Numerical evaluation of the Unified Transform
Method
Mason Brewer
Seattle University
March 19, 2015
Acknowledgements
I
Center for Undergraduate Research in Mathematics (CURM)
I
Dr. Katie Oliveras and Dr. Eric Bahuaud
I
The Eigenseminar research group
Presentation Outline
Partial Differential Equations (PDEs)
Solutions to the heat equation on the whole line
Solutions to the heat equation on the half line
Numerics of the UTM Solution
Partial Differential Equations (PDEs)
Let q be a function of two variables.
I
Heat equation, q = q(x, t)
∂q
∂2q
=
−→ qt = qxx .
∂t
∂x2
Partial Differential Equations (PDEs)
Let q be a function of two variables.
I
Heat equation, q = q(x, t)
∂q
∂2q
=
−→ qt = qxx .
∂t
∂x2
I
Laplace’s equation, q = q(x, y)
qxx + qyy = 0.
Partial Differential Equations (PDEs)
Let q be a function of two variables.
I
Heat equation, q = q(x, t)
∂q
∂2q
=
−→ qt = qxx .
∂t
∂x2
I
Laplace’s equation, q = q(x, y)
qxx + qyy = 0.
I
Other examples, q = q(x, t)
2
qt = e−x qx ,
qqt = qxxx .
Partial Differential Equations (PDEs)
Let q be a function of two variables.
I
Heat equation, q = q(x, t)
∂q
∂2q
=
−→ qt = qxx .
∂t
∂x2
I
Laplace’s equation, q = q(x, y)
qxx + qyy = 0.
I
Other examples, q = q(x, t)
XXX−x
2 qt =
eX
XqX
x,
h
(
(
h
(
h
(qh
qq(
t =
xxx
h
(
h.
Boundary and initial conditions
Time dependent PDEs may have initial conditions, and
boundary conditions.
Initial condition: q(x, 0) = q0 (x)
Boundary condition: q(a, t) where a ∈ ∂D
t
q(x, T )
t=T
D
x
t
q(x, T )
t=T
D
One-dimensional domains
There are two one-dimensional domains of interest.
Whole Line −∞ < x < ∞
Half Line 0 ≤ x < ∞
Presentation Outline
Partial Differential Equations (PDEs)
Solutions to the heat equation on the whole line
Solutions to the heat equation on the half line
Numerics of the UTM Solution
The Unified Transform Method (UTM)
I
Relatively new.
I
Solves linear time dependent PDEs on the whole line, half
line, and interval where classical methods fail.
I
Obtains integral solutions.
I
Can be extrapolated to more difficult problems.
The heat equation on the wholeq(x,line
T)
t
t=T
qt = qxxD,
Domain, D:
x
−∞ < x < ∞,
t
Boundary condition:
Initial condition:
0 ≤ t ≤ T,
q(x, T )
t=T
q(x, t) −→ 0
as
D
x −→ ±∞,
x
q(x, 0)
q(x, 0) = q0 (x).
t
q(x, T )
t=T
q(x, t) → 0 as x → −∞
D
q(x, t) → 0 as x → ∞
x
q(x, 0)
q(x, T )
The Fourier transform (preliminaries)
The Fourier transform is a tool for solving PDEs on the whole line.
Fourier transform:
F(q(x, t)) = qˆ(k, t) =
ˆ
∞
−∞
Inverse Fourier transform:
F
−1
e−ikx q(x, t)dx,
1
(ˆ
q (k, t)) = q(x, t) =
2π
ˆ
∞
eikx qˆ(k, t)dk.
−∞
Green’s theorem (preliminaries)
Let C be a positively oriented, peicewise smooth, simple closed
curve in a plane, and let D be the region bounded by C. If L and
M are functions of x and y defined on an open region containing
D with continuous partial derivatives, then the following equality
exists.
˛
¨ ∂L
∂M
−
dxdy
(Ldx + M dy) =
∂x
∂y
C
D
The Unified Transform Method (UTM)
Consider the heat equation on the whole line:
qt = qxx ,
−∞ < x < ∞,
0 ≤ t ≤ T.
q(x, T )
The Unified Transform Method (UTM)
t
t=T
Consider the heat equation onD the whole line:
x
qt = qxx ,
−∞ < x < ∞,
q(x, T )
t
0 ≤ t ≤ T.
