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Many times in mathematics, as for example when we find the solution of an ODE, we can not claim uniqueness just by construction, instead we have to use a theorem.

The reasoning behind this is that even if we found a solution and the solution appears to be unique from the point of view of the method used, how can we be sure there is no another method which provides another solution?

Now my question: sometimes the following type of argument is accepted as valid. For example, a simple differential equation like this one $y'(x)=x^2$ with $y(0)=0$, we say that $y(x)=\frac{1}{3}x^3$ is the only solution (without the use of an uniqueness theorem, I think is because of the Fundamental Theorem of Calculus).

How can be sure now that there is not another way to solve the equation that provides another solution (without using the uniqueness theorem of course)?

When is this type of argument valid and when not?

Down in some of the answers some people say that the reason in the previous example for uniqueness is because the anti-derivative is unique, which sounds reasonable.

But the Laplace transform of a function that satisfies certain conditions is also unique. And we can't say that the solution of an ODE is unique only beause it was calculated through the Laplace transform method.

Ambesh
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    When is what argument valid? I see no arguments above to validate. – Math Gems Apr 20 '13 at 22:54
  • @MathGems uniqueness without using an uniqueness theorem as in the example. – Ambesh Apr 20 '13 at 22:54
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    Without knowing the uniqueness theorem, what logical basis do you have for believing that the solution should be unique? Does the same reasoning tell you that prime factorizations of integers is unique? – Math Gems Apr 20 '13 at 22:57
  • @MathGems Well, I'm not using my intuition. I have seen teachers do that. I have a very reputed teacher that claims uniqueness on construction basis sometimes. And that is driving me nuts. – Ambesh Apr 20 '13 at 22:59
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    For the specific case you are asking, and similar questions, we do use a uniqueness theorem all the time: If two functions $f$ and $g$ have the same derivative, their difference is constant. We may not bother to mention it all the time, just as we do not bother to mention that arithmetic is commutative every time we have integers $a,b$, and rewrite $ab$ as $ba$. Anyway, the result that if $f'=g'$ then $f,g$ differ by a constant tells you that if you find any function, by any method whatsoever, such that $f'(x)=x^2$, then any other $y$ with $y'=x^2$ will be $f(x)+c$. – Andrés E. Caicedo Apr 20 '13 at 23:05
  • @AndresCaicedo Thank you. I have added two lines to the question that address this issue. – Ambesh Apr 21 '13 at 09:27

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It's unique because it is a ODE with separable variables, and its solution is found via separation of the variables and integration. The function obtained via integration is unique except for the arbitrary constant, which is uniquely determined by the initial condition.

EDIT: I must correct this: let's suppose you have an ODE which is variable-separable that is defined by $$ h(y)y'=g(x) $$ You can find a solution via integration as usual: $$ \int h(y)dy=\int g(x)dx $$ The function is given inplicitly by the above equation. Let's suppose that $y(a)=b$ is the initial condition. If $y'(a)\neq 0$ then $y$ is locally invertible (as per the Inverse Function theorem) and you can find a local unique solution $y(x)$ which can be extended to a maximal conected domain; in this case it is correct to say that the solution obtained via the separation method is unique without recurring to the Existence & Uniqueness of solutions.

An important remark: the Existence and Uniqueness Theorem for ODEs works fine for equations defined as $y'(x)=f(y,x)$, where $|f|$ is limited on a neighbourhood of the initial condition $(a,y(a))$. In the example cited on the comments, $f$ is something like $$ f(x,y)=\dfrac{\hat{f}(x)}{y} $$ for some $\hat{f}$ so if $y(0)=0$, $|f|$ is not limited around $(0,0)$ and uniqueness cannot be guaranteed.

