Solutions of differential equations using transforms
Process:
Take transform of equation and boundary/initial conditions
in one variable.
Derivatives are turned into multiplication operators.
Solve (hopefully easier) problem in k variable.
Inverse transform to recover solution, often as a
convolution integral.
Ordinary differential equations: example 1
− u 00 + u = f (x),
lim u(x) = 0.
|x|→∞
Ordinary differential equations: example 1
− u 00 + u = f (x),
lim u(x) = 0.
|x|→∞
Transform using the derivative rule, giving
k 2 û(k ) + û(k ) = f̂ (k ).
Just an algebraic equation, whose solution is
û(k ) =
f̂ (k )
.
1 + k2
Ordinary differential equations: example 1
− u 00 + u = f (x),
lim u(x) = 0.
|x|→∞
Transform using the derivative rule, giving
k 2 û(k ) + û(k ) = f̂ (k ).
Just an algebraic equation, whose solution is
û(k ) =
f̂ (k )
.
1 + k2
Inverse transform of product of f̂ (k ) and 1/(1 + k 2 ) is
convolution:
∨
Z
1
1 ∞ −|x−y |
u(x) = f (x) ∗
=
e
f (y )dy .
2 −∞
1 + k2
But where was far field condition used?
Ordinary differential equations: example 2
Example 2. The Airy equation is
u 00 − xu = 0,
lim u(x) = 0.
|x|→∞
Ordinary differential equations: example 2
Example 2. The Airy equation is
u 00 − xu = 0,
lim u(x) = 0.
|x|→∞
Transform leads to
−k 2 û(k ) − i û 0 (k ) = 0.
Ordinary differential equations: example 2
Example 2. The Airy equation is
u 00 − xu = 0,
lim u(x) = 0.
|x|→∞
Transform leads to
−k 2 û(k ) − i û 0 (k ) = 0.
Solve by separation of variables: d û/û = ik 2 dk integrates to
û(k ) = Ceik
3 /3
.
Ordinary differential equations: example 2
Example 2. The Airy equation is
u 00 − xu = 0,
lim u(x) = 0.
|x|→∞
Transform leads to
−k 2 û(k ) − i û 0 (k ) = 0.
Solve by separation of variables: d û/û = ik 2 dk integrates to
û(k ) = Ceik
3 /3
.
Inverse transform is
u(x) =
C
2π
Z
∞
exp(i[kx + k 3 /3])dk .
−∞
With the choice C = 1 get the Airy function.
Partial differential equations, example 1
Laplace equation in upper half plane:
uxx + uyy = 0,
−∞ < x < ∞,
u(x, 0) = g(x),
y > 0,
lim u(x, y ) = 0.
y →∞
Partial differential equations, example 1
Laplace equation in upper half plane:
uxx + uyy = 0,
−∞ < x < ∞,
u(x, 0) = g(x),
y > 0,
lim u(x, y ) = 0.
y →∞
Transform in the x variable only:
Z ∞
U(k , y ) =
e−ikx u(x, y )dx.
−∞
Note y -derivatives commute with the Fourier transform in x.
−k 2 U + Uyy = 0,
U(k , 0) = ĝ(k ),
lim U(k , y ) = 0.
y →∞
Partial differential equations, example 1, cont.
Now solve ODEs
−k 2 U + Uyy = 0,
U(k , 0) = ĝ(k ),
lim U(k , y ) = 0.
y →∞
Partial differential equations, example 1, cont.
Now solve ODEs
−k 2 U + Uyy = 0,
U(k , 0) = ĝ(k ),
lim U(k , y ) = 0.
y →∞
General solution is U = c1 (k )e+|k |y + c2 (k )e−|k |y . Using
boundary conditions,
U(k , y ) = ĝ(k )e−|k |y .
Partial differential equations, example 1, cont.
Now solve ODEs
−k 2 U + Uyy = 0,
U(k , 0) = ĝ(k ),
lim U(k , y ) = 0.
y →∞
General solution is U = c1 (k )e+|k |y + c2 (k )e−|k |y . Using
boundary conditions,
U(k , y ) = ĝ(k )e−|k |y .
Inverse transform using convolution and exponential formulas
∨
y
−|k |y
u(x, y ) =g(x) ∗ e
= g(x) ∗
π(x 2 + y 2 )
Z ∞
1
yg(x0 )
=
dx0 .
