# Use Mean Value Theorem to Determine if an Initial Value Problem Has a Solution (Lipschitz)

## 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!

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

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

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

• Thanks for the advice; I posted a full solution below. Do you mind helping me check for inconsistencies and mistakes? – Anthony Krivonos Sep 19 '18 at 16:31
• $F(t,y)$ is continous over the interval $[a,b]$. hmm... but $(t,y) \in \mathbb{R}^2$ and $[a,b] \subset \mathbb{R}$. – Siong Thye Goh Sep 19 '18 at 16:39

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 $$\mathbb R$$.