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Initial Value Problem (IVP) - Definition, Properties,
and Examples
Definition
An Initial Value Problem (IVP) is a differential equation together with a set of initial conditions that specify the value of the unknown function (and possibly its derivatives) at a particular point.
Find y(x) such that y′(x) = f (x, y(x)), y(x 0 ) = y 0
- y′(x) = derivative of y - f (x, y) = given function - y(x 0 ) = y 0 = initial condition Higher-order IVP: y(n)^ = f (x, y, y′,... , y(n−1)), y(x 0 ) = y 0 , y′(x 0 ) = y 1 ,... , y(n−1)(x 0 ) = yn− 1 —
Properties of IVP
- Existence: If f (x, y) is continuous in a region around (x 0 , y 0 ), a solution exists locally.
- Uniqueness: If f (x, y) satisfies a Lipschitz condition in y, the solution is unique.
- Maximal Interval: The solution exists on the largest interval allowed before blow-up or singularity.
- Continuous Dependence: Solution depends continuously on the initial condition.
- Linear IVPs: Solutions can be superposed (linear combination) for linear ODEs.
- Nonlinear IVPs: May have finite-time blow-up, oscillatory solutions, or bounded solutions depending on f (x, y). —
Solved Examples
Example 1 Solve: y′^ = xy, y(0) = 1 Solution: - Separable: dyy = xdx - Integrate: ln |y| = x 22 + C - IC: y(0) = 1 =⇒ C = 0 - Final Solution: y = ex^2 /^2 Example 2 Solve: y′^ − y = ex, y(0) = 0 Solution: - Linear ODE: integrating factor μ(x) = e−^ R^ −^1 dx^ = ex^ - (e−xy)′^ = 1 =⇒ e−xy = x + C - IC: y(0) = 0 =⇒ C = 0 - Final Solution: y = xex
Example 3 Solve: y′^ = y^2 , y(0) = 1 Solution: - Separable: dyy 2 = dx =⇒ − (^1) y = x + C - IC: y(0) = 1 =⇒ C = −1 - Final Solution: y = (^1) −^1 x - Note: solution blows up at x = 1 → maximal interval [0, 1) Example 4 Solve second-order IVP: y′′^ − 3 y′^ + 2y = 0, y(0) = 1, y′(0) = 0 Solution: - Characteristic equation: r^2 − 3 r + 2 = 0 =⇒ r = 1, 2 - General solution: y = C 1 ex^ + C 2 e^2 x^ - IC: C 1 + C 2 = 1, C 1 + 2C 2 = 0 =⇒ C 1 = 2, C 2 = − 1
- Final Solution: y = 2ex^ − e^2 x Example 5 Solve: y′^ + xy = 0, y(0) = 1 Solution: - Linear ODE: integrating factor 1 μ(x) = eR^ xdx^ = ex^2 /^2 - Solution: y = μ(x) =^ e−x^2 /^2 - Properties:^ global existence, monotone decreasing, always positive
Fundamental Existence and Uniqueness Theorem for
IVPs
Problem Setting
Consider the first-order Initial Value Problem (IVP):
y′^ = f (x, y), y(x 0 ) = y 0
- f : D ⊂ R × R → R - (x 0 , y 0 ) ∈ D We are interested in conditions under which a solution exists and is unique near x 0. —
Fundamental Existence Theorem (Peano)
Theorem (Existence) If f (x, y) is continuous in a rectangle R = {(x, y) : |x − x 0 | ≤ a, |y − y 0 | ≤ b} then there exists at least one solution y(x) of the IVP on an interval |x − x 0 | ≤ h ≤ a for some h > 0.
5 Extremely Hard CSIR-NET MSQs on Lipschitz Con-
dition
Question 1 Consider the IVP: y′^ = y^1 /^3 , y(0) = 0 Which of the following statements are true?