Marra
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  • @Gustavo_Marra But how do we know that is really unique? Why integration is different than another method? For example using the Laplace transform to find the solution of an ODE. – Ambesh Apr 20 '13 at 23:00
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    Comes from the uniqueness of the primitive (antiderivative) of a function modulo an adding constant. Also see the Fundamental Theorem of Calculus. – Marra Apr 20 '13 at 23:03
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    @MykeArya uniqueness can be proven by the mean value theorem, which shows that a function defined on an interval with derivative zero is a constant. Then you take the difference of any solution to an ODE with the solution you solve for and use linearity of the derivative to show that that difference has derivative zero. The initial condition identifies the constant. – Chris Janjigian Apr 20 '13 at 23:42
  • And of course the unspoken hypothesis is that the domain is the set of all real numbers (or some interval containing $0$). – Lubin Apr 21 '13 at 02:21
  • @GustavoMarra But still, The Laplace transform of a function that satisfies certain conditions is also unique. But we can't say that the solution of an ODE is unique only beause it was calculated through the Laplace transform (I'm going to add this to the question itself to make my doubt more clear). – Ambesh Apr 21 '13 at 07:51
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    I must be missing something: the equation $2y\frac{dy}{dx}=4x^3$, $y(0)=0$ is separable and has four solutions on $\mathbb{R}$ (two possibilities each side of the origin). – Shane O Rourke Apr 21 '13 at 09:13
  • @ShaneORourke Which are the four solutions? – Ambesh Apr 21 '13 at 09:25
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    As well as $y_1=x^2$ and $y_2=-x^2$, there is $y_3=\left{\begin{array}{rl}x^2 & x<0\ -x^2 & x\geq 0\end{array}\right.$ and $y_4=-y_3$. – Shane O Rourke Apr 21 '13 at 09:36
  • @ShaneORourke Yes but still this case is different from my example because you have $y$ outside de derivative. So it's not solved just by regular integration. My question still stands. – Ambesh Apr 21 '13 at 13:51
  • Many times this questions don't get good response as if they are trivial. But sometimes nobody can answer them with precission. – Ambesh Apr 21 '13 at 13:52
  • @ShaneORourke we are talking about linearly independent solutions. – Marra Apr 21 '13 at 15:33
  • @MykeArya, do you know the formal definition of the Laplace Transform? – Marra Apr 21 '13 at 15:46
  • @GustavoMarra The functions $y_1$ and $y_3$ above are linearly independent (viewed as functions defined on $\mathbb{R}$). – Shane O Rourke Apr 21 '13 at 16:51
  • @MykeArya I agree that the example you gave in your question is of a different sort. (And for example, Chris Ganjigian's comment settles the uniqueness for equations of the form $\frac{dy}{dx}=f(x)$.) However, my example shows that things get a little trickier for more general separable equations. – Shane O Rourke Apr 21 '13 at 16:59
  • @ShaneORourke you see that in your 4-solution argument, what you have in fact are two suggestions for solutions, not four. I'm still think about it, gimme a minute. – Marra Apr 21 '13 at 17:24
  • Ok, I think I got it: when using the separation of variables method in the equation v$h(y)y'=g(x)$, the solution $y=y(x)$ is given implicitly by $\int h(y)dy=\int g(x)dx$. Since in your example one would have $y(x)^2=x^4+C$ as the implicit function that defines the ODE and $y(x)^2=x^4$ the implicit function that satisfies the equation; since the initial condition is $y(0)=0$, and this is the exact point where $y$ is not locally invertible, you cannot write it as a non-implicit function. I'll change my above answer. – Marra Apr 21 '13 at 17:32
  • @GustavoMarra Yes Gustavo, the standard definition. $F(s)=\int_{0}^{\infty}e^{-xt}f(t),dt$, it is said to be unique. So why the same argument does not apply here? – Ambesh Apr 21 '13 at 18:13
  • The Laplace transform is an isomorphism mapping differential equations into easier algebraic equations; thus if the solution for the algebraic equation obtained is unique, the inverse Laplace transform is unique. – Marra Apr 21 '13 at 18:29
  • @GustavoMarra So as the algebraic equation has an unique solution, then so does the differential equation. But this argument is wrong and my question is why. – Ambesh Apr 21 '13 at 18:54
  • Why is it wrong? – Marra Apr 21 '13 at 18:58
  • @GustavoMarra Because in many cases the derivative of a solution obtained through the Laplace transform method is a solution. And also because otherwise we wouldn't have to work so hard to prove uniqueness Picard's theorem. – Ambesh Apr 21 '13 at 19:01
  • I would like to see a formal proof that the bijection of the Laplace Transform is not enough to guarantee uniqueness based on this example you mentioned; also note that you can apply the Laplace Transform for functions defined in $\mathbb{R}^+$ only; also note that Picard's theorem is infinitely more general (and has more applications) when you see it in a multivariable context. – Marra Apr 21 '13 at 19:07
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Although they may not be explicitly mentioned, there are in fact uniqueness theorems that are at the foundation of such "unique by construction" arguments. Let's consider your specific example of solutions of a nonhomogenous differential equation. It is simply a special case of the ubiquitous linear principle that the general solution of a nonhomogeneous linear equation is given by adding any particular solution to the general solution of the associated homogeneous equation. More explicitly, if $\rm\:D\:$ is a linear map then one easily proves

Lemma $\ \ $ If $\rm\ D\:f_1\ =\ g\ $ then $\rm\ D\:f_2\ =\ g\ \iff\ 0\ =\ D\:f_1 - D\:f_2\ =\ D\:(f_1-f_2)$

Therefore $\rm\ D^{-1}(g)\ =\ f_1 +\ ker\ D\ =\:\: $ particular + homogeneous solution, as in linear algebra.

In particular $\rm\ \ \int g\ =\ f_1 +\ c,\ $ for $\rm\ c\in ker\ \dfrac{d}{d\:x}\: =\:\: $ constants w.r.t. the derivation $\rm\ D\: =\: \dfrac{d}{d\:x}\:.$

Compare this to $\rm \ x\, =\, 3\, +\, 5\, \mathbb Z,\:$ the solution of $\rm\ 2\: x\ \equiv\ 6\pmod{10}\:,\: $ with particular solution $\rm x \equiv 3\:,\: $ and homogeneous solution: $\rm\ 2\: x\:\equiv 0\pmod{10}\iff 10\:|\:2\:x\iff 5\:|\:x\iff x\in 5\ \mathbb Z\:.$

Community
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Math Gems
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Based on your reactions to the other answers, I think your problem might be with the form of such arguments.

The general idea is this: we want to prove the uniqueness of a particular gadget satisfying certain properties. From those properties, we deduce that it has other properties, and eventually we show that an any gadget which satisfies all of these properties must be our particular gadget. Maybe there are other ways to approach the problem or other properties of the gadget which we haven't used, but in any case the gadget is unique.

Analogously (?), a detective might determine that a master thief acted alone before determining the thief's identity.

Adam Saltz
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