# Delta function

#### matematikawan

I'm working on discretizing pde with boundary and initial conditions. The assumption of the method is that functions must be at least twice differentiable. I have an intial condition given by the following function
$$u(x,0)=f(x)=\left\{\begin{array}{cc}x,&\mbox{ } 0 \leq x <1\\2-x, & \mbox{ } 1 \leq x <2 \end{array}\right.$$

I can calculate the first derivative as
$$f'(x)=\left\{\begin{array}{cc}1,&\mbox{ } 0 \leq x <1\\-1, & \mbox{ } 1 \leq x <2 \end{array}\right.$$
Nevermind that the derivative at x=1 is not correct. We use discrete variable.

What about the second derivative. Is it equal to $$\delta (x-1)$$ or zero? How do we discretize a delta function?

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#### g_edgar

"Must be at least twice differentiable" ... this means the second derivative must be an actual function, not a generalized function like the delta function. But to answer your question, yes the second derivative is a delta function. It's whatever you have to integrate to recover the previous derivative.

#### defunc

Why do you differentiate the initial conditions? Simply use it as it is, which is continuous. No need for a delta functional.

Anyway, when a delta functional appears in the pde, you will have to solve it in some weak sense.

#### matematikawan

"Must be at least twice differentiable" ... this means the second derivative must be an actual function, not a generalized function like the delta function. But to answer your question, yes the second derivative is a delta function. It's whatever you have to integrate to recover the previous derivative.
It's clear that f"(x) is not zero. Is it a Dirac delta function or an impulse function ? A Dirac delta function would be difficult to write a program code because it involves infinity.

As you pointed out, we need to recover the previous derivative. How is this possible? As I see it (for unit impulse)
$$\int_0^x \delta(t-1) dt =\left\{\begin{array}{cc}0,&\mbox{ } 0 \leq x <1\\1, & \mbox{ } 1 \leq x <2 \end{array}\right.$$

I do not know how to do it for Dirac delta function.

#### matematikawan

Why do you differentiate the initial conditions? Simply use it as it is, which is continuous. No need for a delta functional.

Anyway, when a delta functional appears in the pde, you will have to solve it in some weak sense.
I'm trying to solve a simple wave equation utt=uxx. One method that I read from a paper is to use discretize iteration to approximate the solution (the paper claim the method works for Burger equation and Sine-Gordon equation)

$$\ddot{U}_{s+1,i} = U''_{s,i}$$

I need the initial u"(x,0) to proceed.

#### Arturgower

$$f''(x) = -2 \delta(x-1)$$. Discrete delta is a box with area equal to 1, with a width that depends on the size of your step.