Boundary value problems — shooting method, finite difference: Exercises

Two symbols recur; pin them down once so no line uses an undefined mark:

Level 1 — Recognition

L1.1 — IVP or BVP?

Classify each. State whether you'd march (IVP) or aim / puzzle-solve (BVP), and why.

(a) $y^{''} = - y, y (0) = 1, y^{'} (0) = 0.$
(b) $y^{''} = - y, y (0) = 1, y (π /2) = 0.$

Recall Solution

What decides it: count where the conditions live. Both at the same point $x = 0$ ⇒ IVP (you know $y$ and $y^{'}$ there, so you can step forward). One at each end ⇒ BVP.

(a) Both conditions at $x = 0$ ⇒ IVP. You know the whole state at the start, so march forward with e.g. RK4.
(b) $y (0)$ at the left end, $y (π /2)$ at the right end ⇒ two-point BVP. You cannot march (the starting slope is missing) — use shooting or finite difference.

L1.2 — Read off the stencil coefficients

For the linear BVP written in standard form $y^{''} = p (x) y^{'} + q (x) y + r (x)$ , the finite-difference row at interior node $i$ is $(1 + \frac{h}{2} p_{i}) y_{i - 1} - (2 + h^{2} q_{i}) y_{i} + (1 - \frac{h}{2} p_{i}) y_{i + 1} = h^{2} r_{i} .$ For $y^{''} = 3 y^{'} - 2 y + x$ with $h = 0.1$ , write the three coefficients (lower, diagonal, upper) and the right-hand side at a node where $x_{i} = 0.4$ .

Recall Solution

Identify $p = 3, q = - 2, r = x$ so $r_{i} = x_{i} = 0.4$ .

lower $= 1 + \frac{h}{2} p = 1 + \frac{0.1}{2} (3) = 1 + 0.15 = 1.15 .$
diagonal $= - (2 + h^{2} q) = - (2 + 0.01 (- 2)) = - (2 - 0.02) = - 1.98 .$
upper $= 1 - \frac{h}{2} p = 1 - 0.15 = 0.85 .$
RHS $= h^{2} r_{i} = 0.01 \times 0.4 = 0.004 .$

Notice lower $\neq =$ upper — the asymmetry is exactly the first-derivative (convection) term $p$ .

Level 2 — Application

L2.1 — One secant shooting step

A BVP has landing-miss function $ϕ (s) = y (b; s) - β$ . Two shots gave $ϕ (s_{0}) = ϕ (1) = - 0.4, ϕ (s_{1}) = ϕ (2) = 0.6.$ Do one secant update to produce $s_{2}$ .

Recall Solution

Why secant: we can't cheaply differentiate $ϕ$ (each evaluation is a full IVP solve), so we approximate $ϕ$ by the straight line through the last two points and read off where that line crosses zero. $s_{2} = s_{1} - ϕ (s_{1}) \frac{s _{1} - s _{0}}{ϕ ( s _{1} ) - ϕ ( s _{0} )} = 2 - 0.6 \cdot \frac{2 - 1}{0.6 - ( - 0.4 )} .$ $s_{2} = 2 - 0.6 \cdot \frac{1}{1.0} = 2 - 0.6 = 1.4 .$ Sign check: $ϕ$ changed sign between $s_{0}$ and $s_{1}$ , so the root lies between $1$ and $2$ — and $1.4$ does. Good.

L2.2 — Linear shooting in exactly two shots

Solve the linear BVP $y^{''} = y, y (0) = 0, y (2) = sinh 2$ by the linear-interpolation shortcut. Use shots $s_{0} = 0$ and $s_{1} = 1$ . Target $β = sinh 2 \approx 3.6269$ .

Recall Solution

Why two shots suffice: for a linear ODE the landing value $y (b; s)$ is a straight-line function of $s$ , so $ϕ (s)$ is a line — two points fix it exactly, no iteration.

Shot $s_{0} = 0$ : IVP $y^{''} = y, y (0) = 0, y^{'} (0) = 0 \Rightarrow y \equiv 0$ , so $y (2; 0) = 0$ .
Shot $s_{1} = 1$ : by linearity $y = sinh x$ , so $y (2; 1) = sinh 2 \approx 3.6269$ . Interpolate to hit $β$ : $s^{⋆} = s_{0} + (s_{1} - s_{0}) \frac{β - y ( 2 ; s _{0} )}{y ( 2 ; s _{1} ) - y ( 2 ; s _{0} )} = 0 + (1 - 0) \frac{3.6269 - 0}{3.6269 - 0} = 1.0 .$ The exact starting slope of $sinh x$ is $cosh 0 = 1$ . ✓ Recovered exactly in two shots.

