MHB Friedrich's Problem 3a: Asymptotic Relation and Exact Solution in ODEs

[Alone]

I have a question regarding problem 3a on page 9 in Omalley's Singular Perturbation Methods for ODEs, regarding what he called "Friedrich's problem".
I am not sure how did they get the asymptotic relation: $x(t,\epsilon) \sim (\exp(1-t)-\exp(1-t/\epsilon))$ as $\epsilon \to 0$ uniformly in $t \in [0,1]$.
I get that the solution is
\begin{gather}\nonumber x(t,\epsilon)=(1/(\exp((-1+\sqrt{1-4\epsilon})/(2\epsilon))-\exp((-1-\sqrt{1-4\epsilon})/(2\epsilon)))\\ \nonumber (\exp(((-1+\sqrt{1-4\epsilon})/(2\epsilon))t)-\exp(((-1-\sqrt{1-4\epsilon})/(2\epsilon))t)).
\end{gather}
Now if I am not mistaken I need to show that: $\lim_{\epsilon \to 0, \\ t \in [0,1]} x(t,\epsilon)/(e^{1-t}-e^{1-t/\epsilon})=1$
The part of exponents with $-\sqrt{1-4\epsilon}$ vanishes when $\epsilon \to 0$, and $e^{-t/\epsilon}\to 0$ as $\epsilon \to 0$, but other than that I don't see how to show that the limit approaches 1.
Obviously there's l'HOPITAL there, but I don't see how many times should I use l'HOPITAL?

For those who don't have the book I'll iterate the problem:
3a.
Consider the two-point problem:
$$\epsilon \ddot{x}+\dot{x}+x=0 , t\in [0,1], $$
$$x(0)=0, x(1)=1$$
Determine the exact solution and show that:
$x(t,\epsilon)\sim e^{1-t}-e^{1-t/\epsilon}$ as $\epsilon \to 0$ uniformly in $t\in [0,1]$.

Any pointers on how to compute the limit?

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[Carla1985]

I have a question regarding problem 3a on page 9 in Omalley's Singular Perturbation Methods for ODEs, regarding what he called "Friedrich's problem".
I am not sure how did they get the asymptotic relation: $x(t,\epsilon) \sim (\exp(1-t)-\exp(1-t/\epsilon))$ as $\epsilon \to 0$ uniformly in $t \in [0,1]$.
I get that the solution is
\begin{gather}\nonumber x(t,\epsilon)=(1/(\exp((-1+\sqrt{1-4\epsilon})/(2\epsilon))-\exp((-1-\sqrt{1-4\epsilon})/(2\epsilon)))\\ \nonumber (\exp(((-1+\sqrt{1-4\epsilon})/(2\epsilon))t)-\exp(((-1-\sqrt{1-4\epsilon})/(2\epsilon))t)).
\end{gather}
Now if I am not mistaken I need to show that: $\lim_{\epsilon \to 0, \\ t \in [0,1]} x(t,\epsilon)/(e^{1-t}-e^{1-t/\epsilon})=1$
The part of exponents with $-\sqrt{1-4\epsilon}$ vanishes when $\epsilon \to 0$, and $e^{-t/\epsilon}\to 0$ as $\epsilon \to 0$, but other than that I don't see how to show that the limit approaches 1.
Obviously there's l'HOPITAL there, but I don't see how many times should I use l'HOPITAL?

For those who don't have the book I'll iterate the problem:
3a.
Consider the two-point problem:
$$\epsilon \ddot{x}+\dot{x}+x=0 , t\in [0,1], $$
$$x(0)=0, x(1)=1$$
Determine the exact solution and show that:
$x(t,\epsilon)\sim e^{1-t}-e^{1-t/\epsilon}$ as $\epsilon \to 0$ uniformly in $t\in [0,1]$.

Any pointers on how to compute the limit?

I think when they say 'determine the exact solution' they mean the composite asymptotic solution. I will give you a brief outline of what is required. As $\epsilon$ is multiplying the highest order term we have a singular perturbation problem so have multiple layers.

For the outer layer no rescaling is required. Substitute in the approximation $x(t)\approx x_0+\epsilon x_1+...$. This gives a leading order problem as
$$x_0'+x_0=0$$
Solving this, with the initial condition of the RHS of the interval gives the solution $x_0(t)=\exp(1-t)$.

Now for the inner layer. Rescale by setting $\tau=t/\epsilon^\alpha$ and substitute in asymptotic expansions (we now have $X(\tau)$ rather than $x(t)$) and use the chain rule. This gives
$$\epsilon^{1-2\alpha}(X_0+...)''+\epsilon^{-\alpha}(X_0+...)'+(X_0+...)=0$$

Balancing the coefficients tells us that $\alpha=1$, so we have a leading order problem
$$X_0''+X_0'=0$$
Solving this with the LHS initial condition gives the solution $X_0(\tau)=A(1-\exp(-\tau))$.

Now we need to match these. The matching condition says that
$$\lim_{t\rightarrow 0} x_0(t)=\lim_{\tau\rightarrow\infty} X_0(\tau)$$
which we calculate to give $A=\exp(1)$ so the inner solution is $X_0(\tau)=\exp(1)-\exp(1-\tau)$.

Together we get the composite solution
$$x\approx x_0(t)+X_0(t/\epsilon)-x_0(0)=\exp(1-t)+\exp(1)-\exp(1-t/\epsilon)-exp(1)=\exp(1-t)-\exp(1-t/\epsilon)$$
as required.

I have skipped over a good bit of the workings here so you will still need to learn the intricacies of singular perturbation methods. The book 'Introduction to the foundations of applied mathematics' by Mark Holmes has similar examples and explains it very well also. I actually just used the book to learn this methods myself. Hope that helps.

Carla

 

[Alone]

Well in the end I solved it with the exact solution, I had mistake in my previous calculations, but the asymptotic checks true.

And you can solve this question with the exact solution of the ODE.

 

[Alone]

Here's the solution if someone wants to check it.
Notice that $\lim_{\epsilon \to 0^+ , t \in [0,1]} \exp(-t/\epsilon)=0$
You just need to expand $\sqrt{1-4\epsilon}=1-2\epsilon+\mathcal{O}(\epsilon^2)$, and then calculate the limit:
$x(t,\epsilon)$ becomes after letting $\epsilon \to 0^+$
\begin{gather}(1/(\exp(-1)))(\exp(-t)).
\end{gather}
which is the same as: $\exp(1-t)$.