# I Chain Rule and acceleration as a function of two variables

#### fog37

Hello Forum,
When the force $F$ and its resulting acceleration $a$ have the most general form, the acceleration can depend on the position $x$, time $t$ and speed $v$. Newton's second law is given by $\frac {d^2x}{d^2t}= a(x,t,v)$.
When the acceleration is only a function of position, i.e. $a(x)$, we can use the chain rule $\frac {dv}{dt} = \frac {dv} {dx} v$ and Newton's 2nd law becomes $$v dv = a(x) dx$$ and we can solve for $v(x)$.
But what if the acceleration is instead a function of both time $t$ and position $x$, i.e. $\frac {d^2x}{d^2t}= a(x,t)$? What kind of approach do we use? What kind of chain rule can we use? How do we solve for the speed $v$?

Thanks!

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#### Delta2

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When the acceleration is given as a function of $x=x(t)$ and $t$ and (perhaps $v$) that is $a=a(x,t,v)$ then this method of the chain rule and separation of variables simply wont work and I don't think that there is a similar method based on a generalized chain rule and separation of variables that would work. It will still be $vdv=a(x,t,v)dx$ but when we gonna integrate $\int a(x,t,v)dx$ we cannot treat $t,v$ as constants with respect to $x$ and this is where our problem lies and why this method wont work. We simply have to look for other methods.

If for example it is given that $a=5x+t^2+v$ then we ll have the ODE $\frac{d^2x}{dt^2}=5x+t^2+\frac{dx}{dt}$ or simply $x''(t)-x'(t)-5x(t)-t^2=0$ and this ODE it just cannot be solved by separation of variables. It can be solved with Laplace Transform for example.

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#### fog37

Thank you. That makes sense. I guess not all differential equations can be solved analytically to provide a nice solution equation.
What does it mean when I hear that a differential equation can be solved explicitly or implicitly?

All equations can always be solved numerically, correct?

#### Delta2

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All ODEs can be solved numerically but it doesn't mean that the numerical solution is correct or that converges to the analytical solution. This issue is know as the numerical stability of the solution (or the algorithm that calculates the numerical solution).

Explicit solution to a ODE is when the solution is given as a function $y=y(x)$. For example $y=ce^{kx}$ is an explicit solution of the ODE $y'=ky$.

Implicit solution is when the solution is given in perplexed form, that is as a general equation that involves $y$ and $x$ (but not any of the derivatives of $y$, otherwise it would still be an ODE), that is as $g(x,y)=0$ where $g$ a known function. In the above example, such an equation would be $lny-kx-c_1=0$.

Or as another example we might be able to prove that the solution $y=y(x)$ of a specific ODE, satisfies the equation $g(x,y)=y^3x^2-y^2+x\sin y+5=0$

#### fog37

Thanks Delta2. All clear on explicit/implicit solution.

You made me curious about the fact that a numerical solution can always be found but it may not be correct. What does it mean that it may not be correct? If it is a solution that solves the ODE, why wouldn't it be correct?

As far as the solution being converging, that should simply mean that the converging solution is "close" to the actual solution. The more converging the closer it is to the actual solution. How do we make it more converging?

#### Delta2

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Well strictly speaking, numerical solutions almost never are absolutely correct, in most of the cases they are approximately correct. For example we might have an ODE where the exacty analytical solution y(t) is such that $y(0)=\pi$. The numerical solution it just cant return as result that $y(0)=\pi$ it will be instead something like $y(0)=3.14149$ that is an approximation near $\pi$. On the contrary it might be that the analytical solution has $y(0)=1$ and it might be the case that the numerical solution also returns as result that $y(0)=1$. In this case the numerical solution is correct at the point $x_0=0$.

The converging of the solution has to do with how the numerical solution changes as we change various parameters in the algorithm/program that finds the numerical solution. A specific program maybe give converging solutions for some class of ODEs (for example linear ODEs with constant coefficients) but may give non converging solution for some other class of ODEs (for example linear ODEs with non constant coefficients). The converging of the solutions is something that depends both on the specific program we use and the specific ODE we trying to solve.

#### fog37

Thank you for the discussion. In general, with ODEs and PDEs, we would always try to find the solution over a set of points and not just at a single point, correct?

Also, is it even possible that a physical problem governed by a certain ODE or PDE does not have an actual solution (either analytical or numerical)? Or, like you said, at least a numerical solution always exist?

