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How I can solve an infinite system of equations?

enter image description here

Assume nMax = 5

term[n_][u_, v_] := 
 (Sqrt[32]/Pi)  Sqrt[ (Cosh[v] - Cos[u])]( A[n] LegendreP[n - 1/2, Cosh[v]]-n LegendreQ[n - 1/2, Cosh[v]]) Sin[n u];

partialSum[nMax_][u_, v_] := Sum[term[n][u, v], {n, 0, nMax}];

Differentiating this w.r.t. ν at ν = ν0. Requiring this derivative to be zero. Now how do I determine the A[n] coefficients by solving a system of equations and then after substituting A[n] into the equation for V[v, u], how do I get the result?

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There seems something wrong with your question. First, note that you may differentiate every term separately. E.g.:

term[n_][u_, v_] := 
  D[Sqrt[(Cosh[v] - Cos[u])] A[n] LegendreP[n - 1/2, Cosh[v]] Cos[n u].v];

Then you may get the partial sum for v0 by:

partialSum[nMax_][u_, v0_] := 
  Sum[term[n][u, v], {n, 0, nMax}] /. v -> v0;

This has the form (where fi[u,v0] are expressions from the partial sum):

A[1] f1[u,v0] + A[2] f2[u,v0] + ...

Setting this to zero gives you ONE equation for an infinity of variables!

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  • $\begingroup$ Please explain more... This is a PDE problem. Firstly we might need different eigenfunctions and eigenvalues to satisfy ∂V/∂v=0 at v=v0. Secondly, this is a homogeneous boundary condition. we will need a separate inhomogeneous boundary condition to determine the coefficients of the Fourier series. $\endgroup$ – iman karimipour Sep 13 at 22:02
  • $\begingroup$ You specify: ∂V/∂v=0 at v=v0. So we are left with only 1 variable, v has the value v0 and there are no derivatives left. That means we have an ordinary equation. Now my answer needs a bit explaining. It all depends on the fi[]. If we consider a definite u then we have only 1 equation. However, if u is considered an undetermined variable it all depends if the fi[] are linear dependent or not. If they are not, we have again only one equation. However if the are linear independent, each term must be zero and the term i can be solved for A[i] $\endgroup$ – Daniel Huber Sep 14 at 6:43
  • $\begingroup$ Thanks. Yes, u is determined and for any amount of u in its domain i,e, [0,2pi] we have only one equation to be solved. The system of equation that I want to solve is a similar system that is derived in the following post. math.stackexchange.com/questions/2746660/… $\endgroup$ – iman karimipour Sep 14 at 8:58
  • $\begingroup$ The OP's first comment here is copied from my comment to the OP in the original plotting question, telling the OP why this new question is problematic. But yes, the question is problematic. $\endgroup$ – yawnoc Sep 20 at 9:52
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Taking Daniel's answer a little step further: your problem reduces to find the real numbers $(A_i)_i$ given that $$\forall u\in\mathbb{R},\ \sum_i^\infty A_i f_i(u,v_0) = 0$$

where the $f_i(u,v)$ are given by:

 f[n_][u_,v_] := (1/(2 Sqrt[-Cos[u] + Cosh[v]])) Cos[n u] Csch[v]
               * ((-1 + (1 + 2 n) Cos[u] Cosh[v] -  2n Cosh[v]^2) 
               * LegendreP[-(1/2) + n, Cosh[v]] - (1 + 2 n) (Cos[u] - Cosh[v])
               * LegendreP[1/2 + n,  Cosh[v]])

A geometric interpretation of that is that you are looking for a vector $[A_1,\dots, A_\infty]$ in an infinite-dimensional space that is orthogonal to the vector $[f_1(u,v_0), \dots, f_\infty(u,v_0)]$, no matter the value of $u$. So the solution is likely to rely on the structure of the space covered by $[f_1(u,v_0), \dots, f_\infty(u,v_0)]$ when $u$ covers $\mathbb{R}$.

Now, if think it will be very difficult to find a (non-zero) solution numerically for the following reason. Note that I am absolutely not proving anything, just explaining an intuition. If the A[i] were a solution for a given v0, they would be in particular solution to any linear system of equations of the form:

\begin{align} \sum_i^\infty A_i f_i(u_1,v_0) &= 0\\ \sum_i^\infty A_i f_i(u_2,v_0) &= 0\\ \vdots& \end{align} for some sequence $(u_i)_i$. Of course, numerically, you cannot keep the infinite sum; so you might want to truncate it and see why comes out.