π −∞ (x − x0 )2 + y 2
Same formula as obtained by Green’s function methods!
Partial differential equations, example 2
“Transport equation"
ut + cux = 0,
−∞ < x < ∞,
t > 0,
u(x, 0) = f (x).
Partial differential equations, example 2
“Transport equation"
ut + cux = 0,
−∞ < x < ∞,
As before,
Z
∞
U(k , t) =
t > 0,
u(x, 0) = f (x).
e−ikx u(x, t)dx.
−∞
therefore transform in x variables is
Ut + ikcU = 0,
U(k , 0) = f̂ (k ).
Partial differential equations, example 2
“Transport equation"
ut + cux = 0,
−∞ < x < ∞,
As before,
Z
∞
U(k , t) =
t > 0,
u(x, 0) = f (x).
e−ikx u(x, t)dx.
−∞
therefore transform in x variables is
Ut + ikcU = 0,
U(k , 0) = f̂ (k ).
Simple differential equation with solution
U(k , t) = e−ickt f̂ (k ).
Partial differential equations, example 2
“Transport equation"
ut + cux = 0,
−∞ < x < ∞,
As before,
Z
∞
U(k , t) =
t > 0,
u(x, 0) = f (x).
e−ikx u(x, t)dx.
−∞
therefore transform in x variables is
Ut + ikcU = 0,
U(k , 0) = f̂ (k ).
Simple differential equation with solution
U(k , t) = e−ickt f̂ (k ).
Use translation formula f (x − a) = e−iat f̂ (k ) with a = ct,
u(x, t) = f (x − ct).
Partial differential equations, example 3
Consider the wave equation on the real line
utt = uxx ,
−∞ < x < ∞,
t > 0,
u(x, 0) = f (x),
ut (x, 0) = g(x).
Partial differential equations, example 3
Consider the wave equation on the real line
utt = uxx ,
−∞ < x < ∞,
t > 0,
u(x, 0) = f (x),
ut (x, 0) = g(x).
Transforming as before,
Utt + k 2 U = 0,
U(k , 0) = f̂ (k ),
Ut (k , 0) = ĝ(k ).
Partial differential equations, example 3
Consider the wave equation on the real line
utt = uxx ,
−∞ < x < ∞,
t > 0,
u(x, 0) = f (x),
ut (x, 0) = g(x).
Transforming as before,
Utt + k 2 U = 0,
U(k , 0) = f̂ (k ),
Ut (k , 0) = ĝ(k ).
Solution of initial value problem
U(k , t) = f̂ (k ) cos(kt) +
ĝ(k )
sin(kt).
k
Need inverse transform formulas for cos,sin and f̂ /k .
Partial differential equations, example 3, cont.
For cos,sin, write in terms of complex exponentials:
1 iak
1
[e + e−iak ]∨ = [δ(x + a) + δ(x − a)],
2
2
i
i iak
−iak ∨
∨
] = − [δ(x + a) − δ(x − a)].
[sin(ak )] = − [e − e
2
2
[cos(ak )]∨ =
Partial differential equations, example 3, cont.
For cos,sin, write in terms of complex exponentials:
1 iak
1
[e + e−iak ]∨ = [δ(x + a) + δ(x − a)],
2
2
i
i iak
−iak ∨
∨
] = − [δ(x + a) − δ(x − a)].
[sin(ak )] = − [e − e
2
2
[cos(ak )]∨ =
If f (x) is integrable, int. by parts gives
Z
x
0
f (x )dx
−∞
0
∧
Z
∞
=
−∞
e−ikx
Z
x
−∞
f (x 0 )dx 0 dx =
1
ik
Z
∞
−∞
e−ikx f (x)dx.
Partial differential equations, example 3, cont.
For cos,sin, write in terms of complex exponentials:
1 iak
1
[e + e−iak ]∨ = [δ(x + a) + δ(x − a)],
2
2
i
i iak
−iak ∨
∨
] = − [δ(x + a) − δ(x − a)].
[sin(ak )] = − [e − e
2
2
[cos(ak )]∨ =
If f (x) is integrable, int. by parts gives
Z
x
0
f (x )dx
−∞
0
∧
Z
∞
e−ikx
=
−∞
therefore
"
f̂ (k )
ik
Z
x
f (x 0 )dx 0 dx =
−∞
#∨
Z
x
=
1
ik
Z
∞
e−ikx f (x)dx.
−∞
f (x 0 )dx 0 .
