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As part of a solution to the PDE $$-xu_x+yu_y = u + 2x,$$ through the method of characteristics, the author writes

$$\frac{dx}{x} + \frac{dy}{y} = 0.$$

Now, I know where such an equation comes from:

$$\left(\frac{dx}{dt}=-x\right) \land \left(\frac{dy}{dt} = y\right) \implies \frac{dx}{x} = - \frac{dy}{y}$$ where both sides where multiplied by $-dt$ and by the respective variable.

Even though I detest the ambiguous $dx/dt$ notation, these kinds of arguments are standard, and I was wondering if there any definition of $dx$, $dy$, and $dt$ that justifies the step of multiplying by $dt$?

Perhaps through differentials as in Smooth Manifold theory, or with measure theory?

J.G.
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Sam
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  • It’s a symbolic tool, that’s all. Don’t use it if you’re uncomfortable with it. – A rural reader May 23 '24 at 22:03
  • Non-standard analysis provides a rigorous approach, building off of the use of hyperreal numbers, and is consistent with standard calculus by passing through the "shadow" or "standard part" map. https://en.wikipedia.org/wiki/Nonstandard_calculus and https://en.wikipedia.org/wiki/Standard_part_function – Chickenmancer May 23 '24 at 22:54
  • There's a host of related questions already on this site. Can you describe why they all don't satisfy you, so that we can avoid answering something you already know? – Trebor May 24 '24 at 09:14
  • Been awhile since I've done this but wikipedia says 'These integral curves are called the characteristic curves of the original partial differential equation and are given by the Lagrange–Charpit equations[3]' so $dt$ comes from whatever 'Lagrange–Charpit equations' are right? – BCLC May 24 '24 at 10:43

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The symbolism $\frac{dx}{dt}$ is not at all ambiguous. If you don't like it don't use it. But you will be deprived of a lot of facilities established for centuries of mathematic progress. Non-standard analysis should convince you. $$-xu_x+yu_y=u+2x$$ The characteristic ODEs are : $$\frac{dx}{-x}=\frac{dy}{y}=\frac{du}{u+2x}$$ A first characteristic equation comes from solving $\frac{dx}{-x}=\frac{dy}{y}$ : $$xy=c_1$$ A second characteristic equation comes from solving $\frac{dx}{-x}=\frac{du}{u+2x}$ : $$xu+2x^2=c_2$$ The general solution of the PDE expressed on the form of implicit equation $c_2=F(c_1)$ is: $$xu+2x^2=F(xy)$$ where $F$ is an arbitrary fonction (to be determined according to boundary condition).

On explicit form : $$u(x,y)=-2x+\frac{1}{x}F(xy)$$

JJacquelin
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