Wikipedia:Reference desk/Archives/Mathematics/2018 June 28

= June 28 =

Differential equation
I have constructed the following system of differential equations.

$$\frac{\frac{\partial t(x,v)}{\partial v}}{\frac{\partial g(x,v)}{\partial v}}=x$$, $$\frac{\partial t(x,v)}{\partial x}=\frac{1}{v}$$.

I'm primarily interested in the most general form of a solution for $$g(x,v)$$. I'm pretty sure that it's a large family of solutions, and Mathematica can't seem to help me. A form for $$t(x,v)$$ would be nice too; again, I don't anticipate anything other than a very general expression.--Leon (talk) 13:11, 28 June 2018 (UTC)
 * From the second equation, $$t(x,v) = t(0,v) + \frac{x}{v}$$. Let us denote $$A(v) := t(0,v)$$ (this could be any function that is suitably differentiable). Then $$\frac{\partial t(x,v)}{\partial v} = -\frac{x}{v^2} + A'(v)$$. The first equation becomes (assuming everything needed is nonzero at all relevant points) $$\frac{\partial g(x,v)}{\partial v} = -\frac{1}{v^2} + \frac{A'(v)}{x}$$. Then $$g(x,v) = \frac{1}{v} + \frac{A(v)}{x} + B(x)$$ where B can be any function (again, possibly subject to smoothness conditions).
 * Notice that in your original question, the first equation is an obfuscation of a simple differential equation of the kind $$F'(x) = K G'(x)$$; so the solution to that first equation is rather simple, namely that $$t(x,v) = x g(x,v) + B(x)$$. Tigraan Click here to contact me 14:04, 28 June 2018 (UTC)


 * Thanks. However, I fear that I made a small mistake.


 * $$\frac{\frac{\mathrm{d} t(x,v)}{\mathrm{d} v}}{\frac{\mathrm{d} g(x,v)}{\mathrm{d} v}}=x$$, $$\frac{\mathrm{d} t(x,v)}{\mathrm{d} x}=\frac{1}{v}$$ is what I want to solve.


 * I think that $$\frac{\mathrm{d} t(x,v)}{\mathrm{d} v} = \frac{\partial t(x,v)}{\partial v} + \frac{\partial t(x,v)}{\partial x} \frac{\mathrm{d}x}{\mathrm{d}v}$$, with similar results for the other full derivatives. Is there a way of doing this? I'm primarily interested in the general form of $$g(x,v)$$, much as before.--Leon (talk) 10:21, 29 June 2018 (UTC)
 * The derivative $$\frac{dx}{dv}$$ is meaningless without some way of specifying the dependence between x and v.--Jasper Deng (talk) 15:34, 29 June 2018 (UTC)
 * It is a function of $$x$$ and $$v$$. Does that help?--Leon (talk) 16:09, 29 June 2018 (UTC)


 * Put it another way, $$\frac{\mathrm{d}v}{\mathrm{d}x}(x,v)$$ is a general $$C^1$$ function, and I want a general procedure to move from this to $$t(x,v)$$ and $$g(x,v)$$.--Leon (talk) 19:21, 29 June 2018 (UTC)
 * Then there is unlikely to be a general closed-form expression as the resulting differential equation is highly nonlinear, and the existence of $$\frac{dx}{dv}$$, needed to expand the second equation's left hand side, is extremely dependent on the location of the roots of $$\frac{dv}{dx}$$.--Jasper Deng (talk) 19:31, 29 June 2018 (UTC)
 * Okay, here's another system that might help me.


 * What about $$\frac{\mathrm{d} t(x,v)}{\mathrm{d} v}= \frac{1}{v\frac{\mathrm{d}v}{\mathrm{d}x}(x,v)}$$, $$\frac{\mathrm{d} t(x,v)}{\mathrm{d} x}=\frac{1}{v}$$? Is this "solvable" in some sense?--Leon (talk) 19:57, 29 June 2018 (UTC)


 * Perhaps it will help if I give some context: suppose I have a phase portrait for an autonomous mechanical system. $$x$$ and $$v$$ are the phase space coordinates, and the trajectory that starts at $$(x_0,v_0)$$ is entirely determined by the function $$\frac{\mathrm{d}v}{\mathrm{d}x}(x,v)$$. How would I even set up the problem for finding the time between two points on a trajectory?


 * The idea of my function $$g$$ is as follows. By differentiating energy $$E$$ with respect to velocity $$v$$, I get momentum $$p$$. I wanted something similar such that differentiating time $$t$$ with respect to $$g$$ would give position $$x$$. Can this be done?--Leon (talk) 22:53, 29 June 2018 (UTC)
 * Okay, maybe you want to look at the material derivative, which is the correct way to use the total derivative with respect to time.--Jasper Deng (talk) 02:02, 30 June 2018 (UTC)