General solution of heat equation?

[lriuui0x0]

We know

$$
K(x,t) = \frac{1}{\sqrt{4\pi t}}\exp(-\frac{x^2}{4t})
$$

is a solution to the heat equation:

$$
\frac{\partial u}{\partial t} = \frac{\partial^2 u}{\partial x^2}
$$

I would like to ask how to prove:

$$
u(x,t) = \int_{-\infty}^{\infty} K(x-y,t)f(y)dy
$$

is also the solution to the equation, and also:

$$
\lim_{t\to 0^+} u(x,t) = f(x)
$$

 

[Delta2]

The first part is a lot easier than you might think, you just use the fact that K satisfies the heat equation, that is that $$\frac{\partial K}{\partial t}=\frac{\partial^2K}{\partial x^2}$$ and you go ahead to prove that $u(x,t)$ as it is defined also satisfies the heat equation by working on both sides of the heat equation for this specific $u$.
All you use is that in the expression for$u$, because we integrate with respect to$y$which is independent variable with respect to $x$and$t$, the partial derivative with respect to t (or x) passes under the integral sign, that is for example $$\frac{\partial u}{\partial t}=\frac{\partial \int_{-\infty}^{\infty}K(x-y,t)f(y)dy}{\partial t}=\int_{-\infty}^{\infty}\frac{\partial K(x-y,t)}{\partial t} f(y) dy$$ and similar for the first and second partial derivative with respect to x.

 

[wrobel]

the downstairs limit of the integral must be zero. Actually the whole question is contained in textbooks in PDE

 

[lriuui0x0]

Thanks! I found some reference on this. Basically the integral is a solution follows from the limiting case of linear combination of $K(x-y,t)$, and the limit follows from the limit of Gaussian function is delta function.