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

Welcome to the Wikipedia Mathematics Reference Desk Archives
The page you are currently viewing is an archive page. While you can leave answers for any questions shown below, please ask new questions on one of the current reference desk pages.

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 $F'(x)=KG'(x)$; so the solution to that first equation is rather simple, namely that $t(x,v)=xg(x,v)+B(x)$. 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)