Random do my math homework for me thread

[BerenTebogo]

Let's face it, this needed to be made, so I'll start it off.

Let's say I have a 2nd order ODE, say... x'' - kx' = 0 where k is a constant. I want to make a substitution to reduce the order, y = x'. I now have y' - ky = 0. This is going to give me y(n) = e^kn. Basically, exponential growth.

Now this isn't how I want my function to behave. In fact, it's the exact opposite of how I want my function to behave. I want exponential decay.

My question is, given the parameters of my original equation, is there anything I can do to the function to change x'' - kx' = 0 to x'' + kx' = 0. Assume k is a physical constant that cannot be changed.

I'm considering maybe some type of coordinate system change, or something, but I'm really quite stumped as of right now. For the purposes of discussion, assume this represents some aspect of a physical system so the function can only be altered so much. X(t), position and time, let's say.

[Byrthnoth]

Let me preface this by saying it has been a long time since I have taken a math course of any kind.

Still, if k is negative then it is decay? k is a variable standing for a constant. As far as your equation is concerned, you could either substitute k=-j or just accept that k is a negative number.

For instance:

x(n) = e^(-j*n) :: equation for exponential decay
x'(n) = -k*e^(-j*n)
x'(n) = -j*x(n)
x''(n) = j^2*e^(-j*n)
x''(n) = -j*x'(n)
x''(n) + j*x'(n) = 0
k = -j
x''(n) - k*x'(n) = 0 :: your equation

I would take the moral of the problem to be "signs in front of constants are not always important." Also, "wikipedia + doing things in reverse is a good problem solving technique after you graduate and forget math."

[Richybear]

ky = lube
crisis averted

[BerenTebogo]

Let me preface this by saying it has been a long time since I have taken a math course of any kind.

Still, if k is negative then it is decay? k is a variable standing for a constant. As far as your equation is concerned, you could either substitute k=-j or just accept that k is a negative number.

For instance:

x(n) = e^(-j*n) :: equation for exponential decay
x'(n) = -k*e^(-j*n)
x'(n) = -j*x(n)
x''(n) = j^2*e^(-j*n)
x''(n) = -j*x'(n)
x''(n) + j*x'(n) = 0
k = -j
x''(n) - k*x'(n) = 0 :: your equation

I would take the moral of the problem to be "signs in front of constants are not always important." Also, "wikipedia + doing things in reverse is a good problem solving technique after you graduate and forget math."

Yeah, that would be the easy way out, lol. Unfortunately the constant can never be negative. Even a substitution like you proposed isn't going to work. It represents a physical system that can't ever be negative.

Any weird variable substitutions are more than likely going to invalidate the analysis I need to run on the function itself, which is what makes this really hard. I need to find a legal way of transforming the function from blowing up to being bounded, haha.

[Woozie]

This thread is an awesome idea, by the way.

Edit: From what we've gathered so far, we know the general solution is obviously http://latex.codecogs.com/gif.latex?...kt}+c_2e^{-kt}

Sath insists that the physics of the problem does not allow k to be negative, so we're thinking there's some sort of boundary condition that forces c_1=0 or something.

Edit: Nevermind. I was reading

x''-kx instead of x''-kx' so my solution is way off lol

[Byrthnoth]

Yeah, that would be the easy way out, lol. Unfortunately the constant can never be negative. Even a substitution like you proposed isn't going to work. It represents a physical system that can't ever be negative.

Any weird variable substitutions are more than likely going to invalidate the analysis I need to run on the function itself, which is what makes this really hard. I need to find a legal way of transforming the function from blowing up to being bounded, haha.

Could you give the exact problem?

[BerenTebogo]

The entire is a 2nd order PDE in 4 dimensions expressed in augmented spherical coordinates, which won't help too much, but I can give the exact expression for this component of the resulting variable separation.

(-r^2) del^2(Omega)/del(t)^2 + 4M del(Omega)/del(t) = 0

Variable sub y = Omega dot -> Ydot = 1/M(y) -> y = e^(T/M)

M is 1/2 of the root of a given delta function (for radius) which is Delta = r^2 - 2Mr and the roots r = 0 represent a singularity and r = 2M represents the event horizon of a black hole. The resulting y function represents the radius inspiral with respect to time and the location of the event horizon.

[Cadsuane]

I don't know much ODE and even less physics, but yes indeed the solutions for that ODE are precisely those of exponential growth if k is strictly positive. Of course, the 0 function is also a solution in the general family, but this probably doesn't help.

Also this thread is a fantastic idea.

[Ferion]

I don't know much ODE and even less physics, but yes indeed the solutions for that ODE are precisely those of exponential growth if k is strictly positive. Of course, the 0 function is also a solution in the general family, but this probably doesn't help.

Also this thread is a fantastic idea.

Pretty much this. I haven't worked with differential equations since I took the class junior year and I don't know too much about physics so I can't help much with this, but I like this thread a lot.

[Elesirdur]

This is really on topic for me, doing second-order, linear, autonomous, ODEs in my Math for Economics course. I have to ask, why do you need to convert the problem from a second order to a first order in the first place? Is it to remove the issues if you get complex roots? I suppose if there's nothing else in the function, just k multiplied by y' then you will get a negative root.

