I General solution of heat equation?

lriuui0x0
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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:

$$
\lim_{t\to 0^+} u(x,t) = f(x)
$$
 
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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.
 
the downstairs limit of the integral must be zero. Actually the whole question is contained in textbooks in PDE
 
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.
 
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