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This question is probably very elementary but I don't know how to tackle the conversely part of the following result. Let $M(x,y)$ and $N(x,y)$ be two differentiable and homogeneous functions of the same degree $d$ and such that $M(x,y)dx+N(x,y)dy$ is not exact that is: $$ \frac{\partial M}{\partial y}\neq\frac{\partial N}{\partial x} $$. Using Euler's identity on $M$ and $N$:

$$ xM_{x}(x,y)+yM_{y}(x,y)=d\cdot M(x,y)$$

$$ xN_{x}(x,y)+yN_{y}(x,y)=d\cdot N(x,y)$$

Euler's identity comes from Euler's homogeneous function theorem which is appicable in this case since $M$ and $N$ are both homogeneous functions.

I showed that the function

$$\mu(x,y)=\frac{1}{xM(x,y)+yN(x,y)}\tag1$$

will satisfy:

$$ \frac{\partial}{\partial y}\left(\,\mu\cdot M\right)=\frac{\partial}{\partial x}\left(\mu\cdot N\right).\tag2$$

My question: If we suppose that the PDE above is true,then how we can explain where the formula of $\mu(x,y)$ comes from?, that is:

$${If}\quad N(x,y)\mu_{x}-M(x,y)\mu_{y}=(N_{x}-M_{y})\mu,\quad \text{then where the formula} \quad\mu(x,y)=\frac{1}{xM(x,y)+yN(x,y)} \quad \text{comes from ?} $$

I tried to apply the method of characteristics but I don't see it. This result comes from an old edition of Boyce and Di prima.

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  • $\begingroup$ As I read the questoin, you have shown that your particular $\mu$ satisfies the PDE, and you want to show that it is one solution of the PDE. This doesn't make sense to me. Perhaps you could rephrase the question. $\endgroup$ – Ben McKay Oct 29 '17 at 16:02
  • $\begingroup$ @BenMcKay: I woluld like to prove that $\mu(x,y)=\frac{1}{xM(x,y)+yN(x,y)}$. I want to know where the formula for $\mu$ comes from. $\endgroup$ – Hector Blandin Oct 29 '17 at 16:03
  • $\begingroup$ @BenMcKay: I showed if $\mu$ has the formula I wrote above, then the equation is satisfied but starting from the PDE how can I get $\mu$ ? $\endgroup$ – Hector Blandin Oct 29 '17 at 16:05
  • $\begingroup$ @FrancoisZiegler: Yes, exactly. I have to correct the statement. $\endgroup$ – Hector Blandin Oct 29 '17 at 16:05
  • $\begingroup$ @FrancoisZiegler: Thank you, exactly integrating factors are never unique. The thing is where the formula for $\mu$ comes from? I will correct the statement. $\endgroup$ – Hector Blandin Oct 29 '17 at 16:16
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The last display asks whether (1) is the unique solution of (2). It isn’t: try $M=N=x$, $\mu=1/x$.

In fact an integrating factor is never unique: see e.g. Serret (1886, thm 681).

Now if you are asking for heuristics, then e.g. (ibid., §685) “derives” (1) under the Ansatz that $\mu$ is itself homogeneous (of degree $−d−1$).

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  • $\begingroup$ No, nothing. It was a misktake. I completely acdept the answer ! :) There is No problem. $\endgroup$ – Hector Blandin Nov 29 '17 at 0:43
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(From Serret J.A. Cours de Calcul Differentiel Et Integral... Volume 1 book)

Following suggestions from Francois Ziegler, we will show that we can find an homogeneous function $\mu(x,y)$ of some degree $k\in\mathbb{Z}$ such that:

$$(\mu\cdot M)\,dx+(\mu\cdot N)\,dy=0$$

is an exact 1-form using the fact that $M$ and $N$ are both homogeneous functions of the same degree $d$. In effect, let $\mu(x,y)$ be such an homogeneous function of degree $k\in\mathbb{Z}$, then $\mu\cdot M$ will be an homogeneous function of degree $k+d$ and by Euler's homogeneous function theorem we have: $$ x\frac{\partial}{\partial x}(\mu\cdot M)+y\frac{\partial}{\partial y}(\mu\cdot M)=(d+k)\,\mu\cdot M. $$ Since we want $\mu$ to be a factor such that the original equation is exact we must have: $$ \frac{\partial}{\partial x}(\mu\cdot N)=\frac{\partial}{\partial y}(\mu\cdot M) $$ Then $$ x\frac{\partial}{\partial x}(\mu\cdot M)+y\frac{\partial}{\partial x}(\mu\cdot N)=(d+k)\,\mu\cdot M $$ but $$ y\frac{\partial}{\partial x}(\mu\cdot N)=\frac{\partial}{\partial x}(y\,\mu\cdot N) $$ and $$ x\frac{\partial}{\partial x}(\mu\cdot M)=\frac{\partial}{\partial x}(x\,\mu\cdot M)-\mu\cdot M $$ this implies: $$ \frac{\partial}{\partial x}(x\,\mu\cdot M)-\mu\cdot M+\frac{\partial}{\partial x}(y\,\mu\cdot N)=(d+k)\,\mu\cdot M $$ so $$ \frac{\partial}{\partial x}(x\,\mu\cdot M)+\frac{\partial}{\partial x}(y\,\mu\cdot N)=(d+k+1)\,\mu\cdot M $$ then we get: $$ \frac{\partial}{\partial x}(\mu\,(xM+yN))=(d+k+1)\,\mu\cdot M. $$ Let's choose $k=-d-1$, then $$ \frac{\partial}{\partial x}(\mu\,(xM+yN))=0.$$ Similarly, the function $\mu\cdot N$ is homogeneous of degree $k+d$, then again by Euler's homogeneous function theorem we get: $$ x\frac{\partial}{\partial x}(\mu\cdot N)+y\frac{\partial}{\partial y}(\mu\cdot N)=(k+d)\,\mu\cdot N $$ as before we can write the last equation as follows: $$ x\frac{\partial}{\partial y}(\mu\cdot M)+\frac{\partial}{\partial y}(\mu\cdot N\,y)-\mu N=(k+d)\,\mu\cdot N $$ then $$ \frac{\partial}{\partial y}(x\,\mu\cdot M)+\frac{\partial}{\partial y}(\mu\cdot N\,y)=(k+d+1)\,\mu\cdot N=0 $$ so $$\frac{\partial}{\partial y}(\mu(xM+yN))=0$$ Then the expresion $\mu\cdot(xM+yN)$ satisfies both conditions: $$ \frac{\partial}{\partial x}(\mu(xM+yN))=0$$ and $$\frac{\partial}{\partial y}(\mu(xM+yN))=0$$ so $\mu\cdot(xM+yN)$ is any constant, in particular we can say $$ \mu\cdot(xM+yN) = 1 $$ and hence: $$\mu(x,y)=\frac{1}{xM+yN}.$$

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