- There exists at least one solution in a neighborhood of x = 0
- The solution is unique
- y ≡ 0 is a solution
- y = (x/3)^3 /^2 is a solution Solution: - f (y) = y^1 /^3 is continuous at y = 0 → existence guaranteed (Peano) → 1 True - f not Lipschitz at y = 0 → uniqueness fails → 2 False - y ≡ 0 satisfies y′^ = 0 = 0^1 /^3 → 3 True - y = (x/3)^3 /^2 derivative: y′^ = (1/2)(x/3)^1 /^2 ∗ (1/3)∗? → matches y^1 /^3 → 4 True Correct: 1,3, — Question 2 Let the IVP be: y′^ = sin(y), y(0) = π Which statements are correct?
- f (y) = sin(y) satisfies Lipschitz condition for all y ∈ R
- Solution exists and is unique globally
- The solution is bounded for all x
- The solution may blow up in finite time Solution: - f (y) = sin y → derivative w.r.t y is cos y → bounded → Lipschitz on any bounded interval, but not globally Lipschitz → 1 False - ODE is linear in the sense of derivative dependence on y? Nonlinear but continuous → uniqueness locally; global existence → yes, no blow-up → 2 True - y′^ = sin y → |y′| ≤ 1 → solution grows at most linearly → bounded slope → 3 True - No blow-up → 4 False Correct: 2, —
Question 3 Consider IVP: y′^ = |y|, y(0) = − 1 Which of the following are true?
- f (y) = |y| is Lipschitz continuous for all y ∈ R
- Solution exists and is unique globally
- Solution crosses y = 0 in finite time
- Maximal interval of existence is [0, ∞)
Solution: - f (y) = |y| → derivative undefined at y = 0 → not Lipschitz globally → 1 False - But piecewise linear → uniqueness exists until y = 0, after which two solutions possible? Actually yes, y = −e−x^ etc. → check carefully. - Solve: dy/|y| = dx → y(x) = −ex^ → crosses 0? No, stays negative → 3 False - y = −ex exists for all x ≥ 0 → maximal interval [0, ∞) → 4 True Correct: 4
— Question 4 IVP: y′^ = y^2 , y(0) = 1 Which of the following are true?
- f (y) = y^2 satisfies Lipschitz condition near y = 1
- Solution is unique locally
- Solution exists globally
- Solution blows up at finite x
Solution: - f (y) = y^2 → derivative 2y → bounded in neighborhood of y = 1 → Lipschitz locally → 1 True - Local uniqueness guaranteed → 2 True - Global existence fails because solution y = 1/(1−x) blows up at x = 1 → 3 False - Blow-up occurs at x = 1 → 4 True Correct: 1,2,
—
Mathematical Justification By the Mean Value Theorem:
f (x, y 1 ) − f (x, y 2 ) = ∂f∂y (x, ξ)(y 1 − y 2 ), ξ between y 1 and y 2
Taking absolute values:
|f (x, y 1 ) − f (x, y 2 )| = ∂f∂y (x, ξ) |y 1 − y 2 | ≤ L|y 1 − y 2 |
Hence, bounded partial derivative =⇒ Lipschitz continuity.
Important Notes
- Lipschitz in x is not required for IVP uniqueness; only in y.
- If ∂f∂y is continuous in a rectangle, it is automatically bounded on any compact subset =⇒ guarantees local Lipschitz.
- Functions like f (y) = p|y| fail this: derivative is unbounded at y = 0 =⇒ uniqueness fails.
Examples Example 1: ∂f f (x, y) = x + y ∂y = 1^ =⇒^ bounded^ =⇒^ Lipschitz with^ L^ = 1 Example 2: ∂f f (x, y) = y^2
globally∂y^ = 2y^ =⇒^ bounded in any bounded rectangle of^ y^ =⇒^ Lipschitz locally, not Example 3: ∂f f (y) = p|y| ∂y =^2 √^1 |y| =⇒^ unbounded at^ y^ = 0^ =⇒^ not Lipschitz
Summary Table Condition Lipschitz? Notes ∂f∂y ≤ L Yes Sufficient condition f continuous but ∂f∂y unbounded Not guaranteed Example: f (y) = p|y| f (y) linear in y Always Global uniqueness guaranteed
5 Extremely Hard CSIR-NET MSQs on Lipschitz Con-
dition
Question 1 Consider the IVP: y′^ = y^3 , y(0) = 1 Which of the following statements are true?
