0
$\begingroup$

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

$$ \frac{\delta{u}_t}{\delta{t}} = - \mathcal{L}u_t $$

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

$\endgroup$
0
$\begingroup$

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 $a_j(t)=e^{-\lambda_j}a_j(0)$. With $a_j(0)=\phi_j^T u_0$ you arrive at the expression (2).

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.