−∞
That is, integration in x gives a factor of 1/(ik ) in the transform.
Partial differential equations, example 3, cont.
Now for inverse transform of
U(k , t) = f̂ (k ) cos(kt) +
ĝ(k )
sin(kt).
k
Partial differential equations, example 3, cont.
Now for inverse transform of
U(k , t) = f̂ (k ) cos(kt) +
ĝ(k )
sin(kt).
k
First term:
h
i∨
1
f̂ (k ) cos(kt) = f (x) ∗ [δ(x + t) + δ(x − t)]
2
Z
1 ∞
1
=
f (x − y )[δ(y + t) + δ(y − t)]dy = [f (x − t) + f (x + t)].
2 −∞
2
Partial differential equations, example 3, cont.
Now for inverse transform of
U(k , t) = f̂ (k ) cos(kt) +
ĝ(k )
sin(kt).
k
First term:
h
i∨
1
f̂ (k ) cos(kt) = f (x) ∗ [δ(x + t) + δ(x − t)]
2
Z
1 ∞
1
=
f (x − y )[δ(y + t) + δ(y − t)]dy = [f (x − t) + f (x + t)].
2 −∞
2
Second term:
Z
∨
x
[ĝ(k ) sin(kt)/k ] = i
=
=
=
1
2
1
2
1
2
∨
[ĝ(k ) sin(kt)] (x 0 )dx 0
−∞
Z x
−∞
Z x
−∞
Z x
−∞
g(x 0 ) ∗ [δ(x 0 + t) − δ(x 0 − t)]dx 0
Z
∞
g(x 0 − y )[δ(y + t) − δ(y − t)]dydx 0
−∞
g(x 0 + t) − g(x 0 − t)dx 0 .
Partial differential equations, example 3, cont.
Finally, can change variables ξ = x 0 + t or ξ = x 0 − t
Z
x
g(x 0 + t) − g(x 0 − t)dx 0 =
−∞
Z
x+t
Z
x−t
g(ξ)dξ −
−∞
Z x+t
g(ξ)dξ.
=
x−t
g(ξ)dξ
−∞
Partial differential equations, example 3, cont.
Finally, can change variables ξ = x 0 + t or ξ = x 0 − t
Z
x
g(x 0 + t) − g(x 0 − t)dx 0 =
−∞
Z
x+t
Z
x−t
g(ξ)dξ −
−∞
Z x+t
g(ξ)dξ
−∞
g(ξ)dξ.
=
x−t
All together get d’Alembert’s formula
1
1
u(x, t) = [f (x − t) + f (x + t)] +
2
2
Z
x+t
g(ξ)dξ.
x−t
Fundamental solutions
Consider generic, linear, time-dependent equation
ut (x, t) = Lu(x, t), −∞ < x < ∞, u(x, 0) = f (x), lim u(x, t) = 0,
|x|→∞
where L is some operator(e.g. L = ∂ 2 /∂x 2 ).
Fundamental solutions
Consider generic, linear, time-dependent equation
ut (x, t) = Lu(x, t), −∞ < x < ∞, u(x, 0) = f (x), lim u(x, t) = 0,
|x|→∞
where L is some operator(e.g. L = ∂ 2 /∂x 2 ).
The fundamental solution S(x, x0 , t) is a type of Green’s
function, solving
St = Lx S, −∞ < x < ∞, S(x, x0 , 0) = δ(x−x0 ), lim S(x, x0 , t) = 0.
|x|→∞
Fundamental solutions
Consider generic, linear, time-dependent equation
ut (x, t) = Lu(x, t), −∞ < x < ∞, u(x, 0) = f (x), lim u(x, t) = 0,
|x|→∞
where L is some operator(e.g. L = ∂ 2 /∂x 2 ).
The fundamental solution S(x, x0 , t) is a type of Green’s
function, solving
St = Lx S, −∞ < x < ∞, S(x, x0 , 0) = δ(x−x0 ), lim S(x, x0 , t) = 0.
|x|→∞
Initial condition means S limits to a δ-function as t → 0:
Z ∞
Z ∞
lim
S(x, x0 , t)φ(x)dx =
δ(x − x0 )φ(x)dx = φ(x0 ),
t→0 −∞
−∞
Fundamental solutions, cont.
Claim that the initial value problem has solution
Z ∞
u(x, t) =
S(x, x0 , t)f (x0 )dx0 ,
−∞
Fundamental solutions, cont.