When I truncate at nmax and play with nmax, there seem to be no solution other than the zero sequence. You can check that by your own with that, for example:

v0 = 1.;
nmax = 20;
Solve[Table[Sum[A[i]*f[i][u, v0], {i, nmax}] == 0, {u, 2, nmax}], 
 Table[A[i], {i, 2, nmax}]]

Also, the solutions are sensitive to the sequence $(u_i)$ (here, $=(2,3, \dots, n_\text{max})$.

Of course, this sensitivity could vanish when nmax tends to infinity, but good luck to observe that numerically...

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  • $\begingroup$ Thank so much. I am not familiar with Mathematica. Could you please help with finding A[n] coefficients by solving a system of equations? Then after substituting A[n] into the equation for V[v, u], how do I get the result? Thanks $\endgroup$ – iman karimipour Sep 14 at 20:17
  • $\begingroup$ if we neglect cos (nu), how we can solve it from the begining i.e.,V=Sqrt[ (Cosh[v] - Cos[u])] A[n] LegendreP[n - 1/2, Cosh[v]] $\endgroup$ – iman karimipour Sep 15 at 8:30
  • $\begingroup$ When I run it, this error occurs! How I can remove it. Secondly, the amount of A[i] does not obtain!! We have 9 equations and 11 unknowns! General::munfl: 6.62582*10^-168 (-3.8983*10^-168) is too small to represent as a normalized machine number; precision may be lost. $\endgroup$ – iman karimipour Sep 16 at 19:12
  • $\begingroup$ I assume that u,nu,phi are independent coordinates in a 3D space. When we write the partial derivative with respect to nu, it is implied that u,phi remain constant in the derivation. $\endgroup$ – iman karimipour Sep 17 at 20:52
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In my answer to the original plotting question, I made an educated guess (since it was not mentioned in the question) that the Fourier solution $$ V (u, v) = \sum_{n=0}^\infty \sqrt{\cosh v - \cos u} \cdot A_n \cdot P_{n - 1/2} (\cosh v) \cos(nu) $$ was constructed so that the $A_n$ were to be determined by the non-homogeneous boundary condition $V = V_0$ at $v = v_0$.

In the current question, the boundary condition $\partial V/ \partial v = 0$ at $v = v_0$ is homogeneous. Using the same eigenfunctions $P_{n - 1/2} (\cosh v)$ and $\cos(nu)$ will therefore NOT work, as I have already noted in a comment.

In general, when finding a Fourier solution (separation of variables) to a (linear) boundary value problem, the coefficients must determined by a non-homogeneous boundary condition. If the boundary conditions of the boundary value problem are all homogeneous, then the solution is just identically zero.

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  • $\begingroup$ Thanks so much for your reply. I changed the potential V to other different forms. Please see it in my main question. Now in new definition of function V is possible to take ∂V/∂v=0 at v=v0 and find An? Thanks so much. $\endgroup$ – iman karimipour Sep 20 at 16:42
  • $\begingroup$ The problem is that you are still trying to find Fourier coefficients to satisfy a homogeneous boundary condition. This will NOT work. Also please don't modify a question such that it asks something different to existing answers. $\endgroup$ – yawnoc Sep 20 at 20:25
  • $\begingroup$ Thanks so much. It seems that the same problem which solved in this post has a problem!!. math.stackexchange.com/questions/2746660/… Anyway thanks in advance. I want to study another equation which I provided it in the main my question in the previous post. Could you please help with this new equation similar to what you solved before? mathematica.stackexchange.com/questions/229880/… $\endgroup$ – iman karimipour Sep 20 at 21:32
  • $\begingroup$ Could you please inform me the new equation has any solution or not? Thanks $\endgroup$ – iman karimipour Sep 21 at 20:46
  • $\begingroup$ Dear user72028.Could you please help me how I can solve the new form of function V , having two unknown parameters an and bn to find them?Boundary condition is V0 at tau0 and V1 at tau1. Thanks $\endgroup$ – iman karimipour Sep 24 at 7:51

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