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(For convenience, for any functions, only its first instance the x,y dependence will be written out, all subsequent instance the x,y will be suppressed)

I have an ODE $$M(x,y)+N(x,y)\frac{dy}{dx}=0$$

I understood the ODE is inexact when $$\frac{\partial M}{\partial y} \neq \frac{\partial N}{\partial x}$$

I also learnt that for some inexact ODE, it can be made exact if the integrating factor $\mu$ is a function of x or y only, which is determined if the first (resp second) of these is satisfied

$$\frac{\frac{\partial M}{\partial y}-\frac{\partial N}{\partial x}}{M},\frac{\frac{\partial M}{\partial y}-\frac{\partial N}{\partial x}}{N}$$

So that a function

$$H(x,y)=C$$

can be found as the solution to the ODE

However when I tried to do it as a general $\mu (x,y)$ I got something interesting

Start with

$$M(x,y)+N(x,y)\frac{dy}{dx}=0$$

ODE is exact with integrating factor $\mu$ iff

$$\frac{\partial (\mu M)}{\partial y} = \frac{\partial (\mu N)}{\partial x}$$

$$\mu\frac{\partial M}{\partial y}+\frac{\partial \mu}{\partial y}M = \mu\frac{\partial N}{\partial x}+\frac{\partial \mu}{\partial x}N$$

Rearrange

$$\mu\left(\frac{\partial M}{\partial y}-\frac{\partial N}{\partial x}\right) =\frac{\partial \mu}{\partial x}N-\frac{\partial \mu}{\partial y}M$$

Rewrite with del operators

$$\mu \nabla \cdot \begin{pmatrix} M \\ -N \end{pmatrix} =\nabla\mu \cdot \begin{pmatrix} N \\ -M \end{pmatrix}$$

Move everything to the LHS and move the - sign into the vectors

$$\mu \nabla \cdot \begin{pmatrix} M \\ -N \end{pmatrix} + \nabla\mu \cdot \begin{pmatrix} -N \\ M \end{pmatrix}= 0$$

This thing

$$\mu \nabla \cdot \begin{pmatrix} M \\ -N \end{pmatrix} + \nabla\mu \cdot \begin{pmatrix} -N \\ M \end{pmatrix}= 0$$

looks deceptively similar to the divergence product rule, is it actually possible to do something on it and solve it analytically for $\mu$?

================================================================== UPDATE

KittyL have pointed out a mistake I made in the 1st version. It is now corrected as follows

$$\mu \nabla \cdot \begin{pmatrix} \color{red}{-N} \\ \color{red}{M} \end{pmatrix} =\nabla\mu \cdot \begin{pmatrix} N \\ -M \end{pmatrix}$$

which rearranges to

$$\mu \nabla \cdot \begin{pmatrix} N \\ -M \end{pmatrix} +\nabla\mu \cdot \begin{pmatrix} N \\ -M \end{pmatrix}=0$$

Using the divergence product rule backwards

$$ \nabla \cdot \left(\mu\begin{pmatrix} N \\ -M \end{pmatrix}\right)=0$$

Since the starting functions M,N,$\mu$ are all functions of two variables (e.g. x,y) we can say this problem lives in $\mathbb{R}^2$

Applying Divergence theorem in $\mathbb{R}^2$

$$ \iint_A\nabla \cdot \left(\mu\begin{pmatrix} N \\ -M \end{pmatrix}\right)dA=\iint_A 0 dA$$

$$ \oint \mu\begin{pmatrix} N \\ -M \end{pmatrix}\cdot d\vec{l}=0$$

Because divergence theorem holds for all closed surfaces, let's choose our surface to be a circle $x^2+y^2=1$

Parametrising in terms of $\theta$, we obtained

$$ \int_0^{2\pi} \mu(r,\theta)\begin{pmatrix} N(r,\theta) \\ -M(r,\theta) \end{pmatrix}\cdot \frac{\nabla{r^2}}{||\nabla{r^2}||}d\theta=0$$

$$ \int_0^{2\pi} \mu(r,\theta)\begin{pmatrix} N(r,\theta) \\ -M(r,\theta) \end{pmatrix}\cdot \frac{1}{\sqrt{2r}}\begin{pmatrix} \cos\theta \\ \sin\theta \end{pmatrix}d\theta=0$$

$$ \int_0^{2\pi} \frac{\mu(r,\theta)}{\sqrt{2r}}\left( N(r,\theta)\cos\theta-M(r,\theta)\sin\theta\right)d\theta=0$$

Is there a non trivial solution to $\mu$ for this integral?

If so, can $\mu$ be solved analytically/in a closed form?

If no analytic solutions exists in general, what is the most numerically stable and efficient way to approximate $\mu$?

$$ \int_0^{2\pi} \frac{\mu(r,\theta)}{\sqrt{2r}}\left( N(r,\theta)\cos\theta-M(r,\theta)\sin\theta\right)d\theta=0$$

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  • $\begingroup$ This is an interesting discovery. But I think it should be $\mu \nabla \cdot \begin{pmatrix} -N \\ M \end{pmatrix} + \nabla\mu \cdot \begin{pmatrix} -N \\ M \end{pmatrix}= 0$, which is equivalent as the original statement $\frac{\partial (\mu M)}{\partial y} = \frac{\partial (\mu N)}{\partial x}$. $\endgroup$ – KittyL Jan 18 '15 at 12:56
  • $\begingroup$ Yes, thanks for pointing that out, I realise there is a mistake when doing the algebra .With this $$\mu \nabla \cdot \begin{pmatrix} -N \\ M \end{pmatrix} + \nabla\mu \cdot \begin{pmatrix} -N \\ M \end{pmatrix}= 0$$, (which is the same as the original statement) it seemed it can be integrated by Stoke's Theorem, however I am not sure if this is actually useful... $$\int \int_S \mu \begin{pmatrix} -N \\ M \end{pmatrix} \cdot d\mathbf{S}=0$$ $\endgroup$ – Secret Jan 20 '15 at 0:09

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