# A line integral equation popped up when trying to derive Exact ODE integrating factor, can it be solved analytically?

(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

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

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 $\mu$?

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

• 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 $\frac{\partial (\mu M)}{\partial y} = \frac{\partial (\mu N)}{\partial x}$. – KittyL Jan 18 '15 at 12:56
• 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... $$\int \int_S \mu \begin{pmatrix} -N \\ M \end{pmatrix} \cdot d\mathbf{S}=0$$ – Secret Jan 20 '15 at 0:09