Dirac Delta Function as Initial condition for 1D Diffusion PDE: ONE or TWO equations(conditions)? I have 1D diffusion (u(t,x)) PDE with Dirac Delta initial condition.
Question is regarding it's implementation: 
Dirac delta func is formally defined as an encapsulation of 2 conditions: 
1st condn: function takes value 1 at x=0, 
2nd condition:function takes value 0, if x not equal to 0). 
Thus, to solve the above PDE for initial condition u(t=0,x)=Dirac_delta(x), is it necessary to treat this dirac delta function initial condition (for that PDE) as ONE or TWO separate conditions (viz. after substituting for t=0 and x=0 giving u(t=0,x=0)=1 solution giving 1st equation for initial condition & substituting t=0 and x=x(i.e non-zero x) giving u(t=0,x)=0 giving 2nd equation for initial condition)?? 
This means, 1D diffusion PDE with Dirac Delta function (delta(x)) has 1 OR 2 initial conditions, according to for t=0, (x=0 & x not equal to 0)?
Thanks
 A: Your description of the Dirac delta ($\delta_0$) is incorrect. One of possible correct definitions is: $\delta_0$ is the derivative of the Heaviside function $h_0$, which is defined as $h_0(x)=1$ when $x\ge 0$; $h_0(x)=0$ when $x< 0$. The meaning of the derivative is not the classical one; instead, it is a stipulation that the certain integral identities hold: Fundamental theorem of Calculus and integration by parts. In particular, $\int_a^b \delta_0 =h_0(b)-h_0(a)=1$ whenever $a<0<b$.
Now back to counting initial conditions. If we have $u(0,x)=x^2$ for all $x\in\mathbb R$, is this one condition? Or maybe infinitely many, because there's a condition for every $x\in\mathbb R$? I guess we take the position that this is one condition: the distribution of matter represented by $u$ is given at time $t=0$, and that's all we have. 
Another example: $u(0,x)=h_0(x)$ for all $x\in\mathbb R$. This is also one initial condition, we just have another function here instead of $x^2$. Since the function $h_0$ is piecewise defined, the condition $u(0,x)=h_0(x)$ amounts to 
$$u(0,x)=0 \text{ when } x<0 \ \text{ and } \ u(0,x)=1 \text{ when } x\ge 0$$
But this split is just our way of writing $h_0$. It is of no consequence for the mathematical problem. 
Similarly, the initial condition $u(0,\cdot) =\delta_0$ is also one condition, which may appear as two (or three) depending on how $\delta_0$ is presented in an application-oriented text. 
A: This answer is incomplete reference to the Brownian motion "stocastic Calculus" and the same initial conditions we get the normal distribution
