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The 2D heat equation Homogeneous Dirichlet boundary conditions Steady state solutions
The two dimensional heat equation
Ryan C. Daileda
Trinity University
Partial Differential EquationsMarch 6, 2012
Daileda The 2D heat equation
The 2D heat equation Homogeneous Dirichlet boundary conditions Steady state solutions
Physical motivation
Consider a thin rectangular plate made of some thermallyconductive material. Suppose the dimensions of the plate are a× b.
The plate is heated in some way, and then insulated along itstop and bottom.
Our goal is to mathematically model the way thermal energymoves through the plate.
Daileda The 2D heat equation
The 2D heat equation Homogeneous Dirichlet boundary conditions Steady state solutions
We let
u(x , y , t) = temperature of plate at position (x , y) andtime t.
For a fixed t, the height of the surface z = u(x , y , t) gives thetemperature of the plate at time t and position (x , y).
Under ideal assumptions (e.g. uniform density, uniform specificheat, perfect insulation, no internal heat sources etc.) one canshow that u satisfies the two dimensional heat equation
ut = c2∇2u = c2(uxx + uyy) (1)
for 0 < x < a, 0 < y < b.
Daileda The 2D heat equation
The 2D heat equation Homogeneous Dirichlet boundary conditions Steady state solutions
At the edges of the plate we impose some sort of boundaryconditions. The simplest are homogeneous Dirichlet
conditions:
u(0, y , t) = u(a, y , t) = 0, 0 ≤ y ≤ b, t ≥ 0,
u(x , 0, t) = u(x , b, t) = 0, 0 ≤ x ≤ a, t ≥ 0. (2)
Physically, these correspond to holding the temperature along theedges of the plate at 0.
The way the plate is heated initially is given by the initial
condition
u(x , y , 0) = f (x , y), (x , y) ∈ R , (3)
where R = [0, a]× [0, b].
Daileda The 2D heat equation
The 2D heat equation Homogeneous Dirichlet boundary conditions Steady state solutions
Solving the 2D wave equation: homogeneous Dirichlet
boundary conditions
Goal: Write down a solution to the heat equation (1) subject tothe boundary conditions (2) and initial conditions (3).
As usual, we
separate variables to produce simple solutions to (1) and(2), and then
use the principle of superposition to build up a solution thatsatisfies (3) as well.
Daileda The 2D heat equation
The 2D heat equation Homogeneous Dirichlet boundary conditions Steady state solutions
Separation of variables
Assuming thatu(x , y , t) = X (x)Y (y)T (t),
plugging into the heat equation (1) and using the boundaryconditions (2) yields the separated equations
X ′′ − BX = 0, X (0) = 0, X (a) = 0, (4)
Y ′′ − CY = 0, Y (0) = 0, Y (b) = 0, (5)
T ′ − c2(B + C )T = 0. (6)
Daileda The 2D heat equation
The 2D heat equation Homogeneous Dirichlet boundary conditions Steady state solutions
We have already seen that the solutions to (4) and (5) are
Xm(x) = sinµmx , µm =mπ
a, B = −µ2
m
Yn(y) = sin νny , νn =nπ
b, C = −ν2n ,
for m, n ∈ N. Using these values in (6) gives
Tmn(t) = e−λ2mnt ,
where
λmn = c
√
µ2m + ν2n = cπ
√
m2
a2+
n2
b2.
Daileda The 2D heat equation
The 2D heat equation Homogeneous Dirichlet boundary conditions Steady state solutions
Superposition
Assembling these results, we find that for any pair m, n ≥ 1 wehave the normal mode
umn(x , y , t) = Xm(x)Yn(y)Tmn(t) = sinµmx sin νny e−λ2
mnt .
For any choice of constants Amn, by the principle of superpositionwe then have the general solution to (1) and (2)
u(x , y , t) =∞∑
m=1
∞∑
n=1
Amn sinµmx sin νny e−λ2
mnt .
Daileda The 2D heat equation
The 2D heat equation Homogeneous Dirichlet boundary conditions Steady state solutions
Initial conditions
We must determine the values of the coefficients Amn so that oursolution also satisfies the initial condition (3). We need
f (x , y) = u(x , y , 0) =
∞∑
n=1
∞∑
m=1
Amn sinmπ
ax sin
nπ
by
which is just the double Fourier series for f (x , y). We know thatif f is a C 2 function then
Amn =4
ab
∫ a
0
∫ b
0f (x , y) sin
mπ
ax sin
nπ
bdy dx .