Claim that the initial value problem has solution
Z ∞
u(x, t) =
S(x, x0 , t)f (x0 )dx0 ,
−∞
Check:
Z
u(x, 0) = lim
∞
t→0 −∞
Z
∞
S(x, x0 , t)f (x0 )dx0 =
δ(x−x0 )f (x0 )dx0 = f (x).
−∞
Fundamental solutions, cont.
Claim that the initial value problem has solution
Z ∞
u(x, t) =
S(x, x0 , t)f (x0 )dx0 ,
−∞
Check:
Z
u(x, 0) = lim
∞
t→0 −∞
Z
∞
S(x, x0 , t)f (x0 )dx0 =
δ(x−x0 )f (x0 )dx0 = f (x).
−∞
Plugging u into the equation and moving time derivative inside
the integral
Z ∞
Z ∞
ut =
St (x, x0 , t)f (x0 )dx0 =
Lx S(x, x0 , t)f (x0 )dx0 .
−∞
−∞
Fundamental solutions, cont.
Claim that the initial value problem has solution
Z ∞
u(x, t) =
S(x, x0 , t)f (x0 )dx0 ,
−∞
Check:
Z
u(x, 0) = lim
∞
t→0 −∞
Z
∞
S(x, x0 , t)f (x0 )dx0 =
δ(x−x0 )f (x0 )dx0 = f (x).
−∞
Plugging u into the equation and moving time derivative inside
the integral
Z ∞
Z ∞
ut =
St (x, x0 , t)f (x0 )dx0 =
Lx S(x, x0 , t)f (x0 )dx0 .
−∞
−∞
Now move operator outside integral
Z ∞
ut = Lx
S(x, x0 , t)f (x0 )dx0 = Lx u.
−∞
Fundamental solutions using the Fourier transform, example 1
For diffusion equation on the real line, S solves
St = DSxx , −∞ < x < ∞, S(x, x0 , 0) = δ(x−x0 ), lim S(x, x0 , t) = 0.
|x|→∞
Fundamental solutions using the Fourier transform, example 1
For diffusion equation on the real line, S solves
St = DSxx , −∞ < x < ∞, S(x, x0 , 0) = δ(x−x0 ), lim S(x, x0 , t) = 0.
|x|→∞
Take Fourier Rtransform in x by letting
∞
Ŝ(k , x0 , t) = −∞ S(x, x0 , t)e−ikx dx, giving
Fundamental solutions using the Fourier transform, example 1
For diffusion equation on the real line, S solves
St = DSxx , −∞ < x < ∞, S(x, x0 , 0) = δ(x−x0 ), lim S(x, x0 , t) = 0.
|x|→∞
Take Fourier Rtransform in x by letting
∞
Ŝ(k , x0 , t) = −∞ S(x, x0 , t)e−ikx dx, giving
Ŝt = −Dk 2 Ŝ,
Ŝ(k , 0) = e−ix0 k .
Fundamental solutions using the Fourier transform, example 1
For diffusion equation on the real line, S solves
St = DSxx , −∞ < x < ∞, S(x, x0 , 0) = δ(x−x0 ), lim S(x, x0 , t) = 0.
|x|→∞
Take Fourier Rtransform in x by letting
∞
Ŝ(k , x0 , t) = −∞ S(x, x0 , t)e−ikx dx, giving
Ŝt = −Dk 2 Ŝ,
Ŝ(k , 0) = e−ix0 k .
Solution to this ODE
2
Ŝ = e−ix0 k −Dk t .
Fundamental solutions using the Fourier transform, example 1
Inverse transform of
2
Ŝ = e−ix0 k −Dk t .
uses translation, dilation, and Gaussian formulas:
S(x, x0 , t) = √
1
4πDt
2 /(4Dt)
e−(x−x0 )
.
Fundamental solutions using the Fourier transform, example 1
Inverse transform of
2
Ŝ = e−ix0 k −Dk t .
uses translation, dilation, and Gaussian formulas:
S(x, x0 , t) = √
1
4πDt
2 /(4Dt)
e−(x−x0 )
.