L2.3 — FDM with one interior node

Solve $y^{''} = y, y (0) = 0, y (1) = sinh 1$ with $N = 2$ (so $h = 0.5$ , one interior node $x_{1} = 0.5$ ).

Recall Solution

Here $p = 0, q = 1, r = 0$ . The single interior row ( $i = 1$ ): $(1 + 0) y_{0} - (2 + h^{2} \cdot 1) y_{1} + (1 - 0) y_{2} = 0.$ With $h^{2} = 0.25$ : $y_{0} - 2.25 y_{1} + y_{2} = 0$ . Boundary values known: $y_{0} = 0, y_{2} = sinh 1 \approx 1.1752$ . Move them to the right: $- 2.25 y_{1} = - 1.1752 \Rightarrow y_{1} = \frac{1.1752}{2.25} = 0.5223 .$ True value $sinh (0.5) = 0.5211$ ; error $\approx 0.0012$ with just two intervals.

Level 3 — Analysis

L3.1 — Build and solve the tridiagonal system ( $N = 4$ )

Solve $y^{''} = y, y (0) = 0, y (1) = sinh 1$ with $N = 4$ ( $h = 0.25$ ). Set up the $3 \times 3$ tridiagonal system for $y_{1}, y_{2}, y_{3}$ and solve it. Compare $y_{2}$ to the truth.

Recall Solution

$p = 0, q = 1, r = 0$ , $h^{2} = 0.0625$ . Each interior row: $y_{i - 1} - (2 + 0.0625) y_{i} + y_{i + 1} = 0$ , i.e. diagonal $= - 2.0625$ , off-diagonals $= 1$ . Nodes $x_{1} = 0.25, x_{2} = 0.5, x_{3} = 0.75$ . Boundaries $y_{0} = 0$ , $y_{4} = sinh 1 \approx 1.1752$ .

i=1:&\quad -2.0625\,y_1 + y_2 = -y_0 = 0,\\ i=2:&\quad y_1 -2.0625\,y_2 + y_3 = 0,\\ i=3:&\quad y_2 -2.0625\,y_3 = -y_4 = -1.1752. \end{aligned}$$ **Thomas / elimination** (forward sweep): - Row 1: $y_1=y_2/2.0625$. - Substitute into row 2: $\dfrac{y_2}{2.0625}-2.0625\,y_2+y_3=0\Rightarrow(-2.0625+0.4848)y_2+y_3=0\Rightarrow -1.5777\,y_2+y_3=0$, so $y_3=1.5777\,y_2$. - Substitute into row 3: $y_2-2.0625(1.5777\,y_2)=-1.1752\Rightarrow(1-3.2540)y_2=-1.1752\Rightarrow -2.2540\,y_2=-1.1752$. $$y_2=\frac{1.1752}{2.2540}=\mathbf{0.5214}.$$ Back-substitute: $y_1=0.5214/2.0625=\mathbf{0.2528}$, $y_3=1.5777(0.5214)=\mathbf{0.8226}$. **Compare:** $\sinh(0.5)=0.5211$. Error at the midpoint is now $\approx0.0003$ — versus $0.0012$ at $N=2$. Halving $h$ dropped the error by about $4\times$, confirming $O(h^2)$.

Figure — Boundary value problems — shooting method, finite difference

L3.2 — Why the naive secant can miss the bracket

Given $ϕ (1) = - 0.4$ and $ϕ (2) = 0.6$ (same as L2.1), suppose instead the true $ϕ$ is curved: $ϕ (1.4) = + 0.05$ actually. Explain in one line why one linear secant step is not the exact answer here, and what you must do.

Recall Solution

The secant step assumed $ϕ$ was a straight line, so it predicted a root at $s_{2} = 1.4$ . But the real $ϕ (1.4) = + 0.05 \neq = 0$ : the function is curved (the ODE is nonlinear), so a line through two points cannot land the root in one shot. Fix: iterate — use the new pair $(s_{1}, s_{2}) = (2, 1.4)$ … actually the sign-changing pair $(1, 1.4)$ since $ϕ (1) < 0 < ϕ (1.4)$ — and take another secant step, repeating until $∣ ϕ ∣ < tol$ . Exactness in two shots is a linear-only privilege.

Level 4 — Synthesis

L4.1 — FDM with a convection term $p \neq = 0$

Set up and solve $y^{''} = y^{'} - y, y (0) = 0, y (1) = 1$ with $N = 2$ ( $h = 0.5$ , one interior node $x_{1} = 0.5$ ). Note the standard form is $y^{''} = p y^{'} + q y + r$ with $p = 1, q = - 1, r = 0$ .

