Initial value problems (IVP) vs boundary value problems (BVP): Worked examples

This page is the exhaustive case-book for the parent topic on IVPs and BVPs. The parent told you what the distinction is. Here we walk through every kind of outcome a second-order problem can produce, so you never meet a scenario you have not already seen solved.

The scenario matrix

Before defining anything new, here is the map of every cell we must cover. Each row is a class of situation; the last column names the example that hits it.

#	Cell (scenario class)	Key question	Outcome	Example
A	IVP, both data at one point	Wronskian $\neq = 0$ ?	Always unique	Ex 1
B	IVP with $x_{0} \neq = 0$ (the " $t = 0$ trap")	Same point, not zero?	Still unique	Ex 2
C	BVP, $det M \neq = 0$	Matrix invertible?	Unique	Ex 3
D	BVP, $det M = 0$ , target reachable	Compatible RHS?	Infinitely many	Ex 4
E	BVP, $det M = 0$ , target unreachable	Incompatible RHS?	No solution	Ex 5
F	Eigenvalue sweep (which $λ$ break it?)	For which parameter is $det M = 0$ ?	Discrete "resonant" values	Ex 6
G	Degenerate ODE ( $y^{''} = 0$ ) — real-world rod	Straight-line family	Unique linear profile	Ex 7
H	Mixed / Robin BVP (value at one end, slope at other)	Different points still?	BVP, check $det M$	Ex 8
I	Exam twist: same numbers, IVP vs BVP swap	Where do the conditions live?	One unique, one singular	Ex 9

Two symbols recur, both built in the parent — let us re-anchor them in one breath so nothing here is unexplained.

Group 1 — IVPs always work (cells A, B)

Group 2 — BVPs: the three fates (cells C, D, E)

Here we hit the same ODE $y^{''} + y = 0$ three times, changing only the endpoints. This is the heart of the topic: counting conditions is not enough. Look at the figure — it shows why $[0, π]$ is the "dangerous" interval.

Figure — Initial value problems (IVP) vs boundary value problems (BVP)

Recall The fork at

$det M = 0$ When $det M = 0$ , is the answer "no solution" or "infinitely many"? ::: Depends on the target: if the boundary data is compatible with the singular system (RHS in the column space) → infinitely many; if incompatible → none.

Group 3 — The eigenvalue sweep (cell F)

Now vary a parameter and ask: for which values does the BVP go singular? These special values are eigenvalues; the whole subject of Sturm-Liouville Theory is this question asked systematically.

Worked example Example 6 — Cell F: which

$λ$ give nontrivial solutions? ODE: $y^{''} + λ y = 0$ on $[0, π]$ with $y (0) = 0, y (π) = 0$ , and $λ > 0$ . Write $λ = μ^{2}$ .

Forecast: For "most" $λ$ only $y \equiv 0$ works. For which special $λ$ do we suddenly get real curves?

General solution $y = c_{1} cos μx + c_{2} sin μx$ . Why? For $λ = μ^{2} > 0$ the characteristic roots are $\pm i μ$ , giving cos/sin.
$y (0) = c_{1} = 0$ , so $y = c_{2} sin μx$ . Why? Left endpoint kills the cosine part.
$y (π) = c_{2} sin (μ π) = 0$ . For a nontrivial ( $c_{2} \neq = 0$ ) solution we need $sin (μ π) = 0$ . Why? Only then can $c_{2}$ stay free. That means $μ π = nπ \Rightarrow μ = n$ , so $λ_{n} = n^{2}$ , $n = 1, 2, 3, \dots$
Eigenfunctions: $y_{n} = sin (n x)$ .

Verify: For $λ = 4$ ( $n = 2$ ): $y = sin 2 x$ , $y (0) = 0$ ✓, $y (π) = sin 2 π = 0$ ✓ — nontrivial. For $λ = 2$ : $μ = 2$ , $sin (2 π) \approx sin (4.4429) \approx - 0.9633 \neq = 0$ → only $c_{2} = 0$ → trivial solution $y \equiv 0$ . So the discrete list ${1, 4, 9, \dots}$ is exactly the resonant set. (Compare parent Ex 4, which was $λ = 4$ read as a length sweep.)

Group 4 — Degenerate ODE, real-world word problem (cell G)

Worked example Example 7 — Cell G: steady heat in a rod (units check!)

A metal rod of length $L = 2 m$ sits in steady state. Temperature obeys $y^{''} = 0$ (no time change, no source). Ends are held at $y (0) = T_{0} = 10 0^{\circ} C$ and $y (L) = T_{1} = 2 0^{\circ} C$ .

Forecast: Straight line, curve, or something wilder? And what is the temperature at the midpoint $x = 1$ ?

Integrate twice: $y = c_{1} x + c_{2}$ . Why? $y^{''} = 0$ means slope never changes → a straight line; two integrations give two constants.
$y (0) = c_{2} = 100$ . Why? Left end fixes the intercept.
$y (2) = 2 c_{1} + 100 = 20 \Rightarrow c_{1} = \frac{20 - 100}{2} = - 40$ . Why? Right end fixes the slope.
Profile: $y (x) = 100 - 40 x$ (units: $^{\circ}$ C per metre for the slope $- 40$ ).
Midpoint $x = 1$ : $y (1) = 100 - 40 = 6 0^{\circ} C$ .

