# IVP prob involving convolutions

1. Jan 31, 2005

### fourier jr

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 $$F"(x) = \delta (x)$$ where $$\delta (x)$$ 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 $$F(x) = \frac{1}{2}|x|$$. assuming the function $$f \in L^1 (\mathbb{R})$$ has compact support then the integral $$\frac{1}{2} \int_{-\infty}^{\infty} |x-y|f(y) dy$$ 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 $$u(0) = u_0$$ and $$u'(0) = u'_0$$ where $$f \in C^{\infty}([0,\infty))$$ and $$f = \bigcirc (|x|^{-(2+\epsilon)})$$ as $$|x| \rightarrow \infty$$

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