3
$\begingroup$

$y''+\dfrac{\varepsilon y'}{y^2} - y' = 0, \, y(-\infty)=1$ and $y(\infty) = \varepsilon >0$.

Above is an ODE that was in an asymptotic analysis exam. A part of the question asked us to solve it exactly on the domain $(-\infty, \infty)$, where I got into troubles. I wrote the equation as: $$\left(\dfrac{\varepsilon}{y}\right)' = y''-y'=(y'-y)'$$ So we have $\dfrac{\varepsilon}{y} - y'+y = C_0$ and then $\dfrac{ydy}{y^2-yC_0+\varepsilon} = dx$, which would give a closed formula for the general solution. But this is where the trouble is: the antiderivative of the LHS will be very different functions depending on the relationships between $\varepsilon$ and $C_0$; therefore, it would seem like we need to determine $C_0$ to begin with. But that is impossible because of the initial conditions not involving any $y'.$

How should I proceed from here?

$\endgroup$
4
$\begingroup$

The key hypothesis here (slightly hidden, but still) is that $y(x)$ must be defined for every $x$ in the real line, with prescribed limits when $x\to\pm\infty$.

Thus, the denominator $z^2-C_0z+\epsilon$ must be nonzero for every $z$ between $y(-\infty)$ and $y(+\infty)$ and go to zero at the boundaries $z\to y(-\infty)$ and $z\to y(+\infty)$.

One is given $y(-\infty)=1$ hence $C_0=1+\varepsilon$. Thus, $z^2-C_0z+\varepsilon=(z-1)(z-\varepsilon)$ and $$(1-\varepsilon)x=\int_{y(0)}^{y(x)}\frac{(1-\varepsilon)z}{z^2-C_0z+\varepsilon}dz=\int_{y(0)}^{y(x)}\left(\frac1{z-1}-\frac1{z-\varepsilon}\right)dz$$ For every $y(0)$ between $\varepsilon$ and $1$, this is solved by $$e^{(1-\varepsilon) x}\frac{y(0)-1}{y(0)-\varepsilon}=\frac{y(x)-1}{y(x)-\varepsilon}$$ that is, $$y(x)=\frac{\varepsilon(y(0)-1)e^{(1-\varepsilon) x}-(y(0)-\varepsilon)}{(y(0)-1)e^{(1-\varepsilon) x}-(y(0)-\varepsilon)}$$ For example, if $\varepsilon=\frac12$, one gets $$y(x)=\frac{(1-y(0))e^{x/2}+2y(0)-1}{2(1-y(0))e^{x/2}+2y(0)-1}$$ for every $y(0)$ in $(\frac12,1)$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.