Sarahisme

i think i need to solve : [math] Asin(\sqrt(\lambda)) + Bcos(\sqrt(\lambda)) =0 [/math] [math] Asin(\frac{\sqrt(\lambda)}{2}) + Bcos(\frac{\sqrt(\lambda)}{2}) = 0[/math] but i just can't figure out how!

let me see, well i've work out that the general solution is y(x) = Asin(((lambda)^0.5)/x)) + Bcos((((lambda)^0.5)/x)) but i don't see how to use y(1) = 0 and y(2) = 0 because neither cancels either term.... :S

to be specific, its parts © & (d) that i don't know how to do, just can't quite figure out anything reasonable, other than a trivial solution! :S

Hi guys, i was hopeing someone would be able to guide me through this problem. for part (a) i get the transformed equation to be [math] \frac{d^2y}{ds^2}+\frac{4}{s}\frac{dy}{ds} = -\lambda y [/math] i am not 100% sure this is correct, although i have been through the working about 5 times now as for the rest of the question, well, i guess i really need to get part (a) before i can attempt the rest hey? and i don't really get this eigenvalue/vector stuff for functions :S i suppose (a) isnt really my problem, its the rest of the question that is really giving me a hard time! Thanks -Sarah

hmm ok, how does this look..? for the 'appropritate Green's Function' for part (a): The left boundary condition gives: u(x) = x and the right boundary condition gives: v(x) = x - 1 so the wronskian = W(x) = uv' - u'v = x - (x-1) = 1 so u,v are linearly independent. then the greens function is [math] G(x,e) = x(\epsilon - 1) [/tex] for [tex] 0 \leq x \leq \epsilon [/math] [math] G(x,e) = \epsilon(x - 1) [/tex] for [tex] \epsilon \leq x \leq 1 [/math] So the solution to the associated boundary value problem with homogenous conditions is: [math] y(x) = \int_0^1G(x,\epsilon)f(\epsilon)d\epsilon [/math] so then is the solution to the orginal BVP something like [math] y(x) = \int_0^1G(x,\epsilon)f(\epsilon)d\epsilon + A +Bx [/math] ????? this is where i think i get quite lost (maybe before here if i have already made a giant mistake somewhere! )

hey i am having a bit of trouble with this problem, as the course textbook has nothing on doing these problems using green's functions (all we have is a few slides of dodgey lecture notes)... anyways... the problem.. so to start with, for part (a) now from what i can gather, the BVP has a unique solution if and only if the corresponding HBVP (homogeneous boundary value problem) y''=0, y(0)=0, y(1) = 0 has no solution other than the trivial one, y(x) = 0. i guess then the associated HLDE (homogenous linear differential equation) needs to be solved... so y''(0) = 0 so we look for answers of the form y(x) = Ax + B then the left boundary condition (y(0) = 0 ) leads to u(x) = Ax is this the right way to go about things? cheers -Sarah

ok , fair enough but what about the fact that maple gives 2 different answers?

for this bit, when i used convolution integrals i used t^2 instead of t, i.e. [math] f(x) = 2\int_0^xte^tdt - \int_0^xt^2e^{x-t}dt [/math] but i did put it into maple, and it gives the answer in my original post.... so i can see what you did is correct, i just want to know why if you used t^2 instead of t(x-t) you get an almost completely different answer....??

hey this is not so much a problem that i can't do, its problem that seems to have 2 answers, yet i thought laplace transforms are unique.... if i have [math] F(s) = \frac{2s-1}{(s-1)^2s^2} [/math] then if i take the inverse transform it straight from here (either by using maple, or relising that [math] \frac{2s-1}{(s-1)^2s^2} = \frac{s^2 - (s-1)^2}{(s-1)^2s^2} [/math] i get [math] f(t) = -t + te^t [/math] however if i split up F(s) to [math] F(s) = \frac{2s-1}{(s-1)^2s^2} = \frac{2}{(s-1)^2s} - \frac{1}{(s-1)^2s^2} [/math] and then take the inverse transform (either by using maple or convolution integrals), i get [math] f(t) = 2te^t-4e^t+2t+t^2 +4 [/math] can anybody give an insight as to why this is happening? (as even maple seems to get 2 quite different answers....) -Sarah

ah yep i see now, lol, so simple, but so hard to see! thanks for the help guys!

do you mean something like [math] \tau = x^2 [/math] ??

yeah the answer is [math] \sqrt{\pi} Erf (\sqrt{t}) [/math] but i can't seem to get there, this is what i have done so far: [math] \int^t_0e^{-\tau} \frac{1}{\sqrt{\tau}}d\tau [/math] IBP gives: [math] 2e^{-t} \sqrt{t} + 2 \int_0^te^{-\tau}\tau^{1/2} d\tau [/math] but this seems to end up getting me nowhere....

hmmm any ideas how to go about doing this integral (without a computer...)? [math] \int^t_0e^{-\tau} \frac{1}{\sqrt{\tau}}d\tau [/math] i tried Integration by parts but that seems to end up in me going roudn in circles for 5 hours! thanks Sarah

"the minimum photon energy required for process?" i think this is the work function? (someone please correct me if i am wrong)...

i am happy to help, how far have you gotton so far?

yep,, got it all now. thanks Severian.

thanks for all the btw Severian. much appreciated.

yeah, i know what you mean. well i think i get the general idea of it anyway (the normal modes of oscillation stuff.) thanks for all you help once again swansont

do you mean that the system can be made to oscillat ein a precisely periodic manner if the ratio of the characteristic frequencies are a rational number? hmmm i don't quite understand, do that want anything more than that or do they want actual numbers or something??? ahhh!

ok.

how about for (i) [math] r(t) = \frac{1}{\sqrt{2(t_0-t)}} [/math]

ok i get [math] \frac{d\theta}{dt} = -1 [/math] and so [math] \theta = -t + t_0 [/math] but for the dr/dt part i don't quite see how i can get a different answer... if [math] \frac{dr}{dt} = r^3 [/math] then [math] \frac{1}{r^3}dr = dt [/math] right? but integrating [math] \frac{1}{r^3}dr [/math] gives [math] -\frac{1}{2r^2} [/math] .... which seems to be my problem... also when you are say the limits of integration... that should be r = 0 to r = infinity?

a lot of it isnt homework, i just like to do extra work, i enjoy doing most of it too!