wath is the mathematical definition of the eigensolutions for a system of partial differential equations?
98990
A system of partial differential equations can be described as an nxn matrix (lets call it A). The eigenvector (X) will be the n dimensional vector which has the property that there exists some real eigenvalue (L) such that AX=LX.
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http://mathworld.wolfram.com/HarmonicSeries.html
37116
An especially significant property of chaotic systems is great sensitivity to initial conditions.  Move the tiniest amount in x,y in Mandelbrot and you have a radically different outcome.
196036
If you divide 2 to the second equation, you will notice that it is identical to the first one, so they are dependent to each other, and there are infinitely many solutions to this problem.\n\n^_^
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You can refer this book:\n\nDifferential equations : linear, nonlinear, ordinary, parial - A.C. King, J. Billingham, and S.R. Otto.
81655
Yes:  theory of computational systems.
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It's called spectral analysis and you do this type of analysis on an operator on a Hilbert space. It is a generalization of the study of eigenvalues and eigenvectors of a matrix. If you are at a level where this question could come up, you should do the work yourself.
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the first link explains system boundary and the second is an explanation of computer simulations in the context of "conceptualizing a system".  hope this helps
589794
Formulate stability of the system as a function of its energy. Now, one has to show that minimum energy translates into max stability on finding the derivatives.
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The idea is that the y_n form an orthonormal basis so every function can be expanded as a series of the form \nsum a_n y_n.\n\nHow is this known? Typically, the inverse operator of L can be shown to be a compact operator and the general theory of compact operators can be invoked. Some care may be needed with the convergence properties of the series you obtain. Only mean-square convergence (L_2) is guaranteed. \n\nIn any case, the fact that the y_n are eigenfunctions for L and that L is linear makes the determination of the A_n from the B_n much easier. The convergence details can either be ignored or dealt with separately.
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it called the Cayley-Hamilton Theorem, type it in google and its proved in many places;\n\n\nhttp://delta.cs.cinvestav.mx/~mcintosh/comun/summer99/mcintosh/node9.html  <---this one is good :)\n\nLet B = x*I - A, where I is the identity matrix of the same size as A.  \nLet C be the classical adjoint of B, so that BC = (det B)I. Recall \nf(x) = det(B), by definition. We wish to show that f(A) = 0. It \nsuffices to show that f(A)*e(k) = 0 for 1 <= k <= n, where e(k) is the \nk-th unit vector, so fix a value of k in that range.\n\nNow x = A implies that for all j,\n\n   x*e(j) = A*e(j),\n\n             n\n          = SUM A(i,j)*e(i),\n            i=1\n   \n      n                     n\n   x*SUM delta(i,j)*e(i) = SUM A(i,j)*e(i),\n     i=1                   i=1\n\n    n\n   SUM [x*delta(i,j)-A(i,j)]*e(i) = 0,\n   i=1\n\n    n\n   SUM B(i,j)*e(i) = 0,\n   i=1\n\n           n\n   C(j,k)*SUM B(i,j)*e(i) = 0,\n          i=1\n\n    n\n   SUM B(i,j)*C(j,k)*e(i) = 0.\n   i=1\n\nThen, summing over all j, reversing the order of summation, and using\nthe definition of the classical adjoint,\n\n    n   n\n   SUM SUM B(i,j)*C(j,k)*e(i) = 0,\n   j=1 i=1\n\n    n    n\n   SUM [SUM B(i,j)*C(j,k)]*e(i) = 0,\n   i=1  j=1\n\n    n\n   SUM [det(B)*delta(i,k)]*e(i) = 0,\n   i=1\n\n    n\n   SUM [f(x)*delta(i,k)]*e(i) = 0,\n   i=1\n\n    n\n   SUM [f(A)*delta(i,k)]*e(i) = 0,\n   i=1\n\n         n\n   f(A)*SUM [delta(i,k)]*e(i) = 0,\n        i=1\n\n   f(A)*e(k) = 0.\n\nThis proves that f(A) annihilates all the unit vectors, so must be\nthe zero matrix, and f(A) = 0.\n\n\nsecond one taken off the dr math website, but the link above might be easier to understand.
