Inner solution and outer solution question

Question

Find the 2 term outer solution and one term inner solution for (using matched expansions)

$$ (1+\epsilon)x^2y'=\epsilon((1-\epsilon)xy^2-(1+\epsilon)x+y^3+2\epsilon y^2), \space \space \space \space y(1)=1$$

where $0<\epsilon <<1$ and $x > 0$

Also find a one term uniformly valid solution.

Thanks in advance

Answer 1

The outer solution is just a normal perturbation problem, let $y(x)=y_0(x)+\epsilon y_1(x)+O(\epsilon^2)$ and look at the $O(\epsilon^0)$ and $O(\epsilon)$ equations. At leading order, you get $$ x^2y_0'=0 $$ with initial condition $y_0(1)=1$. Since $x>0$, this is just $y_0'=0$, so we get $y_0=1$. Then, the $O(\epsilon)$ equation is $$ x^2y_0'+x^2y_1'=xy_0^2-x+y_0^3 $$ with initial condition $y_1(1)=0$. The equation simplifies to $$ y_1'=\frac{1}{x^2} $$ and so $y_1=-1/(3x^3)+A$, and the initial condition gives $A=1/3$.

So the two-term outer solution is $$ y=1+\epsilon\left(\frac{x^3-1}{3x^3}\right)+O(\epsilon^2).$$

Looking at the equation, we see that when $x$ is very small, the equation will change quite a lot. So we expect a boundary layer around $x=0$. Introduce a stretched coordinate $X=x\epsilon^{-\alpha}$ and let $Y(X)=y(x)$. The equation becomes $$\epsilon^{\alpha}(1+\epsilon)X^2Y'=\epsilon^{1+\alpha}(1-\epsilon)XY^2-\epsilon^{1+\alpha}(1+\epsilon)X+\epsilon Y^3+2\epsilon^2Y^2,$$ and so we use dominant balance to find the $\alpha=1$. The equation can now be written $$(1+\epsilon)X^2Y'=\epsilon(1-\epsilon)XY^2-\epsilon(1+\epsilon)X+Y^3+2\epsilon Y^2.$$ Now, write a series for $Y$, $Y=Y_0+\epsilon Y_1+O(\epsilon^2),$ and the leading order equation gives $$ X^2Y_0'=Y_0^3. $$ This is a separable ODE, so it can be integrated to give $$-\frac{1}{2}Y_0^{-2}=-X^{-1}+B$$ which rearranges to $$ Y_0=\sqrt{\frac{X}{2-2BX}}.$$ To determine $B$, we have to do asymptotic matching, $$\lim_{X\rightarrow\infty}Y_0(X)=\lim_{x\rightarrow0^+}y_0(x)\Rightarrow\sqrt{\frac{1}{-2B}}=1,$$ so $B=-1/2$, and $$Y_0=\sqrt{\frac{X}{2+X}}.$$

The uniform approximation is given by $y_{uniform}=y_0(x)+Y_0(X)$ plus a matching constant. In this case, the matching constant is chosen so that $y_{uniform}(1)=1$, so it is $-Y_0(1/\epsilon)=-1/\sqrt{1+2\epsilon}$. The one-term uniform solution is therefore $$ y_{uniform}(x)=1+\sqrt{\frac{x/\epsilon}{2+x/\epsilon}}-\frac{1}{\sqrt{1+2\epsilon}}.$$

Now we can check the results, by solving the equation numerically, and plotting our approximations. The code is

dx=0.01; %// grid spacing
xr=1:dx:3; %// have to split into two solution because initial condition is
xl=1:-dx:0; %// in the middle of the domain

epsilon=0.1;

dudt=@(x,y)epsilon*((1-epsilon)*x.*y.^2-(1+epsilon)*x+y.^3+2*epsilon*y.^2)./...
    ((1+epsilon)*x.^2); %// y'(x)=dudt(y,x)
[xr,yr]=ode45(dudt,xr,1); %// solve for x>=1
[xl,yl]=ode45(dudt,xl,1); %// solve for x<=1
X=[xl(end:-1:1);xr]; %// combine solutions into a single vector
Y=[yl(end:-1:1);yr];
plot(X,Y,X,ones(size(X)),X,1+epsilon*((1-X.^(-3))/3),... %// numeric, y_0, y_0+epsilon*y_1
    X,sqrt(X/epsilon./(2+X/epsilon)),... %// Y_0
    X,sqrt(X/epsilon./(2+X/epsilon))+1-1/sqrt(1+2*epsilon)) %// y_uniform
legend('Numeric','One term outer','Two term outer','One term inner','One term uniform')
axis([0 3 0.5 1.2])

and the resulting plot looks like

so you can see that although the inner and outer approximations don't look that great (they get better away from $x=1$), the uniform approximation is very good.