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).



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linear-algebra heat-equation laplacian




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    up vote
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    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).



    enter image description here




    linear-algebra heat-equation laplacian




    share|cite|improve this question





















      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).



      enter image description here




      linear-algebra heat-equation laplacian




      share|cite|improve this question











      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).



      enter image description here





      linear-algebra heat-equation laplacian





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      asked Jul 30 at 6:04









      learner

      10613




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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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            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

























              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























                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













                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













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                answered Jul 30 at 6:23









                Kusma

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