- f (y) = y^3 is Lipschitz continuous for all y ∈ R
- Solution exists locally around x = 0
- Solution exists globally for all x ≥ 0
- Solution blows up at finite x Solution: - ∂f∂y = 3y^2 → unbounded for large y → not globally Lipschitz → 1 False - Continuous function → Peano theorem → local existence → 2 True - Solve separable: dy/y^3 = dx =⇒ y = √ 11 − 2 x → blow-up at x = 1/ 2 → 3 False, 4 True Correct: 2, — Question 2 Let the IVP be: y′^ = sin(y), y(0) = π Which statements are correct?
- f (y) = sin(y) satisfies Lipschitz condition for all y ∈ R
- Solution exists and is unique locally
- Solution exists globally for all x
- Solution may blow up in finite x Solution: - ∂f∂y = cos y → bounded → Lipschitz locally, but not globally (cosine is bounded, so actually Lipschitz everywhere) → 1 True - Continuity + Lipschitz → uniqueness locally → 2 True - |y′| ≤ 1 → solution slope bounded → global existence → 3 True - No blow-up → 4 False Correct: 1,2, —
Question 5 Consider IVP: y′^ = y ln |y|, y(0) = 1 Which statements are correct?
- f (y) = y ln |y| is Lipschitz near y = 1
- Solution is unique locally
- Solution exists globally for all x
- Solution may blow up at finite x Solution: - ∂f∂y = ln |y| + 1 → bounded near y = 1 → Lipschitz locally → 1 True
- Local uniqueness → 2 True - Solve separable: dy/(y ln y) = dx =⇒ ln(ln y) = x+C =⇒ y = eex → blows up as x → ∞, global existence possible but unbounded growth → 3 False, 4 True Correct: 1,2,
Gr¨onwall’s Lemma
- Statement (Integral Form) Let u(x), v(x) be continuous, non-negative functions on [x 0 , X], and α(x) continu- ous, non-negative. If u(x) ≤ v(x) +
Z (^) x x 0 α(t)u(t)dt,^ ∀x^ ∈^ [x^0 , X], then u(x) ≤ v(x) +
Z (^) x x 0 v(t)α(t) exp
Z^ x t^ α(s)ds
dt. In particular, if v(x) = v 0 is constant: u(x) ≤ v 0 exp
Z^ x x 0 α(t)dt
- Differential Form If u(x) satisfies u′(x) ≤ α(x)u(x) + β(x), u(x 0 ) = u 0 , with continuous α(x), β(x), then u(x) ≤ u 0 e
R (^) xx 0 α(s)ds^ +
Z (^) x x 0 β(t)e
R (^) tx α(s)dsdt.
- Intuition
- Gr¨onwall’s Lemma controls the growth of solutions of integral inequalities.
- Shows that if the growth of u(x) is bounded by an integral of itself, u(x) cannot blow up faster than an exponential.
- Fundamental tool in uniqueness proofs and stability estimates in ODEs.
- Applications
- Uniqueness of IVP: If y 1 , y 2 solve y′^ = f (x, y), then |y 1 (x) − y 2 (x)| ≤
Z (^) x x 0 L|y^1 (t)^ −^ y^2 (t)|dt^ =⇒^ y^1 ≡^ y^2.
- Stability Estimates: Estimate deviation of solutions under small pertur- bations.
- Numerical Analysis: Bound errors in Euler or Runge-Kutta methods.
