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The Question:

Given that $J_0(x)$ satisfies

$$x\frac{d^2J_0}{dx^2}+\frac{dJ_0}{dx}+xJ_0=0 \qquad J_0(0)=1 \qquad \frac{dJ_0}{dx}(0)=0$$

Show that the Laplace Transform $\bar{J_0}(p)$ of $J_0$ is given by

$$\bar{J_0}(p) = \frac{1}{\sqrt{1+p^2}}$$

My Attempt:

By applying Laplace Transform to both sides of the differential equation, and using JUST the boundary condition $J_0(0)=1$, I arrived at

$$(p^2+1) \frac{d\bar{J_0}}{dp} + p\bar{J_0}=0$$

Solving this gives

$$\bar{J_0}(p)=\frac{C}{\sqrt{1+p^2}}$$

However, I am unsure how to use the OTHER boundary condition, that $J'_0(0)=0$, to determine the constant $C$.

Any hints?

EDIT:

In the answer it says to "observe that $\dfrac{d\bar J_0}{dp}(\infty)=0$", but I have no idea where that comes from.

glowstonetrees
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1 Answers1

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I dont recall the name of the theorem but it says that

$$ \lim_{x \to \infty} f(x) = \lim_{s \to 0} s\mathcal{L}(f)(s) $$

Setting $f = J'$ then $$ \mathcal{L}(J') = s\mathcal{L}(J)(s)- \mathcal{L}(J)(0) = \frac{Cs}{\sqrt{1+s^2}} - 1 $$

The above unnamned theorem implies $C = 1$.

Edit

Found it, its called Final value theorem and there is a wiki page on it.

Olba12
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    $$\begin{align} \lim_{s\to 0}(sF(s))&=-\lim_{s\to 0},\int_0^\infty f(t)\frac{de^{-st}}{dt},dt\\ &=\lim_{s\to0}\left(-\left.\left(f(t)e^{-st}\right)\right|_{t=0}^{t\to\infty}+\int_0^\infty f'(t)e^{-st},dt\right)\\ &=f(0)+\int_0^\infty f'(t),dt\\ &=f(\infty) \end{align}$$ – Mark Viola Jun 02 '18 at 22:17