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April 3

How do I do this

I have a partial differential equation to solve, and all I would like to find out is the step by step process of going about it, so I can work it out myself. We have u*u_x+u_t = 2, where u_x is the partial derivative of u with respect to x, and such. The initial conditions are given as u(x, 0 ) = -x. We are first asked to find the Characteristic curves. What I did was use the chain rule, and went about as follows : To begin with, let us look at the Chain Rule to see if we can make the equation above fit the pattern it gives. For this we have : du/dt=δu/δx*dx/dt + δu/δt*dt/dt, and so, to give it a solution more conducive to this, we make sure that the u in the problem is analogous to the dx/dt in this variation of the Transport Convection Equation, both being the coefficients of δu/δx , and the 2 from the Right Hand Side of the problem is like the du/dt in the Chain Rule expression, and it is obvious that dt/dt=1.

	So  dx/dt*δu/δx +  δu/δt = du/dt  compared to  u*δu/δx + δu/δt=2  makes u=dx/dt,  and du/dt=2. For a start, this means that along the characteristic lines, u changes for each second with a slope of two, where if we integrate  du/dt=2, and get ∫(du/dt) dt=∫2dt, we end up with u=2t+A , with A some constant, or function not dependent upon t, and at t=0, because the initial condition states that u(x,0)= -x, then A= -x, while for u=dx/dt, this implies that x=ut+B, with B a constant of integration or function not depending on t, since dx/dt= d/dt(ut)=u. I also went on to add : From the information above, since u=2t+A,∀t∈R∶t ≥0, and it was found that A= -x, then u=2t-x. But since it is also known that x=ut+B, and if we rearrange u=2t-x, we can also shew that x=2t-u, then ut+B= 2t-u.

Now out of this, which are the characteristic curves ? Are they the lines x=ut+B, and u=2t-x, or have I totally mucked this up ? Should I have instead used the Lagrange Charpit equations, or a change of variables with t=tau and x=xi ? We are required next to find the solution u in terms of the original variables x and t, so I guess I should change variables, but then I am fuzzy on how to relate this to the chain rule equation. We are also to go on to sketch this, and I assume these are lines in the plane. We are also asked to Consider this same initial value problem but with a new initial condition that u(x, 0) = f(x), where f_0(x) is bounded, then to find those characteristic curves and from the Jacobian, determine the earliest possible time at which the solution will break down. We are also informed that it is not possible to find an explicit form of u as a function of x and t but it is possible to find an implicit solution F(x, t, u) = 0, and we are asked to find it, and I have been looking at YouTube videos all weekend to work out the procedures for solving PDE'S like these, but I am still not sure how to proceed. Finally, we are asked to consider the initial value problem u_t - u^2*u_x = 0, with initial conditions u(x, 0) = g(x) = { negative half, if x less than or equal to zero ; 1 if x between zero and 1 exclusive ; and one half when x is greater than or equal to one } From this we are asked to sketch or plot those base characteristics, then find the solution, as well as determine in which region it is single-valued. Next we sketch or plot the solution for t = 0, 2, 4, and finally find the shock solution for this system.

I have some ideas, but I just need some hints to put it all together. Thank You Chris the Russian Christopher Lilly 08:01, 3 April 2017 (UTC).[reply]

Why should it be so complicated? This is a linear partial differential equation of the first order and should be solved by the method of characteristics. Ruslik_Zero 18:26, 3 April 2017 (UTC)[reply]

@Ruslik0: Unless I'm missing something, his equation as-is is nonlinear: his equation is

u{\frac {\partial u}{\partial x}}+{\frac {\partial u}{\partial t}}=2

and we have a nonlinear term on the left.--Jasper Deng (talk) 08:18, 4 April 2017 (UTC)[reply]

@Jasper Deng: You are right it is quasi-linear but the method still applies. Characteristics:

{\frac {dx}{u}}={\frac {du}{2}}\longleftrightarrow x-{\frac {u^{2}}{4}}=C_{1},

{\frac {dt}{1}}={\frac {du}{2}}\longleftrightarrow t-{\frac {u}{2}}=C_{2}.

The general solution:

F(x-{\frac {u^{2}}{4}},t-{\frac {u}{2}})=0,

where

F

is an arbitrary function. Ruslik_Zero 18:12, 5 April 2017 (UTC)[reply]

For the initial condition

u+x=0

at

t=0

the solution is (I am leaving details for the requester to hash out themselves):

u={\frac {x+t^{2}-2t}{t-1}}={\frac {x-1}{t-1}}+t-1

Ruslik_Zero 19:07, 5 April 2017 (UTC)[reply]

