Scientific Computing - Quiz 3 - Question 1-3

Problem:

Solve the simple boundary value problem
$y′′=−4y$ with $y(0)=0$, $y′(1)=\cos(2)$

(Use your above work and solution to continue with this question)

With boundary value problem considered previously
$y′′=−4y$ with $y(0)=0$, $y′(1)=\cos(2)$

Consider instead

$y′′=−4y$ with $y(0) = 0$, $y′(0) = A$ and use a bisection to compute the iterative values of $A_1 = 0.5$ that converge to the solution using a shooting algorithm (NOTE: use your known solution in the interval $t \in [0,1]$. Here assume that $A_1 = 0.5$ (too low) and $A_2 = 2$ (too high) to compute (a) $A_3$, (b) $A_4$ and (c) $A_5$.

Solution:

Taking Laplace transform on both side gives

$\begin{eqnarray*} \mathcal{L}(y'') &=& \mathcal{L}(-4y) \\ s\mathcal{L}(y') - y'(0) &=& -4\mathcal{L}(y) \\ s(s\mathcal{L}(y) - y(0))- y'(0) &=& -4\mathcal{L}(y) \\ s(s\mathcal{L}(y))- y'(0) &=& -4\mathcal{L}(y) \\ s^2\mathcal{L}(y) + 4\mathcal{L}(y) &=& y'(0) \\ \mathcal{L}(y) &=& \frac{y'(0)}{s^2 + 4} \\ \mathcal{L}(y) &=& \frac{y'(0)}{2}\frac{2}{s^2 + 4} \\ y(t) &=& \frac{y'(0)}{2}\sin(2t) \\ \end{eqnarray*}$

Now we see the general form of $y$, using the boundary condition we can solve for the unknown $y'(0)$ as follow:

$\begin{eqnarray*} y'(t) &=& y'(0)\cos(2t) \\ y'(1) &=& y'(0)\cos(2) \\ \cos(2) &=& y'(0)\cos(2) \\ y'(0) &=& 1 \\ y(t) &=& \frac{1}{2}\sin(2t) \\ \end{eqnarray*}$

The rest is just bisection.

Note that $y'(1) = Acos(2)$

$A$	$y'(1)$	goal	$y'(1)$ is …	$A$ is …
0.5000	-0.2081	-0.4161	too high	too low
2.0000	-0.8323	-0.4161	too low	too high
1.2500	-0.5202	-0.4161	too low	too high
0.8750	-0.3641	-0.4161	too high	too low
1.0625	-0.4422	-0.4161	too low	too high

Note that the question is confusing me saying $A = 0.5$ is too low while $y'(1)$ when $A = 0.5$ is actually too high :(

Also be very careful with the requirement for the number of digits.