Heat Equation on Unbounded Intervals

MATH 467 Partial Differential Equations

J. Robert Buchanan

Department of Mathematics

Spring 2016

Objectives

In this lesson we will learn about: ▶ the fundamental solution to the heat equation, ▶ solutions to the heat equation for 0 ≤ x < ∞, and ▶ solutions to the heat equation for −∞ < x < ∞.

Fundamental Solution

For t > 0 define the function

U(x, t) = \frac{1}{\sqrt{4πkt}}e^{−x^{2}/(4kt)}.

Remarks:

▶ U(x, t) is related to the probability density function for a

normally distributed random variable. ▶ While defined only for t > 0, the limit as t → 0^{+} exists. ▶ U(x, t) solves the heat equation.

Graph

lim_{t→0^{+}}U(x, t)

If x ≠ 0 then

lim_{t→0^{+}}U(x, t) = lim_{t→0^{+}} \frac{1}{\sqrt{4πkt}}e^{−x^{2}/(4kt)} = 0.

If x = 0 then lim_{t→0^{+}}U(0, t) = lim_{t→0^{+}} \frac{1}{\sqrt{4πkt}} = ∞.

lim_{t→0^{+}}U(x, t)

If x ≠ 0 then

lim_{t→0^{+}}U(x, t) = lim_{t→0^{+}} \frac{1}{\sqrt{4πkt}}e^{−x^{2}/(4kt)} = 0.

If x = 0 then lim_{t→0^{+}}U(0, t) = lim_{t→0^{+}} \frac{1}{\sqrt{4πkt}} = ∞.

Justification

Suppose x ≠ 0 then

lim_{t→0^{+}}U(x, t) = lim_{t→0^{+}} \frac{1}{\sqrt{4πkt}}e^{−x^{2}/(4kt)}
= lim_{t→0^{+}} \frac{1/\sqrt{t}}{\sqrt{4πk}e^{x^{2}/(4kt)}} (indeterminate ∞/∞)
= lim_{t→0^{+}} \frac{−1/(2t^{3/2})}{−\frac{x^{2}}{4kt^{2}} \sqrt{4πk}e^{x^{2}/(4kt)}}
= lim_{t→0^{+}} \frac{1}{\frac{x^{2}}{4kt^{1/2}} \sqrt{πk}e^{x^{2}/(4kt)}}
= 0.

Area Under the Curve

Assume that for fixed t > 0 the improper integral

∫_{−∞}^{∞} U(x, t) dx

converges. Find the value of the integral.

Solution

IfS = ∫_{−∞}^{∞} U(x, t) dx
S^{2} = (∫_{−∞}^{∞} U(x, t) dx)(∫_{−∞}^{∞} U(y, t) dy)
= \frac{1}{4πkt} ∫_{−∞}^{∞} e^{−x^{2}/(4kt)}dx ∫_{−∞}^{∞} e^{−y^{2}/(4kt)}dy
= \frac{1}{4πkt} ∫_{−∞}^{∞} ∫_{−∞}^{∞} e^{−(x^{2}+y^{2})/(4kt)}dx dy
= \frac{1}{4πkt} ∫_{0}^{2π} ∫_{0}^{∞} re^{−r^{2}/(4kt)}dr dθ
= \frac{1}{2kt} ∫_{0}^{∞} re^{−r^{2}/(4kt)}dr
S^{2} = 1 ⟹ S = 1.

Dirac Delta Function

Since ▶ for x ≠ 0, lim_{t→0^{+}}U(x, t) = 0 and ▶ ∫_{−∞}^{∞} U(x, t) dx = 1 for all t > 0

then

lim_{t→0^{+}} \frac{1}{\sqrt{4πkt}}e^{−x^{2}/(4kt)} = δ(x)

the Dirac delta function.

