# Analytic solution of this, Advice.

• Clausius2
In summary, the equation has an asymptotic behavior of G\sim As^{\sqrt{2}} and G(s)\sim se^{-s}} for small and large s respectively. The equation is a Riccatti equation, and finding a closed form solution seems to be a difficult task.
Clausius2
Gold Member
I am looking for the analytic solution of this ODE (if it were one):

$$s^2G''+sG'-(1+s^2+s\text{coth}(s))G=-4s^2e^{-s}$$

I have solved this equation numerically, it only gives one physically realizable configuration rejecting conveniently one of the homogenous solutions. I don't have those solutions BUT I have the asymptotic behavior of $$G(s)$$, which turns out to be $$G\sim As^{\sqrt{2}}$$ and $$G(s)\sim se^{-s}}$$ for small and large $$s$$ respectively, where A is a coefficient that I have worked out by means of a linear shooting.

When writting it on Maple in OdeAdvisor, it says to me that it is a Linear ODE (easy thing to know) and with Linear Symmetries. Does this last thing have something to do with Lie Symmetries?. May this ODE be solvable employing that theory, I don't have a clue.

Thanks.

$$u^{\prime \prime} +\lambda u^{\prime} +\lambda^{\prime}u=\gamma e^{\int \eta ds}$$

and make the substitution

$$u = G(s) e^{\int \eta ds}$$

This will give us the equation

$$G^{\prime \prime}(s) + (2 \eta + \lambda) G^{\prime}(s) + (\eta^{\prime} + \eta^2 + \lambda \eta + \lambda^{\prime}) G(s) = \gamma$$which if we set

$$\lambda = \frac{1}{s} - 2 \eta$$

and

$$\gamma = -4 e^{-s}$$

gives us

$$G^{\prime \prime}(s) + \frac{G^{\prime}(s)}{s} + ( \frac{\eta}{s} - \eta^{\prime} - \eta^2 - \frac{1}{s^2}) G(s) = -4e^{-s}$$

which we can see is similar to your equation once we divide it by $$s^2$$.

$$G^{\prime \prime}(s) + \frac{G^{\prime}(s)}{s} - ( \frac{1}{s^2} + 1 + \frac{\coth{(s)}}{s}) G(s) = -4e^{-s}$$Now the next bit is to match the $$G(s)$$ term with your equation. So,

$$\frac{\eta}{s} - \eta^{\prime} - \eta^2 - \frac{1}{s^2} = -\frac{1}{s^2} -1 - \frac{\coth{(s)}}{s}$$

Hence we need to choose an $$\eta$$ that fits. If we try substituting

$$\eta = \coth{(s)} \pm \frac{\sqrt{2}}{s}$$

we get, with some fiddling,

$$\frac{2(1 - \sqrt{2})}{s} \coth{(s)} = \frac{2(1 - \sqrt{2})}{s^2}$$

or

$$\frac{2(1 + \sqrt{2})}{s} \coth{(s)} = \frac{2(1 + \sqrt{2})}{s^2}$$

So, for small s, where we can make the assumption that

$$\coth{(s)} \simeq \frac{1}{s}$$

we therefore have a solution.
The only thing left is to solve the very first equation.
By inspection we can see that it can be integrated directly to yield

$$u^{\prime} + \lambda u = \int \gamma e^{\int \eta ds} ds$$

which is straightforward to solve for u, and hence G.

For large, positive s we can use the same method, except in our choice of $$\eta$$. If we choose

$$\eta = \coth{(s)} - 2$$

and plug this into

$$\frac{\eta}{s} - \eta^{\prime} - \eta^2 - \frac{1}{s^2} = -\frac{1}{s^2} -1 - \frac{\coth{(s)}}{s}$$

and rearrange, we get

$$(\frac{2}{s} + 4) \coth{(s)} = \frac{2}{s} + 4$$

which as s tends to (positive) infinity, coth(s) tends to 1, and this becomes true.

You can then solve for u and G as above.

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will get back to this later...

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Mathew! THANKS A LOT MAN!. Keep on posting, I will keep track of every line. At first sight it seems you don't find a closed form though...it's an asymptotic behavior isn't it?.

Hey man. I re-did all the steps, and it pretty much works. The only problem is that finding an exact solution of:

$$\eta^2+\eta'-\eta/s=1+coth(s)/s$$

seems a pretty difficult task. It's a Riccatti equation. It would be nice to have it for closing the problem. I will mess around with some books to look for a solution (Maple doesn't say anything as always), and also I am going to see how good is your asymptotic approximation compared with the one (coming from Frobenius at s=0 and approximating $$scoth(s)\sim s$$ for large s) that I had before. Where did you find this stuff?. I mean, you have the book where this stuff can be found?.

Thanks.

Sorry, no book - I've just been fiddling around with it, seeing what comes up.

Yeah - for the general case I can't find a solution (one that doesn't involve an approximation), but I'll keep looking.

## 1. What is an analytic solution?

An analytic solution is a mathematical expression or formula that gives the exact solution to a problem, without the need for numerical approximation or iteration. It is often derived from a set of equations and can be used to solve a wide range of problems in science and engineering.

## 2. How is an analytic solution different from a numerical solution?

An analytic solution provides an exact solution to a problem, while a numerical solution involves approximating the solution using methods such as iteration or interpolation. Analytic solutions are often preferred because they are more accurate and can be derived directly from the equations describing the problem.

## 3. What types of problems can be solved using analytic solutions?

Analytic solutions can be used to solve a variety of problems in mathematics, physics, chemistry, engineering, and other scientific fields. These can include differential equations, optimization problems, and other complex mathematical models.

## 4. What are the benefits of using analytic solutions?

Using analytic solutions can save time and resources in problem-solving, as they provide an exact solution without the need for time-consuming numerical calculations. They also provide a deeper understanding of the problem and can be used to validate the results obtained from numerical methods.

## 5. What advice can you give for finding an analytic solution to a problem?

First, make sure to carefully define the problem and understand the underlying equations. Then, try to simplify the problem by breaking it down into smaller, more manageable parts. Utilize mathematical tools such as algebra, calculus, and trigonometry to manipulate the equations and find a solution. It can also be helpful to consult with other experts or references in the field for guidance.

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