How Do You Solve the 1D Heat Equation with Trigonometric Initial Conditions?

Click For Summary
The discussion revolves around solving the 1D heat equation with specified boundary and initial conditions. The proposed solution form is a series involving sine functions and exponential decay, but the user encounters an issue where all coefficients a_n equal zero. Another participant suggests expressing the initial condition as a sine series and questions the relationship between coefficients a_n and b_n. The conversation emphasizes the need to correctly represent the initial condition in a sine series to resolve the issue with the coefficients. Understanding the relationship between the two series is crucial for finding the correct solution.
prolix
Messages
2
Reaction score
0
problem

u_t=u_xx, x is in [0,1], t>0

with
u(0,t)=u(1,t)=0, t>0
u(x,0)=sin(pi*x)-sin(3*pi*x), x is in (0,1)


i think its solution is of the form

u(x,t)=sigma(n=1 to infinity){a_n*sin(n*pi*x)*exp(-n^2*pi^2*t)

where a_n=2*integral(0 to 1){ (sin(pi*x)-sin(3*pi*x)) * sin(n*pi*x) }

but i have a_n = 0, for all n..

i don't know where is my mistake..
 
Physics news on Phys.org
Can you write your initial condition as a sin series?

u(x,0)=\sum b_n sin(n \pi x)

How does this compare to your sin series for

u(x,t)

Can you relate a_n to b_n?
 

Similar threads

  • · Replies 7 ·
Replies
7
Views
3K
  • · Replies 11 ·
Replies
11
Views
4K
Replies
1
Views
2K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 3 ·
Replies
3
Views
3K
  • · Replies 1 ·
Replies
1
Views
3K
  • · Replies 5 ·
Replies
5
Views
3K