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Problem

Determine whether the initial value problem $$ y' = \cos(t + y) $$ given $ y(t_0) = y_0$ has a unique solution defined on all of $ \mathbb{R} $.

Hint: Use the mean value theorem.

Attempt

Let $$ F(t, y) = \cos(t + y).$$ Let $ [a, b] $ be a particular, but arbitrarily chosen interval. Then $$ F_y(t, y) = − \sin(t + y). $$ Let $y_1$ and $y_2 \in \mathbb{R}$. Then, by the mean value theorem there is a $y_0$ between $y_1$ and $y_2$ and also $$| F(t, y_1) − F(t, y_2) | \leq k|y_1 - y_2|.$$

Notes

I got to the point of $$| \cos(t + y_1) − \cos(t + y_2) | \leq k|y_1 - y_2|, $$ but am having trouble simplifying the LHS to look like the RHS, getting a particular value $\leq k$.

Thanks!

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By mean value theorem, there exists a $q$ in between $t+y_1$ and $t+y_2$ such that

$$\cos(t+y_1) - \cos(t+y_2) = -\sin(q)((t+y_1-(t+y_2))$$

$$|\cos(t+y_1) - \cos(t+y_2)| = |-\sin(q)||((t+y_1-(t+y_2))|\le |y_1-y_2|$$

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  • $\begingroup$ Thanks for the advice; I posted a full solution below. Do you mind helping me check for inconsistencies and mistakes? $\endgroup$ – Anthony Krivonos Sep 19 '18 at 16:31
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    $\begingroup$ $F(t,y)$ is continous over the interval $[a,b]$. hmm... but $(t,y) \in \mathbb{R}^2$ and $[a,b] \subset \mathbb{R}$. $\endgroup$ – Siong Thye Goh Sep 19 '18 at 16:39
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Let $ F(t, y) = y' = cos(t + y)$, and $(t_0, y_0)$ be an interior point on the interval $[a,b]$. Since $t \in [a, b]$ and $y \in \mathbb R$, we know that $F(t,y)$ is continuous over the interval $[a,b]$ and differentiable over the interval $(a, b)$.

The mean value theorem (MVT) states that, given an arbitrary function $f(t)$ that is continuous over a closed interval $[a_0, b_0]$ and differentiable over an open interval $(a_0, b_0)$, there exists a value $c_0$ such that

$$ f'(c_0) = \frac{f(b_0)-f(a_0)}{b_0-a_0}. $$

We can apply the MVT to show that $ F(t,y) $ is Lipschitz continuous. Taking an arbitrary $ y_1, y_2 \in \mathbb R $, we must show that there exists a $ y_3 $ between $y_1$ and $y_2$ with $t \in [a,b]$ such that

$$ F_y(t, y_3) = \frac{F(t, y_2)-F(t, y_1)}{y_2-y_1}. $$

Differentiating $F(t,y)$ by $y$ gives us $F_y(t,y) = -sin(t + y)$. Substituting $F_y(t,y)$ into the LHS gives us

$$ -sin(t + y_3) = \frac{F(t, y_2)-F(t, y_1)}{y_2-y_1} $$ $$ -sin(t + y_3)(y_2-y_1) = F(t, y_2)-F(t, y_1) $$ $$ |-sin(t + y_3)(y_2-y_1)| = |F(t, y_2)-F(t, y_1)| $$ $$ |-sin(t + y_3)||y_2-y_1| = |F(t, y_2)-F(t, y_1)| $$ $$ |sin(t + y_3)||y_1-y_2| = |F(t, y_1)-F(t, y_2)| $$

Since the range of $ f(x) = sin(x) $ is $ [-1, 1] $, we know that $ sin(t + y_3) $ on $ \mathbb R $ has a maximum of $1$ and a minimum of $-1$. According to the definition of Lipschitz continuity, for $F(t,y)$ to be closed continuous, we must take a $k > 0$ to satisfy $$ |F(t,y_1) - F(t,y_2)| \leq k |y_1 - y_2|. $$ Thus, $$ |F(t, y_1)-F(t, y_2)| = |sin(t + y_3)||(y_1-y_2)| \leq 1|y_1-y_2| $$ where, in this case, $ k = 1 $. By the global existence and uniqueness theorem, there is a unique solution for $F(t,y)$ on the arbitrary domain $ t \in [a,b] $, and so the solution is defined on $ \mathbb R $.

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