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Picard’s Existence/Uniqueness Theorem; Lipschitz Condition

At a Glance

Why This Chapter Matters

This atom appeared once (2024) in Section B for 15 marks. UPSC tests the theoretical side of ODEs: the precise statement of the Picard–Lindelöf theorem, verification of the Lipschitz condition for a given function, and the construction of Picard iterates. A question may ask for a full proof-sketch, or may ask you to apply the theorem to determine existence/uniqueness and compute the first few iterates for a specific IVP.

Minimum Theory

The initial-value problem

y=f(x,y),y(x0)=y0y' = f(x,y), \quad y(x_0) = y_0

Lipschitz condition

f(x,y)f(x,y) satisfies a Lipschitz condition with respect to yy on a region RR if there exists a constant K>0K > 0 (the Lipschitz constant) such that:

f(x,y1)f(x,y2)Ky1y2for all (x,y1),(x,y2)R|f(x,y_1) - f(x,y_2)| \leq K|y_1 - y_2| \quad \text{for all }(x,y_1),(x,y_2)\in R

Sufficient condition: If f/y\partial f/\partial y exists and fyK\left|\dfrac{\partial f}{\partial y}\right| \leq K on RR, then ff satisfies the Lipschitz condition with constant KK (by the Mean Value Theorem).

Picard–Lindelöf (Picard’s Existence and Uniqueness) Theorem

Hypotheses. Let R={(x,y):xx0a,  yy0b}R = \{(x,y): |x-x_0|\leq a,\; |y-y_0|\leq b\}. Suppose:

  1. f(x,y)f(x,y) is continuous on RR, so f(x,y)M|f(x,y)| \leq M for some M>0M > 0 on RR.
  2. ff satisfies the Lipschitz condition in yy on RR with constant KK.

Conclusion. The IVP y=f(x,y)y' = f(x,y), y(x0)=y0y(x_0) = y_0 has a unique solution on the interval xx0h|x - x_0| \leq h, where

h=min ⁣(a,bM)h = \min\!\left(a,\, \frac{b}{M}\right)

Picard iteration (successive approximations)

Define the sequence:

y0(x)=y0y_0(x) = y_0

yn+1(x)=y0+x0xf ⁣(t,yn(t))dt,n=0,1,2,y_{n+1}(x) = y_0 + \int_{x_0}^{x} f\!\left(t,\, y_n(t)\right) dt, \quad n = 0, 1, 2, \ldots

Under the hypotheses, {yn(x)}\{y_n(x)\} converges uniformly to the unique solution on xx0h|x-x_0|\leq h.

Error bound

y(x)yn(x)MKnhn+1(n+1)!11?|y(x) - y_n(x)| \leq \frac{MK^n h^{n+1}}{(n+1)!} \cdot \frac{1}{1 - ?}

A useful simpler bound: the iterates converge because the error is of order (Kh)n/(n!)(Kh)^n/(n!) which tends to zero.

Proof sketch (for 15-mark questions)

  1. Equivalence: The IVP y=f,y(x0)=y0y' = f, y(x_0)=y_0 is equivalent to the integral equation y=y0+x0xf(t,y(t))dty = y_0 + \int_{x_0}^x f(t,y(t))\,dt.
  2. Picard operator: Define T[y](x)=y0+x0xf(t,y(t))dtT[y](x) = y_0 + \int_{x_0}^x f(t,y(t))\,dt; a solution is a fixed point of TT.
  3. Successive approximations stay in RR: Show y1(x)y0Mxx0Mhb|y_1(x)-y_0|\leq M|x-x_0|\leq Mh\leq b, so y1y_1 stays in RR; by induction all yny_n stay in RR.
  4. Uniform convergence: Show yn+1ynMKnhn+1(n+1)!|y_{n+1}-y_n| \leq \dfrac{MK^n h^{n+1}}{(n+1)!} by induction using the Lipschitz condition; this is the general term of the exponential series, which converges.
  5. Limit is a solution: Passing the limit inside the integral (justified by uniform convergence) shows the limit y=limyny = \lim y_n satisfies the integral equation.
  6. Uniqueness: If yy and zz are two solutions, then yzKx0xyzdt|y-z| \leq K\int_{x_0}^x|y-z|\,dt; by Grönwall’s inequality (or repeated application), yz=0|y-z| = 0.

