Solve initial-value problem for heat equation and find relaxation time

[mmo115]

Homework Statement

Solve the initial-value problem for the heat equation u_t = K[tex]\nabla[/tex]²u in the column 0< x < L₁, 0< y < L₂ with the boundary conditions u(0,y;t)=0, u_x(L₁,y,t)=0, u(x,0;t)=0, u_y(x,L₂;t)=0 and the initial condition u(x,y;0)=1. Find the relaxation time.

Can anyone please explain how to get this solution.. I really don't understand how to arrive at the solution. I'm hoping that if i can learn specifically how to do this i can apply it to other similar problems. I have searched around the internet for some information, but it seems like most of it is using actual situations as opposed to theoretical. i mean the book doesn't even explain what relaxation time is or how to derive it.. =/

Homework Equations

We are told that :

where m,n=1 to infinity
u(x,y;t)= [tex]\Sigma[/tex] B_mnsin (m*pi*x / L₁) sin (n*pi*y)/L₂) * e^-lambda_mnKt

We can use this to solve initial-value problems for the heat equation.

The Attempt at a Solution

I really don't get how to solve for the B_mn... the book really doesn't give a good explanation.
The solution is u(x,y;t) = 4/pi² [tex]\Sigma[/tex]_m,n=1 [ sin[(m-(1/2))([tex]\pi[/tex]x/L₁)] / (m-1/2) ] * [ sin[n-1/2)(pi y/L₂)\ / (n-1/2) ] * [ e^-lambda_mnKt ]

lambda_mn= (m-1/2)² (pi/L₁)² + (n-1/2)²(pi/L₁)²

relaxation time = (4/pi² K)[L1²L2²/L1²+L2²)]

 

[Nate810]

I would really appreciate it if someone could help me understand how to get this solution. Thank you very much in advance.

 

[thomate1]

I understand the frustration of not being able to fully understand a solution or derivation. However, it is important to keep in mind that the heat equation is a complex mathematical concept and it takes time and practice to fully grasp it. With that being said, I will try my best to explain the solution and how to derive it.

Firstly, let's discuss the initial-value problem for the heat equation. The heat equation is a partial differential equation that describes the distribution of heat in a given region over time. In this problem, we are given a rectangular region with dimensions L1 and L2, and we are trying to find the temperature distribution u(x,y;t) at any point within this region.

To solve this problem, we need to use the general solution for the heat equation, which is given by:

u(x,y;t)= \Sigma Bmnsin (m*pi*x / L1) sin (n*pi*y/L2) * e^-lambdamnKt

Where m and n are integers that represent the modes of vibration in the x and y directions, respectively. Bmn is a constant that we need to solve for, and lambdamn is a constant that determines the rate of decay of the temperature. K is the thermal diffusivity constant.

Now, to apply this solution to our specific initial-value problem, we need to use the boundary conditions and the initial condition. The boundary conditions represent the temperature at the edges of the region, while the initial condition represents the temperature at time t = 0.

Using the boundary conditions, we can see that u(0,y;t)=0 and u(x,0;t)=0, which means that the temperature at the edges of the region is always 0. This allows us to eliminate all the terms in the general solution that contain sin(m*pi*x/L1) and sin(n*pi*y/L2). This leaves us with the following solution:

u(x,y;t)= \Sigma Bmnsin (m*pi*x / L1) sin (n*pi*y/L2) * e^-lambdamnKt

Now, we can use the initial condition u(x,y;0)=1 to solve for the constant Bmn. This involves plugging in x=0, y=0, and t=0 into the solution and solving for Bmn. After some algebraic manipulation, we get:

Bmn = (4/L1L2) * int(0,L