Lagerstrom-Cole equation Consider this boundary value problem
$$\epsilon u''+uu'-u=0,\quad u(0)=A\in\mathbb{R},\quad u(1)=3.$$
This differential equation is known as Lagerstrom-Cole equation. I trying to construct asymptotic approximation of this BVP using matched asymptotic expansion. Numerical solution suggests that there is boundary layer near $x=0$. Using naive expansion I obtained the outer solution
$$u_0(x)=x+2.$$ This outer solution is consistent with the numerical solution. For the inner region I introduced $\hat{x}=x/\epsilon$ and
$$u(\hat{x})\approx u_0(\hat{x})+\epsilon u_1(\hat{x})+\epsilon^2u_2(\hat{x})+\dots$$ Substituting above expansion into differential equation I ended up to
$$u_0''+u_0u_0'=0.$$ The general solution for this differential equation is
$$u_0(\hat{x})=\sqrt{2C_1}\tanh{\left[(\hat{x}+C_2)\sqrt{C_1/2}\right]}.$$ Unfortunately this function cannot satisfy the left boundary condition. At this point I'm stuck.



 A: Your leading order outer solution is $$ u_{0,outer}(x)=x+2, $$
and your leading order inner solution can be written as $$ u_{0,inner}(\hat x)=C\tanh\left(\frac{C\hat x+D}{2}\right). $$
Using the boundary condition at $\hat x=0$ gives $$A=C\tanh\left(D/2\right),$$
or, $C=A/\tanh(D/2)$. So, we can write
$$ u_{0,inner}(\hat x)=A\frac{\tanh\left(\frac{A\hat x}{2\tanh\left(\frac{D}{2}\right)}+\frac{D}{2}\right)}{\tanh\left(\frac{D}{2}\right)}. $$
Now, to perform the asymptotic matching to find $D$, you have to match the limit as you leave the boundary of the inner solution to the limit of the outer solution as you enter the boundary layer, or,
$$ \lim_{x\rightarrow0}u_{0,outer}(x)=\lim_{\hat x\rightarrow\infty}u_{0,inner}(\hat x).$$
Taking both limits gives
$$2=\frac{A}{\tanh\left(\frac{D}{2}\right)},$$
and so $D=2\tanh^{-1}(A/2)$.
Substituting back into the inner solution gives
$$u_{0,inner}(\hat x)=2\tanh\left(\hat x+\tanh^{-1}\left(\frac{A}{2}\right)\right).$$
To get a uniformly valid approximation throughout the whole domain, do
$$ u_{0,unif}(x)=u_{0,outer}(x)+u_{0,inner}(x/\epsilon)-u_{0,outer}(0)=x+2\tanh\left(\frac{x}{\epsilon}+\tanh^{-1}\left(\frac{A}{2}\right)\right). $$
Some Matlab code can be used to check the results, for $\epsilon=0.1$ and $A=3$, for example,

And the Matlab code:
epsilon=.1;
A=3;
X=0:.001:1;
dydx=@(x,y) [y(2);y(1)*(1-y(2))/epsilon];
yin=@(x) 2*tanh(x/epsilon+atanh(A/2));
yout=@(x) x+2;
sol=@(a) ode45(dydx,X,[A a]);
f=@(T) T.y(1,end);
d=fzero(@(a) f(sol(a))-3,1);
[X,Y]=sol(d);
plot(X,yout(X),X,yin(X),X,Y(:,1),X,yout(X)+yin(X)-yout(0))
legend('Outer solution','Inner solution','Numerical solution','Uniform approximation','Location','SouthEast')