In a different (physics) thread I asked a question about Newton's 2nd law: $$F(x,t,v)=ma= m \frac{d^2x}{d^2 t}$$ where the force $F$ has a general dependence.
If the force $F$ is solely dependent on position $x$, it means that it does not vary with time at a specific location $x$. An object under the influence of a position dependent force will be changing its positions in time and will be experiencing, from its own perspective, a time-varying force.
For example, the gravitational force $F(y)= \frac {G m1 m2}{y^2}$ solely depends on the position y (altitude) but it can parametrically converted into a force that depends on time $t$ alone: once we find the object's position y(t), we can take plug it into $F(y)= \frac {G m1 m2}{y^2}$ and transform the force $F$ into a function $F(t)$. But this newly obtained function $F(t)$ is specific to the function $y(t)$ which was obtained for a specific set of ICs while the force expression $F(y)= \frac {G m1 m2}{y^2}$ represents the true and general structure of the force to plug into the differential equation above. Is that correct? So when we state that a force is either time-dependent or position dependent, we should never look at the situation from the perspective of the moving object. Otherwise, even a force that is only time dependent $F(x)$ can be re-written as a time-dependent force...

Would you agree?

#### PeroK

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Yes, I would agree. Although, it seems to me not a complex issue. For example, assume it rains in London from noon until 6pm; but is dry outside London. If you leave London at 3pm, then at 3pm the rain stops for you. But, it's still raining in London. The rain (at least for those 6 hours) is position-dependent, but not time dependent. But, if you move around, you can move from wet areas to dry areas. For you, the rain might be on and off.

#### fog37

I agree it can be a matter of perspective but it surely rains in London regardless of me being there or not so that is the true fact to consider.

In the case of the force function $F$ to use in Newton's 2nd law, it seems that a force expression that depends only on position or only on time would make a big difference in terms of finding the solution and the type of solution.

What would an example of a physical force $F(x,t)$ that changes both in time and in position? Do you have anything or any situation in mind? I guess the damping force on a spring would be an example of that...

#### Delta2

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Thank you for the discussion. In general, with ODEs and PDEs, we would always try to find the solution over a set of points and not just at a single point, correct?
Yes both analytical and numerical solutions are over a set of points. Analytical solutions are over continuous intervals $[a,b]$ or over whole $\mathbb{R}$ while numerical solutions are over a discrete set of points ${x_i}$ $i=1,2,...,n$ for some positive integer $n$ (this is one of the parameters we enter to programs). The numerical solutions consist of values $y(x_i)$ that the program computes . Due to the finite precision we can have on computers , the points $x_i$ are essentially always rational that is $x_i\in \mathbb{Q}$, and also the values computed $y(x_i)\in \mathbb{Q}$. So if the exact analytical solution for a specific $x_i$, $Y(x_i)$ is an irrational number like $\pi$, the solution computed by the program, that is $y(x_i)$ can only be a good approximation of the true value $Y(x_i)$.

Also, is it even possible that a physical problem governed by a certain ODE or PDE does not have an actual solution (either analytical or numerical)? Or, like you said, at least a numerical solution always exist?
An ODE or PDE that describes a physical problem should always have an exact solution. However this exact solution might not be possible to be found by analytical means, so an analytical solution might not exist. A numerical solution always exist.
In a different (physics) thread I asked a question about Newton's 2nd law: $$F(x,t,v)=ma= m \frac{d^2x}{d^2 t}$$ where the force $F$ has a general dependence.
If the force $F$ is solely dependent on position $x$, it means that it does not vary with time at a specific location $x$. An object under the influence of a position dependent force will be changing its positions in time and will be experiencing, from its own perspective, a time-varying force.
For example, the gravitational force $F(y)= \frac {G m1 m2}{y^2}$ solely depends on the position y (altitude) but it can parametrically converted into a force that depends on time $t$ alone: once we find the object's position y(t), we can take plug it into $F(y)= \frac {G m1 m2}{y^2}$ and transform the force $F$ into a function $F(t)$. But this newly obtained function $F(t)$ is specific to the function $y(t)$ which was obtained for a specific set of ICs while the force expression $F(y)= \frac {G m1 m2}{y^2}$ represents the true and general structure of the force to plug into the differential equation above. Is that correct? So when we state that a force is either time-dependent or position dependent, we should never look at the situation from the perspective of the moving object. Otherwise, even a force that is only time dependent $F(x)$ can be re-written as a time-dependent force...

Would you agree?
You are right about this, $F(y)$ describes the true and general structure of the force and depends only on the gravitational or electrostatic or magnetostatic field, while $F(y(t))$ depends on the orbit $y(t)$ of the moving object . However i believe under certain condition there can be a 1 to 1 correspondence between a force $F(y)$ and the orbit $y(t)$ that the object takes due to the force $F(y)$.

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