Logarithmic Differentiation (1 of 2)

U(x, t) = \frac{1}{\sqrt{4πkt}}e^{−x^{2}/(4kt)}
lnU = −\frac{1}{2} ln(4πkt) − \frac{x^{2}}{4kt}
\frac{∂}{∂t} [lnU] = \frac{∂}{∂t} [−\frac{1}{2} ln(4πkt) − \frac{x^{2}}{4kt}]
\frac{U_{t}}{U} = −\frac{1}{2t} + \frac{x^{2}}{4kt^{2}}
U_{t} = \frac{1}{\sqrt{4πkt}}e^{−x^{2}/(4kt)}(−\frac{1}{2t} + \frac{x^{2}}{4kt^{2}})

U(x, t) = \frac{1}{\sqrt{4πkt}}e^{−x^{2}/(4kt)}
lnU = −\frac{1}{2} ln(4πkt) − \frac{x^{2}}{4kt}
\frac{∂}{∂t} [lnU] = \frac{∂}{∂t} [−\frac{1}{2} ln(4πkt) − \frac{x^{2}}{4kt}]
\frac{U_{t}}{U} = −\frac{1}{2t} + \frac{x^{2}}{4kt^{2}}
U_{t} = \frac{1}{\sqrt{4πkt}}e^{−x^{2}/(4kt)}(−\frac{1}{2t} + \frac{x^{2}}{4kt^{2}})

U(x, t) = \frac{1}{\sqrt{4πkt}}e^{−x^{2}/(4kt)}
\frac{∂}{∂x} [lnU] = \frac{∂}{∂x} [−\frac{1}{2} ln(4πkt) − \frac{x^{2}}{4kt}]
\frac{U_{x}}{U} = −\frac{x}{2kt}
U_{x} = −U (\frac{x}{2kt})
U_{xx} = −U_{x} (\frac{x}{2kt}) − U (\frac{1}{2kt})
= U (\frac{x^{2}}{4k^{2}t^{2}}) − U (\frac{1}{2kt})
U_{xx} = \frac{1}{\sqrt{4πkt}}e^{−x^{2}/(4kt)}(\frac{x^{2}}{4k^{2}t^{2}} − \frac{1}{2kt})

Solution to the Heat Equation

U_{t}(x, t) = \frac{1}{4\sqrt{kπt}}e^{−x^{2}/(4kt)}(\frac{x^{2}}{2kt^{2}} − \frac{1}{t})
U_{xx}(x, t) = \frac{1}{4k\sqrt{kπt}}e^{−x^{2}/(4kt)}(\frac{x^{2}}{2kt^{2}} − \frac{1}{t})

and thus U_{t} = kU_{xx}. Remark: since the fundamental solution is defined for −∞ < x < ∞, no boundary conditions need be considered. If u(x, 0) = f (x) is an initial condition defined on −∞ < x < ∞, how do we form a solution to the IVP?

Solution to the Heat Equation

U_{t}(x, t) = \frac{1}{4\sqrt{kπt}}e^{−x^{2}/(4kt)}(\frac{x^{2}}{2kt^{2}} − \frac{1}{t})
U_{xx}(x, t) = \frac{1}{4k\sqrt{kπt}}e^{−x^{2}/(4kt)}(\frac{x^{2}}{2kt^{2}} − \frac{1}{t})

and thus U_{t} = kU_{xx}. Remark: since the fundamental solution is defined for −∞ < x < ∞, no boundary conditions need be considered. If u(x, 0) = f (x) is an initial condition defined on −∞ < x < ∞, how do we form a solution to the IVP?

Solving the IVP

Theorem

Consider the initial value problem

u_{t} = k u_{xx} for −∞ < x < ∞ and t > 0
u(x, 0) = f (x), for −∞ < x < ∞.

If f (x) is continuous and if ∫_{−∞}^{∞} |f (x)| dx converges, then the piecewise defined function

u(x, t) = \{ \begin{matrix}∫_{−∞}^{∞} U(x − y, t)f (y) dy & if t > 0, \\ f (x) & if t = 0 \end{matrix}

solves the heat equation and satisfies the initial condition in the sense that

lim_{(x,t)→(x_{0},0^{+})}u(x, t) = f (x_{0}).

