Asymptotic solution of an equation containing an integral

[umby]

I am interested in finding $x$ which solves the following equation in the asymptotic limit of $a\to 0$ and $b\to 0$


$$\int_0^{\frac{\pi }{2}} \exp \left(-\frac{1}{2} \left(\frac{1}{2} \pi a^2 b^2 \left(\cos ^2(\theta )+\frac{1}{\cos ^2(\theta )}-2\right)+\frac{x^2 \cos ^2(\theta )}{\frac{1}{2} \pi a^2 b^2}+2 x \left(1-\cos ^2(\theta )\right)\right)\right) \, d\theta =\frac{\pi a}{2}$$
Any suggestion?

 

[blamocur]

I am interested in finding $x$ which solves the following equation in the asymptotic limit of $a\to 0$ and $b\to 0$


$$\int_0^{\frac{\pi }{2}} \exp \left(-\frac{1}{2} \left(\frac{1}{2} \pi a^2 b^2 \left(\cos ^2(\theta )+\frac{1}{\cos ^2(\theta )}-2\right)+\frac{x^2 \cos ^2(\theta )}{\frac{1}{2} \pi a^2 b^2}+2 x \left(1-\cos ^2(\theta )\right)\right)\right) \, d\theta =\frac{\pi a}{2}$$
Any suggestion?

Ouch!
The only approach which comes to mind is to write software to search for a solution.

 

[umby]

I try to reformulate more correctly the problem.

The equation $$\frac{2}{\pi a} \int_0^{\frac{\pi }{2}} \exp \left(-\frac{\left(\cos (\theta ) \left(\pi a^2 b^2-2 x\right)-\pi a^2 b^2 \sec (\theta )\right)^2}{4 \pi a^2 b^2}\right) \, d\theta=1$$defines, at least theoretically, $x$ as a function of $a$ and $b$, $x=f(a,b)$. All values are supposed to be real. Is it possible to find the asymptotic form of $x=f(a,b)$ in the limit of

1) $a\to 0$ and $b\to 0$;

2) $a\to 1$ and $b\to 0$.

I tried the followings for the first case ($a\to 0$ and $b\to 0$).

Substitute the integrand function with its Taylor series about $a= 0$ and $b=0$, thus the equation becomes:$$\frac{2}{\pi a} \int_0^{\frac{\pi }{2}} \exp \left(x \left(\cos ^2(\theta )-1\right)-\frac{x^2 \cos ^2(\theta )}{\pi a^2 b^2}\right) \, d\theta=1$$
Solve the integral using the modified Bessel function $I_0$

$$\frac{2}{\pi a} \frac{1}{2} \pi e^{-\frac{1}{2} x \left(\frac{x}{\pi a^2 b^2}+1\right)} I_0\left(\frac{1}{2} x \left(\frac{x}{a^2 b^2 \pi }-1\right)\right)=1$$
In the limit of $a\to 0$ and $b\to 0$

$$\frac{x}{\pi a^2 b^2}+1=\frac{x}{\pi a^2 b^2}-1\to \frac{x}{\pi a^2 b^2}$$ then last equation becomes

$$\frac{1}{a} e^{-\frac{x^2}{2 \left(\pi a^2 b^2\right)}} I_0\left(\frac{x^2}{2 \left(a^2 b^2 \pi \right)}\right)=1$$
which becomes using the asymptotic form for for $I_0$ for large real $z$

$$I_0(z)=\frac{\exp (z)}{\sqrt{2 \pi z}}$$
$$\frac{e^{-\frac{x^2}{2 \left(\pi a^2 b^2\right)}} e^{\frac{x^2}{2 \left(\pi a^2 b^2\right)}}}{a \sqrt{\frac{2 \pi x^2}{2 \left(\pi a^2 b^2\right)}}}=1$$
Simplifying:

$$\sqrt{\frac{x^2}{a^2 b^2}}=\frac{1}{a}$$
which has solution

$$x=\pm b$$
Is it correct this reasoning? Any help for the second case? Thanks in advance

 

[blamocur]

Substitute the integrand function with its Taylor series about a=0 and b=0

Taylor series for which variables? a,b,x or $\theta$ ? Also, are you replacing the whole integrand function or just the argument of the exponent?

 

[umby]

Taylor series for which variables? a,b,x or $\theta$ ? Also, are you replacing the whole integrand function or just the argument of the exponent?

Taylor series with respect to $a$ and $b$ about $0$, replacing the whole integrand function.

 

[blamocur]

The integrand in your formula can be represented as

$$f(a,b) = e^{-\left(pa^2b^2 + \frac{q}{a^2b^2}+r \right)}$$where $p,q,r$ depend on $\theta$ and $x$. But I can't recognize the Taylor expansion of $f$ in your subsequent formulae. The integrand in your post #3 looks slightly simpler, i.e., $f(a,b) = e^{-\left(\frac{q}{a^2b^2}+r \right)}$, but I still don't see a correct Taylor expansion either.

BTW, the expressions in your original post and post #3 don't look identical to me -- which one do you need?

