Scribd
Upload
159 views3 pages

Steady State-2D-Heat Equation: Kalpana Mahalingam March 2020

This document discusses solving the 2D steady-state heat equation using separation of variables. It considers the heat equation on a rectangular region with Dirichlet boundary conditions. Using separation of variables, it obtains solutions of the form F(x)G(y) and applies the boundary conditions to determine coefficients. The general solution is a Fourier series involving sin(nx/a)sinh(ny/a) terms, with coefficients determined from the boundary data.

Uploaded by

vanaj123
Copyright
© © All Rights Reserved
We take content rights seriously. If you suspect this is your content, claim it here.
Available Formats
Download as PDF, TXT or read online on Scribd
159 views3 pages

Steady State-2D-Heat Equation: Kalpana Mahalingam March 2020

This document discusses solving the 2D steady-state heat equation using separation of variables. It considers the heat equation on a rectangular region with Dirichlet boundary conditions. Using separation of variables, it obtains solutions of the form F(x)G(y) and applies the boundary conditions to determine coefficients. The general solution is a Fourier series involving sin(nx/a)sinh(ny/a) terms, with coefficients determined from the boundary data.

Uploaded by

vanaj123
Copyright
© © All Rights Reserved
We take content rights seriously. If you suspect this is your content, claim it here.
Available Formats
Download as PDF, TXT or read online on Scribd
You are on page 1/ 3

Steady State- 2D-Heat Equation

Kalpana Mahalingam
March 2020

In this section, we study the solutions of 2D heat equation(steady


state-2D heat equation) that do not vary with time, also called as
2D-Laplacian.
2 ∂ 2u ∂ 2u
∇ u= 2+ 2 (1)
∂x ∂y
A heat problem then consists of this PDE to be considered in some re-
gion R of the xy-plane and a given boundary condition on the boundary
curve C of R. This is a boundary value problem (BVP). If u is pre-
scribed on C (“Dirichlet boundary condition”) then the BVP is called
First BVP or Dirichlet Problem.

Dirichlet Problem on a rectangle :

We consider a Dirichlet problem for Laplace’s equation (1) in a


rectangle R, assuming that the temperature equals a given function
f (x) on the upper side and 0 on the other three sides of the rectangle
and 0 ≤ x ≤ a, 0 ≤ y ≤ b.

Separation of Variables :

1
K.Mahalingam 2D-Laplacian

We use separation of variables technique to solve the given problem.


Let,
u(x, y) = F (x)G(y)
Then, the equation (1) becomes,
00 00
00 00 F (x) G (y)
F (x)G(y) + F (x)G (y) = 0, =⇒ =− =k
F (x) G(y)
where k is a constant.
We now obtain two ODE’s
00 00
F (x) − kF (x) = 0, G (y) + kG(y) = 0

Solving, for F , we get,


√ √
kx − kx
F (x) = Ae + Be

Substituting the boundary conditions F (0) = F (a) = 0, we get,


√ √
F (0) = A + B = 0, F (a) = A(e ka − e− ka
)=0
 nπ 2
=⇒ k =
a
and hence,  nπx 
F (x) = Fn (x) = sin
a
Now solving for G similarly, we get
nπy nπy
Gn (y) = An e a + Bn e− a

Now applying the boundary condition u = 0 on the lower side of rect-


angle, we get Gn (0) = 0 which implies,

Gn (0) = An + Bn = 0, =⇒ An = −Bn
K.Mahalingam 2D-Laplacian

Now applying the condition u(x, b) = f (x) implies that G(b) = f (x)
and substituting in Gn , we get
nπb nπb nπy
Gn (b) = An (e a − An e− a ) = 2An sinh
a
Hence, the general solution of the 2D-Laplacian is
 nπx   nπy 
∗
u(x, y) = Fn (x)Gn (y) = An sin sinh
a a
Since, the given equation is linear, a linear combination of solutions is
also a solution and hence,
X  nπx   nπy 
∗
u(x, y) = An sin sinh
n≥0
a a

We now apply the last boundary condition u(x, b) = f (x) implies


G(b) = f (x). Then,
X  nπx   nπb 
u(x, b) = f (x) = A∗n sin sinh
n≥0
a a

and hence, by Fourier series,


 nπb  2 Z a  nπx 
∗
An sinh = f (x) sin dx
a a 0 a
Hence, X  nπx   nπy 
u(x, y) = A∗n sin sinh
n≥0
a a
with Z a
2  nπx 
A∗n =   f (x) sin dx
a sinh nπb 0 a
a

You might also like