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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 $\frac{\partial t(x,v)}{\partial v}=-\frac{x}{v^2}+A^{\prime }(v)$. The first equation becomes (assuming everything needed is nonzero at all relevant points) $\frac{\partial g(x,v)}{\partial v}=-\frac{1}{v^2}+\frac{A^{\prime }(v)}{x}$. 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 $F^{\prime }(x)=K{G}^{\prime }(x)$; 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)