Proof (1 of 7)

Suppose t > 0 and differentiate inside the integral

u_{t}(x, t) = ∫_{−∞}^{∞} \frac{1}{4\sqrt{kπt}}e^{−(x−y)^{2}/(4kt)}(\frac{(x − y)^{2}}{2kt^{2}} − \frac{1}{t}) f (y) dy

and

u_{xx}(x, t) = ∫_{−∞}^{∞} \frac{1}{4k\sqrt{kπt}}e^{−(x−y)^{2}/(4kt)}(\frac{(x − y)^{2}}{2kt^{2}} − \frac{1}{t}) f (y) dy

and thus u_{t} = k u_{xx}.

Assume t > 0 and fix x_{0} ∈ ℝ.

|u(x, t) − f (x_{0})| = |\frac{1}{\sqrt{4πkt}} ∫_{−∞}^{∞} e^{−(x−y)^{2}/(4kt)}f (y) dy − f (x_{0})|
≤ \frac{1}{\sqrt{4πkt}} ∫_{−∞}^{∞} e^{−(x−y)^{2}/(4kt)}|f (y) − f (x_{0})| dy

▶ For all ϵ > 0, there exists a δ > 0 such that if |z − x_{0}| ≤ 2δ

then |f (z) − f (x_{0})| ≤ ϵ since f is assumed continuous at x_{0}. ▶ Suppose x and y are such that |x − x_{0}| ≤ δ and |x − y| ≤ δ

then |y − x_{0}| ≤ 2δ and consequently |f (y) − f (x_{0})| ≤ ϵ.

\frac{1}{\sqrt{4πkt}} ∫_{x−δ}^{x+δ} e^{−(x−y)^{2}/(4kt)}|f (y) − f (x_{0})| dy ≤ ϵ

Assume t > 0 and fix x_{0} ∈ ℝ.

|u(x, t) − f (x_{0})| = |\frac{1}{\sqrt{4πkt}} ∫_{−∞}^{∞} e^{−(x−y)^{2}/(4kt)}f (y) dy − f (x_{0})|
≤ \frac{1}{\sqrt{4πkt}} ∫_{−∞}^{∞} e^{−(x−y)^{2}/(4kt)}|f (y) − f (x_{0})| dy

▶ For all ϵ > 0, there exists a δ > 0 such that if |z − x_{0}| ≤ 2δ

then |f (z) − f (x_{0})| ≤ ϵ since f is assumed continuous at x_{0}. ▶ Suppose x and y are such that |x − x_{0}| ≤ δ and |x − y| ≤ δ

then |y − x_{0}| ≤ 2δ and consequently |f (y) − f (x_{0})| ≤ ϵ.

\frac{1}{\sqrt{4πkt}} ∫_{x−δ}^{x+δ} e^{−(x−y)^{2}/(4kt)}|f (y) − f (x_{0})| dy ≤ ϵ

Proof (3 of 7)

Using the triangle inequality,

\frac{1}{\sqrt{4πkt}} ∫_{x+δ}^{∞} e^{−(x−y)^{2}/(4kt)}|f (y) − f (x_{0})| dy

≤ \frac{1}{\sqrt{4πkt}} ∫_{x+δ}^{∞} e^{−(x−y)^{2}/(4kt)}(|f (y)| + |f (x_{0})|) dy
= \frac{1}{\sqrt{4πkt}} ∫_{δ}^{∞} e^{−z^{2}/(4kt)}|f (x + z)| dz + \frac{|f (x_{0})|}{\sqrt{4πkt}} ∫_{δ}^{∞} e^{−z^{2}/(4kt)}dz

if z = y − x.

Proof (4 of 7)

The first integral satisfies the inequality,

\frac{1}{\sqrt{4πkt}} ∫_{δ}^{∞} e^{−z^{2}/(4kt)}|f (x + z)| dz

≤ \frac{1}{\sqrt{4πkt}}e^{−δ^{2}/(4kt)}∫_{δ}^{∞} |f (x + z)| dz
≤ \frac{1}{\sqrt{4πkt}}e^{−δ^{2}/(4kt)}∫_{−∞}^{∞} |f (u)| du.

