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

In summary, the conversation discusses solving an initial-value problem for the heat equation in a specified column with given boundary and initial conditions. The solution is found using the equation provided and involves solving for Bmn. The relaxation time is also determined to be (4/pi^2 K)[L1^2 L2^2/(L1^2 + L2^2)]. The individual is seeking further understanding of the solution and how to apply it to similar problems.
  • #1
mmo115
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Homework Statement


Solve the initial-value problem for the heat equation ut = K[tex]\nabla[/tex]2u in the column 0< x < L1, 0< y < L2 with the boundary conditions u(0,y;t)=0, ux(L1,y,t)=0, u(x,0;t)=0, uy(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] Bmnsin (m*pi*x / L1) sin (n*pi*y)/L2) * e^-lambdamnKt

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 Bmn... 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^-lambdamnKt ]

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

relaxation time = (4/pi2 K)[L12L22/L12+L22)]
 
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  • #2
I would really appreciate it if someone could help me understand how to get this solution. Thank you very much in advance.
 
  • #3


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
 

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

What is the heat equation?

The heat equation is a mathematical model that describes how heat diffuses through a medium over time. It is a partial differential equation that relates the change in temperature at a point in space to the heat flux and thermal conductivity of the medium.

What is an initial-value problem?

An initial-value problem is a type of mathematical problem where the values of a function and its derivatives at a given point are known, and the goal is to find the function that satisfies a certain differential equation. In the context of the heat equation, the initial values would represent the initial temperature distribution in the medium.

How do you solve an initial-value problem for the heat equation?

To solve an initial-value problem for the heat equation, you would need to use a method called separation of variables. This involves breaking down the equation into simpler parts and solving for each variable separately. The resulting solution would then need to be checked against the initial conditions to ensure it satisfies the problem.

What is the relaxation time in the context of the heat equation?

The relaxation time in the heat equation refers to the time it takes for the temperature distribution in the medium to reach a steady state. In other words, it is the time it takes for the temperature at any point in the medium to stop changing as heat diffuses through it. This value is an important parameter in understanding the behavior of the heat equation.

How is the relaxation time calculated?

The relaxation time can be calculated using the thermal diffusivity of the medium, which is a measure of how quickly heat diffuses through it. The formula for calculating the relaxation time is t = L^2/α, where t is the relaxation time, L is the characteristic length scale of the system, and α is the thermal diffusivity. This formula is derived from the solution of the heat equation.

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