# Trying to solve a 2nd order diffy-Q with delta function

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My function:

d2f/dx2 + cf = delta(x)

Condition: f is finite and f(50) = f(-50) = 0

Solution: f = C1exp(cx) + C2exp(-cx)

Due to condition,

f = C1exp(cx) for x<=0 and C2exp(-cx) for x>=0

f(50) = C2exp(-c*50) = 0 = > C2 = 0

Likewise, for C1

I don't know if I might have missed something. Also not sure if I dealt with the delta function correctly. I assumed RHS as zero for the purpose of solving the equation.

Orodruin
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Due to condition,

f = C1exp(cx) for x<=0 and C2exp(-cx) for x>=0
This is wrong. What is your argument for ignoring ##C_2## for ##x \leq 0##? Note that your domain is just ##-50 < x < 50##, not ##-\infty < x < \infty##.

Orodruin
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Also not sure if I dealt with the delta function correctly. I assumed RHS as zero for the purpose of solving the equation.
I would say you have not dealt with it at all. Your solution is now identically zero, which of course means that its derivative is zero. In order for a derivative to be proportional to a delta function, the function needs to have a discontinuity at that point. Thus, in order for a second derivative to be proportional to a delta function, its derivative needs to have a discontinuity at that point.

This is wrong. What is your argument for ignoring ##C_2## for ##x \leq 0##? Note that your domain is just ##-50 < x < 50##, not ##-\infty < x < \infty##.

Sorry! Should have specified, delta(x) is a point particle source. f measures the current. By the way if you are wondering why it looks so strange, I was hoping to simplify it. Sorry if it didn't work out as intended.

I would say you have not dealt with it at all. Your solution is now identically zero, which of course means that its derivative is zero. In order for a derivative to be proportional to a delta function, the function needs to have a discontinuity at that point. Thus, in order for a second derivative to be proportional to a delta function, its derivative needs to have a discontinuity at that point.

Again, sorry. It was in relation to the delta being a source. I assumed it likely would be better off as zero due to that. Wouldn't I just specify the discontinuity at exactly x = 0 in this case and leave the rest be?

Orodruin
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I am fully aware of what a delta function (distribution actually) is. What the interpretation of f is is irrelevant for the mathematical problem.

Wouldn't I just specify the discontinuity at exactly x = 0 in this case and leave the rest be?
Yes, but your proposed solition does not have the appropriate discontinuous derivative at x=0, so it must be wrong.

I suggest you reread my first post and reply to its actual content.

Okay, it doesn't need to be bound if the domain is only -50 to 50. That makes sense mathematically.

What if it's charge distribution from a charge source and an insulating material lies at those points? It would decay, wouldn't it? I think I should at least know the physical implications.

Orodruin
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