3
$\begingroup$

I have learned Peano existence theorem and Uniqueness of solutions to IVPs, but I don't understand what does that mean by $y'=f(x,y)$, $y(x_0)=y_0$ has a solution in $(x_0-d,x_0+d)$ for some $d>0$.

  • What is this thm trying to show?
  • How to understand its uniqueness?

I have a hw question following:

Prove that if $f$ is continuous on $\Bbb R× \Bbb C$ and locally Lipschitz in the second argument, and if $x_0 \in \Bbb R$, $y_0 \in \Bbb C$, then there exists an interval $(a; b)$ containing $x_0$ such that the following holds.

  • A solution to IVP $y' = f(x; y)$ and$ y(x_0) = y_0$ exists on $(a; b)$, and
  • if a solution $\tilde y$ to the IVP exists on some open interval $I$ containing $x_0$, then $I\subset(a; b)$ and $\tilde y = y$ on $I$. That is, (a; b) is the largest interval of existence and the solution is unique on it.

(Hint: Show that if $F$ is the family of all couples (open interval containing $x_0$, solution on it), then any two of these solutions coincide on the intersection of their intervals of definition. Conclude that then a solution can be defined on the union of these intervals (show it is an open interval) such that it coincides with each solution in F where the latter is defined.)

I just don't understand how is the hint related to the thm.

I know I have learned this part poorly, I really appreciate if you can let me understand the thm.

$\endgroup$
1
$\begingroup$

What you are actually asked to do is to prove the existence of the maximal interval on which the solution to your initial value problem exists. You have at your disposal the existence and uniqueness theorem, which says that given the problem $$ \dot y=f(x,y), \quad y(x_0)=x_0 $$ and assuming that $f$ is continuous in $x$ and Lipschitz continuous in $y$ then there exists a unique solution to this problem defined on some local interval $(x_0-\epsilon,t_0+\epsilon)$, where $\epsilon$ can be estimated from the properties of $f$. Again, the theorem is essentially local and a very natural question to ask what if we would like to extend our interval?

The solution to your problem can be given in the following form (following the hint):

Lemma: Suppose that two solutions $\phi_1$ and $\phi_2$ to your problem are defined on the same open interval $I$ containing $x_0$. Then $\phi_1(x)=\phi_2(x)$ for all $x\in I$.

Proof: First, we know that $\phi_1(x)=\phi_2(x)$ for $x$ in $|x-x_0|<\epsilon$ due to existence and uniqueness theorem, which is an open interval. For different $\epsilon$ we'll have different open intervals, but their union is also open interval (union of any number of open sets is open). Let me denote this union as $I^*$. Assume $I^*\neq I$, then there must be an end point of $I^*$ that belong to $I$, I denote it as $x_1\in I$. By continuity $\phi_1(x_1)=\phi_2(x_1)$, therefore, by the same existence and uniqueness theorem there must an open interval $|x-x_1|<\epsilon_1$ (call it $I'$) where $\phi_1(x)=\phi_2(x)$, hence $\phi_1(x)=\phi_2(x)$ in $I^*\cup I'$ which is larger than $I^*$, and hence $I^*$ must be equal to $I$.

This lemma implies that for each $y_0$ there exists a maximal open interval $(\alpha,\beta)$, containing $x_0$, on which there is a unique solution $\phi$ with $\phi(x_0)=y_0$.

$\endgroup$
0
$\begingroup$

Not all differential equations have unique solutions, for example $D^2f=0$ admits solutions $f=ax+b$, and even if we form $f(0)=c$, this still leaves $a$ undetermined. The point of uniqueness is, that if you have an equation of the form $Df=F(x, f)$, the solutions only have one degree of freedom, which can be choosen by determining the function at any point. So namely, if we have have two functions, $f,g$, such that $Df=F(x, f)$, $Dg=F(x, g)$, and $f(0)=g(0)$, then $f(x)=g(x)$. This shouldn't be too suprising though, since the change in $f$ is determined by the value of $f$ and $x$, so the change at $0$ is the same for $f$ and $g$, so they are changing at the same rate, and equal to the same value at zero, so they are never gonna be different!

Additionally, not all differential equations are solvable. For example, in higher dimensions, the set $\partial_x f=0$, $\partial_y f=x^2$ cannot be solvable, since $2x=\partial_x\partial_yf=\partial_y\partial_xf=0$. There is also an equation on $\mathbb{C}$ that is not solvable, $\partial_{x}f+ix\partial_{y}f=F$, where $F$ is a specific, but hard to write down function. The existence states that when we are in one dimension, these higher dimension effects that can make an equations unsolvable disappear, and that all such equations are solvable.

Now your hint talks about both generating functions on intervals, and then showing that they coincide on there intersections, which concides with using existence and then uniqueness. This is actually a remarkably common yoga in mathematics. The important notion is that if a function solves a differential equation on a cover of open intervals, then it solves it everywhere!

$\endgroup$
  • $\begingroup$ This is all outright wrong. $\endgroup$ – Artem Oct 26 '15 at 12:53
  • 1
    $\begingroup$ @Artem In what way? You didn't explain why? If you have a error bring it forward and I will correct it delete the post. $\endgroup$ – Pax Kivimae Oct 26 '15 at 13:46
  • $\begingroup$ The whole answer is one error. First two lines: equation $f''=0$ has a perfectly fine unique solution given two initial conditions, the point of uniqueness is actually very different, why you need to bring up the condition $f(0)=c$ is not clear. You should remove your narrative not to confuse people who know nothing about the subject. $\endgroup$ – Artem Oct 26 '15 at 14:10
  • $\begingroup$ @Artem The theorem mentioned in the question is specifically about first order ODEs. The requirement for uniqueness is explicitly $f(x_0)=y_0$, as is used in the statment of the theorem, which is why I mentioned that, since it is more likely than not what OP mean by an IVP. The fact that constraints on higher deriavatives restrict higher equations is ussally taught after this, atleast in classes I have graded. $\endgroup$ – Pax Kivimae Oct 26 '15 at 14:27
  • $\begingroup$ Your last comment is also false. $\endgroup$ – Artem Oct 26 '15 at 14:41

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.