Matching inner and outer solutions gives me zero

Question

I'm looking to find inner and outer solutions to this ODE and to do some matching $$\epsilon \ddot y + (1+t) \dot y = 0 \;\;\;\;\; y(0) = 0 \;\; \dot y (0) = \frac1\epsilon$$

So I understand I need to convert to a new time variable $\tau = \frac t\epsilon$ ($\dot y$ and $y'$ are $\frac{dy}{dt}$ and $\frac{dy}{d\tau}$ respectively), which I plug in to get

$$y''+(1+\epsilon \tau)y' =0 \;\;\;\;\; y(0) = 0 \;\; y' (0) = 1$$

So if I plug in $y^i=y_0^i + \epsilon y_1^i$ to the $\tau$ equation and solve I get $y^i = 1-e^{-\tau}+\epsilon(1-e^{-\tau}(1-\tau-\frac{\tau^2}2))$.

When I plug in $y^o=y_0^o + \epsilon y_1^o$ to the $t$ equation and solve ignoring the boundary conditions I just get a constant. So $y^o = c+\epsilon d$.

Now to apply Van Dyke matching rules. Express $y^i$ in terms of t to get $y^{io}=1-e^{-\epsilon t}+\epsilon(1-e^{-\epsilon t})$ plus higher order terms. $1-e^{-\epsilon t}$ goes to zero, so the whole thing is zero. $y^{oi}=y^{o}$ since it's constant, so equating $y^{oi}=y^{io}$ I just get zero.

Something seems wrong. Is zero really the best approximation available here? Where did I go wrong?

Answer 1

Your inner solution is in $τ$, it should read $$ y^i(τ) = 1-e^{-τ}+ϵ\Bigl(1-e^{-τ}\bigl(1-τ-\frac{τ^2}2\bigr)\Bigr). $$ Then the matching rules tell you to equate $y^i(\infty)$ with $y^o(0)$ which gives the constant function $y^o(t)=1+ϵ$.

As $e^{-τ}τ^2\le 4e^{-τ/2}$ the difference to the constant will be lower than second order in $ϵ$ for $τ>-2\lnϵ$ or $t>2ϵ|\lnϵ|$.

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(Correction mar30) After trying to validate that formula graphically and recomputing, you must have made some sign errors in the inner solution. The correct solution should read $$ y^i(τ)=1-e^{-τ}-ϵ\Bigl(1-e^{-τ}\bigl(1+τ+\tfrac12τ^2\bigr)\Bigr) $$ giving the constant for the outer solution as $y^i(∞)=1-ϵ\equiv y^o$. The numerical solution has an asymptote close to but below $1-ϵ+3ϵ^2$. (The next order boundary is above $1-ϵ+3ϵ^2-20ϵ^3$.)

For $ϵ=0.1$ the following diagram shows the numerical solution, the above first-order approximation and the bounds for the asymptote.

Answer 2

So far so good, I can see how you come up with the time scale change to the new variable $\tau$ which leads to the following differential equation (DE) :

$y'' + (1 + \epsilon \tau)y' = 0\quad y(0)=0\quad y'(0) = 1\quad$

which agrees with your DE.

I have not checked the rest to see if there are any errors though.