2
$\begingroup$

I've some doubts about initial values problems involving differential equation with absolute values.

For example if I have a differential equation like $y'=|x+1|$ with initial condition $y(3)=-2$, since $3>-1$ I can trascurate the absolute value and solve $y'=x+1$, is it correct?

But if the condition is for instance $y(-1)=2$ then I must consider the two different cases? That is, $y'=x+1$ if $x>-1$ and $y'=-x-1$ if $x<-1$

Is this the right way to solve this kind of problems?

$\endgroup$
1
$\begingroup$

You have two ODEs: $$y_1' = x + 1\tag{1}$$ which holds for $x \ge -1$, and $$y_2' = -x-1\tag{2}$$ which holds for $x \le -1$. The solutions combine to give the solution $y$ to $y' = |x + 1|$: $$y(x) = \begin{cases} y_1(x) & x \ge -1\\y_2(x) & x \le -1\end{cases}$$ Since $y$ is continuous, it is required that $y_1(-1) = y_2(-1)$.

In your first example, ODE (1) has initial value $y_1(3) = -2$, which uniquely defines $y_1(x)$ for $x \in [-1, \infty)$. ODE (2) is then uniquely defined by the continuity condition $y_2(-1) = y_1(-1)$.

In your second example, both (1) and (2) have initial value $y_1(-1) = y_2(-1) = 2$, which automatically meets the continuity condition.

In either case, in order to get the full solution, you have to examine both sides of $x = -1$.

$\endgroup$
1
$\begingroup$

The equation $y'(x)=|x+1|$ does not require ODE theory since it does not involve $y$. It is just an equation that can be integrated. Integrating both sides over $x \in [t_0, t]$ gives:

$$ y(t) - y(t_0) = \int_{t_0}^t |x+1|dx $$

You likely intended to consider equations like $y'+y=|x+1|$ or $y'=|y+1|$.


The equation $y'=|y+1|$ is more interesting: If $y(0)=-1$ then $y(t)=-1$ for all $t \geq 0$. If $y(0)>-1$ then $y(t)>-1$ for all $t \geq 0$ (since solutions with different initial conditions cannot cross). So you can use $y'=y+1$ for that case. The case $y(0)<-1$ is similar.

$\endgroup$

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.