Daileda The 2D heat equation
The 2D heat equation Homogeneous Dirichlet boundary conditions Steady state solutions
Conclusion
Theorem
Suppose that f (x , y) is a C 2 function on the rectangle
[0, a]× [0, b]. The solution to the heat equation (1) withhomogeneous Dirichlet boundary conditions (2) and initial
conditions (3) is given by
u(x , y , t) =∞∑
m=1
∞∑
n=1
Amn sinµmx sin νny e−λ2
mnt ,
where µm =mπ
a, νn =
nπ
b, λmn = c
√
µ2m + ν2n , and
Amn =4
ab
∫ a
0
∫ b
0f (x , y) sin
mπ
ax sin
nπ
by dy dx .
Daileda The 2D heat equation
The 2D heat equation Homogeneous Dirichlet boundary conditions Steady state solutions
Remarks:
We have not actually verified that this solution is unique, i.e.that this is the only solution to problem.
We will prove uniqueness later using the maximum principle.
Example
A 2× 2 square plate with c = 1/3 is heated in such a way that the
temperature in the lower half is 50, while the temperature in the
upper half is 0. After that, it is insulated laterally, and the
temperature at its edges is held at 0. Find an expression that gives
the temperature in the plate for t > 0.
Daileda The 2D heat equation
The 2D heat equation Homogeneous Dirichlet boundary conditions Steady state solutions
We must solve the heat equation problem (1) - (3) with
f (x , y) =
{
50 if y ≤ 1,
0 if y > 1.
The coefficients in the solution are
Amn =4
2 · 2
∫ 2
0
∫ 2
0f (x , y) sin
mπ
2x sin
nπ
2y dy dx
= 50
∫ 2
0sin
mπ
2x dx
∫ 1
0sin
nπ
2y dy
= 50
(
2(1 + (−1)m+1)
πm
)(
2(1− cos nπ2 )
πn
)
=200
π2
(1 + (−1)m+1)(1− cos nπ2 )
mn.
Daileda The 2D heat equation
The 2D heat equation Homogeneous Dirichlet boundary conditions Steady state solutions
Since
λmn =π
3
√
m2
4+
n2
4=
π
6
√
m2 + n2
the solution is
u(x , y , t) =200
π2
∞∑
m=1
∞∑
n=1
(
(1 + (−1)m+1)(1− cos nπ2 )
mnsin
mπ
2x
× sinnπ
2y e−π2(m2+n2)t/36
)
.
Daileda The 2D heat equation
The 2D heat equation Homogeneous Dirichlet boundary conditions Steady state solutions
Steady state solutions
To deal with inhomogeneous boundary conditions in heatproblems, one must study the solutions of the heat equationthat do not vary with time.
These are the steady state solutions. They satisfy
ut = 0.
In the 1D case, the heat equation for steady states becomes
uxx = 0.
The solutions are simply straight lines.
Daileda The 2D heat equation
The 2D heat equation Homogeneous Dirichlet boundary conditions Steady state solutions
Laplace’s equation
In the 2D case, we see that steady states must solve
∇2u = uxx + uyy = 0. (7)
This is Laplace’s equation.
Solutions to Laplace’s equation are called harmonic
functions.
See assignment 1 for examples of harmonic functions.
Daileda The 2D heat equation
The 2D heat equation Homogeneous Dirichlet boundary conditions Steady state solutions
We want to find steady state solutions to (7) that satisfy theDirichlet boundary conditions
u(x , 0) = f1(x), u(x , b) = f2(x), 0 < x < a (8)
u(0, y) = g1(x), u(a, y) = g2(y), 0 < y < b (9)
The problem of finding a solution to (7) - (9) is known as theDirichlet problem.
We will begin by assuming f1 = g1 = g2 = 0.
The general solution to the Dirichlet problem will be obtainedby superposition.
