fundamental solution for the heat equation in a rod, $0It is known that the function, called fundamental solution, $K(x,t)=\frac{e^{-\frac{x^2}{4t}}}{\sqrt{4 \pi t}}$ satisfies the equation $u_t=u_{xx}$. I want to find the solution of the problem 
D.E: $u_t=u_{xx}$, $0<x<1$, $0<t$,
BCs: $u(0,t)=u(1,t)=0$, $0<t$,
IC: $u(x,0)=f(x)$, $0<x<1$.
By using seperation of variables I can find the solution easily. But I want to find the solution of the problem in terms of the fundamental solution, $K(x,t)$. In the book where I saw the problem the solution given as 
$u(x,t)=\int_0^1\big\{\sum_{m=-\infty}^{m=\infty}[K(x-(\xi+2m),t)-K(x+(\xi-2m),t)]\big\}f(\eta)d\eta$,
and gives a hint as
Hint: Extend $f$ as an odd function to $-1<x<0$. Then extend $f$ as periodic with period $2$ to $-\infty<x<\infty$ and solve the initial-value problem for the periodic extension.
My Question: Could anyone please help me solve this problem or give me some hints? Please, I couldn't find the odd periodic extension of the funtion $f$.   
 A: it will help if we have a specific $f$ that is defined on $0 < x < 1.$ so let us take the simplest. that is $f(x) = 1, 0 < x < 1.$ we will extend $f$ to $-1 < x < 0$ and we will do so in way that makes the new function odd. only way to do that is to define $f(-x) = -f(x) = - 1, 0 < x < 1.$ now we have $$f(x) = \begin{cases} 1 & \text{ if  0 < x < 1}\\-1 &\text{ if $-1 < x < 0$}  \end{cases}$$
you can extend this function to $1 < x < 3$ by defining $f(x) = f(x - 2), 1 < x < 3.$ and to $(0, \infty)$ by defining $f(x) = f(x - 2k)$ and $k$ is quotient when $x$ is divided by $2$ so that $0< x - 2k< 1.$and extend $f$ to $(-\infty, 0)$ by defining $f(-x) = -f(x).$
now we have a function still called $f$ that is defined on $(-\infty, \infty),$ is periodic of period $2,$  and is an odd function. consider $u(x,t)$  defined by 
$$u(x,t) = \int_{ -\infty}^\infty f(\xi)K(x-\xi, t)\, d \xi.\tag 1$$ verify that $u_t = u_{xx}$ and $u(x,0) = f(x).$
using periodicity of $f,$ we can write break up the integral 
$$\begin{align}u(x,t) &= \int_{ -\infty}^\infty f(\xi)K(x-\xi, t)\, d \xi \\
&= \cdots +\int_{-1}^1f(\xi)K(x-\xi, t)\, d \xi +\int_{1}^3 f(\xi)K(x-\xi, t)\, d \xi + \cdots\\
&=\cdots+\int_{-1}^1f(\xi)K(x-\xi, t)\, d \xi +\int_{0}^1 f(\xi)K(x+2-\xi, t)\, d \xi \\
&=\cdots+\int_{-1}^0f(\xi)K(x-\xi, t)\, d \xi +\int_{0}^1 f(\xi)K(x-\xi, t)\, d \xi \\
&+\int_{-1}^0f(\xi)K(x+2-\xi, t)\, d \xi +\int_{0}^1 f(\xi)K(x+2-\xi, t)\, d \xi \\
&=\cdots+\int_{0}^1 \left(f(\xi) -f(-\xi) \right)K(x-\xi, t)\, d \xi \\& +\int_{0}^1 \left(f(\xi) - f(-\xi)\right)K(x+2-\xi, t)\, d \xi \\
&=\int_0^1  \left(f(\xi) - f(-\xi)\right)\left\{ \cdots+K(x-\xi, t)+K(x+2-\xi, t) +\cdots \right\} d \xi  
\end{align}$$
