Asymptotic form of the function doesn't satisfy original differential equation

Question

Modified Bessel function of the first kind $I_0$ is defined as a solution of the differential equation $$ \frac{d^2y}{dx^2}+\frac{1}{x}\frac{dy}{dx}-y=0. $$ For large arguments it has the asymptotic form $$I_0(x)\rightarrow \frac{e^x}{ \sqrt{2\pi x}}.$$ Substituting it back into the original equation we get $$\sqrt{2}e^{x}\frac{1}{8\sqrt{\pi}x^{5/2}}=0,$$ which is clearly nonsence. What am I missing ?

In the comments I was told that what is important is the ratio of the computed LHS and original approximation. Could you please tell why is that so and provide some reference? Is there some general way to know if some function is asymptotic approximation to the solution given the computed LHS ?

Answer 1

Modified Bessel function of the first kind $I_0$ is defined as a solution of the differential equation $(1)$. $$ \frac{d^2 I_0(x) }{dx^2}+\frac{1}{x}\frac{dI_0(x)}{dx}-I_0(x)=0. \tag 1$$ For large arguments it has the asymptotic form : $$I_0(x)\rightarrow f(x)=\frac{e^x}{ \sqrt{2\pi x}} \tag 2$$ $f(x)$ is not solution of Eq.$(1)$ because $f(x)\neq I_0(x)$

So, it should be nonsence to expect that substituting it back into the original equation would get $0.$

In fact, $f(x)$ is solution of the equation : $$ \frac{d^2f(x)}{dx^2}+\frac{1}{x}\frac{df(x)}{dx}-\left(1+\frac{1}{4x^2}\right)f(x)=0. \tag 3$$ In comparing Eq.$(3)$ to Eq.$(1)$, we see that they are close one from the other.

Of course, $\frac{1}{4x^2}$ is small compared to $1$.

But $\frac{f(x)}{4x^2}=\frac{e^x}{ \sqrt{2\pi}\:x^{5/2}} $ is not small due to the exponential.

Nevertheless, $\frac{f(x)}{4x^2}$ is small compared to $f(x)$.

This means that the relative deviation is small, while the absolute deviation is large.

This is strengthen by the asymptotic series expansion : $$I_0(x)\sim\frac{e^x}{ \sqrt{2\pi x}}\left(1+\frac{1}{8x^2} +\frac{9}{128x^4}+...\right)$$ "Asymptotic series" doesn't mean that the absolute deviation tend to $0$. It means that the relative deviation tends to $0$.

If we compare the graphs of $I_0(x)$ and $f(x)=\frac{e^x}{ \sqrt{2\pi x}}$ in a range of large $x$, they cannot be distinguished one from the other because, by necessity, the scale of graphing is very large. The absolute difference between both is not visible at large scale, even with a large absolute difference.

Answer 2

Using the known form of the asymptote, consider $$ u =\ln(e^{-x}\sqrt xy(x))= -x +\frac12\ln x +\ln y(x) $$ Then $$ u'(x)=-1+\frac1{2x}+\frac{y'(x)}{y(x)} $$ and \begin{align} u''(x)&=-\frac1{2x^2}+\frac{y''(x)}{y(x)}-\frac{y'(x)^2}{y(x)^2}\\ &=-\frac1{2x^2}+\frac{y-\frac{y'}{x}}{y}-\left(u'+1-\frac1{2x}\right)^2\\ &=-\frac1{2x^2}+1-\frac{u'}{x}-\frac1x+\frac1{2x^2}-u'^2-1-\frac1{4x^2}-2u'+\frac{u'}{x}+\frac1x\\ &=-u'(x)^2-\frac1{4x^2} \end{align} For large $x$ the last term is small, now try to argue that $e^u$ has to be considered as relatively constant.