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

• random_soldier
Sorry.In summary, the function d2f/dx2 + cf = delta(x) has a solution of f = C1exp(cx) + C2exp(-cx) with the condition that f is finite and f(50) = f(-50) = 0. The solution can be further simplified by setting C2 = 0 and specifying a discontinuity at x = 0 for the delta function. However, the proposed solution does not have the appropriate discontinuous derivative at x=0 and may not accurately represent the physical situation being described. Further clarification of the problem is needed.

#### random_soldier

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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.

random_soldier said:
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##.

random_soldier said:
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.

Orodruin said:
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.

Orodruin said:
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?

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

random_soldier said:
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.

I am sorry, but you are not making any sense whatsoever. You need to specify the problem you are actually wanting to solve mathematically. It is not at all clear what physical situation you are trying to describe. You need to give the exact problem statement as required by the homework template (which you deleted).

Nevermind. I'll post it in the other forum, then. It's not for math but I was thought math would suit it better.