There exists a δ_{1} > 0 so that if 0 < t < δ_{1} then

\frac{1}{\sqrt{4πkt}} ∫_{δ}^{∞} e^{−z^{2}/(4kt)}|f (x + z)| dz ≤ ϵ.

Proof (4 of 7)

The first integral satisfies the inequality,

\frac{1}{\sqrt{4πkt}} ∫_{δ}^{∞} e^{−z^{2}/(4kt)}|f (x + z)| dz

≤ \frac{1}{\sqrt{4πkt}}e^{−δ^{2}/(4kt)}∫_{δ}^{∞} |f (x + z)| dz
≤ \frac{1}{\sqrt{4πkt}}e^{−δ^{2}/(4kt)}∫_{−∞}^{∞} |f (u)| du.

There exists a δ_{1} > 0 so that if 0 < t < δ_{1} then

\frac{1}{\sqrt{4πkt}} ∫_{δ}^{∞} e^{−z^{2}/(4kt)}|f (x + z)| dz ≤ ϵ.

Proof (5 of 7)

The second integral satisfies the inequality,

\frac{|f (x_{0})|}{\sqrt{4πkt}} ∫_{δ}^{∞} e^{−z^{2}/(4kt)}dz ≤ \frac{|f (x_{0})|}{\sqrt{4πkt}} ∫_{δ}^{∞} e^{−zδ/(4kt)}dz
= \frac{|f (x_{0})|\sqrt{4kt}}{δ\sqrt{π}}e^{−δ^{2}/(4kt)}.

Since

lim_{t→0^{+}} \begin{matrix}√ & \overline{4kt} \\ √ & \overline{π} \end{matrix}e^{−δ^{2}/(4kt)} = 0

there exists a δ_{2} > 0 such that if 0 < t < δ_{2} then

\frac{|f (x_{0})|}{\sqrt{4πkt}} ∫_{δ}^{∞} e^{−z^{2}/(4kt)}dz ≤ ϵ.

Proof (6 of 7)

Define δ′ = min{δ_{1}, δ_{2}} then for 0 < t < δ′

\frac{1}{\sqrt{4πkt}} ∫_{x+δ}^{∞} e^{−(x−y)^{2}/(4kt)}|f (y) − f (x_{0})| dy ≤ 2ϵ.

Similarly we can show

\frac{1}{\sqrt{4πkt}} ∫_{−∞}^{x−δ} e^{−(x−y)^{2}/(4kt)}|f (y) − f (x_{0})| dy ≤ 2ϵ.

Proof (6 of 7)

Define δ′ = min{δ_{1}, δ_{2}} then for 0 < t < δ′

\frac{1}{\sqrt{4πkt}} ∫_{x+δ}^{∞} e^{−(x−y)^{2}/(4kt)}|f (y) − f (x_{0})| dy ≤ 2ϵ.

Similarly we can show

\frac{1}{\sqrt{4πkt}} ∫_{−∞}^{x−δ} e^{−(x−y)^{2}/(4kt)}|f (y) − f (x_{0})| dy ≤ 2ϵ.

Proof (7 of 7)

Suppose 0 < t < δ′ then

|u(x, t) − f (x_{0})| ≤ \frac{1}{\sqrt{4πkt}} ∫_{−∞}^{∞} e^{−(x−y)^{2}/(4kt)}|f (y) − f (x_{0})| dy
= \frac{1}{\sqrt{4πkt}} ∫_{−∞}^{x−δ′} e^{−(x−y)^{2}/(4kt)}|f (y) − f (x_{0})| dy

+ \frac{1}{\sqrt{4πkt}} ∫_{x−δ′}^{x+δ′} e^{−(x−y)^{2}/(4kt)}|f (y) − f (x_{0})| dy

+ \frac{1}{\sqrt{4πkt}} ∫_{x+δ′}^{∞} e^{−(x−y)^{2}/(4kt)}|f (y) − f (x_{0})| dy

≤ 2ϵ + ϵ + 2ϵ = 5ϵ.

Since ϵ > 0 is arbitrary then lim_{(x,t)→(x_{0},0^{+})}|u(x, t) − f (x_{0})| = 0.