Yes, it is non linear because of the u term as a coefficient of the partial derivative of u with respect to x, and Thank You. What I did next was to make the Chain Rule, where du/dtau=δu/δx*dx/dtau + δu/δt*dt/dtau, then I got u=dx/dtau, 2=du/dtau, and dt/dtau=1, since these matched up with their analogous coefficients in the original equation, then I said because u=dx/dτ, ∫udτ= ∫dx/dτ dτ → x= uτ+ ξ → ξ = x-uτ, and we also let t=τ . In addition, since du/dτ=2, then ∫du/dτ dτ =∫2dτ → u=2τ+ ξ → ξ = u-2τ, but then I have two expressions for xi ( ξ ), where one equals x-uτ, and the other is u-2τ, yet I am not sure if that is right, or what to do next. I need to find the characteristic curves, then solve to find u(x, t ).. Chris the Russian Christopher Lilly 01:48, 5 April 2017 (UTC)[reply]

@Christopher1968: If you don't mind, please consider using LaTeX markup, as it's that much more difficult to read your comments with only text representations.--Jasper Deng (talk) 02:08, 5 April 2017 (UTC)[reply]

Okay, I shall look at that and try to put the question again, using that to make the derivatives clearer, thank You. Chris the Russian Christopher Lilly 03:01, 5 April 2017 (UTC)[reply]

Evaluating an integral

How much is $I(t,t_{0})=\int _{\tau =0}^{t}\left((\tau +t_{0})(t-\tau )\right)^{-1/2}d\tau$ ? (t, t0 both nonnegative reals)

On the one hand, WolframAlpha seems to think that $I=2\arctan {\sqrt {t/t_{0}}}$ .

On the other hand, when I do the computations by hand (develop the product, change variables), I end up finding $I=\int _{u={\frac {t_{0}-t}{t_{0}+t}}}^{1}{\frac {du}{\sqrt {1-u^{2}}}}={\frac {\pi }{2}}-\arcsin {\frac {t_{0}-t}{t_{0}+t}}$ (EDIT 07:22, 4 April 2017 (UTC): arcsin(1) is pi/2, not pi/4...)

Both expressions look compatible on limit cases, but WA's numerical evaluation tells me they are not equal (example with t0=1). Tigraan^{Click here to contact me} 16:33, 3 April 2017 (UTC)[reply]

@Tigraan: Indeed, they are not equal for

t_{0}=t=1

. As for your change of variables, could you please elaborate on the particular change of variables you used? --Jasper Deng (talk) 18:11, 3 April 2017 (UTC)[reply]

Did you mean to write π/2 ? --catslash (talk) 18:29, 3 April 2017 (UTC)[reply]

@Catslash: He's off by at least a sign, since plotting the two results as functions of two variables (so in 3D space) gives two graphs that aren't parallel.--Jasper Deng (talk) 18:54, 3 April 2017 (UTC)[reply]

In the special case where $t_{0}=0,t=1$ , this is equivalent to $\mathrm {B} (1/2,1/2)$ which has value $\pi$ , which shows that @Tigraan:'s value is incorrect.--Jasper Deng (talk) 20:43, 3 April 2017 (UTC)[reply]

Note that

\cos \left({\frac {\pi }{2}}-\arcsin \left({\frac {t_{0}-t}{t_{0}+t}}\right)\right)={\frac {t_{0}-t}{t_{0}+t}}

and

\tan \left({\frac {\theta }{2}}\right)={\sqrt {\frac {1-\cos(\theta )}{1+\cos(\theta )}}}

(tangent_half-angle_formula) --catslash (talk) 22:28, 3 April 2017 (UTC)[reply]

I had made a plotting mistake earlier. The answers do turn out to be equivalent when the mistake is corrected to

\pi /2

.--Jasper Deng (talk) 23:27, 3 April 2017 (UTC)[reply]

Yeah, that is a π/2 instead of π/4, corrected. Maybe the integral is correct, but I still fail to see how the arctan of a square root can be equal to the arccos of a fraction (loosely speaking).

The calculation goes by reformulating

(\tau +t_{0})(t-\tau )=\left({\frac {t_{0}+t}{2}}\right)^{2}-\left(\tau +{\frac {t_{0}-t}{2}}\right)^{2}=\left({\frac {t_{0}+t}{2}}\right)^{2}\left(1-\left(1+{\frac {2(\tau -t)}{t_{0}+t}}\right)^{2}\right)

, the change of variables at that point is then

u=1+{\frac {2(\tau -t)}{t_{0}+t}}

which gives the integral in u I wrote above, of which arcsin is a primitive. Tigraan^{Click here to contact me} 07:22, 4 April 2017 (UTC)[reply]

@Tigraan: Catslash's identities establish it when

t_{0},t\geq 0

.--Jasper Deng (talk) 08:01, 4 April 2017 (UTC)[reply]

Reverse Gaussian correlation inequality

If I have a set of correlated numeric attributes for objects, GCI claims that the probability of an object having all attributes in the 95% range per attribute is equal or greater than the product of each individual probability being in the 95% range for each individual attribute. Is it valid to reverse this? May I make a valid claim that the probability of being outside the 95% for a specific attribute is lowered if the object is within the 95% range for another correlated attribute? Specific application: If a patient is deemed healthy based on X measurable attributes, the probability that the patient will be unhealthy for another attribute is less than the probability of being unhealthy based solely on that attribute. 209.149.113.5 (talk) 17:33, 3 April 2017 (UTC)[reply]

How quickly can a computer solve quintic equations?

