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IVP prob involving convolutions

  1. Jan 31, 2005 #1
    there's an example in the text that we're supposed to use to solve this problem. the example solves the ODE u" = f, & to find the fundamental solution F(x) we want to solve [tex]F"(x) = \delta (x)[/tex] where [tex]\delta (x)[/tex] is the Dirac delta function. the Heaviside function satisfies (H(x)+c)' = delta(x) for any c but for convenience use c = -1/2 & solve F(x) = 1/2 for x>0 or F(x) = -1/2 for x<0 & integrate to get the FUNDAMENTAL SOLUTION [tex]F(x) = \frac{1}{2}|x|[/tex]. assuming the function [tex]f \in L^1 (\mathbb{R})[/tex] has compact support then the integral [tex]\frac{1}{2} \int_{-\infty}^{\infty} |x-y|f(y) dy[/tex] converges and defines a solution of u" = f.

    now to the problem:
    a) use use the fundamental solution (above) to solve the initial value problem u" = f for x>0 with [tex]u(0) = u_0[/tex] and [tex]u'(0) = u'_0[/tex] where [tex]f \in C^{\infty}([0,\infty))[/tex] and [tex]f = \bigcirc (|x|^{-(2+\epsilon)}) [/tex] as [tex]|x| \rightarrow \infty[/tex]

    i've used the convolution property that [tex]F' * f(0) = F * f'(0) = u'(0)[/tex] on [tex]u(x) = F * f(x) = \int_{\mathbb{R}^n} F(x-y)f(y) dy[/tex] to get [tex]u_0 = u(0) = F * f(0) = \frac{1}{2} \int_{-\infty}^{\infty} |y|f(y) dy[/tex] & [tex]u'_0 = u'(0) = F' * f(0) = \frac{1}{2} \int_{-\infty}^{\infty}f(y) dy[/tex] (since x>0) but not sure how to pick an f so that [tex]f = \bigcirc (|x|^{-(2+\epsilon)}) [/tex] as [tex]|x| \rightarrow \infty[/tex]
     
  2. jcsd
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