# Finding a composite solution to an ODE (boundary layer problem)

Given $\epsilon \frac{d^2u}{dt^2}-a(t)\frac{du}{dt}+b(t)u=0$, where $a(t)>0$, $u(0)=1$, $u(1)=1$, and assuming that the boundary layer is at $t=1$, and the boundary layer variable is $T=\frac{1-t}{e^\nu}$, how does one find the composite solution to this ODE? I'm completely lost because there's just virtually no theory on this in the course notes, and the example shown is very difficult to extrapolate from. Unfortunately, I haven't also had much success in finding this on the internet. I don't even know which books would be useful to consult for this sort of problems.

Here's my attempt:

• Outer solution:

$u(t)=u_{out}(t)+O(\epsilon)$, $\epsilon \frac {d^2u_{out}}{dt^2}-a(t)\frac{du_{out}}{dt}+b(t)u_{out}=0$, $a(t)\frac{du_{out}}{dt}-b(t)u_{out}=0$.

$\implies \int\frac{du_{out}}{u_{out}}=\int\frac{b(t)}{a(t)}dt \implies u_{out}=Ae^{\int\frac{b(t)}{a(t)}dt}$. $A=e^{-\int\frac{b(1)}{a(1)}dt}\implies u_{out}=e^{\int(\frac{b(t)}{a(t)}-\frac{b(1)}{a(1)})dt}$.

• Inner solution:

Substitute $T$ into the equation and get:

$\epsilon^{\nu^2+1}u''+a(T)\epsilon^\nu u'+b(T)u=0$, where $\nu^2+1=\nu$, and this equation has complex roots.

How do we go from here? I'd appreciate some hints.

For your outer solution, don't scale the independent variable, but write $u$ as an asymptotic series in powers of $\epsilon$: $$u_{outer}=u_0+\epsilon u_1+\epsilon^2 u_2+O(\epsilon^3)$$

Substitute this into the original equation, $$\epsilon u_0''+\epsilon^2 u_1''-au_0'-a\epsilon u_1'-a\epsilon^2 u_2'+bu_0+\epsilon bu_1+\epsilon^2 bu_2+O(\epsilon^3)=0.$$

You can pull out equations corresponding to each power of $\epsilon$, for $O(1)$, $$-au_0'+bu_0=0,$$ at $O(\epsilon)$, $$u_0''-au_1'+bu_1=0,$$ and so on.

You have a boundary layer at $t=1$, so the outer solution will satisfy the $t=0$ boundary condition, and so $u_0+\epsilon u_1+\epsilon^2 u_2+O(\epsilon^3)=1$ which means $u_0(0)=1$ and $u_i(0)=0$ for all $i>0$.

You can homogeneous equations with an integrating factor, and the use variation of parameters to obtain the full solution of the correction terms.

For the inner solution, you have to rescale $t$, like you did, as $t=1-\epsilon^\alpha T$, and write $U(T)=u(t)$, which gives $$\epsilon^{1-2\alpha}U''-\epsilon^{-\alpha}a(T)U'+b(T)U=0.$$ You have to use the method of dominant balance to determine what $\alpha$ should be. You want $\alpha>0$ so that you have actual scaling, and you need the two (or more) largest terms in the equation to balance. You get $1-2\alpha=-\alpha$ so $\alpha=1$, and the equation in the boundary layer is $$U''-a(T)U'+\epsilon b(T)U=0.$$ Again you can solve this easily enough (power series in $\epsilon$ for $U$ again).

You will have one constant left over from the inner solution, and you work it out by requiring $$\lim_{t\rightarrow1}u_0(t)=\lim_{T\rightarrow\infty}U_0(T)$$ and solving for the constant.