# Sufficient condition for integrating factor to be a function of one variable and other theorems involved in the solution

Suppose I have an inexact differential equation (please forgive me for my sloppy use of differentials etc.)

$$0 = A(x,y)dy+B(x,y)dx.$$

Now my understanding is that we employ a few theorems to solve this problem:

(1) We assert that some integrating factor $$\mu$$ exists for which $$\frac{\partial(\mu A)}{\partial x}=\frac{\partial(\mu B)}{\partial y}$$. Is it obvious that such a $$\mu$$ exists or am I missing something?

(2) It is straightforward to show that if $$\mu$$ is only a function of one of $$x$$ or $$y$$ then it has a form given by some integral. For example, (WLOG, I now take the case that $$\mu = \mu(x)$$), given that (1) is true, I have $$\mu \frac{\partial B}{\partial y} = A\frac{d\mu}{dx}+\mu \frac{\partial A}{\partial x},$$ from which we see that $$\frac{d\mu}{dx}=\frac{\mu}{A}\left(\frac{\partial B}{\partial y} - \frac{\partial A}{\partial x} \right) \equiv \mu f$$ where in the last step I've defined $$f = \frac{1}{A}\left(\frac{\partial B}{\partial y} - \frac{\partial A}{\partial x} \right)$$ (and indeed it is only a function of $$x$$ by our hypothesis that $$\mu$$ is only a function of $$x$$). We can easily proceed from here to compute $$\mu$$ and to solve for an in general implicit relationship between $$y$$ and $$x$$ via the theory of exact differential equations. No problem there. However, it's also stated in my textbook (Riley, Hobson, and Bence, 3rd edition) that $$f$$ can give a sufficient condition for when $$\mu$$ will be only a function of $$x$$. That is, it says that if I compute $$f$$ and find that $$f(x,y) = f(x)$$ then I can conclude that $$\mu$$ will be only a function of $$x$$. Is this obvious?

In summary, I'm hoping someone can provide a proof of the two theorems I've stated and/or a resource to such a proof.

• W. Walter, Ordinary Differential Equations, GTM 182, Chapter I.§3 (p. 39 ff.) is a reference for Euler multipliers. Commented Jan 2, 2023 at 20:54

In the following, we (mostly directly) describe results outlined in "Differential Equations with Applications and Historical Notes (3ed)" by George F. Simmons (Pages 74 to 77 and Pages 8-9).

$$\textbf{Claim 1}$$: The equation \begin{align}\label{eqn::1}\tag{1} B(x,y) dx + A(x,y) dy = 0 \end{align} always has an integrating factor if it has a general solution (to be defined later) $$f(x,y) = \xi$$ for some scalar $$\xi.$$

$$\textit{Proof}$$. Suppose that (\ref{eqn::1}) has the general solution \begin{align}\label{eqn::2}\tag{2} f(x,y) = \xi \end{align} for some $$\xi \in \mathbb{R}.$$ Then one may compute the total derivative of $$f,$$ which eliminates $$\xi,$$ and yields \begin{align}\label{eqn::3}\tag{3} df = \frac{\partial f}{\partial x} dx + \frac{\partial f}{\partial y} dy = 0. \end{align} By (\ref{eqn::1}) and (\ref{eqn::3}), one has \begin{align}\label{eqn::4}\tag{4} \frac{dy}{dx} = - \frac{B(x,y)}{A(x,y)} = - \frac{\frac{\partial f}{\partial x}}{\frac{\partial f}{\partial y}}. \end{align} Therefore, \begin{align}\label{eqn::5}\tag{5} \frac{\frac{\partial f}{\partial x}}{B(x,y)} = \frac{\frac{\partial f}{\partial y}}{A(x,y)} \end{align} Set $$\mu(x,y) = \frac{\frac{\partial f}{\partial x}}{B(x,y)} = \frac{\frac{\partial f}{\partial y}}{A(x,y)}.$$ This implies that $$\frac{\partial f}{\partial x} = \mu(x,y) B(x,y)$$ and $$\frac{\partial f}{\partial y} = \mu(x,y) A(x,y).$$ Hence, multiplying (\ref{eqn::1}) by $$\mu(x,y)$$ yields \begin{align}\label{eqn::6}\tag{6} \mu(x,y) B(x,y) dx + \mu(x,y) A(x,y) dy = 0 \end{align} or, equivalently \begin{align}\label{eqn::7}\tag{7} \frac{\partial f}{\partial x} dx + \frac{\partial f}{\partial y} dy = 0, \end{align} which is exact. Thus, if (\ref{eqn::1}) has a general solution, then it has at least one integrating factor. In fact, (\ref{eqn::1}) as arbitrarily many integrating factors in this case. Observe that for any mapping $$T(f)$$ of $$f,$$ one has \begin{align}\label{eqn::8}\tag{8} \mu(x,y) T (f) (B(x,y) dx + A(x,y) dy) &= T(f) ( \mu(x,y) B(x,y) dx + \mu(x,y) A(x,y) dy )\\ &= T(f) df \\ &= d \left[ \int T(f) df \right], \end{align} in which case, $$\mu(x,y) T(f)$$ is also an integrating factor for (\ref{eqn::1}).

