Showing unique solution to an Initial Value Problem

Question

I am trying to show that the equation $$dy/dt = \cos^2(y)$$ has a unique solution when $$0 \le t \le1$$ and $$y(0) = \pi/4$$ I don't believe I can just solve this by integrating both sides like I can with simple IVPs, yet it seems from some searching on the web that the "Lipschitz condition" can be used to demonstrate that this equation has a unique solution. I'm finding it very difficult to understand what the Lipschitz condition means, and would be very grateful both for an explanation of what it means conceptually as well as how it can be used to solve this problem.

Answer 1

Don't fret when you see a $y$ inside parentheses in a differential equation!

The equation

$$\dfrac{dy}{dt}=\cos^2(y)$$

looks nonlinear but is actually 'trivially' integrable.

As long as a ODE is of the form

$$y'(x)=A(y(x))\cdot B(x)$$

its integrable (regardless of whether your variables are $y(x)$, $x{t}$ or anything else). Or more generally

$$y^{(n)}(x)=A(y(x))\cdot B(x)$$

where $y^{(n)}$ is the $n$-th derivative of $y$.

Solution by separation of variables

$$\dfrac{dy}{dx}=\cos^2(y)$$ $$\dfrac{dy}{cos^2(y)}=dx$$ $$\int \dfrac{dy}{cos^2(y)}=\int dx$$

Now you can use $u$-substitution to get the antiderivative of $\frac{1}{cos^2(x)}$ or remember what is the derivative of $\tan(x)$.

$$\tan(y)=x+C \hspace{10mm} C\in \mathbb{R}$$

Explicitly:

$$y=\arctan(x+C)$$

To find the IVP you need to plug in $0$ and $\pi/4$, solve for $C$ and put that $C$ value back to the solution.

Let's prove the solution by the Lipschitz continuity

This condition is usually introduced when learning about solution of systems of differential equations. The theorem states that if there is a positive real number $K$ and $K\cdot |x_1-x_2|$ is greater or equal to $|f(x_1)-f(x_2)|$, then $f(x)$ is Lipschitz continous on set $X$ for any $x_1,x_1\in X$.

Let's dissect this:

Imagine you have a function and you draw a secant of it (a line that crosses the function twice) you take. The absolute value of the absolute value of its steepness is essentially the number $K$:

$$\dfrac{|f(x_1)-f(x_2)|}{|x_1-x_2|}\leq K$$

If $x_2\to x_1$ then this is just the absolute value of its derivative at $x_1$:

$$\lim\limits_{x_2 \to x_1}\dfrac{|f(x_1)-f(x_2)|}{|x_1-x_2|}=\left|\dfrac{d\hspace{1mm}f(x_1)}{dx}\right|$$

I'm sure you can imagine a number that has a greater value than the steepness of a function's derivative. The reason you see a cone at the Wikipedia page of it is because we took the absolute value of the differences of $x_1$ and $x_2$, thus always having a steeper $K$.

$$\dfrac{d\hspace{1mm}f(x_1)}{dx}\leq K$$

If there was no greater or equal $K$ the function wouldn't really be well-defined.

If $x_1\neq x_2$ we have more problems. Imagine the function

$$f(x)=x^2$$

We know that its steepness continously grows, thus we cannot find an upper bound where the function won't cross the cone. This function is globally not Lipschitz continous, it's not Lipschitz continous on its entire domain.

Another example would be

$$f(x)=\sqrt{x}$$ on the region $[0,1]$ as we get closer to $0$ the steepness of the function grows to $\infty$ thus we'll always have an intersection.

Let our function be

$$y(t)=\tan(t)$$

on the domain $[0,1]$

From the graph of $tan(t)$ it's clearly visible that the function is Lipschitz continous (locally) on the domain.

If you want prove it:

$$\dfrac{|tan(t_1)-tan(t_2)|}{|t_1-t_2|}\leq K$$

$$\exists K\in\mathbb{R}^+ \hspace{2mm}\forall t_1, t_2 \in \mathbb{R}$$

Ironically the famous universal solution of linear DE, the exponential function is not Lipschitz continous. It is analytic though.

Additional fun exercise: Try solving

$$\frac{dy}{dt}=cos^2(t)$$

Compare it to your original problem. Is there a connection between the two.