Fourier Series Solution of 1-D Heat Flow

Curtis15
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Homework Statement




Length of rod = 1

Initial Conditions: u(x,0)=sin(πx)

Boundary conditions: u(0,t)=0 and u(1,t)=5.


Alright I am supposed to find the temperature at all times, but I am curious about the setup of the problem itself.

When x = 1, the boundary condition says that u = 5.

When t = 0, the initial condition says that u = sin(x∏).

So u(1,0) is supposed to equal what exactly? The boundary says it should be 5, but the initial condition says that sin(∏) = 0, so what would the answer be. I feel like this is contradictory but people are saying that it isn't and I am an idiot.

I have asked this somewhere else and got responses just saying this was a stupid question.


Thanks for any help
 
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You're right, it is contradictory. You should ask for clarification at your instructor.

A possible way to interpret the problem is that the ##u(1,t)=5## condition only holds for large ##t## and not for all ##t##.
 
micromass said:
You're right, it is contradictory. You should ask for clarification at your instructor.

A possible way to interpret the problem is that the ##u(1,t)=5## condition only holds for large ##t## and not for all ##t##.

You have no idea how much I appreciate this response. Thank you very much!
 
@curtis15: You didn't specify for what ##t## your boundary conditions ##u(0,t)=0,\ u(1,t)=5## apply. A reasonable interpretation would be for ##t>0##. Or you could think of the bar having the initial temperature distribution ##u(x,0) = \sin(\pi x)## suddenly inserted into a situation with those boundary conditions. I don't think there is anything contradictory here and working the problem should be straightforward.

[Edit, added later:] Another reasonable interpretation is to assume the initial condition ##u(x,0) = \sin(\pi x)## holds for ##0 < x < 1##. That works even better intuitively.
 
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Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...
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