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Understanding the method of Green's function

  1. Feb 25, 2016 #1
    I'm trying to understand the derivation for methods of Greens functions for PDEs but I can't get my head around some parts. I'm starting to feel comfortable with the method itself but I want to understand why it works.

    The thing I have problem with is quite crucial and it is the following:

    I have the general problem ##-\Delta u=0## in the upper half plane and some boundary condition ##g(x)## on the ##x##-axis itself. Why is it enough to solve ##-\Delta u=0## for ##y>0## and ##u(x,o)=\delta(x)## and then use the convolution to get the solution for the original problem? I have the whole derivation in front of me but I can't get a intuitive feeling about what is going on with the delta function, how can one just replace the boundary condition with a Dirac's delta and still get a solution that works for an arbitrary ##g(x)##?
     
    Last edited: Feb 25, 2016
  2. jcsd
  3. Feb 25, 2016 #2

    Twigg

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    Gold Member

    Convolution is like superposition with time delay and/or spatial separation. For example, consider a damped spring. Suppose the spring loses all its energy and sticks after 10 seconds, but someone gives it a quick push of 5 newtons every 15 seconds. As you'd expect, the spring will oscillate and damp out in exactly the same way starting the moment it gets pushed until 10 seconds later, then it sits still for 5 seconds, and then repeats with the next push. The driving force in this case is a periodic delta function, ## F(t) = \sum 5\delta(t - 15n) ## for all positive integers n. As it turns out, in this special case, the solution repeats periodically too: ## x(t) = \sum Aexp(-\beta (t - 15n))cos(\omega (t-15n)) ##. The idea is that you're taking the spring's intrinsic response to an instantaneous outside impulse (aka Dirac delta spike) and integrating over all the spikes. That "intrinsic response" is the Greens function. The trick is that the driving force, which more generally is just an inhomogenous term in an autonomous linear differential equation, doesn't have to be spread out like in my example. Any function can be decomposed into an infinite progression of delta spikes, as in: ## f(x) = \int f(x') \delta(x - x') dx ##. Think of that integral as an infinite sum over delta spikes located at point x' with height f(x').

    For your particular example, think about what that scenario would look like. You're solving Laplace's equation, so imagine that u is a steady state temperature distribution. Solving ## u(x,0) = \delta(x) ## is like solving for the temperature distribution you would see if you only fixed the temperature at the point (0,0). By taking the convolution, you're taking those temperature distributions, centering them on a boundary point (x',0), scaling by the magnitude g(x'), and summing up over all the points on that boundary (i.e., integrating over x').
     
  4. Feb 26, 2016 #3
    Thank you! This helped a lot! Now I can actually understand what's going on here.
     
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