Just curious: where do these formulae come from? Your post does not look like a typical homework assighment.

 

[umby]

The correct problem is the following one, probably I did some error using Latex.


The equation $$\frac{2}{\pi r} \int_0^{\frac{\pi }{2}} \exp \left(-\frac{1}{2} \left(\frac{1}{2} \pi a^2 b^2 \left(\cos ^2(\theta )+\frac{1}{\cos ^2(\theta )}-2\right)+\frac{x^2 \cos ^2(\theta )}{\frac{1}{2} \pi a^2 b^2}+2 x \left(1-\cos ^2(\theta )\right)\right)\right) \, d\theta=1$$
defines, at least theoretically, $x$ as a function of $a$ and $b$, $x=f(a,b)$. All values are supposed to be real. Is it possible to find the asymptotic form of $x=f(a,b)$ in the limit of


1) $a\to 0$ and $b\to 0$;


2) $a\to 1$ and $b\to 0$.


I tried the followings for the first case ($a\to 0$ and $b\to 0$).


Substitute the integrand function with its Taylor series about $a= 0$ and $b=0$,

$$\exp \left(-\frac{1}{2} \left(\frac{1}{2} \pi a^2 b^2 \left(\cos ^2(\theta )+\frac{1}{\cos ^2(\theta )}-2\right)+\frac{x^2 \cos ^2(\theta )}{\frac{1}{2} \pi a^2 b^2}+2 x \left(1-\cos ^2(\theta )\right)\right)\right) = \exp \left(\frac{-\frac{x^2 \cos ^2(\theta )}{\pi b^2}+O\left(b^2\right)}{a^2}+x \left(\cos ^2(\theta )-1\right)+O\left(a^2\right)\right)$$
thus the equation becomes:$$\frac{2}{\pi a} \int_0^{\frac{\pi }{2}} \exp \left(x \left(\cos ^2(\theta )-1\right)-\frac{x^2 \cos ^2(\theta )}{\pi a^2 b^2}\right) \, d\theta=1$$

Solve the integral using the modified Bessel function $I_0$


$$\frac{2}{\pi a} \frac{1}{2} \pi e^{-\frac{1}{2} x \left(\frac{x}{\pi a^2 b^2}+1\right)} I_0\left(\frac{1}{2} x \left(\frac{x}{a^2 b^2 \pi }-1\right)\right)=1$$

In the limit of $a\to 0$ and $b\to 0$


$$\frac{x}{\pi a^2 b^2}+1=\frac{x}{\pi a^2 b^2}-1\to \frac{x}{\pi a^2 b^2}$$ then last equation becomes


$$\frac{1}{a} e^{-\frac{x^2}{2 \left(\pi a^2 b^2\right)}} I_0\left(\frac{x^2}{2 \left(a^2 b^2 \pi \right)}\right)=1$$

which becomes using the asymptotic form for for $I_0$ for large real $z$


$$I_0(z)=\frac{\exp (z)}{\sqrt{2 \pi z}}$$

$$\frac{e^{-\frac{x^2}{2 \left(\pi a^2 b^2\right)}} e^{\frac{x^2}{2 \left(\pi a^2 b^2\right)}}}{a \sqrt{\frac{2 \pi x^2}{2 \left(\pi a^2 b^2\right)}}}=1$$
and so on, like in the previous post...

The integrand in your formula can be represented as

$$f(a,b) = e^{-\left(pa^2b^2 + \frac{q}{a^2b^2}+r \right)}$$where $p,q,r$ depend on $\theta$ and $x$. But I can't recognize the Taylor expansion of $f$ in your subsequent formulae. The integrand in your post #3 looks slightly simpler, i.e., $f(a,b) = e^{-\left(\frac{q}{a^2b^2}+r \right)}$, but I still don't see a correct Taylor expansion either.

BTW, the expressions in your original post and post #3 don't look identical to me -- which one do you need?

Just curious: where do these formulae come from? Your post does not look like a typical homework assighment.

Thanks for your interest. The correct formula is the first one, in the original post, see below. It comes from a heat transfer problem where a gaussian heat source is applied to the surface of a semi-infinite medium. The solution should represent the intercept of a generic isotherm with the x axis. I will try to check the Taylor expansion.

 

[umby]

The integrand in your formula can be represented as

$$f(a,b) = e^{-\left(pa^2b^2 + \frac{q}{a^2b^2}+r \right)}$$where $p,q,r$ depend on $\theta$ and $x$. But I can't recognize the Taylor expansion of $f$ in your subsequent formulae. The integrand in your post #3 looks slightly simpler, i.e., $f(a,b) = e^{-\left(\frac{q}{a^2b^2}+r \right)}$, but I still don't see a correct Taylor expansion either.

BTW, the expressions in your original post and post #3 don't look identical to me -- which one do you need?

Just curious: where do these formulae come from? Your post does not look like a typical homework assighment.

You, are right, the Taylor expansion is wrong. Further, I cannot do the series expansion of the integrand function because it is not defined for $a=0$ and $b=0$.