Uniqueness

Theorem

Consider the initial value problem with conditions imposed as x → ±∞,

u_{t} = ku_{xx}, for −∞ < x < ∞ and t > 0
u(x, 0) = f (x), for −∞ < x < ∞
lim_{x→±∞}( max_{0≤t≤T}|u(x, t)|) = 0

for −∞ < x < ∞, t > 0, and T > 0. If f (x) is continuous, if

lim_{x→±∞}f (x) = 0, and if ∫_{−∞}^{∞} |f (x)| dx converges, then

u(x, t) = \{ \begin{matrix}∫_{−∞}^{∞} U(x − y, t)f (y) dy & if t > 0, \\ f (x) & if t = 0 \end{matrix}

is the unique, continuous solution to the initial value problem above on (−∞, ∞) × [0, ∞).

Example

Find the unique solution to the following initial boundary value problem.

u_{t} = u_{xx}, for −∞ < x < ∞ and t > 0
u(x, 0) = e^{−x^{2}} cos x, for −∞ < x < ∞
lim_{x→±∞}( max_{0≤t≤T}|u(x, t)|) = 0

Solution (1 of )

Since k = 1 then

u(x, t) = ∫_{−∞}^{∞} U(x − y, t)e^{−y^{2}} cos y dy
= ∫_{−∞}^{∞} \frac{1}{\sqrt{4πt}}e^{−(x−y)^{2}/(4t)}e^{−y^{2}} Re (e^{iy}) dy
= \frac{1}{\sqrt{4πt}}e^{−x^{2}/(4t)}∫_{−∞}^{∞} e^{(2xy−y^{2})/(4t)}e^{−y^{2}} Re (e^{iy}) dy
= U(x, t) ∫_{−∞}^{∞} e^{(2xy−[1+4t]y^{2})/(4t)}Re (e^{iy}) dy
= U(x, t) ∫_{−∞}^{∞} Re (e^{(2xy−[1+4t]y^{2})/(4t)}e^{iy}) dy
u(x, t) = U(x, t) Re (∫_{−∞}^{∞} e^{([2x+i4t]y−[1+4t]y^{2})/(4t)}dy)

Solution (1 of )

Since k = 1 then

u(x, t) = ∫_{−∞}^{∞} U(x − y, t)e^{−y^{2}} cos y dy
= ∫_{−∞}^{∞} \frac{1}{\sqrt{4πt}}e^{−(x−y)^{2}/(4t)}e^{−y^{2}} Re (e^{iy}) dy
= \frac{1}{\sqrt{4πt}}e^{−x^{2}/(4t)}∫_{−∞}^{∞} e^{(2xy−y^{2})/(4t)}e^{−y^{2}} Re (e^{iy}) dy
= U(x, t) ∫_{−∞}^{∞} e^{(2xy−[1+4t]y^{2})/(4t)}Re (e^{iy}) dy
= U(x, t) ∫_{−∞}^{∞} Re (e^{(2xy−[1+4t]y^{2})/(4t)}e^{iy}) dy
u(x, t) = U(x, t) Re (∫_{−∞}^{∞} e^{([2x+i4t]y−[1+4t]y^{2})/(4t)}dy)

Solution (1 of )

Since k = 1 then

u(x, t) = ∫_{−∞}^{∞} U(x − y, t)e^{−y^{2}} cos y dy
= ∫_{−∞}^{∞} \frac{1}{\sqrt{4πt}}e^{−(x−y)^{2}/(4t)}e^{−y^{2}} Re (e^{iy}) dy
= \frac{1}{\sqrt{4πt}}e^{−x^{2}/(4t)}∫_{−∞}^{∞} e^{(2xy−y^{2})/(4t)}e^{−y^{2}} Re (e^{iy}) dy
= U(x, t) ∫_{−∞}^{∞} e^{(2xy−[1+4t]y^{2})/(4t)}Re (e^{iy}) dy
= U(x, t) ∫_{−∞}^{∞} Re (e^{(2xy−[1+4t]y^{2})/(4t)}e^{iy}) dy
u(x, t) = U(x, t) Re (∫_{−∞}^{∞} e^{([2x+i4t]y−[1+4t]y^{2})/(4t)}dy)

