I wanna solve this, I have three equations and the needed boundary conditions as the picture shows, but after trying so many methods I still could not solve it, the picture has attached, note that the initial conditions have been chosen arbitrarily, they can change
code
could you help, please
I used the shooting method, and I expect to get some function of f g and h with respect to z, considering the boundary conditions
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The following is the plot of a curve f(r), where r is the radial coordinate, and plotted for different values of a parameter as shown:
However, I don't know the functional form of the curve and I am interested to find the same. Are there any numerical methods which can be used to find the functional form of f(r) in terms of the radial coordinate and the parameter?
I had found a solution of the problem based on the suggestion by ja72 to use the Eureqa software which churns through the data to create accurate predictive models using evolutionary search algorithm.
In the question, the different curves corresponds to different values of . So, initially I obtained the best fit equation for different values of and found that the following model equation is suitable for my purpose:
Then, I repeated the process for a large number of values of and calculated the values of the four functions for different values of and then individually fitted these four functions. The following are the results that I obtained:
N.B.: Eureqa gave several other better fitting formulas than those mentioned in the answer. But the formulas that I mentioned are sufficiently accurate for my purpose and have minimum complexity.
A blind curve fit without an underlying model is a dangerous thing.
You need to have an understanding of the physical model behind the data to create a successful fit. The reason is that if r is distance and the best fit curve uses r^0.4072 for example, that dimension raised to a decimal power bears no meaning and it hides any underlying assumptions.Like some other dimension l not included in the model, whereas only the dimensionless quantity (r/l) would make sense to raise to the decimal power.
From a function analysis standpoint
These curves are not the result of any standard math function. Well I am not that familiar with bessel functions, gamma functions and legendre polynomials. But none of the standard functions you find in a scientific calculator jumps out here.
If r is assumed to be dimensionless, then you try to match the asymptotic behavior when r -> 0 and when r -> ∞. The would be the baseline curve. To me it does not look hyperbolic, but rather close to 1/LN(1+r).
So change the variables make g=1/LN(1+r) and plot f(r) against g(r) and see what that looks like. Then try another round of curve fitting in the new curves ... and so on.
Nobody can answer this question
Nobody else could effectively answer this question but you, because a) you have the data, and b) you need to make assumptions about what region is important or not, and what is acceptable deviation.
While I was reading the blog of Colah,
In the diagram we can clearly see that zt is going to
~ht and not rt
But the equations say otherwise. Isn’t this supposed to be zt*ht-1 And not rt*ht-1.
Please correct me if I’m wrong.
I see this is somehow old, however, if you still haven't figured it out and care, or for any other person who would end up here, the answer is that the figure and equations are consistent. Note that, the operator (x) in the diagram (the pink circle with an X in it) is the Hadamard product, which is an element-wise multiplication between two tensors of the same size. In the equations, this operator is illustrated by * (usually it is represented by a circle and a dot at its center). ~h_t is the output of the tanh operator. The tanh operator receives a linear combination of the input at time t, x_t, and the result of the Hadamard product between r_t and h_{t-1}. Note that r_t should have already been updated by passing the linear combination of x_t and h_{t-1} through a sigmoid. I hope the reset is clear.
I have the following question on a practice exam:
I need to use MATLAB to solve it. The problem is, I have not seen a problem like this before and I'm struggling to get started.
I have my 1x1 grid, split into 10x10. I know I can calculate the whole bottom row besides the corners using 1/10 * x*2. I also know I can calculate the entire right row using (1/10)(1+t)^2. However, I cannot figure out how to get enough points to be able to fill in the values for the entire grid. I know it must have something to do with the partial derivatives given in the problem, but I'm not quite sure where they come into play (especially the u_x equation). Can someone help me get a start here?
I don't need the whole solution. Once I have enough points I can easily write a matlab program to solve the rest. Really, I think I just need the x=0 axis solved, then I just fill in the middle of the grid.
I have calculated the bottom row, minus the two corners, to be 0.001, 0.004, 0.009, 0.016, 0.025, 0.036, 0.049, 0.064, 0.081. And similarly, the entire right row is trival to calculate using the given boundry condition. I just can't piece together where to go from there.
Edit: the third boundry condition equation was mistyped. it should read:
u_x(0,t) = 1/5t, NOT u(0,t) = 1/5t
First realise that the equation you have to solve is the linear wave equation, and the numerical scheme you are given can be rewritten as
( u^(n+1)_m - 2u^n_m + u^(n-1)_m )/k^2 = ( u^n_(m-1) - 2u^n_m + u^n_(m+1) )/h^2
where k is the time step and h is the delta x in space.
The reformulated numerical scheme makes clear that the left- and right-hand sides are the second order centred finite difference approximations of u_tt and u_xx respectively.
To solve the problem numerically, however, you need to use the form given to you because it is the explicit update formula that you need to implement numerically: it gives you the solution at time n+1 as a function of the previous two times n and n-1. You need to start from the initial condition and march the solution in time.
