Differential equation with a trigonometric coefficient for an Ultrasonic Horn

Question

I'm trying to solve a differential equation for the modeling of an ultrasonic horn in wich his form is a catenary, and his differential equation for a wave involves a hyperbolic tangent. I have the solution for the differential equation, but I need to do the "steps by steps". I had transformed the trigonometric coefficient into a polynomial but I keep having problems with the series in the development of the solution (I'm using series for solving this) so if someone can give me a hand, because I don't know if I'm doing a good approach to the equation, I would be very grateful. Here is the equation: $$\ v''+\frac{2k}{r}tanh(\frac{kx}{r})v'+\frac{w^2}{c^2}v=0$$

and here is transformed into a polynomial

$$ \frac{k^2}{r^2}(1+u)^2 v''+\frac{w^2}{c^2}v=0 $$

thanks in advance and I wait for your answers if someone knows. ($k, r, w$ and c are constants)

Answer 1

Starting from the second form:

Divide the whole equation by the constant coefficient of $v^{\prime\prime}$ to get $(1+u)^2v^{\prime\prime} + \left( \frac{rw}{kc} \right)^2v = (u^2+2u+1)v^{\prime\prime} + \omega^2v = 0$. Let $v = \sum_{i=0}^\infty a_iu^i$, and differentiation term-by-term gives $v^{\prime\prime} = \sum_{i=2}^\infty i(i-1)a_iu^{i-2}$. Substituting into the equation gives $\sum_{i=2}^\infty i(i-1)a_iu^i + \sum_{i=2}^\infty 2i(i-1)a_iu^{i-1} + \sum_{i=2}^\infty i(i-1)a_iu^{i-2} + \sum_{i=0}^\infty \omega^2a_iu^i = 0$. We want to shift the sums so that they all have same power of $u$ in them, which will let us factor it out of the sum. So, we have $\sum_{i=2}^\infty i(i-1)a_iu^i + \sum_{i=1}^\infty 2i(i+1)a_{i+1}u^{i} + \sum_{i=0}^\infty (i+1)(i+2)a_{i+2}u^{i} + \sum_{i=0}^\infty \omega^2a_iu^i = \sum_{i=2}^\infty i(i-1)a_iu^i + \sum_{i=2}^\infty 2i(i+1)a_{i+1}u^{i} + \sum_{i=0}^\infty (i+1)(i+2)a_{i+2}u^{i} + \sum_{i=2}^\infty \omega^2a_iu^i + (6a_3 + 4a_2+\omega^2a_1)u + (\omega^2a_0+2a_2) = (6a_3 + 4a_2+\omega^2a_1)u + (\omega^2a_0+2a_2) + \sum_{i=2}^\infty u^i[\left(i(i-1) + \omega^2\right)a_i + 2i(i+1)a_{i+1} + (i+1)(i+2)a_{i+2} ] = 0$.

From here, you get that each coefficient of $u^i$ must be $0$, which gives you a recurrence relation which you can use to get the terms $a_i$ given some constants $a_0$ and $a_1$.

Answer 2

Define $y(x)$ such that $v(x) = y(x)/\cosh(kx/r)$. (EDIT: see below for why you might want to do this.) Then we have $$ v'(x) = \frac{1}{\cosh (kx/r)} y'(x) - \frac{k}{r} \frac{\sinh (kx/r)}{\cosh^2 (kx/r)} y(x) = \frac{1}{\cosh (kx/r)} \left[ y'(x) - \frac{k}{r} \tanh \left( \frac{kx}{r} \right) y(x)\right], $$ and (after a fair amount of algebra), $$ v''(x) = \frac{1}{\cosh (kx/r)} \left[ y''(x) - \frac{2k}{r} \tanh \left( \frac{kx}{r} \right) y'(x) + \frac{k^2}{r^2} \left( 1 - 2 \,\mathrm{sech}^2 (kx/r) \right) y(x)\right]. $$ Thus, $$ v''(x) + \frac{2k}{r} \tanh \left(\frac{2k}{r} \right) v'(x) = \frac{1}{\cosh(kx/r)} \left[ y''(x) + \frac{k^2}{r^2} \left( 1 - 2 \,\mathrm{sech}^2 (kx/r) - 2 \tanh^2 (kx/r) \right) y(x)\right] \\ = \frac{1}{\cosh(kx/r)} \left[ y''(x) - \frac{k^2}{r^2} y(x)\right], $$ since $\tanh^2(x) + \mathrm{sech}^2(x) = 1$. Plugging this into the original ODE we have $$ v''+\frac{2k}{r}\tanh\left(\frac{kx}{r}\right)v'+\frac{w^2}{c^2}v = \frac{1}{\cosh(kx/r)} \left[ y''(x) + \left( \frac{w^2}{c^2} - \frac{k^2}{r^2} \right) y(x) \right] = 0. $$ Since $\cosh(kx/r) \neq 0$, we can conclude that $$ y''(x) + \left( \frac{w^2}{c^2} - \frac{k^2}{r^2} \right) y(x) = 0, $$ which is an ODE that you can probably solve in your sleep.

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In general, if you have an ODE of the form $$ y'' + p(x) y' + q(x) y = 0, $$ you can always make a substitution of the form $y(x) = f(x) z(x)$ to rewrite it in terms of another function $z$. After some algebra you get $$ z'' + \left( p' + 2\frac{f'}{f} \right) z' + \left( \frac{f''}{f} + p \frac{f'}{f} + q \right) z = 0. $$ A reasonable guess for $f(x)$ would be that function that causes the coefficient of $z'$ to vanish, i.e., $$ \frac{2f'}{f} = - p(x) \quad \rightarrow \quad f = \exp \left[ - \frac{1}{2} \int p(x) dx \right]. $$ Note that if we plug in $p(x) = \frac{2k}{r} \tanh (2kx/r)$, we get $f(x) = 1/\cosh(kx/r)$. So that's nice.

In general, the remaining term proportional to $z$ will still be present in the transformed ODE, and will probably be some really ugly function of $x$. We lucked out in this particular equation in that the quantity $$ \frac{f''}{f} + \frac{f'}{f} p + q $$ reduced down to a constant. So this trick won't always work; still, it's probably worth a shot if you have no idea how to solve a second-order ODE.

Note that this trick is similar to the use of integrating factors for first-order ODEs; in that case, it's guaranteed to work because you don't have another term in the ODE to worry about.

Answer 3

Appropriate changes of function and variable leads to a second order linear ODE easy to solve :

If $\lambda>1$ change the exponential to sinusoidal functions.