Solution (1 of )

Since k = 1 then

u(x, t) = ∫_{−∞}^{∞} U(x − y, t)e^{−y^{2}} cos y dy
= ∫_{−∞}^{∞} \frac{1}{\sqrt{4πt}}e^{−(x−y)^{2}/(4t)}e^{−y^{2}} Re (e^{iy}) dy
= \frac{1}{\sqrt{4πt}}e^{−x^{2}/(4t)}∫_{−∞}^{∞} e^{(2xy−y^{2})/(4t)}e^{−y^{2}} Re (e^{iy}) dy
= U(x, t) ∫_{−∞}^{∞} e^{(2xy−[1+4t]y^{2})/(4t)}Re (e^{iy}) dy
= U(x, t) ∫_{−∞}^{∞} Re (e^{(2xy−[1+4t]y^{2})/(4t)}e^{iy}) dy
u(x, t) = U(x, t) Re (∫_{−∞}^{∞} e^{([2x+i4t]y−[1+4t]y^{2})/(4t)}dy)

Solution (1 of )

Since k = 1 then

u(x, t) = ∫_{−∞}^{∞} U(x − y, t)e^{−y^{2}} cos y dy
= ∫_{−∞}^{∞} \frac{1}{\sqrt{4πt}}e^{−(x−y)^{2}/(4t)}e^{−y^{2}} Re (e^{iy}) dy
= \frac{1}{\sqrt{4πt}}e^{−x^{2}/(4t)}∫_{−∞}^{∞} e^{(2xy−y^{2})/(4t)}e^{−y^{2}} Re (e^{iy}) dy
= U(x, t) ∫_{−∞}^{∞} e^{(2xy−[1+4t]y^{2})/(4t)}Re (e^{iy}) dy
= U(x, t) ∫_{−∞}^{∞} Re (e^{(2xy−[1+4t]y^{2})/(4t)}e^{iy}) dy
u(x, t) = U(x, t) Re (∫_{−∞}^{∞} e^{([2x+i4t]y−[1+4t]y^{2})/(4t)}dy)

Solution (2 of )

u(x, t) = U(x, t) Re (∫_{−∞}^{∞} e^{([2x+i4t]y−[1+4t]y^{2})/(4t)}dy)

Complete the square in the exponent.

u(x, t) = U(x, t) Re (∫_{−∞}^{∞} e^{− \frac{1+4t}{4t}(y^{2}− \frac{2x+i4t}{1+4t}y)}dy)
= U(x, t) Re (∫_{−∞}^{∞} e^{− \frac{1+4t}{4t} (y^{2}− \frac{2x+i4t}{1+4t}y+[\frac{x+i2t}{1+4t}]^{2}−[\frac{x+i2t}{1+4t}]^{2})}dy)
= U(x, t) Re (e^{\frac{1+4t}{4t}} ^{[^{\frac{x+i2t}{1+4t}}]}^{2} ∫_{−∞}^{∞} e^{− \frac{1+4t}{4t}(y− \frac{x+i2t}{1+4t})^{2}} dy)

Solution (3 of )

u(x, t) = U(x, t) Re (e^{\frac{(x+i2t)^{2}}{4t(1+4t)}} ∫_{−∞}^{∞} e^{− \frac{1+4t}{2t}(y− \frac{x+i2t}{1+4t})^{2}/2} dy)

Substitute z = \sqrt{\frac{1 + 4t}{2t}} (y − \frac{x + i2t}{1 + 4t}).

u(x, t) = U(x, t) Re (e^{\frac{(x+i2t)^{2}}{4t(1+4t)}} \sqrt{\frac{2t}{1 + 4t}} ∫_{−∞}^{∞} e^{−z^{2}/2} dz)

= U(x, t) Re (e^{\frac{(x+i2t)^{2}}{4t(1+4t)}} \sqrt{\frac{4πt}{1 + 4t}} )

Homework

▶ Read Section x.y ▶ Exercises: