# Derivative with respect to Time

• Derivatives and Bank Collapse
• But just as much as it is easy to find the derivative of a given quantity, so it is difficult to find the integral of a given derivative. Moreover, sometimes we cannot say with certainty whether the integral of a given quantity can be found or not. (Johann Bernoulli)
• Among all of the mathematical disciplines the theory of differential equations is the most important… It furnishes the explanation of all those elementary manifestations of nature which involve time. (Sophus Lie)

We are now ready to give a formal definition of the derivate $\dot u(t)$ of a function $u(t)$ depending on time t. We use here use the “dot-age” notation of Newton to be compared with Leibniz’ notation $\frac{du}{dt}$ or $Du(t)$.

Definition: A function $u: I\rightarrow R$ defined on an interval $I=(a,b)$, is said to be differentiable in I with derivative $\dot u: I\rightarrow R$ if for some positive constant $C_u$

• $\vert u(t+dt)-u(t) -\dot u(t)*dt\vert \le C_u\vert dt\vert^2\quad\mbox{for }t,t+dt\in I$.

A differentiable function $u(t)$ is locally close to a linear function in t in the sense that $u(t+dt)\approx u(t)+\dot u(t)*dt$ up to a quadratic term in $\vert dt\vert$.

A differentiable function is Lipschitz continuous, since it is locally close to a linear function and a linear function is Lipschitz continuous.

We note the following connection between Lipschitz continuity and the size of the derivative.

Theorem: If $u:I\rightarrow R$ is differentiable with $\vert \dot u(t)\vert\le L$ for $t\in I$, then $u:I\rightarrow R$ is Lipschitz continuous with Lipschitz constant L.

Proof: This result should be intuitively obvious: It is like saying that if your velocity is never bigger than 1 km/hour, then it is impossible to travel a distance longer than 1 km in an hour. Right? If you hesitate, consider the follwoing formal proof:

Given $t,s\in I$, we are supposed to show that

• $\vert u(t)-u(s)\vert \le L*\vert t-s\vert$.

To this end let us note that it is sufficient to prove that for any given $\epsilon >0$,

• $\vert u(t)-u(s)\vert \le (L+\epsilon )*\vert t-s\vert$ .

Now choose first $\epsilon >0$ and then dt such that $C_udt\le \epsilon$ and $(t-s) =n*dt$ for some natural number n, assuming $t>s$. We then have splitting the interval $(s,t)$ into $n$ subintervals of length $dt$: $(s,s+dt)$, $(s+dt,s+2dt)$,….$(s+(n-1)dt, s+ndt)$, and using the definition of $\dot u$ on each subinterval:

• $\vert u(s+(m+1)dt)-u(s+mdt)-\dot u(s+mdt)*dt\vert\le C_udt^2 \quad\mbox{for } m=0,...,n-1$,

that is, using that $\vert \dot u(s+mdt)\vert \le L$,

• $\vert u(s+(m+1)dt)-u(s+mdt)\vert\le L*dt + C_u*dt^2$.

By the triangle inequality we now have since $n*dt =t-s =\vert t-s\vert$,

• $\vert u(t)-u(s)\vert \le \vert u(s+dt)-u(s)\vert +\vert u(s+2dt)-u(s+dt)\vert +...\vert u(s+(n-1)dt-u(t)\vert$
• $\le n(L*dt +C_u*dt^2)\le L*\vert t-s\vert + C_u*dt*\vert t-s\vert=(L+\epsilon )*\vert t-s\vert$,

which we wanted to show.

#### Example: Kink

The function $u(t)=\vert t\vert$ is Lipschitz continous (in particular at $t=0$), but is not differentiable at $t=0$ because it is not close to a linear function for $t$ close to 0 (since it has a kink).

#### To Remember:

• A Lipshitz continuous function is locally close to a constant function.
• A differentiable function is locally close to a linear function

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