Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Is there any theoretical basis for employing differential methods like separation of variables in solving differential equations? As we all know, differential is formally defined as a linear transformation, so how can it be said that it's only infinitesimal and be summed up on both sides of the equation? It's not an exact explanation, let alone an exact proof.

share|improve this question
When you are trying to find a solution, you can pretty much do whatver you like, including dancing a bit in honor of Tlaloc, the God of Rain. Now, when you have found your solution, you need to check that it is a solution: then we become a little pickier... –  Mariano Suárez-Alvarez Dec 18 '11 at 2:17
I don't understand what is supposed to be ambiguous here. –  Michael Hardy Dec 18 '11 at 2:22
In any case, behind separation of variables is the fact that functions which are finite sums of products $f(x)g(y)$ with variables separated are dense in most sensible spaces of functions of two variables. For example, one can prettify this statement into an isomorphism $L^2([0,1]\times[0,1])\cong L^2([0,1])\otimes L^2([0,1])$, and so on. –  Mariano Suárez-Alvarez Dec 18 '11 at 2:22
You might find this helpful. –  Mike Spivey Dec 18 '11 at 3:58
add comment

3 Answers

up vote 2 down vote accepted

$$ f(x) = g(y)\;\frac{dy}{dx}$$ where $f$ and $g$ are known functions, and we want $y$ as a function of $x$. So $$ f(x)\;dx = g(y)\;dy $$ $$ F(x) = G(y) + C$$ $$ F(x) - C = G(y) $$ $$ G^{-1}(F(x) - C) = y $$

The question is how to justify this rigorously.

The answer must be the chain rule, and the question is how to show that.

If $x$ and $y$ both depend on some parameter $t$, then $$ f(x) \frac{dx}{dt} = g(y)\frac{dy}{dt}, $$ $$ F\;'(x)\frac{dx}{dt} = G'(y)\frac{dy}{dt}.\tag{1} $$ If $t$ happens to be $x$, this is simply the original differential equation.

The meaning of $f(x)\;dx = g(y)\;dy$ could be taken to be that no matter what variable $t$ parametrizes the graph in a smooth fashion, (1) must hold. And the chain rule tells us that $$ \frac{dy/dt}{dx/dt} = \frac{dy}{dx}. $$

Another P.O.V. might be to say that $$ f(x)\;\Delta x \approx g(y)\;\Delta y $$ and then examine limits as $\Delta x\to0$.

share|improve this answer
add comment

Assume you want to solve the Cauchy problem: $$\tag{C} \begin{cases} y^\prime (x)= f(y(x))\ g(x) &\text{, for } x\in I \\ y(x_0)=y_0\end{cases}$$ where $I\subseteq \mathbb{R}$ an interval, $x_0\in I$ an interior point and $y_0\in \mathbb{R}$. Assume also you got existence and uniqueness for solution of (C) and some monotonicity and regularity from a priori analysis (e.g., you know any solution of (C) is strictly increasing and $C^1$). If $f(y_0)\neq 0$ then, by continuity, there is a complete neighbourhood of $x_0$ contained in $I$ s.t. $f(y(x))\neq 0$, hence you can divide both LH and RH sides in the ODE by $f(y(x))$ to get: $$\frac{y^\prime (x)}{f(y(x))} =g(x)\; .$$ Now, let $x>x_0$. You integrate both sides of the previous equation over $[0,x]$ and, since $\tau =y(t)$ is an admissible change of variable, you gain: $$\int_{y_0}^{y(x)} \frac{1}{f(\tau)}\ \text{d} \tau =\int_{x_0}^x\frac{y^\prime (t)}{f(y(t))}\ \text{d} t =\int_{x_0}^x g(t)\text{d} t\; ;$$ if you call $\Phi,G$ some antiderivative of $1/f,g$, then you rewrite previous equation as: $$\Phi (y(x)) =G(x)-G(x_0)+\Phi (y_0)\; ,$$ i.e. $$\Phi (y(x))=G(x)+c$$ with $c$ a suitable additive constant.

Notice that the latter equation is the usual way to present the solution of a separable ODE in implicit form.