Observe that the solution is assigned on the boundaries of the domain (x=0 and x=1), so the values of the discretized solution u^(n)_0 and u^(n)_10 are known for any n (t=n*k). At the nth time step your unknown is the vector [u^(n+1)_1, u^(n+1)_2, ..., u^(n+1)_9].
Observe also that to use the update formula to find the solution at the n+1 step, requires the knowledge of the solution at two previous steps. So, how do you start from n=0 if you need information from two previous times? This is where the initial conditions come into play.
You have the solution at n=0 (t=0), but you also have u_t at t=0. These two pieces of information combined can give you both u^0 and u^1 and get you started.
I would use the following start-up scheme:
u^0_m = u(h*m,0) // initial condition on u
(u^2_m - u^0_m)/(2k) = u_t(h*m,0) // initial condition on u_t
that combined with the numerical scheme used with n=1 gives you everything you need to define a linear system for both u^1_m and u^2_m for m=1,...,9.
To summarize:
--use the start-up scheme to find solution at n=1 and n=2 simultaneously.
--from there on march in time using the numerical scheme you are given.
If you are completely lost check out things like: finite difference schemes, finite difference schemes for advection equations, finite difference schemes for hyperbolic equations, time marching.
EDITING:
For the boundary condition on u_x you typically use the ghost cell method:
Introduce a ghost cell at m=-1, i.e. a fictitious (or auxiliary) grid point that is used to deal with boundary condition, but that is not part of the solution.
The first node m=0 is back into your unknown vector, i.e. you are now working with [u_0 u_1 ... u_9].
Use the left side boundary condition to close the system.
Specifically, by writing down the centered approx of the boundary condition
u^n_(1) - u^n_(-1) = 2*h*u_x(0,k*n)
The above equation allows you to express the solution on the ghost node in terms on the solution on an internal, real node. Therefore you can apply the time-marching numerical scheme (the one you are given) to the m=0 node. (The numerical scheme applied to m=0 would contain contributions from the m=-1 ghost node, but now you have that expressed in terms of the m=1 node.)
The question basically says it all. I would like to add that lets suppose I have an image, a photograph and I wish to calculate its mathematical function, so that when I input x and y pixel value, it returns a vector consisting of R,G,B values at that x,y point. Therefore I can use a for loop to construct the whole image by just that function. I am not asking about the whole solution or algorithm here, but just that if this thing is possible, which direction should I take to go about doing this. Reference to relevant papers would be really nice.
Thanks
Azmuh
Yes, it is absolutely always possible. Basically, if you choose some points, there is always (an infinity of) smooth explicit functions (that is nice functions) which value on the points is exactly the one you choose.
For example, you can have a look at http://en.wikipedia.org/wiki/Lagrange_polynomial or http://en.wikipedia.org/wiki/Trigonometric_interpolation. They are two different methods to compute an explicit function which pass exactly by the data points you have. So you can apply those methods for your image, seen as a set of data points, and separately for R, G, and B.
At the end, you get one simple function explicitly (a polynomial or a trigonometric series, depending on what you chose), and you can compute its values where you want.
However, note that I would definitely not recommend to use those methods to effectively retrieve the data. Indeed, the functions you get are absolutely not optimized (that is with a veeeery high degree (for a n×m image, each color will have a degree nm-1), very high coefficients) and furthermore will have extremely large values between your original points (look for Runge's phenomenon).
This is not possible in general... Imagine an image that has been generated by random values for each pixel. You can't find a mathematical expression that will give you the value of a pixel given its 2d coordinates.
Now it may be possible for some images that have been generated using a function. In that case, it's not a problem specific to image processing, it's get back the function from some points of the function (in your case, you have all the points). It's exactly the same thing as extrapolating a curve from a set of points when you trace a graph in excel. The more points you have, the more precise the function you wind will be.
Look for information about Regression analysis. I can't help you much but there are some algorithms that exist.
Does anyone know how to prescribe boundary conditions of like u[t,0,y]==u[t,1,1-y] in Mathematica using NDSolve... It always complains that the arguments of the dependent variable should literally match the independent variable.
Thanks in advance.
This symmetry condition can probably be recast in the form Derivative[0,1][u][x,1/2]==0. Of course, more information on the problem would be helpful.
Edit in response to rcollyer:
The algebraic identity f(x)=f(1-x) for all x in (0,1) implies a geometric symmetry: the graph of f will be symmetric about the line x=1/2. Now draw the graph of such a function; if it is differentiable, you will find that f'(1/2)=0.
Now, I don't know for sure that the OP's problem can be recast this way; it rather depends on the specifics of the problem. This situation frequently arises when dealing with PDEs on the disk where the function u is a function of polar coordinates r and theta. If the disk represents a clamped drum, then perhaps you've got u(1,t)=0. But, what of u(0,t)? If the function is symmetric and smooth, then u_x(0,t)=0 is a reasonable condition.