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  1. R

    I need to verify Bessel function expension.

    Equation 2 is taking an arbitrary function f(x) and writing it as an expansion of Bessel functions of order p. In order to do this, f(x) can't blow up at the origin (or else you would have to include Bessel functions of the 2nd kind in your expansion). The Bessel functions of any order 'p'...
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    Generic question on boundary conditions

    I never thought about it that way, that analytic continuation is possible because of a differential equation (the Cauchy-Riemann conditions), but that's precisely what allows two variables (x,y) to be treated just like one (z=x+iy) without a disc in R^2 but only a curve in Z: the presence of...
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    Generic question on boundary conditions

    Analytic continuation can be done on a line. For example, f(z)=1/(1-z) is the analytic continuation into the complex plane z (except the pole at z=1) of the power series f(x)=1+x+x^2+...on the real interval (-1,1). It's true that the curve can't determine the gradient of u on the curve, but...
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    Generic question on boundary conditions

    I think you only need a line to do analytic continuation. It's the basis for the Schwartz reflection principle. Disregarding complex functions, you only need the values of a real, differentiable function f(x) on a small continuous interval, say [0,\epsilon], and then the entire function is...
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    Generic question on boundary conditions

    A partial differential equation requires boundary conditions. Consider a 2-dimensional problem, where the variables are 'x' and 'y'. The boundary is the line x=0 and you are given all sorts of information about the function on that line. If you are given just the values of the function on the...
  6. R

    Laplacian of f equals zero and spherical harmonics equation

    Actually, Wikipedia has the answer: http://en.wikipedia.org/wiki/Associated_Legendre_function They talk about 'm' and 'l' needing to be integers for the solution not to be singular on [-1,1]
  7. R

    Laplacian of f equals zero and spherical harmonics equation

    The differential equations are: \frac{1}{\Phi}\frac{d^2 \Phi}{d \phi^2}=-m^2 and (1-x^2)\frac{d^2\Theta}{dx^2}-2x\frac{d\Theta}{dx}+[l(l+1)-\frac{m^2}{1-x^2}]\Theta=0 where x=cos(\theta) . The solution of the first equation requires that "m" is an integer for the solution e^{im\phi} to...
  8. R

    Help doing differential equations and laplace

    If you want to find the inverse Laplace transform of \frac{s-a}{(s+1)^2+1} then find the poles which are at s=-1 \pm i so the sum of the residues of e^{sx}\frac{(s-a)}{(s+1-i)(s+1+i)} are: e^{sx}\frac{(s-a)(s+1-i)}{(s+1-i)(s+1+i)}|_{s=-1+i}+...
  9. R

    Boundary conditions

    But it ought to be true that if you were also given all the 2nd derivatives at the origin, along with the first derivatives and the function value at the origin, then the solution of the 2nd order PDE would be completely determined? The question then becomes how do you find the 2nd...
  10. R

    Boundary conditions

    If you have the value of a function of many variables, and its 1st-derivatives, at a single point, and a 2nd-order partial differential equation, then haven't you determined the entire function? You can use a Taylor expansion about that point to build the entire function because you have the...
  11. R

    Unsolvable differential equations?

    Re: unsolveable differential equations? Would it be safe to say that a differential equation has at least a number of constants equal to the order? So for your example you have a first order equation having two constants, and two is greater than one. So for example a 2nd order differential...
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    Unsolvable differential equations?

    Re: unsolveable differential equations? oops. In general, is the number of constants equal to the highest order derivative? But can you really see vertical asymptotes from a graph? If you have y=x^2, then at x=1000 all the slopes look vertical. Anyways, the differential equation I chose is...
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    Unsolvable differential equations?

    Suppose you have the differential equation: dg/dt=g^3 for a function g(t). I got that the solution works out to be: g(t)=\pm \left( \frac{1}{-2t}\right)^{1/2} Does this mean that the original differential equation has no solution for t>0, since you can't have a negative in a...
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