- Properties
- Non-negativity required: u(x), v(x), α(x) ≥ 0
- Exponential bound appears naturally
- Useful in comparison principles: If u(x) ≤ v(x), then bound on v(x) bounds u(x)
- Can be used for both local and global solution estimates
- Solved Examples Example 1: u(x) ≤ 2 +
Z (^) x 0 3 u(t)dt By Gr¨onwall: u(x) ≤ 2 e^3 x Example 2 (Differential Form):
u′(x) ≤ 2 u(x) + 1, u(0) = 0
Solution bound:
u(x) ≤
Z (^) x 0 e
R (^) tx 2 ds dt =
Z (^) x 0 e
2(x−t)dt =^1 2 (e
2 x (^) − 1)
- Solved Examples Example 1: y′^ = x + y, y(0) = 1, R = [− 1 , 1] × [0, 2] ∂f∂y = 1 (bounded) =⇒ Lipschitz satisfied =⇒ Existence uniqueness guaranteed. Solution using integrating factor: y(x) = ex^ + x Example 2: y′^ = y^2 , y(0) = 1, R = [− 1 , 1] × [0, 2] ∂f∂y = 2y (bounded in rectangle) =⇒ Lipschitz locally Solution:
y(x) = (^1) −^1 x exists and unique on [0, 1)
- Notes
- The rectangle ensures local existence; the solution may be extended as long as it stays inside.
- The theorem generalizes to higher dimensions (systems of ODEs).
- Links directly with Lipschitz condition and Gr¨onwall’s Lemma for uniqueness proofs.
Picard Existence Theorem (Conceptual Version)
- Initial Value Problem Consider the IVP: ( y′^ = f (x, y), (x, y) ∈ R ⊂ R × R, y(x 0 ) = y 0 , where R = [x 0 − a, x 0 + a] × [y 0 − b, y 0 + b] is a rectangle in the xy-plane.
- Theorem Statement Existence: If f (x, y) is continuous on a rectangle R, then there exists at least one solution y(x) of the IVP on some interval [x 0 − h, x 0 + h] ⊂ [x 0 − a, x 0 + a]. Uniqueness: If f additionally satisfies a Lipschitz condition in y: ∂f ∂y ≤^ L^ on^ R, then the solution is unique in that interval.
- Conditions for Existence and Uniqueness
- f (x, y) is continuous on R =⇒ existence of solution
- ∂f /∂y exists and bounded on R =⇒ uniqueness
- Interval of existence depends on the size of rectangle and bounds of f
- Intuition
- Continuity of f ensures derivative does not blow up → a solution exists
- Lipschitz condition controls slope changes → solution is unique
- Geometrically: slope field is well-behaved, solution curves cannot cross
Rule: Interval of Solution in Rectangle Consider the IVP: y′^ = f (x, y), y(x 0 ) = y 0 on rectangle R = {(x, y) : |x − x 0 | ≤ a, |y − y 0 | ≤ b}.
- Check Continuity: Ensure f (x, y) is continuous in R =⇒ existence of solution.
- Compute Lipschitz Constant: L = (^) (x,ymax)∈R^ ∂f∂y. If L finite =⇒ unique- ness in rectangle.
- Maximum Interval: h = min
a, Mb
, M = (^) (x,ymax)∈R |f (x, y)| =⇒ solution exists uniquely in [x 0 − h, x 0 + h].
- Notes:
- If f continuous globally, solution may extend beyond rectangle.
- If ∂f /∂y unbounded, uniqueness may fail.
- If solution reaches boundary, interval cannot be extended without fur- ther analysis.
