Delta function

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
[tex]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.[/tex]

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

What about the second derivative. Is it equal to [tex]\delta (x-1) [/tex] or zero? How do we discretize a delta function?
 
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"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.
 
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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.
 
"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.
https://www.physicsforums.com/showthread.php?t=309469

As you pointed out, we need to recover the previous derivative. How is this possible? As I see it (for unit impulse)
[tex]\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.[/tex]

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

[tex] \ddot{U}_{s+1,i} = U''_{s,i} [/tex]

I need the initial u"(x,0) to proceed.
 
[tex]f''(x) = -2 \delta(x-1) [/tex]. Discrete delta is a box with area equal to 1, with a width that depends on the size of your step.
 

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