It follows that the solution to ut = Duxx and u(x, 0) = f (x) is
Z ∞
f (x )
2
√ 0 e−(x−x0 ) /(4Dt) dx0 .
u(x, t) =
4πDt
−∞
Fundamental solutions using the Fourier transform, example 2
Linearized Korteweg - de Vries (KdV) equation:
ut = −uxxx ,
u(x, 0) = f (x),
lim u(x, t) = 0.
|x|→∞
Fundamental solutions using the Fourier transform, example 2
Linearized Korteweg - de Vries (KdV) equation:
ut = −uxxx ,
u(x, 0) = f (x),
lim u(x, t) = 0.
|x|→∞
Fundamental solution solves
St = −Sxxx ,
S(x, x0 , 0) = δ(x − x0 ), lim S(x, x0 , t) = 0.
|x|→∞
Fundamental solutions using the Fourier transform, example 2
Linearized Korteweg - de Vries (KdV) equation:
ut = −uxxx ,
u(x, 0) = f (x),
lim u(x, t) = 0.
|x|→∞
Fundamental solution solves
St = −Sxxx ,
S(x, x0 , 0) = δ(x − x0 ), lim S(x, x0 , t) = 0.
|x|→∞
Transforming
Ŝt = ik 3 Ŝ,
Ŝ(k , 0) = e−ix0 k ,
3
whose solution is Ŝ(k , x0 , t) = e−ix0 k eik t .
Fundamental solutions using the Fourier transform, example 2
Linearized Korteweg - de Vries (KdV) equation:
ut = −uxxx ,
u(x, 0) = f (x),
lim u(x, t) = 0.
|x|→∞
Fundamental solution solves
St = −Sxxx ,
S(x, x0 , 0) = δ(x − x0 ), lim S(x, x0 , t) = 0.
|x|→∞
Transforming
Ŝt = ik 3 Ŝ,
Ŝ(k , 0) = e−ix0 k ,
3
whose solution is Ŝ(k , x0 , t) = e−ix0 k eik t .
3
Recall transform of Airy function Ai(x) is eik /3 , therefore
h
i∨ h
i∨
3
3
S(x, x0 , t) = e−ix0 k eik t = ei(k /a) /3 (x − x0 )
= aAi a(x − x0 ) , a ≡ (3t)−1/3 .
Fundamental solutions using the Fourier transform, example 2
Linearized Korteweg - de Vries (KdV) equation:
ut = −uxxx ,
u(x, 0) = f (x),
lim u(x, t) = 0.
|x|→∞
Fundamental solution solves
St = −Sxxx ,
S(x, x0 , 0) = δ(x − x0 ), lim S(x, x0 , t) = 0.
|x|→∞
Transforming
Ŝ(k , 0) = e−ix0 k ,
Ŝt = ik 3 Ŝ,
3
whose solution is Ŝ(k , x0 , t) = e−ix0 k eik t .
3
Recall transform of Airy function Ai(x) is eik /3 , therefore
h
i∨ h
i∨
3
3
S(x, x0 , t) = e−ix0 k eik t = ei(k /a) /3 (x − x0 )
= aAi a(x − x0 ) , a ≡ (3t)−1/3 .
Solution to original equation:
1
u(x, t) =
(3t)1/3
Z
∞
Ai
−∞
x − x0
(3t)1/3
f (x0 )dx0 .
The method of images for fundamental solutions
For solutions on half-line x > 0, can’t use Fourier transform
directly.
The method of images for fundamental solutions
For solutions on half-line x > 0, can’t use Fourier transform
directly.
Fundamental solution must satisfy boundary condition at
x =0
The method of images for fundamental solutions
For solutions on half-line x > 0, can’t use Fourier transform
directly.
Fundamental solution must satisfy boundary condition at
x =0
Inspiration: method of images. If S∞ (x; x0 , t) is the
fundamental solution for the whole line, then:
Odd reflection S = S∞ (x; x0 , t) − S∞ (x; −x0 , t) gives
S(0, x0 , t) = 0.
Even reflection S = S∞ (x; x0 , t) + S∞ (x; −x0 , t) gives
Sx (0, x0 , t) = 0.