Question 3 IVP: y′^ = (^1) −y x, y(0) = 2
Rectangle: |x| ≤ 0. 5 , |y − 2 | ≤ 1 Options:
- ∂f /∂y = 1/(1 − x) ≤ 2
- Maximal interval: h ≤ min(a, b/L)
- Solution exists up to x = 1
- Solution becomes unbounded as x → 1 − Solution:
- L = 1/(1 − 0 .5) = 2
- Max interval h = b/L = 1/ 2
- Solution blows up as x → 1 − Answer: 1,2,
Question 4 IVP: y′^ = p1 + y^2 + x, y(0) = 0 Rectangle: |x| ≤ 0. 5 , |y| ≤ 1 Options:
- f continuous =⇒ existence
- Lipschitz in y fails globally
- Maximal interval h ≤ min(a, b/L) ≈ 0. 5
- Solution may escape rectangle Solution:
- ∂f /∂y = y/p1 + y^2 ≤ 0. 707 =⇒ Lipschitz locally
- Max interval h = min(0. 5 , 1 / 0 .707) = 0. 5
- Solution may escape rectangle outside interval Answer: 1,3,
Question 5 IVP: y′^ = y
1 + x^2 ,^ y(0) = 0 Rectangle: |x| ≤ 1 , |y| ≤ 1 Options:
- Lipschitz: L = 2
- Maximal interval: h = b/L = 0. 5
- Continuous globally =⇒ solution may extend
- Uniqueness holds locally Solution:
- ∂f /∂y = 2y/(1 + x^2 ) ≤ 2 → Lipschitz
- Max interval h = b/L = 0. 5
- f continuous globally → solution may extend
- Uniqueness locally Answer: 1,2,3,
Maximal Interval of Solution in Rectangle IVP: y′^ = f (x, y), y(x 0 ) = y 0 Rectangle: R = {(x, y) : |x − x 0 | ≤ a, |y − y 0 | ≤ b}
- Compute M = max(x,y)∈R |f (x, y)|
- Maximal interval: h = min
a, Mb
- Solution exists uniquely in [x 0 − h, x 0 + h]
- Notes:
- If f continuous globally, solution may extend beyond rectangle
- If ∂f /∂y unbounded, uniqueness may fail
- If solution reaches boundary, interval cannot be extended
Question 3 IVP: y′^ = (^1) −y x, y(0) = 2
Rectangle: |x| ≤ 0. 5 , |y − 2 | ≤ 1 Options:
- ∂f /∂y = 1/(1 − x) ≤ 2
- Maximal interval h = b/L = 0. 167
- Solution exists up to x = 1
- Solution blows up as x → 1 − Solution: L = 1/(1 − 0 .5) = 2, h = b/L = 1/ 6 ≈ 0 .167. Solution becomes unbounded at x = 1. Answer: 1,2,
Question 4 IVP: y′^ = p1 + y^2 + x, y(0) = 0 Rectangle: |x| ≤ 0. 5 , |y| ≤ 1 Options:
- f continuous =⇒ existence
- Lipschitz fails globally
- Maximal interval h = 0. 5
- Solution may escape rectangle outside interval Solution: ∂f /∂y = y/p1 + y^2 ≤ 0 .707, h = min(a, b/M ) = 0.5. All correct except global Lipschitz fails. Answer: 1,3,
Question 5 IVP: y′^ = y
1 + x^2 ,^ y(0) = 0 Rectangle: |x| ≤ 1 , |y| ≤ 1 Options:
- Lipschitz: L = 2
- Maximal interval h = 0. 5
- Continuous globally =⇒ solution may extend
- Uniqueness holds locally Solution: ∂f /∂y = 2y/(1 + x^2 ) ≤ 2, h = b/L = 0.5, solution may extend globally, uniqueness holds. Answer: 1,2,3,
Global Existence and Uniqueness Theorem Consider the IVP: y′^ = f (x, y), y(x 0 ) = y 0 Theorem: If f (x, y) is continuous in x for all x ∈ R and globally Lipschitz in y:
|f (x, y 1 ) − f (x, y 2 )| ≤ L|y 1 − y 2 |, ∀x, y 1 , y 2 ∈ R,
then the IVP has a unique solution for all x ∈ R. Notes:
- If f grows faster than linear in y (e.g., yp, p > 1), solution may blow up in finite time → global existence fails.
- Linear ODEs always satisfy global Lipschitz → global existence and unique- ness.
- Nonlinear but bounded growth in y → solution exists globally.
Example 1
y′^ = 2y + x, y(0) = 1 Solution: ∂f /∂y = 2 (bounded) → global Lipschitz. Answer: Global existence and uniqueness for all x ∈ R.