Recall Solution

Coefficients at the single interior row:

lower $= 1 + \frac{h}{2} p = 1 + \frac{0.5}{2} (1) = 1 + 0.25 = 1.25$ ,
diagonal $= - (2 + h^{2} q) = - (2 + 0.25 (- 1)) = - (2 - 0.25) = - 1.75$ ,
upper $= 1 - \frac{h}{2} p = 1 - 0.25 = 0.75$ ,
RHS $= h^{2} r = 0$ .

Row: $1.25 y_{0} - 1.75 y_{1} + 0.75 y_{2} = 0$ . Boundaries $y_{0} = 0, y_{2} = 1$ : $- 1.75 y_{1} + 0.75 (1) = 0 \Rightarrow y_{1} = \frac{0.75}{1.75} = 0.4286 .$ The asymmetry lower $= 1.25 \neq =$ upper $= 0.75$ is the fingerprint of the convection term $p = 1$ : information leans in the flow direction.

L4.2 — Order-of-accuracy audit

Two FDM runs of the same BVP give midpoint errors $E (h = 0.25) = 0.0012$ and $E (h = 0.125) = 0.0003$ . Estimate the observed order $p$ from $p \approx lo g_{2} \frac{E ( h )}{E ( h /2 )} .$ Is the scheme performing as designed?

Recall Solution

Why $lo g_{2}$ : if $E \propto h^{p}$ , then halving $h$ multiplies error by $2^{- p}$ ; the ratio $E (h) / E (h /2) = 2^{p}$ , and $lo g_{2}$ of that recovers $p$ . $p \approx lo g_{2} \frac{0.0012}{0.0003} = lo g_{2} 4 = 2.0 .$ The central-difference scheme is designed for $O (h^{2})$ , and the measured order is $2$ — it performs exactly as designed. (See Truncation error and order of accuracy.)

Level 5 — Mastery

L5.1 — Choose the method and justify

You must solve a stiff, nonlinear BVP on a long domain $[0, 50]$ where a tiny slope error at $x = 0$ grows exponentially. Which method — shooting or FDM — do you pick, and give two concrete reasons.

Recall Solution

Pick finite difference. Reasons:

Stability on long/stiff domains: shooting marches an IVP forward, so a small error in the guessed slope $s$ is amplified exponentially across $[0, 50]$ — the very definition of the difficulty here. FDM solves all nodes simultaneously, keeping errors local rather than propagating.
No sensitive root-find: $ϕ (s)$ for a stiff problem is wildly steep, so secant/Newton on $s$ is ill-conditioned. FDM turns the problem into a (banded) algebraic system; for the nonlinear part you apply Newton to the whole system (Jacobian is tridiagonal-ish, still cheap via Tridiagonal systems — Thomas algorithm).

Trade-off acknowledged: FDM needs a nonlinear solve per Newton step, but its stability wins decisively when shooting blows up.

L5.2 — Full pipeline, one clean pass

Solve $y^{''} = y, y (0) = 0, y (1) = sinh 1$ by shooting with RK-free reasoning: because the ODE is linear you may use the exact IVP solution $y (x; s) = s sinh x$ . Shoot $s_{0} = 0.5$ and $s_{1} = 1.5$ , then linearly interpolate to the exact slope, and finally state $y (0.5)$ from the recovered solution.

Recall Solution

Why $y = s sinh x$ : for $y^{''} = y, y (0) = 0, y^{'} (0) = s$ , the unique IVP solution is $s sinh x$ (it satisfies the ODE, $y (0) = 0$ , and $y^{'} (0) = s cosh 0 = s$ ).

Shot $s_{0} = 0.5$ : $y (1; 0.5) = 0.5 sinh 1 = 0.5 (1.1752) = 0.5876$ .
Shot $s_{1} = 1.5$ : $y (1; 1.5) = 1.5 sinh 1 = 1.5 (1.1752) = 1.7628$ .
Target $β = sinh 1 = 1.1752$ . Interpolate: $s^{⋆} = 0.5 + (1.5 - 0.5) \frac{1.1752 - 0.5876}{1.7628 - 0.5876} = 0.5 + 1 \cdot \frac{0.5876}{1.1752} = 0.5 + 0.5 = 1.0 .$ With $s^{⋆} = 1$ : $y (x) = sinh x$ , so $y (0.5) = sinh (0.5) = 0.5211$ — the exact midpoint value (shooting recovered the true solution because the ODE is linear).

Recall One-line self-test

Central second-difference stencil for $y^{''}$ ::: $\frac{y _{i + 1} - 2 y _{i} + y _{i - 1}}{h ^{2}}$ , accurate to $O (h^{2})$ . Landing-miss function in shooting ::: $ϕ (s) = y (b; s) - β$ ; solve $ϕ (s) = 0$ . Why linear BVP shoots exact in 2 tries ::: $ϕ (s)$ is a straight line, and 2 points fix a line. Observed order from two errors ::: $p = lo g_{2} (E (h) / E (h /2))$ .