Also note the following: if $$\mu$$ is an integrating factor for (\ref{eqn::1}), then, by the exactness of (\ref{eqn::6}), we have (suppressing the arguments of the functions for convenience) \begin{align}\label{eqn::9}\tag{9} \frac{\partial (\mu B)}{\partial y} = \frac{\partial (\mu A)}{\partial x} &\iff \mu \frac{\partial B}{\partial y} + M \frac{\partial \mu}{\partial y} = \mu \frac{\partial A}{\partial x} + A \frac{\partial \mu}{\partial x} \\ &\iff \frac{1}{\mu} \left[ A \frac{\partial \mu}{\partial x} - B \frac{\partial \mu}{\partial y} \right] = \frac{\partial B}{\partial y} - \frac{\partial A}{\partial x}. \end{align}

$$\textbf{Claim 2}$$:

(i) If $$\mu(x,y) \equiv \mu(x),$$ then there exists some function $$L(x)$$ such that $$\mu(x) = e^{\int L(x) dx}.$$

(ii) If $$\frac{\frac{\partial B}{\partial y} - \frac{\partial A}{\partial x}}{A(x,y)} = H(x),$$ then there exists a function of $$x,$$ say $$\hat{\mu}(x),$$ such that $$\hat{\mu}$$ is an integrating factor for (\ref{eqn::1}). (simply define $$\hat{\mu}(x) = e^{\int H(x) dx}$$ and check that it obeys the aforementioned requirements for an integrating factor).

$$\textit{Proof of}$$ (i). Suppose that $$\mu(x,y) \equiv \mu(x).$$ Then (\ref{eqn::9}) yields \begin{align}\label{eqn::10}\tag{10} \frac{1}{\mu} \frac{\partial \mu}{\partial x} = \frac{\frac{\partial B}{\partial y} - \frac{\partial A}{\partial x}}{A(x,y)}. \end{align} Set $$L(x) = \frac{\frac{\partial B}{\partial y} - \frac{\partial A}{\partial x}}{A(x,y)}.$$ Then we have \begin{align}\label{eqn::11}\tag{11} \frac{1}{\mu} \frac{\partial \mu}{\partial x} = L(x) &\iff \frac{d (\log \, \mu)}{dx} = L(x) \\ &\iff \log \, \mu = \int L(x) \, dx \\ &\iff \mu = e^{\int L(x) \, dx}. \end{align} Note that this line of logic can be done in the opposite direction to yield (ii). That is, assume (ii), then $$\hat{\mu}(x) = e^{\int H(x) dx}$$ depends only on $$x$$ and also satisfies (\ref{eqn::9}).

$$\textbf{What is a General Solution}$$?

• Theorem (Banach-Picard): If $$f(x,y)$$ and $$\partial f / \partial y$$ are continuous functions on a closed rectangle $$R,$$ then for every point $$(a,b) \in \textrm{int} \, R,$$ there passes a unique integral curve of the equation $$dy / dx = f(x,y).$$

If one fixes $$a$$ in the context of the theorem above, then the integral curve passing through $$(a,b)$$ is a one-parameter family of curves, parametrized by $$b;$$ i.e., it is a family of functions of the form $$\{ y_{\alpha}(x) \}_{\alpha \in I}$$ (here, $$I$$ is a (possibly uncountable) indexing set dictating the space of the parameters; in our case $$I = \mathbb{R}$$).

Thus, the integral curve that passes through the point $$(a,b)$$ corresponds to some value $$\alpha$$ for which $$b = y_{\alpha}(a).$$ If we denote this critical $$\alpha$$ by $$\alpha_0,$$ then $$y(x) = y_{\alpha}(x)$$ is called the general solution of $$dy / dx = f(x,y)$$ and $$y(x) = y_{\alpha_0}(x)$$ is called the particular solution of $$dy / dx = f(x,y).$$

A more thorough description of results for differential equations of this type can be found in "Differential Calculus" by Henri Cartan (Section 3.10., Pages 144 to 147).

• Thank you for your answer. I am about to leave the house but will try to review this tonight or early tomorrow.
– EE18
Commented Jan 2, 2023 at 22:04
• I am reading now. I think below (2) you mean to write $\xi$ rather than $\alpha$.
– EE18
Commented Jan 3, 2023 at 19:20
• I'm afraid I still don't follow how claim (ii) follows. In particular, if $\frac{\frac{\partial B}{\partial y} - \frac{\partial A}{\partial x}}{A(x,y)} = H(x)$ then from your last expression in (9) I conclude that $\frac{1}{\mu} \left[ \frac{\partial \mu}{\partial x} - B/A \frac{\partial \mu}{\partial y} \right] = H(x)$, but why can't this be (I suppose your are trying to say that the LHS cannot be a funciton of only $x$ here, but I don't see it.
– EE18
Commented Jan 3, 2023 at 19:39
• Yes. More specifically, I mean that if $\frac{ \frac{\partial B}{\partial y} - \frac{\partial A}{\partial x} }{A(x,y)} = H(x),$ then one can define an arbitrary function of $x,$ say $\hat{\mu}(x)$ via the integration in (11). In which case, one can then show that $\hat{\mu}(x)$ satisfies (9) (along with the other aforementioned properties of an integrating factor). This yields an existence of such an integrating factor. I should be careful though in my phrasing, in that, it is not necessarily the same one (but we know this to be true by (8)) Commented Jan 4, 2023 at 2:08
• I've made edits to reflect this. Commented Jan 4, 2023 at 2:13