=T
We can rewrite the above in divergence t form:
D
(Eq)t − (E(ikq + qx ))x = 0,
q(x, 0)
x
−∞ < x < ∞,
2
where E = eikx+k t .
q(x, T )
t
t=T
q(x, t) → 0 as x → −∞
D
q(x, t) → 0 as x → ∞
x
q(x, 0)
q(x, T )
t
Double integral
over all of D,
t=T
¨
q(0, t) = g (t)
q(x, t) → 0 as x → ∞
D
x
(Eq)
− (E(ikq + qx ))
x dxdt = 0.
q(x,t0) = q (x)
0
D
0
0 ≤ t ≤ T,
The Unified Transform Method (UTM)
Now applying Green’s theorem to D,
˛
(Eq) dx + (E(ikq + qx )) dt = 0.
C
The Unified Transform Method (UTM)
Now applying Green’s theorem to D,
˛
(Eq) dx + (E(ikq + qx )) dt = 0.
C
We can evaluate this boundary integral,
ˆ
ˆ
∞
Eq0 (x)dx|t=0 +
−∞
ˆ
−
T
E(ikq + qx )dt|x=∞
0
∞
−∞
Eq(x, t)dx|t=T −
ˆ
T
E(ikq + qx )dt|x=−∞ = 0.
0
The Unified Transform Method (UTM)
Now applying Green’s theorem to D,
˛
(Eq) dx + (E(ikq + qx )) dt = 0.
C
We can evaluate this boundary integral,
ˆ
X
ˆX
TX
XX X
Eq0 (x)dx|t=0 +
E(ikq + X
qxX
)dt|
Xx=∞
XX
−∞
0
X
ˆX
ˆ ∞
T XX
X
X
−
Eq(x, t)dx|t=T −
E(ikq
qxX
)dt|
= 0.
+ X
Xx=−∞
XXX
−∞
0
∞
The Unified Transform Method (UTM)
The remaining terms are
ˆ ∞
ˆ
e−ikx q0 (x)dx −
−∞
∞
e−ikx ek
2T
q(x, T )dx = 0,
−∞
or
qˆ0 (k) − ek
2T
qˆ(k, T ) = 0.
Solving for q(x, T ) yields
1
q(x, T ) =
2π
ˆ
∞
eikx e−k
2T
qˆ0 (k)dk.
−∞
This is true for any t = T , so our solution is
ˆ ∞
1
2
q(x, t) =
eikx−k t qˆ0 (k)dk.
2π −∞
Presentation Outline
Partial Differential Equations (PDEs)
Solutions to the heat equation on the whole line
Solutions to the heat equation on the half line
Numerics of the UTM Solution
D
The heat equation on the half line
x
q(x, T )
t
qt = qxx ,
Domain, D:
t=T
D
0 ≤ x < ∞,
0 ≤ t ≤ T,
x
q(x, 0)
Boundary conditions:
q(0, t) = q(x,
g0 (t)
T)
t
q(x, t) →condition:
0 as x → −∞
Initial
q(x, t) −→ 0
x −→ +∞,t = T
as
D
q(x, 0) = q0 (x).
q(x, t) → 0 as x → ∞
x
q(x, 0)
t
q(x, T )
t=T
q(0, t) = g0 (t)
D
q(x, t) → 0 as x → ∞
x
q(x, 0) = q0 (x)
General Approach of UTM
I
Assume you have a solution to the PDE.
I
Evaluate along the boundary of the domain using Green’s
theorem to obtain a solution.
I
Extend k into the complex plane to get rid of extra boundary
conditions.
UTM solution to the Heat Equation on the half line
The solution is
ˆ ∞
1
2
e−k t+ikx qˆ0 (k)dk
2π −∞
ˆ
1
2
−
e−k t+ikx (ˆ
q0 (−k) + 2ik˜
g0 (k 2 , t))dk,
2π ∂D+
q(x, t) =
where
ˆ
2
g˜0 (k , t) =
0
contains our boundary condition.
t
2 0
ek t g0 (t0 )dt0
Presentation Outline
Partial Differential Equations (PDEs)
Solutions to the heat equation on the whole line
Solutions to the heat equation on the half line
Numerics of the UTM Solution
Contour Integral in the complex plane
In our solution, k is complex and so must be parametrized.
(
s + is, s ≥ 0,
k=
s − is, s ≤ 0,
This is the
boundary of D+ .
Contour Integral in the complex plane cont.
Parameterizing k yields the solution of the form:
ˆ ∞
1
2
q(x, t) =
e−k t+ikx qˆ0 (k)dk
2π −∞
ˆ
1 + i ∞ −k2 t+ikx
e
(ˆ
q0 (−k) + 2ik˜
g0 (k 2 , t))ds,
−
2π −∞
Where k = s + i|s|.
Qualities of Numerical Solutions
I
High degree of oscillations.
I
Different parameterizations have different rates of
convergence.
Results
qt = qxx
g0 (t) = 0
q0 (x) = xe−x
2
Thank you!