01 - Picard's Existence Theorem

Picard's Existence Theorem

Let $f:R\to \mathbb{R}$ be a function defined on rectangle

$$R:=\{(x,y):|x-a|\le h,|y-b|\le k\}$$

Which satisfies

• $P(i)$ - $(a)$: $f$ is continuous in $R$ with $|f(x,y)|\le M$ for all $(x,y)\in R$
• $P(i)$ - $(b)$: $Mh\le k$
• $P(ii)$: $f$ satisfies Lipschitz Condition in $R$

Then IVP

$$y^{\prime }(x)=f(x,y(x))\text{ with }y(a)=b$$

has a unique solution $y$ on interval $[a-h,a+h]$

Picard’s Theorem doesn’t give the best (largest) interval where existence/uniqueness holds

Proof - Existence - Via Successive Approximation

Successive Approximation (Iteration)

Start of with constant function $y_0(x)$

$$y_0(x)=b\forall x\in [a-h,a+h]$$

Then for $n=0,1,\cdots$ then

$$y_{n+1}(x):=b+{\int }_a^{x}f(t,y_n(t))dt$$

Link to original

Claim 1: Each $y_n(x)$ is well-defined and continuous on $[a-h,a+h]$ which satisfies

$$|y_n(x)-b|\le k\text{ for all }x\in |a-h,a+h|$$

Claim 1

True for $n=0$ as it is constant
For $t\in [a-h,a+h]$ then $(t,y_n(t))\in R$ so evaluate $f$ at this point

As $y_n:[a-h,a+h]\to [b-k,b+k]$ is continuous then

$$t\mapsto (t,y_n(t))\text{ is a continious function}$$

from $[a-h,a+h]$ to the rectangle $R$
As $f$ is continuous function from $R$ to $\mathbb{R}$ then

$$\text{composition }[a-h,a+h]\ni t\mapsto f(t,y_n(t))\in \mathbb{R}\text{ is continious }$$

Hence it is integrable

So next iterate $y_{n+1}$ is well-defined function from $[a-h,a+h]\to \mathbb{R}$
So by properties of integration it is differentiable hence continuous

Hence we have

$$|y_{n+1}(x)-b|\le \left|{\int }_a^x|f(t,y_n(t))|dt\right|\le \left|{\int }_a^{x}Mdt\right|=M|x-a|\le Mh\le k$$

for every $x\in [a-h,a+h]$
PG 12