Bounded Solution of the Heat PDE: Is u Necessarily the Heat Kernel?

[yetar]

Lets say we have a solution u, to the cauchy problem of the heat PDE:
u_t-laplacian(u) = 0
u(x, 0) = f(x)

u is a bounded solution, meaning:
u<=C*e^(a*|x|^2)
Where C and a are constant.

Then, does u is necesseraly the following solution:

u = integral of (K(x, y, t)*f(y))

Where K is the heat kernel?

Thanks in advance.

 

[Clausius2]

Lets say we have a solution u, to the cauchy problem of the heat PDE:
u_t-laplacian(u) = 0
u(x, 0) = f(x)

u is a bounded solution, meaning:
u<=C*e^(a*|x|^2)
Where C and a are constant.

Then, does u is necesseraly the following solution:

u = integral of (K(x, y, t)*f(y))

Where K is the heat kernel?

Thanks in advance.

Use the Green's Formula (integral form of the Lagrange's Identity) for the Heat Operator, and realize that u can be represented as your integral plus a term involving the initial condition.

 

[ehrenfest]

I am confused about the usage of the word "kernel" in this context. Where does the term "heat kernel" come from? Is kernel being used in the sense of "all elements that get mapped to 0" by some function or in some other sense? Another PDE example is the "Poisson kernel". I've also seen the "Dirichlet kernel" in analysis. Is there a relation between these "kernels"?

 

[jostpuur]

I am confused about the usage of the word "kernel" in this context. Where does the term "heat kernel" come from? Is kernel being used in the sense of "all elements that get mapped to 0" by some function or in some other sense?

In other sense. It is common to call functions, that are used in convolutions, kernels. For example if you have an operator

$$T:L^{\infty}([0,1],\mathbb{C})\to L^{\infty}([0,1],\mathbb{C}),\quad (Tf)(x) = \int\limits_0^1 dy\; f(y)K(x,y),$$

where

$$K:[0,1]\times [0,1]\to\mathbb{C}$$

is some bounded function, then $K$ is called the kernel of this integral operator.