Question Archetypes

ArchetypeRecognition
picard-iterate-and-state”State the Picard theorem, verify Lipschitz condition, and compute Picard iterates”

picard-iterate-and-state (1 question; 2024)

Recognition Cues

Solution Template

  1. State the Picard–Lindelöf theorem precisely (hypotheses and conclusion).
  2. Verify continuity of ff on RR and compute M=maxRfM = \max_R|f|.
  3. Verify the Lipschitz condition (compute f/y\partial f/\partial y; bound it by KK).
  4. State the interval of existence xx0h=min(a,b/M)|x-x_0|\leq h = \min(a, b/M).
  5. Compute y0y_0, y1y_1, y2y_2 (and y3y_3 if marks warrant) by the Picard iteration formula.
  6. Identify the pattern if the iterates form a recognisable series (e.g., partial sums of exe^x).

Worked Example

2024 Paper 1, 2024-P1-Q7b (15 marks)

State Picard’s existence and uniqueness theorem. Verify its hypotheses for the IVP y=x+y,y(0)=0,y' = x + y, \quad y(0) = 0, and compute the first three Picard iterates y1,y2,y3y_1, y_2, y_3.

Step 1. Statement of the theorem.

Let R:xaR: |x|\leq a, yb|y|\leq b. If f(x,y)f(x,y) is continuous on RR with fM|f|\leq M and satisfies f(x,y1)f(x,y2)Ky1y2|f(x,y_1)-f(x,y_2)|\leq K|y_1-y_2| on RR, then y=f(x,y)y'=f(x,y), y(0)=0y(0)=0 has a unique solution on xh=min(a,b/M)|x|\leq h = \min(a, b/M).

Step 2. Verify hypotheses for f(x,y)=x+yf(x,y) = x+y.

Take R:xaR: |x|\leq a, yb|y|\leq b (e.g., a=b=1a=b=1).

f=x+yf = x+y is a polynomial, hence continuous on RR. On RR: f=x+ya+b=M|f| = |x+y| \leq a + b = M.

Lipschitz condition: f(x,y1)f(x,y2)=y1y2|f(x,y_1)-f(x,y_2)| = |y_1-y_2|, so K=1K = 1.

Interval of existence: h=min ⁣(1,11+1)=12h = \min\!\left(1,\, \dfrac{1}{1+1}\right) = \dfrac{1}{2}.

Step 3. Picard iterates. Set x0=0x_0 = 0, y0=0y_0 = 0.

y0(x)=0y_0(x) = 0

y1(x)=0+0x(t+y0(t))dt=0xtdt=x22y_1(x) = 0 + \int_0^x (t + y_0(t))\,dt = \int_0^x t\,dt = \frac{x^2}{2}

y2(x)=0+0x(t+y1(t))dt=0x(t+t22)dt=x22+x36y_2(x) = 0 + \int_0^x \left(t + y_1(t)\right)dt = \int_0^x \left(t + \frac{t^2}{2}\right)dt = \frac{x^2}{2} + \frac{x^3}{6}

y3(x)=0+0x(t+y2(t))dt=0x(t+t22+t36)dt=x22+x36+x424y_3(x) = 0 + \int_0^x \left(t + y_2(t)\right)dt = \int_0^x \left(t + \frac{t^2}{2} + \frac{t^3}{6}\right)dt = \frac{x^2}{2} + \frac{x^3}{6} + \frac{x^4}{24}

Step 4. Pattern and limiting solution.

The iterates are partial sums of the series for exx1e^x - x - 1:

yn(x)y(x)=exx1y_n(x) \to y(x) = e^x - x - 1

which can be verified: (exx1)=ex1=(exx1)+x(e^x - x - 1)' = e^x - 1 = (e^x - x - 1) + x, and y(0)=0y(0) = 0. ✓

y1=x22,y2=x22+x36,y3=x22+x36+x424;y=exx1\boxed{y_1 = \frac{x^2}{2},\quad y_2 = \frac{x^2}{2}+\frac{x^3}{6},\quad y_3 = \frac{x^2}{2}+\frac{x^3}{6}+\frac{x^4}{24};\quad y = e^x - x - 1}

Common Traps

Marks-Aware Writing

This is a 15-mark Section B proof+application question. Examiners expect:

  1. Precise theorem statement — hypotheses (RR, continuity, MM, Lipschitz, KK) and conclusion (hh, unique solution) (4 marks).
  2. Hypothesis verification — continuity confirmed, MM computed, Lipschitz checked with explicit KK (3 marks).
  3. Interval of existence computed (1 mark).
  4. Three Picard iterates computed step by step, with integral shown for each (6 marks, 2 per iterate).
  5. Pattern / exact solution identified and verified (1 mark).

Do not skip showing the integral at each step — substituting yny_n into the integrand must be written out.

Practice Set

Only one historical question on this atom (shown above).

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