1
$\begingroup$

I asked this question but no one answered this...

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.

$\endgroup$
1
$\begingroup$

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

$$ y_2(x) = \frac{1}{12} \left(3 x-6 x^2+4 x^3-x^4\right). $$

$\endgroup$
  • $\begingroup$ 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 $y_0$? $\endgroup$ – Teddy Frei May 9 '14 at 15:54
  • $\begingroup$ 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? $\endgroup$ – Antonio Vargas May 11 '14 at 15:51

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.