Is it hot?

Calculus Level 5

The solution to the heat equation u ( x , t ) t = u ( x , t ) x x u(x, t)_t = u(x, t)_{xx} is given as

u ( x , t ) = m = 1 A m e π 2 m 2 t sin ( π m x ) u(x, t) = \sum_{m=1}^{\infty} A_m e^{- \pi^2 m^2 t} \sin(\pi m x)

Considering a rod of total length 1 such that the temperature at the extremities is kept constant and equal to 0 and at the beginning of the heat change the temperature at any point on the bar is given by u ( x , 0 ) = x u(x, 0) = x . Then what is A 1 \left \lfloor A_1 \right \rfloor ?


Notations:

  • u ( x 1 , x 2 , . . . , x n ) x k , u(x_1, x_2, ..., x_n)_{x_k}, for 1 k n 1 \leq k \leq n denotes for the partial derivative of u u with respect to x k . x_k.
  • \lfloor \cdot \rfloor denotes the floor function .


The answer is 0.

This section requires Javascript.
You are seeing this because something didn't load right. We suggest you, (a) try refreshing the page, (b) enabling javascript if it is disabled on your browser and, finally, (c) loading the non-javascript version of this page . We're sorry about the hassle.

1 solution

Hosam Hajjir
Jan 3, 2019

This is just a Fourier Series problem, we have the following formula for that,

A 1 = 2 T 0 T f ( x ) s i n ( π x ) d x = 0 2 x s i n ( π x ) d x = 2 π A_1 = \dfrac{2}{T} \displaystyle \int_{0}^{T} f(x) sin(\pi x ) dx = \displaystyle \int_{0}^{2} x sin(\pi x ) dx = \dfrac{2}{\pi}

Therefore, A 1 = 0 \lfloor A_1 \rfloor = 0

0 pending reports

×

Problem Loading...

Note Loading...

Set Loading...