Now by "decay" do you mean that it must converge towards the steady state? In this case your SS is zero (not undefined), so any value of k less than zero would do the trick. So in that case you could define y=-x', which would get you your negative sign on the k coefficient so you can get the solution x(t)=Ce^-kt. However, I don't really have a framework to understand these questions from a physics perspective, so I'm not sure if this would work for your problem.

[Cadsuane]

This is really on topic for me, doing second-order, linear, autonomous, ODEs in my Math for Economics course. I have to ask, why do you need to convert the problem from a second order to a first order in the first place? Is it to remove the issues if you get complex roots? I suppose if there's nothing else in the function, just k multiplied by y' then you will get a negative root.

Basically if a relationship between a function's first and second derivative defines that function or family of functions, then that's the same as saying a relationship between a function and it's first derivative defines a family of it's first anti-derivatives.

... Maybe that's not actually clearer.

[Woozie]

I was talking to Sath on facebook for a bit, and he says that k cannot be negative, and that his solution must be exponentially decaying. So the only conclusion is that this ODE does not actually model his system. He says he's going to ask his professor about it.

[Kaylia]

Sath, can you post the whole problem. I have the feeling the answer lies in spatial coordinate being reversed (ie: a spring slowly getting back to its initial position which implies a negative x').

[edit]
Actually, nvm, it would also change the sign of the acceleration.

My question is, given the parameters of my original equation, is there anything I can do to the function to change x'' - kx' = 0 to x'' + kx' = 0. Assume k is a physical constant that cannot be changed.

Both term are linearly independent, and x+a cannot be equal to x-a unless a = 0, which is not the case here.


Also, this thread should be titled "predetermined mathematics and physics question", because screw randomness.

[BerenTebogo]

The thing is that not all PDEs can be expressed in terms of analytic solutions, that's just a harsh reality. In fact, as I understand it, something like 80-90% of them cannot be. It may entirely be the case that I cannot do anything to this function that will yield me the solution I need to construct a general analytic solution.

Like has been suggested, and has been done (by me and others), you can model a exp(-iwt) function much earlier on in the separation and solve for a specific frequency of the inspiral and hope that there are no instabilities. The thing is, that you can pick random w values forever and never actually be able to see where the instabilities occur. Given that this is a black hole dynamics problem, physical instabilities aren't as easily able to be hand-waved out as they are in, say, kinematics.

I'll keep the question open in case anyone can think of anything to do. Maybe a transform to a different vector space with different conditions? Don't know if this is even legal with physical systems. Maybe a change of coordinates, who knows. I've been leaning towards it not being solvable for a while, as has my professor. Guess we'll see!

Also, fuck randomness.

[Osede]

well, if it can be solved analytically, then finite difference/numerical method is the way lol

[BerenTebogo]

Yeah, there are a couple of accepted approaches for numerical methods. Thing is, like, my entire math and physics programs at my school are computational and numerical centered and I really like analytical. I do all the computational stuff, I hit matlab, mathematica, python, etc hard, but I like the beauty that comes from an analytic solution. I think the way that most (if not all) 2nd order periodic PDEs resolve to a fourier series is just fantastic. It's one of the things that really clicked with me when I took PDEs.

I'm okay with there being no analytical solution because there are ongoing numerical method applications to the same project I can take up and over the summer I'm getting paid to do research into some area of computational math so I could easily do that. I just would love, love love love, if I could somehow find a way to resolve this.

I've been thinking a lot lately of what it must have been like to be Laguerre, Heun, Bessel, who could see patterns in functions, or thought 'hey, what if I change my separation constant to l(l+1)...wonder if that will make AWESOME THINGS HAPPEN'. I love my some analytical PDEs haha.

[Woozie]

The thing is that not all PDEs can be expressed in terms of analytic solutions, that's just a harsh reality. In fact, as I understand it, something like 80-90% of them cannot be.

More like 100% of them. Well, not quite, but very very very close to it.

[Mojo]

The sign of the coefficient k determines the behavior of the time domain solution for this type of a problem. I don't think that you can re-write the expression in any way that will change this in the time domain. I'm not sure what application you're looking for with this, but you may want to consider domain transforms, if you're looking for more convenient ways to analyze solutions to this. I'll revisit this later, don't have too much time right now.

[Cadsuane]

So I'm taking some economics and physics this semestre at a fairly low level and it's really crashing down on me how differently mathematics is approached outside of pure math, which is my major. In particular I remember attending a physics colloquium way back in the fall where this Waterloo physics PhD made some funny remarks about the different ways physicists and mathematicians approach problems, which incidentally is a constant source of jokes from my professors, who apparently have nothing but disdain for this. ie.:

"So the limit of derivatives is indeed the derivative of the limits in this series of infinite functions, provided you have uniform convergence which is a sufficient but not necessary condition. Of course if you did calculus like a physicist, all of this would be entirely too confusing and you'd do it without checking."

I know this is good natured trash talking between departments, but I was thinking about doing a write up or two on an interesting theoretical subject that maybe you guys are already familiar with but maybe were given a very application oriented treatment of, or maybe something weird like Galois groups.

[oldoldman]

I have a stupid chemistry question that I just want to make sure I'm getting right. The question is What is the primary natural force that is responsible for Hund's Rule? A) The electrical charge of the electron B) The Heisenberg Uncertainty principle C) Magnetism D) The nuclear mass E) None of these

It's A right? Since it's caused by the repulsion between electrons?