Recent content by Set Abominae

But then doesn't u_{t}-u_{x}+2u = 2f? :confused:

So I would guess that they are of the form u(x,t)=f(x+t)+g(x-t), where if \eta=x-t, then g'(\eta)=g(\eta), much like ODE's when you add the solutions... But I've tried plugging that back in and I'm not sure it works... Sorry if I'm missing the point, I've only just started studying these...

I tried a coordinate change of x^{'}=t-x , t^{'}=t+x and tried looking for solutions of the form u(x,t)=v(x^{'},t^{'}). This reduces the problem to \frac{\partial u}{\partial t}(x,t) - \frac{\partial u}{\partial x}(x,t) = 2\frac{\partial v}{\partial x^{'}}(x^{'},t^{'}). Now I think I have to...

Hi there, I'm trying to find all solutions of: \frac{\partial u}{\partial t}(x,t)-\frac{\partial u}{\partial x}(x,t)=-2u(x,t) I know that one solution is u(x,t)=Ae^{x-t}, and any solution of \frac{\partial u}{\partial t}(x,t)-\frac{\partial u}{\partial x}(x,t)=0 is of the form...

I should have mentioned, I was considering a>1. Thanks.

Hi there, I'm trying to show that \sum\frac{log(r)}{r^{a}} (r going from 1 to infinity) converges using the ratio test, but I can't seem to deal with the log(r+1)/log(r) term. Any help would be much appreciated. Thanks.

Yes, I now have: 1 unstable fixed point x=0 for h<\frac{mg}{k}+\frac{1}{2} 1 fixed point x=0 for h=\frac{mg}{k}+\frac{1}{2}, and system is structurally unstable. 1 stable fixed point x=0 and 2 unstable fixed points at x=\pm\sqrt{h-\frac{mg}{k}-\frac{1}{2}} for h>\frac{mg}{k}+\frac{1}{2} I...

f(x)=\frac{k}{2}(x^{2}+(h-x^{2})^{2}+mgx^{2} \Rightarrow f'(x)=2kx^{3}+kx-2hkx+2mgx Solve f'(x)=0 to give x=0 for h\leq\frac{mg}{k}+\frac{1}{2}, and x=0, x=\pm\sqrt{h-\frac{mg}{k}-\frac{1}{2}} for h>\frac{mg}{k}+\frac{1}{2} Now, f''(x)=6kx^{2}+k+2mg-2hk h<\frac{mg}{k}+\frac{1}{2}...

I forgot about that! Upon further thought, I would imagine that I would subtract the negative gpe, so be adding it to the elastic energy to give f(x). I say this by considering energy conservation - assuming we don't have any stupidly big oscillations, the higher the bead gets, the larger mgh...

But I also get the same result of 3 unstable fixed points when I subtract the g.p.e. from the elastic energy... :confused:

Actually, thinking about it, I would imagine that I would subtract the g.p.e from the elastic energy to get f(x) (though I'm cautious about doing so, as the question says f(x)=elastic energy + g.p.e.). Another reason for doing this is that if I sum them to get f(x), when I start drawing my...

It's done like that by definition, as far as I am aware. I don't really know why it's defined like that though...:rolleyes:

The question states that: When the elastic has length d its elastic energy is \frac{k}{2} d^{2}, where d> 0... So when the bead is at (x,x^{2}), we have that: d = \sqrt{x^{2}+(h-x^{2})^{2}} (It'a a question from a math assignment, so its probably not totally accurate...) So that will...

Hi there. I have a bead of mass m, which slides down a frictionless parabolic wire in the form y=x^2, and is attached by elastic to the point (0,h), and I want to write down total energy f(x) (= elastic energy + gravitational potential energy) (no mention of kinetic energy...) of the bead at...

Hi everyone. I have a system with two strictly positive parameters, for which we have a single equilibrium point or 3 equilibrium points if a certain inequality holds involving the two parameters. I'm struggling to identify whether or not this is an example of a cusp catastrophe or not...