IVP prob involving convolutions

[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 $$|x| \rightarrow \infty$$

 

[Nate810]

b) can someone please explain how to get the step-function (Heaviside) solution? The Heaviside function solution can be found by solving the ODE u" = f, with the initial conditions u(0) = u_0 and u'(0) = u'_0. To do this, we define the function H(x) = 0 for x<0 and H(x) = 1 for x>0, and then solve F"(x) = \delta (x) where \delta (x) is the Dirac delta function. We use c = -1/2 so that (H(x)+c)' = \delta(x), and then integrate to get the fundamental solution F(x) = \frac{1}{2}|x|. Then, we use the convolution property, 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, thus giving us the step-function solution.

 

[wais]

The initial value problem given is u" = f with u(0) = u_0 and u'(0) = u'_0, where f is a smooth function and f = \bigcirc (|x|^{-(2+\epsilon)}) as |x| \rightarrow \infty. To solve this problem, we can use the fundamental solution F(x) = \frac{1}{2}|x| as given in the example in the text.

First, we can use the convolution property to find the solution u(x) = F * f(x) = \int_{\mathbb{R}^n} F(x-y)f(y) dy. This gives us u(x) = \frac{1}{2} \int_{-\infty}^{\infty} |x-y|f(y) dy.

Next, we can use the initial conditions to find the constants u_0 and u'_0. Using the convolution property again, we have u_0 = u(0) = F * f(0) = \frac{1}{2} \int_{-\infty}^{\infty} |y|f(y) dy and u'_0 = u'(0) = F' * f(0) = \frac{1}{2} \int_{-\infty}^{\infty}f(y) dy.

Now, we need to choose an appropriate f that satisfies the given condition f = \bigcirc (|x|^{-(2+\epsilon)}) as |x| \rightarrow \infty. This means that f must approach 0 faster than |x|^{-(2+\epsilon)} as |x| \rightarrow \infty. One possible choice could be f(x) = \frac{1}{x^{2+\epsilon}}.

Plugging this into the solution u(x) = \frac{1}{2} \int_{-\infty}^{\infty} |x-y|f(y) dy, we can see that the integral converges and defines a solution of u" = f. Therefore, we have successfully solved the initial value problem using the fundamental solution and an appropriate choice of f.