0
$\begingroup$

Show that $y_1$ and $y_2$ exist, determin the sign and estimate the value of $y_1(3)-y_2(3)$ with $y'=\arctan(y+x+\cos(y)^2)$ and $y_1(0)=1, y_2(0)=2$

I already solved the first question, they exist and are unique because $\arctan(y+x+\cos(y)^2)$ is continuous and its partial derivatives are also continuous. How can i proceed? I don't think that determining an explicit solution is a good path also because i wouldn't know how to do it in this case.

  • 0
    i think a breckat is missing2017-01-23
  • 0
    Do you mean $\cos(x^2)$, $\cos^2 x$, or the entire thing you're taking the arctan of squared2017-01-23
  • 0
    corrrected @Dr.SonnhardGraubner2017-01-23
  • 0
    Do you mean y or y'? It seems like you are asking about a differential equation.2017-01-23
  • 0
    If $\tan y(x)=y(x)+x+\cos(y(x))^2$ how come you can choose $y(0)=1$ and/or $y(0)=2$?2017-01-23
  • 0
    it's a typo i intended y' @martycohen2017-01-23
  • 0
    $y_1 and y_2$ are two separate functions @Did2017-01-23
  • 0
    Yes, now that you have corrected your post, they are.2017-01-23

1 Answers 1

1

Solutions are unique, so that the graphs of $y_1$ and $y_2$ do not cross. Since $y_1(0)

  • 0
    Thank you for your answer, but i didn't really comprehend the last inequality, why did you use 3? is it an arbitrary value?2017-01-23
  • 0
    No, it is an upper bound on the derivative with respect to $y$ of the right hand side of the equation. I have used $$|\cos^2a-\cos^2b|\le2|a-b|.$$2017-01-23
  • 0
    ok, so following this should be: define $z(x) = y_1(x)-y_2(x)$ derivative with respect to x and i get: $\lvert z'(x) \rvert \le 3\int_0^x \lvert z'(t) \rvert dt=3(\lvert z(x)\rvert-\lvert z(0)\rvert) \le 3(\lvert z(x)\rvert+1)$ so applying gronwall's inequality in a=0 t=3 i get $z(3)+1 \le e^9(\lvert z(0)\rvert + 1)$ so $z(3)\le e^9 -1$, my textbooks reports $z(3)\le e^6$ so the answer is correct but not quite accurate, where did i do wrong? Thank you for your precious time, you're really helping me out2017-01-23
  • 0
    Gronwall's lemma gives $y_1(x)-y_2(x)|\le e^{3x}$. This gives an upper bound, which surely will not be optimal. May be there is a better estimate for the Lipschitz constant. For instance, it is not difficult to prove that $y_i$ are increasing. Then a better estimate for the partial derivative is $3/2$.2017-01-23