For example, let us solve: $$\begin{cases} y^\prime (x) =2x\ y(x) &\text{, in } \mathbb{R} \\ y(1)=1\; .\end{cases}$$ Picard-Lindelöf theorem can be used to prove that the problem has locally unique solution, which we can extend to a maximal interval, say $I$. Such a maximal solution $y(x)$ is of class $C^0(I)$; but then $y^\prime (x)=2x\ y(x)$ is also of class $C^0$ hence $y\in C^1(I)$; then again $y^\prime \in C^1(I)$ and $y\in C^2(I)$... Bootstrapping, we see that $y\in C^\infty (I)$. Moreover, $y$ is positive in a neighbourhood of $x_0=1$ because $y^\prime (1)=2\ y(1)=2>0$ and $y, y^\prime$ are continuous. The maximal solution $y$ cannot equal zero anywhere in $I$: in fact if by contradiction there were a point $x_1\in I$ s.t. $y(x_1)=0$, then $y$ would also solve the homogeneous Cauchy problem: $$\begin{cases} y^\prime =x\ y \\ y(x_1)=0\end{cases}$$ which is uniquely solved by $\bar{y}(x)=0$; hence $y(x)=0$ everywhere, but this is a contradiction because $y(1)=1\neq 0$! Therefore $y$ cannot change sign in $I$ and thus $y(x)>0$ everywhere in $I$. Consequently $y$ increases strictly in $I\cap [0,\infty[$ and decreases strictly in $I\cap ]-\infty ,0]$.

Finally, we are in the position to eveluate the solution of (C). We fix $x>1$ and compute: $$x^2 -1 =\int_1^x 2t\ \text{d} t=\int_1^x \frac{y^\prime (t)}{y(t)}\ \text{d} t \stackrel{\tau =y (t)}{=} \int_1^{y(x)} \frac{1}{\tau}\ \text{d} \tau =\ln y(x)\; ,$$ hence: $$\tag{1} y(x)=\exp (x^2-1)$$ for $x>1$. On the other hand, it is easily seen that $y$ as in (1) solves the ODE in (C) also for $x< 1$; therefore (1) gives the maximal solution of (C), which is defined in $I=\mathbb{R}$. As we expected, $y$ is $C^\infty$, positive, strictly decreasing in $]-\infty ,0]$ and strictly increasing in $[0,\infty[$.

share|improve this answer
Nice rigorous solution! The two answers are respectively given by change of variable and chain rule. However, they are different aspects of the same thing! And now we can say that separation of variables is deeply rooted in the chain rule. –  Strin Dec 18 '11 at 3:50
add comment

The explanation I used in my class today was simply that an expression $$P(x,y) \, dx + Q(x,y) \, dy$$ has meaning only inside a line integral, and so a "differential form equation" $$P(x,y) \, dx + Q(x,y) \, dy = 0$$ simply states that for every pair of vectors $\vec{r}_0 = (x_0, y_0)$ and $\vec{r}_1 = (x_1, y_1)$, and every smooth enough curve $C$ from one to the other, we have $$\int_C P(x,y) \, dx + Q(x,y) \, dy = \int_C 0 = c$$ where $c$ is a constant.

This is the key motivation. From it, you can proceed to derive the solution rigorously as follows. Since $C$ is not part of the data, this includes the assurance that the value of the integral must be independent of it, forcing the vector field $(P,Q)$ to be conservative and thus exact, hence the gradient $\nabla F$ of some potential function $F$ satisfying, a fortiori, the equation $$F(x_1, y_1) - F(x_0, y_0) = c.$$ Letting $(x_1, y_1)$ vary (thus, renaming them back to $(x,y)$) and absorbing $F(x_0, y_0)$ into $c$, this gives the implicit solution $$F(x,y) = c.$$ Now, if the field $(P,Q)$ is not exact (as seen by its not being irrotational) then you need an integrating factor to make it exact, and although this changes the differential form equation it doesn't change the associated differential equation $$\frac{dy}{dx} = -\frac{P(x,y)}{Q(x,y)}$$ because the ratio is unchanged if $P$ and $Q$ are scaled. This bit of logic is made rigorous using the chain rule, of course: if $F(x,y) = c$ implicitly defines $y = y(x)$ then $$\frac{dy}{dx} = -\frac{\partial F/\partial x}{\partial F/\partial y} = -\frac{\mu P}{\mu Q}$$ if $\mu$ is the integrating factor.

share|improve this answer
add comment

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.