# General solution of heat equation?

• I
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
Homework Helper
Gold Member
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