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So you assume your solution is u(x,t)=F(x)G(t). For my problem, the conditions given in the beginning are: u(0,t)=u(pi,t)=0 and u(x,0)=sin4x+3sin6x-sin10x. Now, my question is, what is the difference between the boundary conditions and initial conditions and how do you tell them apart? In this question, I know the first set are B.C and the 2nd part are I.C. Now this is a heat flow problem, so naturally though at x=0 and x=pi with t varying, 0 and pi are the ends of the bar which implies boundary conditions. However, what if this wasn't a heat flow problem and x and t weren't used as variables? Another thing is, when you try to solve your ODE of F(x), why do you apply the condition that F(0)=F(pi)=0? Is this because the B.C is saying whatever the value of t is, it doesnt matter and since the solution is of F(x) and G(t), F(0)=F(pi)=0 must be true? Sorry if this kinda lengthy, but I know how to solve the problem, I'm just having trouble understanding some of the theory behind it. Thanks
Your initial conditions in any problem like this (i.e. where time is involved) are the conditions at the beginning - i.e. when t=0 (always). Your boundary conditions refer to conditions on the spatial dimensions here. It MIGHT be the case that sometimes a mathematician may say BOUNDARY conditions, and include time in that definition...you may have to pay attention to the context. This is because we're talking about the boundary of the set on which the function takes its values, and part of that set is time. Here, T x S is the Cartesian cross-product of the set on which your function will take its values: both T and S (time and space) have a starting point.
I don't have your specific differential equation before me, but I would imagine the condition that F(0)=0=F(pi) is just based upon the construct of the problem...i.e., this is just what the person wishing to model this physical problem wishes to insist on. Normally, in the conduction equation, we would use F(0)=0 and F(L)=0 where L is the length of the bar. Since you get trigonometric solutions of this PDE, I'm thinking maybe L is of a length such that F(pi)=F(L)=0.
I just want to add something else...I'm thinking the fact that the ends of the bar are held at temperature 0 is confusing you since you're looking at an equation to do with conduction (and therefore, a transfer of heat from a higher temperature to a lower one). It turns out you can transform the more general case into the special case where the temperature is 0 at both ends. The solution in this scenario is simpler and you can get back to the original problem after solving the easier one. We do this a lot in math.
Is this making sense?
I'm guessing the exercise here is learning how to use the separation of variables method, rather than a physical appraisal of what's going on.
Yeah it is, I don't think we deal specifically with the physical nature of the problems in this class. Though I wouldn't know haha since I'm not taking this class right now, but just learning it on my own so I can skip class later :P
Yeah, learn the method. Apart from mathematics, it's used in physics a great deal (e.g. solving certain types of Schrödinger equation in quantum mechanics, etc.). I think as long as you can understand why sep. of variables works, that's good enough.
And engineering...and chemistry, and economics.
Yeah, well it's usually not too bad except the problem needs to be overly long... How long do you think is a reasonable amount of time to solve one of these pde? 15-20 min?
Well, it depends on the PDE, the boundary conditions, how much sleep you had...
Though this is probably a simpler example, one on our exam would most likely involve fourier series as well :(
Yeah, you could expect Fourier.
Oh yeah, what do you do if you have two eigenvalue and eigenfunctions instead of one?
Are you getting real or complex ones?
Well, not for any question in particular, but I was just curious what would happen if you had 2 lambdas. I know its kinda hard to answer something liek that haha, but like do you get 2 set of solutions that you can then sum up?
Yeah, pretty much. The form will depend on the nature of the roots of the lambdas.
kk, but you can still express them in one set of solution with the principle of superposition right?
I'm signing out. If you have any more questions, post them here and I'll get back to you when I can. Good luck!