75048 Mathematical modeling 1998

Similarity solutions for the heat equation

Harald Hanche-Olsen

                       hanche@math.ntnu.no

Consider the heat equation in one space dimension:

∂_{t}u = ∂_{x}^{2}u.  (1)

Note that the function (x, t) ↦ Bu(Ax, A^{2}t) solves the equation if u does. If u is a non-zero solution satisfying

u(x, t) = A^{−μ}u(Ax, A^{2}t) for all A > 0

then u is called a similarity solution of the heat equation.^{1} By selecting A = x^{−1} one arrives at a representation of a similarity solution in terms of a function of a single variable:^{2}

u(x, t) = x^{μ}v(\frac{x^{2}}{4t} ).  (2)

Now, it is a simple exercise to show that a function u defined in this way solves (1) if and only if v = v(ξ) solves

ξ^{2}v′′ + (ξ^{2} + \frac{2μ + 1}{2}ξ)v′ + \frac{μ(μ − 1)}{4}v = 0.

Two interesting special cases occur for μ ∈ {0, 1}. In these cases, the final term in the above equation drops out, and we are left with a first order separable equation for w = v′, with solution given by

∫ \frac{dw}{w} = − ∫ \frac{ξ^{2} + (μ + \frac{1}{2})ξ}{ξ^{2}} dξ = −(μ + \frac{1}{2}) ln ξ − ξ + constant

1You may verify that, if u(x, t) = B(A)u(Ax, A^{2}t) for all x > 0, t > 0, and A > 0 with B a continuous function of A, we must have B = A^{−μ} for some μ: First show that B(A_{1}A_{2}) = B(A_{1})B(A_{2}).

2I planted the extra factor 4 in the numerator because it does simplify things later. We might also select A = t^{−1/2}, leading to the representation u(x, t) = t^{μ/2}w(x^{2}/4t), which is of course essentially equaivalent. But our current choice turns out to make the calculations a bit easier.

so that

v′(ξ) = w(ξ) = constant · ξ^{−μ−1/2}e^{−ξ}.

This latter expression is easily integrated using the substitution ξ = η^{2}:

∫ ξ^{−μ−1/2}e^{−ξ} dξ = 2 ∫ η^{−2μ}e^{−η^{2}} dη.  (3)

Heating by constant surface temperature: μ = 0. When μ = 0 the above integral is easily evaluated, leading to

v(ξ) = C_{1} erf η + C_{2} = C_{1} erf \sqrt{ξ} + C_{2}.

With u(x, t) = v(x^{2}/4t) we can impose the boundary conditions v(0) = 1, v(∞) = 0, which imply C_{2} = 1 and C_{1} + C_{2} = 0. Thus

u(x, t) = erfc \frac{x}{2\sqrt{t}}

solves (1) with the initial and boundary conditions

u(x, 0) = 0, u(0, t) = 1.

Heating by constant surface heat flow: μ = 1. With μ = 1 we can evaluate (3) using partial integration:

∫ η^{−2}e^{−η^{2}} dη = −η^{−1}e^{−η^{2}} − 2 ∫ e^{−η^{2}} dη = −η^{−1}e^{−η^{2}} − 2 erf η.

Thus we get

v(ξ) = C_{1}(η^{−1}e^{−η^{2}} + 2 erf η) + C_{2} = C_{1}(ξ^{−1/2}e^{−ξ} + 2 erf \sqrt{ξ}) + C_{2}

leading to

u(x, t) = xv(\frac{x^{2}}{4t} ) = 2C_{1}(\sqrt{t}e^{−x^{2}/4t} + x erf \frac{x}{2\sqrt{t}}) + C_{2}x

If we put −C_{2} = 2C_{1} = 1, we then have the solution

u(x, t) = \sqrt{t}e^{−x^{2}/4t} − x erfc \frac{x}{2\sqrt{t}}

3 Similarity solutions for the heat equation which satisfies

u(x, 0) = 0, ∂_{x}u(0, t) = −1

Appendix: The error function. The error function erf and the complementary error function are defined by

erf η = \frac{2}{\sqrt{π}} ∫_{0}^{η} e^{−ζ^{2}} dζ, erfc η = \frac{2}{\sqrt{π}} ∫_{η}^{∞} e^{−ζ^{2}} dζ.

Note that erf η + erfc η = 1, erf 0 = erfc ∞ = 0, erf ∞ = erfc 0 = 1.