Recent content by Jitse Niesen

1. Yes 2. If you set TSPAN (the second argument of ode45) to [-L/2 L/2] then Matlab applies the initial conditions at -L/2 and integrates from x = -L/2 to x = L/2. Since you want to apply the initial conditions at x = 0, you have to set TSPAN to [0 L/2] (as you do below). Then you can compute...

I'm sorry, my fault. I hadn't seen that you were given an initial condition. Yes, the solution \[\ln (x) - \frac{{y^2 }}{x} = c\] is correct (assuming that x > 0, which is the case here). And indeed, the initial condition implies that c = 0. However, the solution \[y = \sqrt {x\ln (x)} \]...

That's just one solution for the ODE. The general solution has a constant in there. The equation is not defined when 2xy=0. So y=0 is not the whole story. No. It means that the equation has a unique solution for y>0 and y<0. Since you only found the solution for y>0, your solution is not...

Silverbabe, search for "Weierstrass substitution" or "trigonometric substitution" in your calculus book, library, the Internet, or Wikipedia, and you'll find out what it means

The equations that you wrote down (the ones you get after discretization) are almost the algorithm. They can be translated directly into code. Have you ever programmed a finite difference method before? If not, do you have an example that you can use to get you started? You don't talk about...

I still don't quite understand what you're doing. So instead of answering directly, let me try to explain how I would linearize that equation and perhaps you find the answer to your question in my explanation. In general, linearization means that you assume that something is so small that...

I'm not sure I can explain where I got the idea from. The first step was to guess that you have to start by using the inequality |a+b| \ge |a| - |b| , because that's one of the few inequalities I know that give a lower bound of the absolute vale of something. Then the trick is to choose a and b...

Start with |e^z-1| = |z + (e^z-z-1)| \ge \ldots

I don't understand what you want to do. As you say, the equation you get after "linearization" is very different from the original equation, so there must be a mistake in this linearization process that you're doing. I would say that the original equation is already linear, so the linearized...

I don't understand you. Are you saying that any local maxima of f (the left-hand side of the inequality) restricted to T (the triangular region from the problem) lies either in the interior or the boundary of T? I agree with that, but I don't see how you want to prove that f restricted to T has...

Because there is a constraint. Even if the function does not have a maximum in the region { a,b,c > 0 }, it may have a maximum in { a,b,c > 0 and a+b+c = 3 }. Of course you can get rid of the constraint by eliminating one of the variables (say a), and instead look for extrema of (3-b-c+b^2)...

(b+c^2)(c+a^2) + 2a(a+b^2)(b+c^2) + \lambda = 0 and the two cyclic permutations of this equation, plus the constraint a+b+c=3 where lambda is a Lagrange multiplier.

arildno, there is also the constraint a+b+c=3 to take into account. matt grime, the procedure you sketch does not seem to be that easy. The equations for finding stationary points in the interior are quite horrible. Even finding the maxima on the boundary is something I'd rather not do by...

\sqrt{x^2+1} is the integrating factor for the equation \frac{dy}{dx} + \frac{xy}{x^2+1} = 0 that you get after you divide the original equation through by x^2 + 1. You have to multiply that equation by \sqrt{x^2+1}, not the original one.