Recent content by Defconist

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    Change of variables to polar coordinates

    Thanks, I've been sudying PDE's few days in a row and I'm really getting tired..I should take a break and yet I can't, it's so interesting..
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    Simple looking but hard to solve nonlinear PDE

    Yes, there is a general solution for any first order nonlinear equation. Method of characteristics, usually this methods requires initial condition, but you can supply something general like u(x,0) = f(x). This way you get either a general solution or something that is pretty close. I myself...
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    Change of variables to polar coordinates

    I thought I grasped coordinate changes well, but now I've run into some problems. Usually I would have some function f(x,y) and transformation equations like s = a*x+b*y . I would apply chain rule and stayed left with new equations in new variables. (old ones get away through...
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    Mathematica An integral that Mathematica won't solve

    Hi, Here is a proper Mathemaica syntax : Integrate[Exp[-i*t*u] (1 - 2 i* t* a)^(-1/2) (1 - 2 i* t* b)^(-1/2), {t, -Infinity, +Infinity}, Assumptions -> {u > 0, a > 0, b > 0}] Indeed it doesn't converge...at all it is computed analytically so it is not just numerical issue. Defconist
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    Method of characteristics for linear PDE's (variable coefficients)

    Oh, I get it. It is a system od ODE's because the in y the second equation is the same as y in the first one... It's easy to see why I missed that. It is a possibility I feared from the very beginning. Anyway, thanks for getting me from this predicament. :)
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    Method of characteristics for linear PDE's (variable coefficients)

    I was going through an inroductory book on PDE's and at one point they proceed with little show of work. I have problem with equation -yu_x + xu_y = u . Characteristics for this equation are x_t = -y, y_t = x, u_t = u . So far it is clear, but now books states that solution of first...
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    Linear Partial Differential Equations Explained

    I don't know of any pictorial way, but this is an interpretation I use. It is useful mainly because it is easily verifiable in a routine "not thinking" way. Of course as you gain confidence studying you will eventually just see this. This from Differential equations, an intoduction by Strauss...
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