Here is my problem:

Consider the boundary-value problem \begin{align*} y''=y^{3}+x, \qquad y(a) = \alpha, \qquad y(b)=\beta, \qquad a \leq x \leq b \end{align*} To use the shooting method to solve this problem, one needs a starting guess for the initial slope $y′(a)$. One way to obtain such a starting guess for the initial slope is, in effect, to do a ”preliminary shooting” in which we take a single step of Euler’s method with $h=b−a$.

(a) Using this approach, write out the resulting algebraic equation for the initial slope.

(b) What starting value for the initial slope results from this approach?

I'm really not sure how to begin here; my idea was to write the 2nd order ODE using a new variable, such as: $y'=u$, $u'=y''=y^3+x$. Then, maybe I could use Euler's method to solve for $u=y'$; however, I'm not sure how to go about doing this. Any help appreciated.


Yes, that is correct. Now the Euler step says that \begin{align} y_1&=y_0+hu_0\\ u_1&=u_0+h(y_0^3+x_0) \end{align} As you can see, the differential equation does not really enter the equation for the value of $y_1$, so that one simply gets $$ u_0=\frac{β-α}{b-a}, $$ the slope of the line connecting $(a,α)$ to $(b,β)$.

Now, to get an initial guess for a multiple shooting or simultaneous collocation method, you could integrate from $(a,α,u_0)$ forward and from $(b,β,u_0)$ backwards with an ODE solver and blend the two solutions via cross-fading, $(1-s)y_0(a+sh)+sy_1(a+sh)$. (Be careful in its use, this formula is or at least was patented).

  • $\begingroup$ So, you solved the top equation for $u_0$, and used the fact that $h=b-a$ so that $y_1=\beta$, is that correct? That makes sense to me; the question in part (b) however confuses me, because it seems to be asking for some particular value... $\endgroup$
    – mXdX
    Apr 8 '20 at 7:49
  • 1
    $\begingroup$ I think the first equation is the answer to a) and the slope formula the answer to b). I do not see what else it could be. $\endgroup$ Apr 8 '20 at 8:00

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