Temperature distribution - PDE's

Click For Summary
SUMMARY

The discussion focuses on solving a partial differential equation (PDE) for temperature distribution in a unit square with specified boundary conditions. The boundary conditions include zero derivative at the sides x=0, x=1, and y=0, with a defined temperature distribution u(x,1)=x along y=1. The solution involves applying Laplace's equation, \nabla^2 u(x, y)= 0, and utilizing the method of separation of variables to express the solution as u(x, y) = X(x)Y(y). The participants emphasize the importance of correctly applying boundary conditions to derive non-trivial solutions.

PREREQUISITES
  • Understanding of partial differential equations (PDEs)
  • Familiarity with Laplace's equation
  • Knowledge of boundary value problems
  • Experience with the method of separation of variables
NEXT STEPS
  • Study the method of separation of variables in depth
  • Learn how to apply boundary conditions in PDEs
  • Explore solutions to Laplace's equation in rectangular domains
  • Investigate non-trivial solutions in boundary value problems
USEFUL FOR

Students and researchers in mathematics, particularly those studying heat distribution problems, as well as educators teaching PDEs and boundary value problems.

Anabelle37
Messages
35
Reaction score
0

Homework Statement



Consider a square of unit length which is insulated on the sides x=0, x=1 and y=0 and has the temperature distribution u(x,1)=x along the side y=1. find the temperature distribution u in the square.

Homework Equations





The Attempt at a Solution



So I have the boundary conditions:
du(0,y)/dx = 0; du(1,y)/dx = 0; du(x,0)/dx = 0; u(x,1)=x (where these are partial derivatives)

I am unsure where to go from here...do i solve laplaces equation? if so I don't know how to put the boundary conditions in terms of the separation of variables method.

please help!
 
Physics news on Phys.org
Yes, the equilibrium distribution of temperature on a plate satisfies [itex]\nabla^2 u(x, y)= 0[/itex].

If you write u(x, y)= X(x)Y(y), then
[tex]\frac{\partial u}{\partial x}= Y(y)\frac{dX}{dx}[/tex]
so
[tex]\frac{\partial u(0, y)}{\partial x}[/tex][tex]= Y(y)\frac{dX(0)}{dx}= 0[/tex]
so either Y(y)= 0 for all y, which gives only the trivial solution, u= 0 for all x and y or
[tex]\frac{dX(0)}{dx}= 0[/tex]
That's your boundary condition on X.

Similarly for the other two.
 
ok thanks,
do i partial du/dx for the other 2? or du/dy?
 

Similar threads

Replies
5
Views
2K
Replies
19
Views
3K
Replies
7
Views
1K
  • · Replies 4 ·
Replies
4
Views
2K
  • · Replies 27 ·
Replies
27
Views
3K
  • · Replies 12 ·
Replies
12
Views
3K
Replies
5
Views
2K
  • · Replies 18 ·
Replies
18
Views
3K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 12 ·
Replies
12
Views
2K