$ \epsilon \frac{d^2 u}{d x^2}+ \frac{du}{dx} -a -2bx=0$ ; 0<x<1 where u(0)=0,and u(1)=1. Find the approximate solutio

Question

Consider the problem
\begin{equation} \tag{*} \label{equ:given eqn} \epsilon \frac{d^2 u}{d x^2}+ \frac{du}{dx} -a -2bx=0 ; \ \ \ 0<x<1 \end{equation} where \begin{equation} \tag{**} \label{equ:given_condition_1} u(0)=0, ~\text{ and }~ u(1)=1 \end{equation} Find the approximate solution

Step 1: Outer Solution

Consider $$ u(x) \sim u_0(x) + \epsilon u_1(x)+ ... $$ Then we obtain $$\epsilon (u''_0(x) + \epsilon u''_1(x)+ ...) + (u'_0(x) + \epsilon u'_1(x)+ ...)=a+2bx $$

The $O(1)$ equation is therefore $$ u'_0(x)=a+2bx \tag1 $$ Using the separation of variables we get the general solution \begin{equation} \tag2 \label{equ:general_outer_solution} u_0(x)= ax+bx^2+C \end{equation} Where $C$ is any arbitrary constant.

Looking at the solution, we have a dilemma because there is only one arbitrary constant but there are two boundary conditions. We have no idea which boundary condition, if any, we should require $u_0(x)$ to satisfy.

Step 2: Boundary Layer

Let us assume that there is a boundary layer at $x=0$, we introduce a boundary layer co-ordinate as \begin{equation} \bar{x}= \frac{x}{\epsilon^\alpha} \tag3 \end{equation} where $\alpha>0$. From the change of variables and the chain rule, we have that \begin{align} \frac{d}{dx}&=\frac{1}{\epsilon^\alpha} \frac{d}{d \bar{x}}\tag{4a} \\ \frac{d^2}{dx^2}&=\frac{1}{\epsilon^{2\alpha}} \frac{d^2}{d^2 \bar{x}}\tag{4b} \end{align} If we let $U(\bar{x})$ denote the solution of the problem when using the boundary layer co-ordinate, then \eqref{equ:given eqn} transforms to \begin{equation} \tag5 \label{equ:boundary_layer} \epsilon^{1-2\alpha} \frac{d^2U}{d\bar{x}^2}+\epsilon^{-\alpha} \frac{dU}{d\bar{x}}=a+2b {\epsilon}^\alpha \bar{x} \end{equation} Then \begin{equation}\tag6 U(0)=0 \end{equation} The boundary condition $x=0$ is included here because the boundary layer is at the left end of the interval.

The appropriate expansion for the boundary-layer solution is now \begin{equation}\tag7 U(\bar{x}) \sim U_0(\bar{x})+ \epsilon^\gamma U_1(\bar{x})+... \end{equation} where $\gamma >0$. In this expansion $\bar{x}$ is held fixed as $\epsilon$ goes to zero. Then \begin{equation}\tag8 \epsilon^{1-2\alpha} ( U''_0(\bar{x})+ ... )+ \epsilon^{-\alpha} ( U'_0(\bar{x}+ ...)= a+2b {\epsilon}^\alpha \bar{x} \end{equation} Now I want to find the value of $\alpha$ by balancing. By balancing First term with the second term I get $\alpha =1$ and by balancing first term with last term we get $\alpha=1/3$. But I am not sure.

Answer 1

I am puzzled as to why the problem asks for an "approximation" when it is very easy to find the exact solution! Because this is a linear equation it is not necessary to use perturbation methods.

The equation is $\epsilon \frac{d^2u}{dx^2}+ \frac{du}{dx}- a= 2bx $. The associated homogeneous equation is $\epsilon \frac{d^2y}{d^2x}+ \frac{du}{dx}- a$ which has characteristic equation $\epsilon r^2+ r- a= 0$, solutions $r= \frac{-1\pm\sqrt{1+ 4a\epsilon}}{2\epsilon}$.

We would then seek a solution to the entire equation of the form $u= Ax+ B$. $\frac{du}{dx}= A$ and $\frac{d^2u}{dx^2}= 0$ so the equation reduces to $A- a= Ax+ B$. So A= 0 and B= -a. $u= -a$ satisfies the equation so the general solution is $u(x)= e^{-x/2\epsilon}\left(C_1e^{\frac{\sqrt{1+ 4a\epsilon}}{2\epsilon}t}+ C_2e^{\frac{\sqrt{1+ 4a\epsilon}}{2\epsilon}t}\right)- a$.