat $u(t)$ at every node but I am unable to get the heat kernel. I would be very thankful if someone could give an overview explanation or some pointers as to how they got from equation (1) to (3).

linear-algebra heat-equation laplacian

share|cite|improve this question

asked Jul 30 at 6:04

learner

10613

add a comment | 

up vote
0
down vote

favorite

up vote
0
down vote

favorite

While reading this paper, I couldn't understand how they derived heat kernel $H_t$ from the heat equation,

$$ fracdeltau_tdeltat = - mathcalLu_t $$

When I take the integration, I can derive the equation of heat $u(t)$ at every node but I am unable to get the heat kernel. I would be very thankful if someone could give an overview explanation or some pointers as to how they got from equation (1) to (3).

linear-algebra heat-equation laplacian

share|cite|improve this question

asked Jul 30 at 6:04

learner

10613

While reading this paper, I couldn't understand how they derived heat kernel $H_t$ from the heat equation,

$$ fracdeltau_tdeltat = - mathcalLu_t $$

When I take the integration, I can derive the equation of heat $u(t)$ at every node but I am unable to get the heat kernel. I would be very thankful if someone could give an overview explanation or some pointers as to how they got from equation (1) to (3).

linear-algebra heat-equation laplacian

share|cite|improve this question

asked Jul 30 at 6:04

learner

10613

share|cite|improve this question

share|cite|improve this question

asked Jul 30 at 6:04

learner

10613

asked Jul 30 at 6:04

learner

10613

asked Jul 30 at 6:04

learner

10613

10613

add a comment | 

add a comment | 

1 Answer
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Write $u_t=sum a_j(t) phi_j$. Then the PDE becomes (using $phi_j$ are eigenfunctions)
$$sum a_j'(t) phi_j = -sum lambda_j a_j(t) phi_j.$$ As $phi_j$ are a basis, $a_j'(t)=-lambda_j a_j$ and so solving the ODE gives $a_j(t)=e^-lambda_ja_j(0)$. With $a_j(0)=phi_j^T u_0$ you arrive at the expression (2).

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answered Jul 30 at 6:23

Kusma

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1 Answer
1

active

oldest

votes

1 Answer
1

active

oldest

votes

active

oldest

votes

active

oldest

votes

up vote
0
down vote



accepted

Write $u_t=sum a_j(t) phi_j$. Then the PDE becomes (using $phi_j$ are eigenfunctions)
$$sum a_j'(t) phi_j = -sum lambda_j a_j(t) phi_j.$$ As $phi_j$ are a basis, $a_j'(t)=-lambda_j a_j$ and so solving the ODE gives $a_j(t)=e^-lambda_ja_j(0)$. With $a_j(0)=phi_j^T u_0$ you arrive at the expression (2).

share|cite|improve this answer

answered Jul 30 at 6:23

Kusma

1,027111

add a comment | 

up vote
0
down vote



accepted

Write $u_t=sum a_j(t) phi_j$. Then the PDE becomes (using $phi_j$ are eigenfunctions)
$$sum a_j'(t) phi_j = -sum lambda_j a_j(t) phi_j.$$ As $phi_j$ are a basis, $a_j'(t)=-lambda_j a_j$ and so solving the ODE gives $a_j(t)=e^-lambda_ja_j(0)$. With $a_j(0)=phi_j^T u_0$ you arrive at the expression (2).

share|cite|improve this answer

answered Jul 30 at 6:23

Kusma

1,027111

add a comment | 

up vote
0
down vote



accepted

up vote
0
down vote



accepted

Write $u_t=sum a_j(t) phi_j$. Then the PDE becomes (using $phi_j$ are eigenfunctions)
$$sum a_j'(t) phi_j = -sum lambda_j a_j(t) phi_j.$$ As $phi_j$ are a basis, $a_j'(t)=-lambda_j a_j$ and so solving the ODE gives $a_j(t)=e^-lambda_ja_j(0)$. With $a_j(0)=phi_j^T u_0$ you arrive at the expression (2).

share|cite|improve this answer

answered Jul 30 at 6:23

Kusma

1,027111

Write $u_t=sum a_j(t) phi_j$. Then the PDE becomes (using $phi_j$ are eigenfunctions)
$$sum a_j'(t) phi_j = -sum lambda_j a_j(t) phi_j.$$ As $phi_j$ are a basis, $a_j'(t)=-lambda_j a_j$ and so solving the ODE gives $a_j(t)=e^-lambda_ja_j(0)$. With $a_j(0)=phi_j^T u_0$ you arrive at the expression (2).

share|cite|improve this answer

answered Jul 30 at 6:23

Kusma

1,027111

share|cite|improve this answer

share|cite|improve this answer

answered Jul 30 at 6:23

Kusma

1,027111

answered Jul 30 at 6:23

Kusma

1,027111

answered Jul 30 at 6:23

Kusma

1,027111

1,027111

add a comment | 

add a comment | 

 

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