# BVP and Perturbation Methods

This is a question on one of my ODE past papers:

You are given the non linear boundary value problem $$y^{\prime\prime}+y^2=0,\ \text{subject to}\ y(0)=\epsilon\ \text{and}\ y(1)=0,$$ where $\epsilon$ is a given constant.

(i) Show that for $\epsilon=0$, a solution is given by $y(x)=0$;
(ii) Pose a perturbation expansion of the form $$y(x)=y_0(x)+\epsilon y_1(x)+\epsilon^2y_2(x)+\cdots$$ for $|\epsilon|\ll1$ and find the first three terms, $y_0(x),y_1(x)$ and $y_2(x)$.

Here is my approach:

(i) If $\epsilon=0$, then $y(0)=y(1)=0$, so substitute $y(x)=0$ and $y^{\prime}=y^{\prime\prime}=0$ into the equation, giving that $y(x)=0$ is a solution.

(ii) First consider the boundary condition: $$y(0)=y_0(0)+\epsilon y_1(0)+\epsilon^2y_2(0)+\cdots=\epsilon,$$ and $$y(1)=y_0(1)+\epsilon y_1(0)+\epsilon^2y_2(1)+\cdots=0.$$ Compare the coefficients of powers of $\epsilon$, we have $$y_0(0)=0,\ y_1(0)=1,\ y_2(0)=0,\ \dots,$$ and $$y_0(1)=y_1(1)=y_2(1)=\cdots=0.$$ Then consider $$y^2=(y_0+\epsilon y_1+\epsilon^2y_2+\cdots)^2=y_0^2+\epsilon(2y_0y_1)+\epsilon^2(y_1^2+2y_0y_2)+\cdots,$$ Substitute the above equation into the original ODE, we get $$y^{\prime\prime}_0+\epsilon y^{\prime\prime}_1+\epsilon^2 y^{\prime\prime}_2+\cdots+y_0^2+\epsilon(2y_0y_1)+\epsilon^2(y_1^2+2y_0y_2)+\cdots=0.$$ Collect the powers of $\epsilon$, we get three homogeneous ODE's $$y^{\prime\prime}_0+y_0^2=0,\\ y_1^{\prime\prime}+2y_0y_1=0,\\ y^{\prime\prime}_2+y_1^2+2y_0y_2=0$$ for the first three terms of $y$.

Now the question comes: Should I use the result in (i) and let $y_0=0$ since $y_0$ satisfies the condition in (i)? If I can't use (i), the perturbation doesn't do anything and I still need to solve the ODE algebraically as the ODE I got for $y_0$ in last the step has the same form as the original one. So how should I solve this perturbation problem?

Thank you.

Yes, use $(i)$ to take $y_0 = 0$. Then $y_1'' = 0$ with $y_1(0) = 1$ and $y_1(1) = 0$, so
$$y_1(x) = 1-x.$$
Further, $y_2'' + (1-x)^2 = 0$ with $y_2(0) = y_2(1) = 0$, so
$$y_2(x) = \frac{1}{12} \left(3 x-6 x^2+4 x^3-x^4\right).$$
• Thanks a lot and I believe that is the solution! Just one more question: why can we use $y_0=0$? Is it the unique solution for $y_0$? – Teddy Frei May 9 '14 at 15:54
• I'm not sure how to show that $y'' + y^2 = 0$ with $y(0) = y(1) = 0$ has a unique solution ($y=0$). Perhaps you could ask a new question for that? – Antonio Vargas May 11 '14 at 15:51