By brute force if that's the best they can do. How powerful a general purpose computer is needed to solve a random quintic equation in under 1 hour (worst case) if the equation is picked from the ones with only integers below z-digits? Would increasing z by 1 take 10 hours or 10⁵ or what? Is this a very parallelizable problem? If under z-digits takes 1 hour how long would sexic equations under z-digits take? How many digits for a quartic, cubic, quadratic or linear equation before a modern PC starts slowing down? (must be a lot?) Sagittarian Milky Way (talk) 18:23, 3 April 2017 (UTC)[reply]

@Sagittarian Milky Way: See root-finding algorithm. Newton's method and fixed-point iteration is generally fast as long as you pick a good starting point.--Jasper Deng (talk) 19:08, 3 April 2017 (UTC)[reply]

You'll need to clarify if you mean a quintic with integer coefficients or one with integer roots. If the former, then the typical solutions will involve irrational numbers, so you have to specify how precisely you want to determine the roots. In general though, such problems can be done very fast on modern computers unless you are considering very large integers. On my copy of Matlab I generated 1000 random quintic polynomials with integer coefficients < 1,000,000 and asked it to find the roots. On my laptop, the average time for finding all roots (including complex roots) per polynomial was 38 microseconds. Dragons flight (talk) 19:19, 3 April 2017 (UTC)[reply]

Pretty sure Matlab doesn't use brute force though. If you really use brute force then every additional digit would take 10 times as long, which is why nobody uses brute force except for trivial cases (e.g. x⁵-1=0). If you use binary search, which some people might consider brute force, then the time would increase by a constant amount for every additional digit, assuming you're just counting arithmetic operations. If you're doing multi-precision operations then the time would increase somewhat faster depending on how you do the arithmetic. In a typical case you'd use binary search to get an approximate value, then start applying Newton's method. Since Newton converges quadratically, the time essentially just depends on how long it takes to do addition and multiplication. --RDBury (talk) 22:01, 3 April 2017 (UTC)[reply]

MATLAB doesn't use brute force. Binary search is the bisection method, but cannot be used for complex roots (since complex numbers are not an ordered field), and requires lower and upper bounds on the roots. From MATLAB's documentation I see that it converts the problem to that of finding eigenvalues of a companion matrix of the polynomial.--Jasper Deng (talk) 23:35, 3 April 2017 (UTC)[reply]

By "brute force" do you mean just plugging in numbers until you hit a solution? Bubba73 ^{You talkin' to me?} 18:43, 6 April 2017 (UTC)[reply]

Yes. Though I know now that there's better ways and the supposedly unsolvable quintic equation is super easy for a computer. Sagittarian Milky Way (talk) 18:54, 6 April 2017 (UTC)[reply]

"Unsolvable" means that there is no closed-form solution, as there is with quartic and below. But there are fast ways to get approximate solutions. Bubba73 ^{You talkin' to me?} 19:13, 6 April 2017 (UTC)[reply]

There are closed form solutions to general quintics, just not in terms of a finite number of basic arithmetic operations (addition, multiplication, root extraction) over the rationals. See Bring radical. --Deacon Vorbis (talk) 21:15, 6 April 2017 (UTC)[reply]

Thanks, I didn't know that. Bubba73 ^{You talkin' to me?} 21:23, 6 April 2017 (UTC)[reply]

Matrix Algebra for ARIMA(1,0,0)

An ARIMA(1,0,0) model can be expressed in the matrix as follows:

   _          _       _           _
  | y_{1}      |     | 1          0|
  | y_{2}      |     |1-\phi=0.68 0|
  | ...        |     |...         0|    _    _
  | y_{n_{1}}  |     |.68         0|   |L     |
  | ---        |   = |---       ---| * |      |
  | y_{n_1}+1} |     |.68         1|   |\delta|
  | y_{n_{1}+2}|     |.68         1|    _    _
  | ...        |     |...       ...|
  | y_{N}      |     |.68         1|
   _          _       _           _

Therefore,

 _    _
|L     |
|      | = (((X^T)*X)^-1)*((X^T)Y)
|\delta|
_     _|

Where X is the Nx2 matrix in the first matrix form above (I goofed around with Latex for too long, pardon my ascii math).

So, doing the math (with software),

   (((X^T)*X)^-1)*((X^T)Y)=0.56.

Question: how do I differentiate L and \delta? Schyler (exquirere bonum ipsum) 21:42, 3 April 2017 (UTC)[reply]