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Consider the initial value problem $\sqrt{\epsilon} \, u'' + u' - u = e^{2t}$ , with $u(0)=1$, and $u'(0)=1/\sqrt{\epsilon}$.

I am trying to use a matched asymptotic expansion to find the leading term of the solution.

First by setting $\epsilon=0$, and using the initial condition $u(0)=1$, we get the outer solution $u_0(t)=c_1 e^t + e^{2t}$.

I set the change of variables $\tau=t/\sqrt{\epsilon}$.

Writing the original equation in terms of the new variable I got:

$U'' + U' -\sqrt{\epsilon} \, U = \sqrt{\epsilon} \, e^{2\tau \epsilon^{1/2}}$

with $U(0)=0$ ; $U'(0)=1$.

By setting $\epsilon = 0 $, we get the boundary layer equation : $U'' + U' = 0$ with same initial conditions.

The solution of this equation is then $U(\tau)= 2-e^{-\tau}$.

Finally matching terms:

$\displaystyle \lim_{\tau \to \infty} U(\tau)=\displaystyle \lim_{t \to 0} u_0(t)$

So that $2=c_1+1$ then $c_1=1$.

Therefore the solution will be $u(t)=U(t/\sqrt{\epsilon}) +u_0(t) - 2= - e^{-t/\sqrt{\epsilon}} + e^{2t} + e^t $, which corresponds to adding the two solutions found earlier and subtracting the common term.

Is my work correct? If so, how come my solution satisfies the first initial condition but not the second?

Any suggestions will be greatly appreciate it. Thanks!

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As the approximation is to the first order in $\sqrtϵ$, you can also only expect to satisfy the initial conditions to first order. And indeed for the combined solution $y'(0)=\frac1{\sqrtϵ}+3=\frac1{\sqrtϵ}(1+3\sqrtϵ)$ has only a first order (relative) error.

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  • $\begingroup$ Thank you for your answer LutzL. I just realized there is a typo in the initial condition for $U$. It should read $U(0)=1$ not $U(0)=0$. In that case I think $U(\tau)=2-e^{-\tau}$ would be correct. $\endgroup$ – user569959 Nov 23 '18 at 17:05

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