  # Sarahisme

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1. Hello peoples, I think this is a trick question... well sort of for part (a) i get that the cosine Fourier Series for f(x) = cos(x) to be: i hope that is ok, but its part (b) that is troubling me... is all that happens as is that the cosine Fourier series of cos(x) goes to 0? i am guessing i am missing some trick to this question? Cheers! Sarah
2. its cool everyone, got the question, its all good!
3. ah ok, yeah that all makes sense now. since you people have been so helpful with this PDE question, if any of you guys have time, i would really love some help with some of my other questions thanks guys! Sarah
4. but don't we also need to satify the initial conditions?
5. hmmm i have no idea where to even start with this problem, i cannot find any examples that are similar or anything like that anywhere! anyone got an idea as to a good first step to take? thanks sarah edit: i tryed something wild and came up with $A(u) = e^{u(x,y)}$ but i didnt use any knowlage of PDE's to work that out....hmmm.... :S
6. hi all, sorry again for the thousands of stupid questions! but heres another one... ok i said for part (a) that the PDE is a hyperbolic type and so has two sets of real characteristics. and that the characteristics are then for part (b) i used the coordinates: which leads to the equation: but then i am stuck... how can i get the 'general solution' from this, i think i am heading in the right direction, but i am not sure. any help would be greatly appreciated! thanks Sarah
7. but then how do you find a singularity when you have an answer of 1 and 0? :S
8. ok i got it now , thanks Bignose
9. lol my head is about to explode! i think this is similar to a previous question i asked but i can't quite get it none the less... now what i did was to following this : then using equations (31) & (32) from that i just plugged in the values and got: u(x,t) = 1 for x > ct and u(x,t) = 0 for 0 < x < ct how does this look to you intelligent mathematically inclined people? :S Sarah
10. how do you know to add that? i see that it works, but at what point do you figure out to add it? that is it doesnt come out in any of the steps i take to solve the problem... does this mean for this problem: that i have to do a similar thing? (i.e. add $e^{x+ct}$ ? i get for that problem that u(x,t) = 1 for x > ct and u(x,t) = 0 for 0 < x < ct but that seems to simple, am i forgetting something completely obvious swansont? :S
11. its ok i think its correct, its some other questions that i am having problems with now!
12. ok, i think i did something like that...i said that part (a) has a solution of: $u(x,t) = f(x+ct)+g(x-ct)-\frac{x^3t}{6c^2}$ then using the initial conditions i solved for f and g... and came up with my answer of $u(x,t) = \frac{1}{6}xt^3$
13. ok. well i plugged my answer for part (a) back in and it works out, although i think what i have got as an answer could/should be simplified... leaving the answer as an integral looks a bit messy. however for part (b) i am totally stumped... how do you work with the BC that u(0,t) = 1 ?? any suggestions?
14. i can't see how to apply seperation of variables (at least for part (a)).... i get this after plugging u(x,t) = X(x)T(t) into the orginal PDE: XT'' = (c^2)X''T + xt but you can't seperate the x's and t's ...? lol i am probably just completely wrong!
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