Singular perturbations. II


### Singular perturbations. II

#### Boundary layer type solution. 2

Consider now the same $Lu=\alpha(x)u$ but $Mu=\beta(x) u'+\gamma(x)u$ where real valued functions $\alpha$, $\beta$ do not vanish. Then we need just one boundary condition: \begin{align} &L_\varepsilon u:= (L+\varepsilon M)u=f, && 0< x< a \label{eq-4.3.1}\\ &u(0)=g. \label{eq-4.3.2} \end{align} As before we can assume that $f=0$ (we can achieve it ignoring boundary condition (\ref{eq-4.3.1}) and looking as in the regular perturbation theory). Then we can try $u=we^{-\phi(x)/\varepsilon}$ and looking for coefficient at $\varepsilon^0$ we get $$\phi'=\alpha(x)/\beta(x), \qquad \phi(0)=0 \label{eq-4.3.3}$$ Since we need $\phi' >0$ we should assume that $$\alpha(x)/\beta(x)>0. \label{eq-4.3.4}$$

**Remark 1.. If $$\alpha(x)/\beta(x)< 0. \label{eq-4.3.5}$$ we should instead of (\ref{eq-4.3.2}) set the condition on the other end $u(a)=g$ and then $\phi(a)=0$.

The rest is the same as in Section 4.2: we look for $w\sim\sum_{n\ge 0}w_n\varepsilon^n$ satisfying $$Pw_n+ Qw_{n-1}=0 \label{eq-4.3.6}$$ with \begin{align} &Pw:= (\beta w'+\gamma w) \label{eq-4.3.7}\\ &Qw=\alpha w \label{eq-4.3.8} \end{align}

#### Boundary layer type solution. 3

Consider now $Lu=\alpha(x)u'+\gamma(x)u$ but $Mu=(\beta(x) u')'$ where real valued functions $\alpha$, $\beta$ do not vanish. There could be also lower order terms in $M$ but they do not change the scheme: \begin{align} &L_\varepsilon u:= (L+\varepsilon M)u=f, && 0< x< a \label{eq-4.3.9}\\ &u(0)=g_1, &&u(a)=g_2. \label{eq-4.3.10} \end{align} Again we can assume that $f=0$. Again we $u=we^{-\phi(x)/\varepsilon}$ and get equation (\ref{eq-4.3.3}) since $\phi'\ne 0$. Under condition (\ref{eq-4.3.3}) we take $\phi(0)=0$ and under condition (\ref{eq-4.3.5}) we take $\phi(a)=0$ which means that as $\varepsilon=0$ should be left for operator $L$ only condition $u(0)=g_1$ or $u(a)=g_2$ respectively.

Finally we get $$Pw_n +Qw_{n-1}=0, \qquad Pw=\alpha w'+\gamma, \label{eq-4.3.11}$$ with the boundary condition $w_n(0)=g_{1n}$ under assumption (\ref{eq-4.3.3}) and $w_n(a)=g_{2n}$ under assumption (\ref{eq-4.3.5}).

Therefore boundary layer type solution exists only near one end. For another end regular perturbation theory takes care. Indeed we take $U\sim \sum_{n\ge 0}U_n \varepsilon ^n$ where $U_n$ are found from \begin{align} &LU_n+Mu_{n-1}=f_n \label{eq-4.3.12}\\ &U_n(b)=g_{*n} \label{eq-4.3.13} \end{align} where $b$ is the other end.