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## 数学代写|常微分方程代写ordinary differential equation代考|Dependence on the initial condition

Usually, in applications several data are only known approximately. If the problem is well-posed, one expects that small changes in the data will result in small changes of the solution. This will be shown in our next theorem. As a preparation we need Gronwall’s inequality.

Lemma $2.7$ (Generalized Gronwall’s inequality). Suppose $\psi(t)$ satisfies
$$\psi(t) \leq \alpha(t)+\int_0^t \beta(s) \psi(s) d s, \quad t \in[0, T]$$
with $\beta(t) \geq 0$. Then
$$\psi(t) \leq \alpha(t)+\int_0^t \alpha(s) \beta(s) \exp \left(\int_s^t \beta(r) d r\right) d s, \quad t \in[0, T] .$$
Moreover, if in addition $\alpha(s) \leq \alpha(t)$ for $s \leq t$, then
$$\psi(t) \leq \alpha(t) \exp \left(\int_0^t \beta(s) d s\right), \quad t \in[0, T] .$$
Proof. Abbreviate $\phi(t)=\exp \left(-\int_0^t \beta(s) d s\right)$. Then one computes
$$\frac{d}{d t} \phi(t) \int_0^t \beta(s) \psi(s) d s=\beta(t) \phi(t)\left(\psi(t)-\int_0^t \beta(s) \psi(s) d s\right) \leq \alpha(t) \beta(t) \phi(t)$$

by our assumption (2.34). Integrating this inequality with respect to $t$ and dividing the resulting equation by $\phi(t)$ shows
$$\int_0^t \beta(s) \psi(s) d s \leq \int_0^t \alpha(s) \beta(s) \frac{\phi(s)}{\phi(t)} d s .$$
Adding $\alpha(t)$ on both sides and using again (2.34) finishes the proof of the first claim. The second is left as an exercise (Problem 2.10).

We will also use the following simple consequence (Problem 2.11). If
$$\psi(t) \leq \alpha+\int_0^t(\beta \psi(s)+\gamma) d s, \quad t \in[0, T]$$
with $\beta \geq 0$, then
$$\psi(t) \leq \alpha \exp (\beta t)+\frac{\gamma}{\beta}(\exp (\beta t)-1), \quad t \in[0, T]$$
Now we can show that our IVP is well-posed.

## 数学代写|常微分方程代写ordinary differential equation代考|Extensibility of solutions

We have already seen that solutions might not exist for all $t \in \mathbb{R}$ even though the differential equation is defined for all $t \in \mathbb{R}$. This raises the question about the maximal interval on which a solution of the IVP (2.11) can be defined.

Suppose that solutions of the IVP (2.11) exist locally and are unique (e.g., $f$ is Lipschitz). Let $\phi_1, \phi_2$ be two solutions of the IVP (2.11) defined on the open intervals $I_1, I_2$, respectively. Let $I=I_1 \cap I_2=\left(T_{-}, T_{+}\right)$and let $\left(t_{-}, t_{+}\right)$be the maximal open interval on which both solutions coincide. I claim that $\left(t_{-}, t_{+}\right)=\left(T_{-}, T_{+}\right)$. In fact, if $t_{+}<T_{+}$, both solutions would also coincide at $t_{+}$by continuity. Next, considering the IVP with initial condition $x\left(t_{+}\right)=\phi_1\left(t_{+}\right)=\phi_2\left(t_{+}\right)$shows that both solutions coincide in a neighborhood of $t_{+}$by Theorem 2.2. This contradicts maximality of $t_{+}$and hence $t_{+}=T_{+}$. Similarly, $t_{-}=T_{-}$.
Moreover, we get a solution
$$\phi(t)= \begin{cases}\phi_1(t), & t \in I_1 \ \phi_2(t), & t \in I_2\end{cases}$$
defined on $I_1 \cup I_2$. In fact, this even extends to an arbitrary number of solutions and in this way we get a (unique) solution defined on some maximal interval.

Theorem 2.12. Suppose the IVP (2.11) has a unique local solution (e.g. the conditions of Theorem 2.5 are satisfied). Then there exists a unique maximal solution defined on some maximal interval $I_{\left(t_0, x_0\right)}=\left(T_{-}\left(t_0, x_0\right), T_{+}\left(t_0, x_0\right)\right)$.
Remark: If we drop the requirement that $f$ is Lipschitz, we still have existence of solutions (see Theorem $2.18$ below), but we already know that we loose uniqueness. Even without uniqueness, two given solutions of the IVP (2.11) can slill be glued Logether al $l_0$ (if necessary) to ublain a sululion defined on $I_1 \cup I_2$. Furthermore, Zorn’s lemma ensures existence of maximal solutions in this case.

Now let us look at how we can tell from a given solution whether an extension exists or not.

# 常微分方程代写

## 数学代写|常微分方程代写ordinary differential equation代考|Dependence on the initial condition

$$\psi(t) \leq \alpha(t)+\int_0^t \beta(s) \psi(s) d s, \quad t \in[0, T]$$

$$\psi(t) \leq \alpha(t)+\int_0^t \alpha(s) \beta(s) \exp \left(\int_s^t \beta(r) d r\right) d s, \quad t \in[0, T] .$$

$$\psi(t) \leq \alpha(t) \exp \left(\int_0^t \beta(s) d s\right), \quad t \in[0, T] .$$

$$\frac{d}{d t} \phi(t) \int_0^t \beta(s) \psi(s) d s=\beta(t) \phi(t)\left(\psi(t)-\int_0^t \beta(s) \psi(s) d s\right) \leq \alpha(t) \beta(t) \phi(t)$$

$$\int_0^t \beta(s) \psi(s) d s \leq \int_0^t \alpha(s) \beta(s) \frac{\phi(s)}{\phi(t)} d s .$$

$$\psi(t) \leq \alpha+\int_0^t(\beta \psi(s)+\gamma) d s, \quad t \in[0, T]$$

$$\psi(t) \leq \alpha \exp (\beta t)+\frac{\gamma}{\beta}(\exp (\beta t)-1), \quad t \in[0, T]$$

## 数学代写|常微分方程代写ordinary differential equation代考|Extensibility of solutions

$$\phi(t)=\left{\phi_1(t), \quad t \in I_1 \phi_2(t), \quad t \in I_2\right.$$

$I_{\left(t_0, x_0\right)}=\left(T_{-}\left(t_0, x_0\right), T_{+}\left(t_0, x_0\right)\right)$

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