

Sarahisme
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its cool, got it! don't worry guys
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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
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its cool everyone, got the question, its all good!
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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
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but don't we also need to satify the initial conditions?
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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 [math] A(u) = e^{u(x,y)} [/math] but i didnt use any knowlage of PDE's to work that out....hmmm.... :S
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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
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but then how do you find a singularity when you have an answer of 1 and 0? :S
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ok i got it now , thanks Bignose
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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
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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 [math] e^{x+ct} [/math] ? 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
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its ok i think its correct, its some other questions that i am having problems with now!
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ok, i think i did something like that...i said that part (a) has a solution of: [math] u(x,t) = f(x+ct)+g(x-ct)-\frac{x^3t}{6c^2} [/math] then using the initial conditions i solved for f and g... and came up with my answer of [math] u(x,t) = \frac{1}{6}xt^3 [/math]
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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?
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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!