Show $\frac{du}{dx}=|u| \ (x\in\mathbb{R}$) has solutions $u=Ae^x$

Question

Show the following ODE has solutions of the form $u=Ae^x$: $$\frac{du}{dx}=|u|, \ x\in\mathbb{R}.$$

My attempt:

I first considered the case where $u>0$. So, \begin{align} \frac{du}{dx}&=u \\ \int \frac{du}{u}&=\int dx \\ \ln|u|&=x+C, \ \ C\in\mathbb{R} \\ u&=Ae^x, \ \ A=e^c\in\mathbb{R} \end{align} Next, for the case where $u<0$ (using separations of variables as above), I get that, $$u=Ae^{-x}.$$ For the final case, $u=0$, $$u=0.$$ Only one of these cases admits the required form. Where have I gone wrong?

Answer 1

$u=0$ is part of the first case for $u=Ae^x$ if you were to say $A \ge 0$. Your first case can actually be $A>0$ because you assumed $u>0$. For your second case you should get

$$\ln|u|=x+C$$

$$\implies \ln(-u)=x+C$$

$$\implies -u=Ae^x$$

$$\implies u=-Ae^x$$

for $A=e^C > 0$

Altogether we can merge the three cases to have $u(x)=Ae^x, x \in \mathbb R, A \in \mathbb R$ where $sgn(A)=sgn(u(x))$

Answer 2

Actually, $u>0$ and $u<0$ are not both possible. Either $u>0$ everywhere or $u<0$ everywhere. Here is the argument: consider the open set $U=\{x: u(x)>0\}$. On this set we have $u(x)=Ae^{x}$. Clearly $A >0$. Let us show that the open set $U$ has no boundary point. If there is a boundary point $x$ then there is a sequence $x_n \in U$ converging to $x$. We then have $u(x) =0$ and $u(x)=\lim u(x_n)$ so $Ae^{x_n} \to 0$, hence $e^{x_n} \to 0$, But this is possible only when $x_n \to -\infty$ contradicting the fact that $x_n \to x$. This proves that $U$ is an open set with no boundary points. This implies that it is either empty or the whole line. Thus the cases $u>0$ and $u<0$ cannot both occur.

Answer 3

Suppose $u$ solves the DE $u'(x)=|u(x)|$ on $\mathbb R.$ Some cases to consider:

1. If $u(x) >0$ on $\mathbb R,$ then $u(x) = Ae^x$ for some $A>0.$ Your proof works here.

2. If $u(x) <0$ on $\mathbb R,$ then $u(x) = Ae^{-x}$ for some $A<0.$

Proof: We have $u'(x)=|u(x)|= -u(x)$ in this case. Thus $u'(x)/u(x)\equiv -1.$ Integrating gives $\ln |u(x)| = -x+c,$ which implies $|u(x)|= e^{-x}e^c.$ Since $|u|=-u,$ we get $u=-e^ce^{-x}.$ Set $A=-e^c$ to see $u$ has the desired form.

3. $u\equiv 0.$ Not much to say here, except that in this case $u$ has the form $Ae^x,$ with $A=0.$

Claim: If $u$ solves the DE, then one of 1,2,3 must hold.

Proof: Suppose $u$ is a solution and $u(a)=0$ for some $a.$ We will show $u\equiv 0.$

Suppose, to reach a contradiction, that $u(b)>0$ for some $b>a.$ Let $c$ be the largest number in $[a,b]$ such that $u(c)=0.$ Then $u$ is nonzero in $(c,b].$ By the IVT, $u>0$ or $u<0$ on $(c,b].$ Since $u(b)>0,$ we must have $u>0$ on this interval. So we have $u>0$ on $(c,b).$ Exactly as you showed in your proof of $1,$ we then have $u(x) = Ae^x$ in $(c,b)$ for some $A>0.$ It follows that $\lim_{x\to c^+}u(x)= Ae^c >0.$ But this limit also equals $u(c)=0$ by the continuity of $u$ at $c.$ This is a contradiction, proving there is no such $b.$

Similarly there is no $b<a$ at which $u$ is positive. Furthermore, using the same ideas, there is no $b\ne a$ at which $u$ is negative. Therefore $u\equiv 0.$

We have shown that if $u$ satisfies the DE and equals $0$ somewhere, it equals $0$ everywhere. By the IVT, any other solution to the DE satisfies either $u>0$ or $u<0$ everywhere, which lands us in cases 1 or 2 above.

There is one point above that needs mention: Why is there a largest $c$ in $[a,b]$ such that $u(c)=0?$ It is because the set $\{x: u(x)=0\}$ is a closed subset of $\mathbb R$ by the continuity of $u.$ I can say more on this if you would like.

By the way, in a comment you said you were hoping to avoid any topology. But the MVT is embedded in the integration results we have used. We also used the IVT. The rigorous proofs of these results, often sidestepped in a first year calculus course, depend on basic topology as well.