D. Andrew S. Rees
• A Fourier Cosine Series example of the solution of Fourier’s equation.
• A quarter-range Fourier Sine Series example.
• Laplace’s equation in rectangular domains.
4.7 A Fourier Cosine Series solutions of Fourier’s equation
Recall that the PDE is, ∂θ
∂t = α
∂2θ
∂x2 .
Example 3. Solve Fourier’s equation where ∂θ/∂x = 0 on x = 0 and x = 1 where the initial
condition is, θ = x at t = 0.
Given the boundary conditions we will let θ = T (t) cos(nπx), where n is either zero or else a positive integer. Hence,
dT
and dT
2 A0 when n = 0.
On reconstructing θ, and superposing all the valid solutions, we have
θ = 1 2 A0 +
Ane −αn2π2t cos(nπx).
Now we apply the Initial Condition that θ = x when t = 0.
x = 1 2 A0 +
n=1
An cos(nπx).
This is a Fourier Cosine Series. The coefficients are given by,
An = 2
Also A0 = 2
θ = 1 2 −
•• 1
x
θ
••
Depicting the above solution of Fourier’s equation. The initial condition is displayed in red. Successive black profiles are at a time-interval of 0.05/απ2. The two blue profiles correspond to t = 0.001/απ2 and t = 0.01/απ2.
A different initial condition: θ = 1 (0 < x < 1 2 ) and θ = 0 (1
2 < x < 1) at t = 0.
θ = 1 2 +
••
Depicting the above solution of Fourier’s equation. The initial condition is displayed in red. Successive black profiles are at a time-interval of 0.05/απ2. The three blue profiles correspond to t = 0.0001/απ2, t = 0.001/απ2
and t = 0.01/απ2.
Another different initial condition: θ = 1 (1 2 < x < 3
4 ) and θ = 0 (otherwise) at t = 0.
θ = 1 4 +
••
Depicting the above solution of Fourier’s equation. The initial condition is displayed in red. Successive black profiles are at a time-interval of 0.05/απ2. The three blue profiles correspond to t = 0.0001/απ2, t = 0.001/απ2
and t = 0.01/απ2.
Yet another different initial condition: θ = 4x(1 − x) at t = 0.
θ = 2 3 −
••
Depicting the above solution of Fourier’s equation. The initial condition is displayed in red. Successive black profiles are at a time-interval of 0.05/απ2. The three blue profiles correspond to t = 0.0001/απ2, t = 0.001/απ2
and t = 0.01/απ2.
Recall that the PDE is, ∂θ
∂t = α
∂2θ
∂x2 .
Given the boundary conditions we will let θ = T (t) sin(nπx/2), where n is a positive odd integer. Hence,
dT
dt = −(αn2π2/4) T =⇒ T = Be−αn2π2t/4.
On reconstructing θ, and superposing all the valid solutions, we have
θ = ∞∑
Bne −αn2π2t/4 sin(nπx/2).
Now we apply the Initial Condition that θ = x − x2 when t = 0.
This gives,
x − x2 = ∞∑
n=1
n odd
Bn sin(nπx/2).
Bn = 2
θ =
e−αn2π2t/4 sinnπx/2.
N.b. the term, sin(nπ/2) oscillates between +1 and −1 as n increases through consecutive odd values.
θ = ∞∑
n = 1, 3, 5, 7, · · · =⇒ m = 0, 1, 2, 3 · · · .
θ = ∞∑
Was that worth the effort? The jury’s out....
The initial condition is θ = x − x2 at t = 0.
θ =
••
Depicting the above solution of Fourier’s equation. The initial condition is displayed in red. Successive black profiles are at a time-interval of 0.1/απ2. The dotted blue profiles correspond to απ2t = 0.0025, 0.005, 0.01, 0.02 and 0.05.
The initial condition is θ = x at t = 0.
θ = ∞∑
••
Depicting the above solution of Fourier’s equation. The initial condition is displayed in red. Successive black profiles are at a time-interval of 0.1/απ2. The dotted blue profiles correspond to απ2t = 0.0025, 0.005, 0.01, 0.02 and 0.05.
Solutions of Laplace’s equation in finite domains.
To be honest, we are talking about rectangles and squares.
Laplace’s equation: ∂2θ
y=1
Separation of Variables — sines in which direction?
In this case: in the x-direction because θ = 0 on both x = 0 and x = 1.
Let θ = Y (y) sinnπx, for n = 1, 2, · · · .
Hence Y ′′ − n2π2Y = 0 =⇒ Y = Aenπy + Be−nπy.
After reconstruction of θ and superposition:
θ(x, t) =
y=1
θ(x, t) = ∞∑
n=1
−nπy )
sinnπx.
In the infinite strip we argued that An = 0, but we cannot do so here.
We have to apply both BCs in the y-direction.
y = 0 : f(x) = ∞∑
∫ 1
0
−nπ = 0.
−nπ = 0.
A pair of simultaneous equations for An and Bn where Cn is known.
Hence An = −Bne −2nπ which means that Bn =
Cn
Cne −2nπ
θ = ∞∑
enπ − e−nπ
y=1
∞∑
y=1
y=1
x=0 x=1θ = 0 y=0
y=1
y=1
y=1
x=0 x=1θ = 0 y=0
y=1
y=1
1 1
1 1
A quarter-range Fourier Sine Series solution