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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)?


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Your question is quite a mess. First, please tell us typically the diffusion PDE. Second, please use TEX when typing the equations. – doraemonpaul Jan 15 '13 at 11:29
up vote 0 down vote accepted

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.

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@5PM..Thanks for the answer. Could you then tell me, how one implements such a Dirac Delta initial condition on 1D Diffusion PDE (u_sub{t}=D*u_sub{xx}, D:diffusion constant, _sub{x} refers to partial derivative taken wrt x)? Is it implemented only for x=0 (assuming it is δ(0,x)) or rather implemented as an integral(δ(0,x)) fashion? I would like to hear if it can be implementated in x=0 fashion, rather than in integral fashion. Hope its clear. – ems Jan 15 '13 at 22:53
@ems Are you familiar with the concept of a fundamental solution? – user53153 Jan 15 '13 at 22:55
@5PM...Yes. Fundamental solution helps finding the solution for any arbitrary initial condition, provided I have the solution calculated using Dirac Delta initial condition, right? I am trying to find the solution to 1D PDE not by analytical means (which I can find in any textbook), but rather using Differential Transform method (numerical approach, basis function based) – ems Jan 15 '13 at 22:58
@ems For numerical computations you need a nascent delta function, such as the piecewise linear $\eta_\epsilon(x)=\epsilon^{-1} \max(0,1-|x|/\epsilon)$, for small $\epsilon>0$. How small you can take it to be will depend on the particulars of your methods. – user53153 Jan 15 '13 at 23:11

This answer is incomplete reference to the Brownian motion "stocastic Calculus" and the same initial conditions we get the normal distribution

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Welcome to MSE! Was this supposed to be a comment. I realize you don't yet have enough reputation, but this area is for answers. Regards – Amzoti May 3 '13 at 5:29

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