← 2024 Paper 1

UPSC 2024 Maths Optional Paper 1 Q7a — Step-by-Step Solution

15 marks · Section B

Picard's Existence/Uniqueness Theorem; Lipschitz Condition · ODEs · Read the full method →

Question

State the uniqueness theorem for the existence of a unique solution of the IVP $\dfrac{dy}{dx}=f(x,y)$, $y(x_0)=y_0$ in the rectangle $R:|x-x_0|\le a,\,|y-y_0|\le b$. Test the existence and uniqueness of the solution of $\dfrac{dy}{dx}=2\sqrt{y},\;y(1)=0$. If more than one solution exists, find all of them.

Technique

State Picard–Lindelöf; check Lipschitz via $\partial f/\partial y$; construct the non-unique solution family by patching.

Solution

Picard–Lindelöf Theorem

Let $f(x,y)$ be continuous on $R:|x-x_0|\le a,\,|y-y_0|\le b$, and satisfy a Lipschitz condition in $y$ on $R$:

for some $L>0$. Let $M=\max_R|f|$ and $h=\min(a,b/M)$. Then the IVP has a unique solution on $|x-x_0|\le h$.

Application to $f(x,y)=2\sqrt{y}$, $y(1)=0$

Continuity: $2\sqrt{y}$ is continuous on $y\ge 0$. Existence is guaranteed by Peano’s theorem.

Lipschitz check: $\partial f/\partial y=1/\sqrt{y}\to\infty$ as $y\to 0^+$. More directly, for $y_1,y_2\ge 0$:

No Lipschitz constant exists on any rectangle containing $(1,0)$. Picard’s theorem does not guarantee uniqueness.

All Solutions

Separating variables for $y>0$: $dy/(2\sqrt{y})=dx\Rightarrow \sqrt{y}=x+C$. With $y(1)=0$: $C=-1$, so one solution is $y=(x-1)^2$ for $x\ge 1$.

But $y\equiv 0$ also satisfies the ODE and $y(1)=0$.

General family. For each $a\ge 1$, define

At the join $x=a$: $y_a$ is continuous (both give $0$) and $y_a'$ is continuous (both give $0$). Each piece satisfies the ODE. And $y_a(1)=0$ since $a\ge 1$.

Answer

There are infinitely many solutions. The case $a=1$ gives $y=(x-1)^2$ for $x>1$; $a\to\infty$ gives $y\equiv 0$.