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http://en.wikipedia.org/wiki/Eigenvalue\n\nI read through this and it made more sense to me this time than it ever has.  You can think of the solutions to a problem as vectors that point in a certain specific direction.  An eigenvector is a solution vector that points along the same direction but has a different scale length compared to the original vector.  That scale length is the eigenvalue/s.
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Let u=ln t, e^u=t, du=dt/t, dt=t du\nThen:\n∫ln² t dt = ∫t u² du = ∫u² e^u du\nIntegrating by parts:\nu² e^u - ∫2u e^u du\nu² e^u - 2u e^u + ∫2e^u du\nu² e^u - 2u e^u + 2e^u + C\nt ln² t - 2t ln t + 2t +C\nAnd we are done.
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wow..that site http://www.algebra.com/services/renderin... is really amazing. You can also check out mupad. It will do simplifications and give answers to integration and much more. There's a demo available at http://www.sciface.com/
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Great question.\n\nMaybe you can work backwards from the characteristic polynomial.  You know the roots so:\n\np(l) = (l+2)(l-1)(l-3)  right?  Now multiply out, the coeff of the linear term is the negative of the trace.  The sum of the eigenvalues is equal to the trace.  The constant term is the determinant.  \n\nor, start with the diagonal matrix and apply a basis transformation to get a non-triangular matrix.  \n\nor, choose 3 orthogonal vectors to call the eigenvectors, then find the transformation of the ....(out of steam)
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s(n) =  [1-r^n]*a/(1-r) + [(1-n)r^n]*d/(1-r) + [1-r^(n-1)]*dr/(1-r)^2\nfor n > 0\n\ns(infinity) = (a + r(d-a))/(1-r)^2\nfor |r| < 1\n\nSee simulation\nhttp://www.geocities.com/derekcowley/public/agp.xls\n\nproof:\n\ncompute the difference equation:\ny(n) = [a + (n - 1)d]*r^(n-1) = a*r^(n-1) + (n-1)d*r^(n-1)\ny(n+1) = [a+nd]*r^n = a*r^n + nd*r^n = y(n)*r  + d*r^n\ny(n+1) = y(n)*r  + d*r^n\n\ngenerate the LTI discrete time system state space:\n\nx2(n) = r^(n-1)\nx2(n+1) = r*x2(n)\nx1(n+1) = r*x1(n) + d*x2(n) \nx1(1) = a\nx2(1) = r\n\nX(n+1) = A*X(n) + B*delta(n)\ny(n+1) = C*X(n)\n\nwhere A is a 2x2 matrix [r,d;0,r]\nB is column vector [a;r]\nC is a row vector [1,0]\ndelta(n) = 1 for n=0, 0 otherwise\n\nCompute the Z-transform:\n\nH(z) = C*[(zI-A)^(-1)]*B = a/(z-r) + dr/(z-r)^2\n\ncompute running sum:\n\nAccumulator U(z) = z/(z-1)\nS(z) = H(z)*U(z) = az/(z-1)(z-r) + drz/(z-1)(z-r)^2\n\nCompute partial sum (partial fraction expansion):\n\nS(z) = e/(z-1) + f/(z-r) + g/(z-r)^2\n\ne = a/(1-r) + dr/(1-r)^2\nf = -ar/(1-r) - dr/(1-r)^2\ng = -dr^2/(1-r)\n\nInverse Z-transform\n\ns(n) = e  + f*r^(n-1) + g*(n-1)r^(n-2) [n>0]\n= see top line above\n\nTo see a spectral analysis of the sequence based on pole/zero locations see\nhttp://www.geocities.com/derekcowley/public/agp2.xls\nYou will need the "Analysis ToolPak" add-in in Excel to properly see this file.
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Actually, you get X''/X=C and Y''/Y=-C where C does not have to be 0. Also, remember that sums of solutions are solutions, so the solutions you get from separation of variables only give a basis for the vector space of solutions, not all the solutions. The specifics are difficult in this forum, but you have to set up the series and use the boundary conditions to find the coefficients of the series.