Picard定理

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DE1notes21-10-06.pdf page 9
常微分方程的柯西问题的存在与唯一性定理


Theorem 1.1. (Picard's existence theorem):
Let $f: R \rightarrow \mathbb{R}$ be a function defined on the rectangle $R:=\{(x, y):|x-a| \leq h,|y-b| \leq k\}$ which satisfies
P(i) : (a) $f$ is continuous in $R$, with bound $M$ (so $|f(x, y)| \leq M$) and (b) $M h \leq k$.
P(ii) : $f$ satisfies a Lipschitz condition in $R$.
Then the IVP
$$
y'(x)=f(x, y(x)) \text { with } y(a)=b .
$$
has a unique solution $y:[a-h, a+h] \rightarrow[b-k, b+k]$.

We will prove the existence of a solution by showing that the iterates $y_{n}$ defined in (1.5) converge as $n \rightarrow \infty$ to a solution $y$ of the IVP and will do this by showing that the series in (1.7) converges. We break the proof into a series of steps:

Claim 1: Each $y_{n}$ is well defined, continuous and $\left|y_{n}(x)-b\right| \leq k$ for $x \in[a-h, a+h]$.
Proof of Claim 1: This is clearly true for $n=0$, so suppose claim is true for some $n \geq 0$. Then for $t \in[a-h, a+h]$ we have that $\left(t, y_{n}(t)\right) \in R$ so as $f$ is defined and continuous on $R$ and as $y_{n}$ is continuous we have that $t \mapsto f\left(t, y_{n}(t)\right)$ is a continuous function on the interval $[a-h, a+h]$. Thus $y_{n+1}$ is well defined and continuous by properties of integration and by P(i)
\begin{aligned}\left|y_{n+1}(x)-b\right| & \leq\left|\int_{a}^{x}\left| f\left(t, y_{n}(t)\right)\right|d t\right| \\ & \leq\left|M \int_{a}^{x} d t\right|=M|x-a| \leq M h \leq k \end{aligned}for every $x ∈ [a − h, a + h]$. Thus the claim is true by induction.

Figure 1.2: successive iterates graphed in $R$.

We next prove:
Claim 2: For $|x-a| \leq h$ and $n \in \mathbb{N}$
$$
\left|e_{n}(x)\right| \leq \frac{L^{n-1} M}{n !}|x-a|^{n}\tag{1.11}
$$
where $L$ is such that the Lipschitz condition (1.9) holds.
We remark that this claim in particular implies that
$$
\left|e_{n}(x)\right| \leq \frac{L^{n-1} M}{n !} h^{n} \text { for all }|x-a| \leq h .\tag{1.12}
$$
which will be the estimate that we will use to show that $\sum e_{n}$ converges uniformly using M-test.

Proof of Claim 2: We recall that the Lipschitz condition P(ii) combined with the fact that the graph of $y_{n}$ is in the rectangle implies that for all $|t-a| \leq h$
$$
\left|f\left(t, y_{n}(t)\right)-f\left(t, y_{n-1}(t)\right)\right| \leq L\left|y_{n}(t)-y_{n-1}(t)\right|=L\left|e_{n}(t)\right|\tag{1.13}
$$
From (1.8) and (1.13) thus
Proof of Claim 2: We recall that the Lipschitz condition P(ii) combined with the fact that the graph of $y_{n}$ is in the rectangle implies that for all $|t-a| \leq h$
$$
\left|f\left(t, y_{n}(t)\right)-f\left(t, y_{n-1}(t)\right)\right| \leq L\left|y_{n}(t)-y_{n-1}(t)\right|=L\left|e_{n}(t)\right| .
$$
From (1.8) and (1.13) thus
\begin{aligned}
\left|e_{n+1}(x)\right| & \leq\left|\int_{a}^{x}\left| f\left(t, y_{n}(t)\right)-f\left(t, y_{n-1}(t)\right) \right| d t\right| \\& \leq L\left|\int_{a}^{x}\left| e_{n}(t)\right|d t\right| .
\end{aligned}
Now we prove (1.11) by induction:
$$
e_{1}(x)=y_{1}(x)-b=\int_{a}^{x} f(t, b) d t
$$
By P(i), $f$ is bounded by $M$ so that
$$
\left|e_{1}(x)\left| \leq\left|\int_{a}^{x}\right| f(t, b)\right|d t\right| \leq M|x-a|
$$
so (1.11) is true for $n=1$. Now suppose that (1.11) is true for $n$, then