- #1

Wledig

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$$p(x)\dfrac{dG}{dx} \Big|_{t-\epsilon}^{t+\epsilon} +\int_{t-\epsilon}^{t+\epsilon} q(x) \;G(x,t) dx = 1$$

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- Thread starter Wledig
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- #1

Wledig

- 69

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$$p(x)\dfrac{dG}{dx} \Big|_{t-\epsilon}^{t+\epsilon} +\int_{t-\epsilon}^{t+\epsilon} q(x) \;G(x,t) dx = 1$$

- #2

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They can, but not if it is to hold as ##\epsilon \to 0##.

- #3

Wledig

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- #4

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No, that would not help. The integral would still go to zero.Couldn't the same be accomplished by letting G have a discontinuity in t?

This is really not a strange thing. There are many examples of sich functions. For example ##f(x) = |x|##.I guess I just find a bit awkward to let G be continuous but not its derivative, not sure if this is a condition that should hold true.

Still, I can relate to your worries and I also never liked this argument although it does its job in the end. My preferred argument, and the one I use in my book, is to write ##G## as a piecewise function ##G(x,x’) = \theta(x-x’)g_1(x) + \theta(x’-x)g_2(x)## and insert this into the differential equation using that ##\theta’## is a delta function. You will need some delta function properties, but once the smoke clears you will obtain exactly the same result.

- #5

Wledig

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You're right, I completely forgot this classic example.This is really not a strange thing. There are many examples of sich functions. For example f(x)=|x|f(x)=|x|f(x) = |x|.

That's actually a much more natural way of thinking about the problem, since the derivative of the step function is Dirac's delta, we're just left with the boundary conditions. I just wonder how to reach the relation below from it though. $$ \lim_{\epsilon \rightarrow 0} \left( \dfrac{dG}{dx} \Big|_{x=t+\epsilon} - \dfrac{dG}{dx} \Big|_{x=t-\epsilon} \right) = \dfrac{1}{p(t)} $$Still, I can relate to your worries and I also never liked this argument although it does its job in the end. My preferred argument, and the one I use in my book, is to write GGG as a piecewise function G(x,x′)=θ(x−x′)g1(x)+θ(x′−x)g2(x)G(x,x′)=θ(x−x′)g1(x)+θ(x′−x)g2(x)G(x,x’) = \theta(x-x’)g_1(x) + \theta(x’-x)g_2(x) and insert this into the differential equation using that θ′θ′\theta’ is a delta function. You will need some delta function properties, but once the smoke clears you will obtain exactly the same result.

- #6

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What do you get when you do the insertion into your differential equation?

- #7

Wledig

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- #8

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It is impossible to know that unless you actually show us what you got.Am I following your line of thought correctly?

- #9

Wledig

- 69

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$$ p[(g''_1 \theta(x-x') + 2g'_1 \delta(x-x') + g_1 \delta(x-x')') + (g''_2 \theta(x'-x) + 2g'_2 \delta(x'-x) + g_2 \delta(x'-x)')] + p'(g'_1 \theta(x-x') + g'_2\theta(x'-x) + g_1 \delta(x-x') + g_2 \delta(x'-x)) + q(g_1 \theta(x-x') + g_2\theta(x'-x)) = \delta(x-x')$$

Now, I looked it up that the derivative of the delta function is ##\frac{\delta}{x}## so I could substitute it there, but it doesn't seem to lead anywhere. I also thought about isolating ##\frac{dG}{dx}## and inserting in the other terms to come up with something, but that didn't seem to work either.

- #10

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