The method of images for fundamental solutions, example
Consider diffusion equation on half line:
ut = Duxx ,
u(x, 0) = f (x), u(0, t) = 0, lim u(x, t) = 0.
x→∞
The method of images for fundamental solutions, example
Consider diffusion equation on half line:
ut = Duxx ,
u(x, 0) = f (x), u(0, t) = 0, lim u(x, t) = 0.
x→∞
Use odd reflection of fundamental
solution for whole line
√
2
S∞ = e−(x−x0 ) /(4Dt) / 4πDt,
S(x, x0 , t) = √
1
4πDt
h
2 /(4Dt)
e−(x−x0 )
2 /(4Dt)
− e−(x+x0 )
i
The method of images for fundamental solutions, example
Consider diffusion equation on half line:
ut = Duxx ,
u(x, 0) = f (x), u(0, t) = 0, lim u(x, t) = 0.
x→∞
Use odd reflection of fundamental
solution for whole line
√
2
S∞ = e−(x−x0 ) /(4Dt) / 4πDt,
S(x, x0 , t) = √
1
4πDt
h
2 /(4Dt)
e−(x−x0 )
2 /(4Dt)
− e−(x+x0 )
i
Therefore the solution u is just
Z ∞
i
f (x ) h −(x−x0 )2 /(4Dt)
2
√ 0
u(x, t) =
e
− e−(x+x0 ) /(4Dt) dx0 .
4πDt
0
The age of the earth
Lord Kelvin: simple model of temperature of earth
u(x, t) at depth x and time t
ut = Duxx , x > 0,
u(x, 0) = U0 , u(0, t) = 0.
Scale chosen so u = 0 on surface; assumes initially
constant temperature (U0 ) throughout the molten earth.
The age of the earth
Lord Kelvin: simple model of temperature of earth
u(x, t) at depth x and time t
ut = Duxx , x > 0,
u(x, 0) = U0 , u(0, t) = 0.
Scale chosen so u = 0 on surface; assumes initially
constant temperature (U0 ) throughout the molten earth.
We found solution
Z ∞
i
2
f (x ) h −(x−x0 )2 /(4Dt)
√ 0
u(x, t) =
e
− e−(x+x0 ) /(4Dt) dx0 .
4πDt
0
The age of the earth
Lord Kelvin: simple model of temperature of earth
u(x, t) at depth x and time t
ut = Duxx , x > 0,
u(x, 0) = U0 , u(0, t) = 0.
Scale chosen so u = 0 on surface; assumes initially
constant temperature (U0 ) throughout the molten earth.
We found solution
Z ∞
i
2
f (x ) h −(x−x0 )2 /(4Dt)
√ 0
u(x, t) =
e
− e−(x+x0 ) /(4Dt) dx0 .
4πDt
0
Temperature gradient µ at surface is therefore
Z ∞
2
U0
1
U0
x0 e−x0 /(4Dt) dx0 = √
.
µ = ux (0, t) = √
4πDt Dt 0
πDt
The age of the earth
Lord Kelvin: simple model of temperature of earth
u(x, t) at depth x and time t
ut = Duxx , x > 0,
u(x, 0) = U0 , u(0, t) = 0.
Scale chosen so u = 0 on surface; assumes initially
constant temperature (U0 ) throughout the molten earth.
We found solution
Z ∞
i
2
f (x ) h −(x−x0 )2 /(4Dt)
√ 0
u(x, t) =
e
− e−(x+x0 ) /(4Dt) dx0 .
4πDt
0
Temperature gradient µ at surface is therefore
Z ∞
2
U0
1
U0
x0 e−x0 /(4Dt) dx0 = √
.
µ = ux (0, t) = √
4πDt Dt 0
πDt
This relates the age of earth t to quantities we can estimate
U0 ≈ melting temp. of iron ≈ 104 C,
D ≈ 10−3 m2 /s,
µ ≈ 10−2 C/m,
The age of the earth
Lord Kelvin: simple model of temperature of earth
u(x, t) at depth x and time t
ut = Duxx , x > 0,
u(x, 0) = U0 , u(0, t) = 0.
Scale chosen so u = 0 on surface; assumes initially
constant temperature (U0 ) throughout the molten earth.
We found solution
Z ∞
i
2
f (x ) h −(x−x0 )2 /(4Dt)
√ 0
u(x, t) =
e
− e−(x+x0 ) /(4Dt) dx0 .
4πDt
0
Temperature gradient µ at surface is therefore
Z ∞
2
U0
1
U0
x0 e−x0 /(4Dt) dx0 = √
.
µ = ux (0, t) = √
4πDt Dt 0
πDt
This relates the age of earth t to quantities we can estimate
U0 ≈ melting temp. of iron ≈ 104 C,
which gives t ≈ 3 × 107 years !!??
D ≈ 10−3 m2 /s,
µ ≈ 10−2 C/m,