Daileda The 2D heat equation
The 2D heat equation Homogeneous Dirichlet boundary conditions Steady state solutions
Solution of the Dirichlet problem on a rectangle
Goal: Solve the boundary value problem
∇2u = 0, 0 < x < a, 0 < y < b,
u(x , 0) = 0, u(x , b) = f2(x), 0 < x < a,
u(0, y) = u(a, y) = 0, 0 < y < b.
Picture:
Daileda The 2D heat equation
The 2D heat equation Homogeneous Dirichlet boundary conditions Steady state solutions
Separation of variables
Setting u(x , y) = X (x)Y (y) leads to
X ′′ + kX = 0 , Y ′′ − kY = 0,
X (0) = X (a) = 0 , Y (0) = 0.
We know the nontrivial solutions for X :
X (x) = Xn(x) = sinµnx , µn =nπ
a, k = µ2
n.
for n ∈ N = {1, 2, 3, . . .}. The corresponding solutions for Y are
Y (y) = Yn(y) = Aneµny + Bne
−µny .
Daileda The 2D heat equation
The 2D heat equation Homogeneous Dirichlet boundary conditions Steady state solutions
The condition Y (0) = 0 yields An = −Bn. Choosing An = 1/2 wefind that
Yn(y) = sinhµny .
This gives the separated solutions
un(x , y) = Xn(x)Yn(y) = sinµnx sinhµny ,
and superposition gives the general solution
u(x , y) =
∞∑
n=1
Bn sinµnx sinhµny .
Daileda The 2D heat equation
The 2D heat equation Homogeneous Dirichlet boundary conditions Steady state solutions
The general solution satisfies the Laplace equation (7) inside therectangle, as well as the three homogeneous boundary conditionson three of its sides (left, right and bottom).
We now determine the values of Bn to get the boundary conditionon the top of the rectangle. This requires
f2(x) = u(x , b) =∞∑
n=1
Bn sinhnπb
asin
nπ
ax ,
which is the Fourier sine series for f2(x) on 0 < x < a.
Appealing to previous results, we can now summarize our findings.
Daileda The 2D heat equation
The 2D heat equation Homogeneous Dirichlet boundary conditions Steady state solutions
Conclusion
Theorem
If f2(x) is piecewise smooth, the solution to the Dirichlet problem
∇2u = 0, 0 < x < a, 0 < y < b,
u(x , 0) = 0, u(x , b) = f2(x), 0 < x < a,
u(0, y) = u(a, y) = 0, 0 < y < b.
is
u(x , y) =∞∑
n=1
Bn sinµnx sinhµny ,
where µn =nπ
aand Bn =
2
a sinh nπba
∫ a
0f2(x) sin
nπ
ax dx .
Daileda The 2D heat equation
The 2D heat equation Homogeneous Dirichlet boundary conditions Steady state solutions
Remarks: As we have noted previously:
We have not proven this solution is unique. This requires themaximum principle.
If we know the sine series expansion for f2(x) then we can usethe relationship
Bn =1
sinh nπba
(nth sine coefficient of f2(x))
to avoid integral computations.
Daileda The 2D heat equation
The 2D heat equation Homogeneous Dirichlet boundary conditions Steady state solutions
Example
Example
Solve the Dirichlet problem on the square [0, 1] × [0, 1], subject tothe boundary conditions
u(x , 0) = 0, u(x , b) = f2(x), 0 < x < a,
u(0, y) = u(a, y) = 0, 0 < y < b.
where
f2(x) =
{
75x if 0 ≤ x ≤ 23 ,
150(1 − x) if 23 < x ≤ 1.
Daileda The 2D heat equation
The 2D heat equation Homogeneous Dirichlet boundary conditions Steady state solutions
We have a = b = 1 and we require the sine series of f2(x). Thegraph of f2(x) is:
According to exercise 2.4.17 (with p = 1, a = 2/3 and h = 50):
f2(x) =450
π2
∞∑
n=1
sin 2nπ3
n2sin nπx .
Daileda The 2D heat equation
The 2D heat equation Homogeneous Dirichlet boundary conditions Steady state solutions
Thus,
Bn =1
sinh nπ
(
450
π2
sin 2nπ3
n2
)
=450
π2
sin 2nπ3
n2 sinh nπ,
and
u(x , y) =450
π2
∞∑
n=1
sin 2nπ3
n2 sinh nπsin nπx sinh nπy .
Daileda The 2D heat equation
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