Verify: $det M = y_{1} (0) y_{2} (L) - y_{2} (0) y_{1} (L)$ with $y_{1} = x, y_{2} = 1$ : $= 0 \cdot 1 - 1 \cdot 2 = - 2 \neq = 0$ → unique ✓. Endpoints: $y (0) = 100$ ✓, $y (2) = 100 - 80 = 20$ ✓. Units: slope $\times$ length $= - 40 \frac{^{\circ} C}{m} \times 2 m = - 8 0^{\circ} C$ , so $100 - 80 = 20$ ✓. This is the steady-state limit of the Heat Equation.

Group 5 — Mixed / Robin BVP (cell H)

A condition can mix a value and a slope — but what makes it a BVP is that they live at different points. Contrast this with the parent's warning: $y (a)$ and $y^{'} (a)$ (same point) would be an IVP.

Worked example Example 8 — Cell H: value at one end, slope at the other

ODE: $y^{''} + y = 0$ , conditions $y (0) = 0, y^{'} (\frac{π}{2}) = 4$ . Different points → genuine BVP.

Forecast: Mixing a value and a derivative — will $det M$ still be checkable? Unique or not?

$y = c_{1} cos x + c_{2} sin x$ ; $y (0) = c_{1} = 0$ . Why? Left endpoint value condition.
$y^{'} (x) = - c_{1} sin x + c_{2} cos x$ ; at $x = \frac{π}{2}$ : $- c_{1} \cdot 1 + c_{2} \cdot 0 = - c_{1}$ . Wait — that is $- c_{1} = 4$ ? Why check carefully? $cos \frac{π}{2} = 0$ kills $c_{2}$ . So the slope condition reads $- c_{1} = 4$ .
But step 1 already gave $c_{1} = 0$ . So $- 0 = 4$ → $0 = 4$ , contradiction → no solution here.

Verify: The mixed BVP matrix uses rows "value at $0$ " and "slope at $π /2$ ": $M = (cos 0 - sin \frac{π}{2} sin 0 cos \frac{π}{2}) = (1 - 1 00)$ , $det M = 1 \cdot 0 - 0 \cdot (- 1) = 0$ → singular, and target $(0, 4)$ is incompatible (second row forces $- c_{1} = 4$ while first forces $c_{1} = 0$ ) → no solution ✓. Lesson: mixed conditions at different points are still a BVP and can be singular exactly like Ex 5.

Group 6 — Exam twist (cell I)

Worked example Example 9 — Cell I: identical numbers, IVP vs BVP swap

Same ODE $y^{''} + y = 0$ and the same two numbers " $0$ and $0$ ", read two ways. (a) IVP: $y (0) = 0, y^{'} (0) = 0$ (both at $x = 0$ ). (b) BVP: $y (0) = 0, y (π) = 0$ (at $x = 0$ and $x = π$ ).

Forecast: Will they give the same answer? (Trap: they look almost identical.)

(a) $y (0) = c_{1} = 0$ ; $y^{'} (0) = c_{2} = 0$ . Both dials forced to zero → unique $y \equiv 0$ . Why unique? Wronskian $= 1 \neq = 0$ , IVPs are always well-posed.
(b) From Ex 4: $c_{1} = 0$ and $c_{2}$ free → $y = c_{2} sin x$ , infinitely many. Why different? The second condition now lives at $x = π$ , where $sin π = 0$ erases $c_{2}$ instead of pinning it.
So the location alone flips "unique trivial" into "infinitely many."

Verify: (a) $W = cos^{2} x + sin^{2} x = 1 \neq = 0$ → unique, and $y \equiv 0$ satisfies $y (0) = y^{'} (0) = 0$ ✓. (b) $det M = cos 0 sin π - sin 0 cos π = 0$ , RHS $(0, 0)$ consistent → infinitely many, each $c_{2} sin x$ hits $y (0) = 0, y (π) = 0$ ✓.

Recall check

Recall Rapid-fire over the whole matrix

Cell C vs D: what single quantity separates "unique" from "infinitely many"? ::: $det M$ — nonzero gives unique, zero pushes you to the none/∞ fork. Cell F: for $y^{''} + λ y = 0$ , $y (0) = y (π) = 0$ , list the eigenvalues. ::: $λ_{n} = n^{2}, n = 1, 2, 3, \dots$ with eigenfunctions $sin (n x)$ . Cell H: are $y (0)$ and $y^{'} (π /2)$ together an IVP or BVP? ::: BVP — different points, even though one is a slope. Cell G: solution of $y^{''} = 0$ , $y (0) = 100, y (2) = 20$ ? ::: $y = 100 - 40 x$ , midpoint $6 0^{\circ}$ C, unique since $det M = - 2 \neq = 0$ . Cell I: same $(0, 0)$ data — why does IVP give one solution but BVP give infinitely many? ::: IVP has both conditions at one point (Wronskian $\neq = 0$ ); BVP reads the second at $x = π$ where $sin π = 0$ frees $c_{2}$ .

Connections

Sturm-Liouville Theory — cell F is the entry point: $det M = 0$ generates the eigenvalues.
Wronskian and Linear Independence — the reason cells A, B never fail.
Picard-Lindelöf Existence and Uniqueness Theorem — the deeper guarantee behind IVP uniqueness.
Heat Equation — cell G is its steady-state limit.
Wave Equation — the same $y^{''} + λ y = 0$ eigenvalues drive standing waves.
Separation of Variables (PDE) — turns PDEs into exactly these BVP eigenvalue problems.
Green's Functions — the systematic inverse of an invertible BVP (cell C).