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Let say we have to solve a given differential equation $$ty''+y'+ty=0$$ $$y(0)=1,\ y'(0)=0$$ (which is Bessel equation with the solution $y=J_0 (t)$, of course) with the Laplace transform. Then we get $$Y(s)=\mathcal{L}(y)(s)=\frac{K}{\sqrt{1+s^2}}$$ Now to get the y, we perform the inverse Laplace transform with Mellin's inverse formula. $$y(t)={K\over 2\pi i}\lim_{T\to\infty}\int_{\gamma-iT}^{\gamma+iT}{e^{st}\over\sqrt{1+s^2}}ds$$ with arbitrary positive number $\gamma$. To decide K, we use the initial condition $y(0)=1$. $$y(0)={K\over 2\pi i}\lim_{T\to\infty}\int_{\gamma-iT}^{\gamma+iT}{1\over\sqrt{1+s^2}}ds$$ In this stage we can use the residue theorem but I want to avoid that because of personal reason. But since $$\int_{\gamma-iT}^{\gamma+iT}{1\over\sqrt{1+s^2}}ds=\ln(\gamma+iT+\sqrt{1+(\gamma+iT)^2})-\ln(\gamma-iT+\sqrt{1+(\gamma-iT)^2})$$, and from $$(\gamma+iT+\sqrt{1+(\gamma+iT)^2})^*=\gamma-iT+\sqrt{1+(\gamma-iT)^2}$$ and $$\ln(z)-\ln(z^*)=2i\arg(z)$$, we get $$1=y(0)={K\over \pi }\lim_{T\to\infty}\arg(\gamma+iT+\sqrt{1+(\gamma+iT)^2})$$ $$={K\over \pi}{\pi\over 2}={K\over 2}$$ So we arrive at $$K=2$$ But since we know that $$\mathcal{L}(J_0(t))(s)={1\over \sqrt{1+s^2}}$$ so $K$ must be 1. This seems to be contradictory. What is the reason of this?

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Have a look in, for example, Apostol's Mathematical Analysis. The edition I looked at was old (5th, 1981), it was exercise 11.39. Here is the formal statement of the inverstion formula for Laplace transforms.

Let $c>0$ be a positive number such that $\int_0^\infty |f(t)|e^{-c t}\,dt$ exists, and define $F(s)=\int_0^\infty e^{-s t}f(t)\,dt$ for $\Re s>c$. Let $t$ be a point and $[t-\delta,t+\delta]$ an interval around it, such that (as in the Fourier integral theorem)

  • $f$ is of bounded variation on $[t-\delta, t+\delta]$,
  • or else both limits $f(t\pm)$ exist and both integrals $$ \int_0^\delta \frac{f(t\pm u)-f(t\pm)}{u}\,du $$ exist.

Then for every $a>c$ we have $$ \frac{f(t+)+f(t-)}{2} = \frac{1}{2\pi} \lim_{T\to+\infty} \int_{-T}^{T} e^{(a+i v)t} F(a+i v)\,dv. $$ (This is copied from Apostol.)

In your case, the formula you use is incomplete, because for the purposes of proving the Laplace transform inversion formula, we say that $f(t)=0$ for all $t<0$. So at $t=0$, $J_0(t+)=1$, but $J_0(t-)=0$, which gives $$ \frac{J_0(0+)+J_0(0-)}{2} = \frac12, $$ which is exactly what you are seeing in your calculation.

The formula given on wikipedia is only correct for $t>0$ and continuous $f(t)$.

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  • $\begingroup$ Oh, thank you sir... This problem had bugged me for 5 months and now it's relieved. $\endgroup$ – generic properties Oct 19 '14 at 9:32

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