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I'm having a hard time trying to solve this coupled pair of differential equations by the perturbation method.

These are the equations:

enter image description here

Where Br you can treat as ε (base solution).

The solution should look like:

enter image description here

Thanks in advance

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Obviously, setting $Br=0$ gives $$w_0''=0\ , w_0(0)=w_0(1)=0$$ which has $w_0(s)=0$ as the only solution. In the first equation $$((1-\beta\,w_0)(u_0'))'=0$$ this results in $u_0''=0$ so that $u_0$ is linear and thus $u_0(s)=1-s$.

Now insert into the equation for $w_1$, $w_1''=-1$, $w_1(0)=w_1(1)=0$ to get $$w_1=-\frac12(s^2-s).$$

In the equation for $u_1$ we get $$((1-\beta\,Br\,w_1)(u_0'+Br\,u_1'))'=O(Br^2)$$ to find $$ 0=u_1''-β(w_1u_0')'=u_1''-βw_1'u_0'=u_1''-β(s-\tfrac12),\\u_1(0)=u_1(1)=0, $$ which can be integrated to $$ u_1=\fracβ{12}(2s^3-3s^2+s). $$

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  • $\begingroup$ That's all very well, but you are getting w1(0) and w1(1) not equal to zero $\endgroup$ – Zygi Orfejas Dec 3 '17 at 20:48
  • $\begingroup$ Ups, that was the wrong key, should have returned to the first of s^2. $w_1=-\frac12s(1-s)$. $\endgroup$ – Lutz Lehmann Dec 3 '17 at 21:13

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