Apologies for any mistakes or inefficiencies in my submission and wording of the question. It is my first time using stackexchange.

The IVP is as follows:

$u' = \cos(u(t))$, subject to the initial conditions $u(0) = 0$, where $0\le t\le 1$.

I separated the variables, solved for the general solution, and applied the initial conditions giving me the solution $u(t) = 2\arctan(e^{t + \log(\tan(\pi))}) - \frac{\pi}{2}$

According to wolfram alpha, however, the correct solution is $u(t) = 2\arctan(\tanh(\frac{t}{2}))$.

I would like to see where my error is as well as how to show well-posedeness of the problem. I have already shown the solution to exist and be unique, however, I don't quite know how to show that it's behaviour changes continuously with respect to the initial conditions(stability).


It is difficult to find where your error is if you do not show your work. However, something is definitely wrong, since $\log(\tan\pi)$ is undefined.

Let's check now the well poshness. Let $u(x,a)$ be the solution with initial value $u(0)=a$. Then \begin{align} |u(x,a)-u(x,b)|&=\Bigl|a-b+\int_0^x\bigl(\cos(u(t,a))-\cos(u(t,b))\bigr)\,dt\Bigr|\\ &\le|a-b|+\int_0^x|\cos(u(t,a))-\cos(u(t,b))|\,dt\\ &\le|a-b|+\int_0^x|u(t,a)-u(t,b)|\,dt. \end{align} Now use Gronwall's inequality.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.