0
$\begingroup$

Hi everyone I need to solve an equation of this type:

$|T(x)'+A(T,x)+B(T,x)| = f(T,x)$

with boundaries conditions.

The absolute value is my problem. Of course without it, the solution of these is well treated in the literature.

After search in the questions I found this: Differential equation with absolute value

So, can I do the same procedure? Or there is another way to solve this? Thanks.

  • 0
    Crosspost on Computational Science: http://scicomp.stackexchange.com/questions/2212/2012-05-14
  • 0
    Cross posting on different sites simultaneously is discouraged. http://meta.stackexchange.com/questions/64068/is-cross-posting-a-question-on-multiple-stack-exchange-sites-permitted-if-the-qu/64073#64073 Wait for a couple of days before deciding to cross-post.2012-05-14
  • 1
    ok I deleted the other one.2012-05-14

1 Answers 1

1

Your equation does not determine $T'$ in terms of $T$ and $x$, so uniqueness of solutions may be a problem. You can say $T'$ is either $f(T,x) - A(T,x) - B(T,x)$ or $-f(T,x) - A(T,x) - B(T,x)$. Presumably there will be one region where it's the first and one region where it's the second, and (assuming $f,A,B$ are continuous) if you want $T'$ to exist everywhere it'll be impossible to switch between one and the other except when $T' = 0$.

  • 0
    Thank you Robert. But can you explain me why you only take two combinations and not more? I mean why not $f(T,x)+A(T,x)−B(T,x)$ for instance ?2012-05-14
  • 0
    $|t|$ is either $t$ or $-t$ (assuming we're talking about real numbers).2012-05-14
  • 0
    Does not matter if A(T,x) and B(T,x) are negatives right?. Thank you very much!2012-05-15
  • 0
    Right. Why would it matter?2012-05-16