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Home , Heat equation

CHAPTER 6

The Equation

We introduce several PDE techniques in the context of the :

The is the heart of the theory of infinite domain prob- lems. The fundamental solution also has to do with bounded domains, when we introduce Green’s functions later.

The Maximum Principle applies to the heat equation in domains bounded in space and . It is an important property of parabolic equations used to deduce a variety of results such as uniqueness of solutions, comparison principles. (E.g., if boundary conditions are changed in a way that suggests intuitively the resulting temperature should be smaller, this can be proved using the maximum principle.)

The Method works analogously to the equation, except that the physical (heat) energy is less interesting than a mathematical energy, which typically decays. As for the wave equation, this leads to straightfor- ward uniqueness results. It is also useful for obtaining estimates on solutions that are part of the existence and regularity theory for parabolic equations.

Initial Boundary Value Problems. We will spend some time describing ex- plicit solutions, expressed as infinite series of functions, of the heat equation plus initial and boundary conditions. There is a general technique frequently referred to as , or as expansions. The development of this technique leads us to an analysis of eigenvalue problems for ordinary and partial differential equations, and to the analysis of .

6.1. The Fundamental Solution

To start with, we consider the heat equation in one space variable, plus time. In this section, we derive the fundamental solution and show how it is used to solve the Cauchy problem:

ut = kuxx, |x| < ∞, t > 0 u(x, 0) = g(x), |x| < ∞

59 60 6. THE HEAT EQUATION

The heat equation has a scale invariance property that is analogous to scale invariance of the wave equation or scalar conservation laws, but the scaling is different.

Let a > 0 be a constant. Under the scaling x → ax, t → a2t the heat equation is unchanged. More precisely, if we introduce the change of variables: t = a2t, x = ax, then the heat equation becomes

ut = kuxx This scale invariance suggests that we seek solutions v depending on the 2 similarity variable x , or on √x . However, there is a property of the heat t t equation we would like to preserve in our similarity solution, that of con- servation of energy. Suppose u is a solution of the heat equation with the R ∞ property that | −∞ u(x, 0) dx| < ∞, and ux(x.t) → 0 as x → ±∞. Then, integrating the PDE, we find d Z ∞ u(x, t) dx = 0, dt −∞ so that the total heat energy is conserved: Z ∞ (1.1) u(x, t) dx = constant. −∞ However, ∞ ∞ Z  x  1 Z v √ dx = t 2 v(y) dy. −∞ t −∞ − 1 This suggests we should scale the v by t 2 : 1  x  (1.2) u(x, t) = √ v √ t t With this scaling, heat is conserved in the sense of (1.1).

Substituting (1.2) into the PDE leads to an ODE for v = v(y), with non- constant coefficients: 1 1 (1.3) kv00(y) + yv0(y) + v(y) = 0 2 2

Since this is a second order equation, we should have two independent solu- tions. First rewrite the ODE as 1 kv00(y) + (yv(y))0 = 0. 2 Thus, 1 kv0(y) + yv(y) = constant. 2 6.1. THE FUNDAMENTAL SOLUTION 61

Since we are really only seeking one solution, it is convenient to set the constant to zero, and write the solution of the homogenous equation:

2 − y v(y) = Ae 4k .

Converting back to (x, t), with y = √x , we obtain the similarity solution t

2 1 − x (1.4) u(x, t) = A√ e 4kt t

Usually, we choose a particular value of A so that the constant in (1.1) is unity; for this choice of constant, we have the fundamental solution of the heat equation: x2 1 − Φ(x, t) = √ e 4kt . 4πkt

n In higher , x ∈ R , the fundamental solution takes a similar form:

1 2 Φ(x, t) = e−|x| /4kt (4πkt)n/2 = φ(r, t), r = |x|, where φ(r, t) = 1 e−r2/4kt. Then u(x, t) = Φ(x, t) satisfies (4πkt)n/2

ut = k∆u, ∆ = div grad (the Laplacian with respect to x).

n−1 Exercise. Show that ∆φ(r, t) = φrr + r φr. Hence,  n − 1  φ = k φ + φ . t rr r r

Properties of the Fundamental Solution Φ(x, t)

1. Φ(x, t) > 0 for all x ∈ R, t > 0 2. Φ is C∞ in (x, t), t > 0 R 3. n Φ(x, t)dx = 1 for all t > 0. R Properties 1 and 2 are obvious; here is the proof of property 3 for n = 1.

Z ∞ Z ∞ 2 1 − x Φ(x, t)dx = √ e 4kt dx −∞ 4πkt −∞ 62 6. THE HEAT EQUATION

∞ Z 2 x dx √1 −y √ √ = π e dy (let y = , dy = ) −∞ 4kt 4kt = 1.

y

1 4π kt

y = Φ (x,t)

0 x

Figure 6.1. Graph of the fundamental solution for the heat equation, with t > 0. .

Therefore, Φ(x, t) is a probability distribution for each t > 0, with interesting dependence on t in the limits t → ∞ and t → 0 :

The area under the graph is 1 for all t > 0, yet as t → ∞, maxx Φ(x, t) → 0; the tail spreads out to maintain R Φ = 1

As t → 0 the maximum (at x = 0) blows up like √1 , but the integral remains t constant. We also observe Φ(x, t) → 0 for x 6= 0, as t → 0 + .

6.2. The Cauchy Problem for the Heat Equation

As for the wave equation, the Cauchy problem is the pure initial value problem, here stated in the one-dimensional case, n = 1:

ut = kuxx, |x| < ∞, t > 0 (a) (2.5) u(x, 0) = g(x), |x| < ∞ (b) 6.2. THE CAUCHY PROBLEM FOR THE HEAT EQUATION 63

Recall that the fundamental solution

2 1 − x Φ(x, t) = √ e 4kt 4πkt satisfies (2.5(a)) for t > 0.

Now Φ(x − y, t) is a solution of (2.5(a)) for all y, by translation invariance: x → x − y does not change the heat equation. Thus, Φ(x − y, t)g(y) is also a solution of (2.5(a)). For later reference, we note that the heat equation is invariant under time translation also.

By linearity and homogeneity of the PDE, we can also take linear combina- tions of solutions. This suggests that Z ∞ (2.6) u(x, t) = Φ(x − y, t)g(y)dy −∞ should also be a solution. Moreover, properties of Φ suggest that as t → 0+, u(x, t) → g(x), since Φ(x − y, t) collapses to zero away from y = x, and blows up at y = x in such a way (i.e., preserving R Φ = 1) that the initial condition is satisfied in the sense u(x, t) → g(x) as t → 0 + .

It is straightforward to check that the integrals for u, ut, uxx all converge provided g ∈ C(R) is bounded. Then Z ∞ ∂Φ Z ∞ ∂2Φ ut = (x − y, t)g(y)dy; uxx = 2 (x − y, t)g(y)dy, −∞ ∂t −∞ ∂x so that u satisfies the PDE for t > 0.

It is more complicated to check that the initial condition is satisfied. We need to show u(x, 0) = g(x). But t = 0 is a singular point for Φ : Φ(x, t) is not defined at t = 0. To get an idea of why limt→0+ u(x, t) = g(x), let’s fix x.

Then, for δ > 0, R ∞ −∞ Φ(x − y, t)g(y)dy R R = (x−y)<δ Φ(x − y, t)g(y)dy + |x−y|>δ Φ(x − y, t)g(y)dy R ≈ |x−y|<δ Φ(x − y, t)g(x)dy By continuity, g(y) ≈ g(x) for y near x, which explains how the first integral is approximately the final line. The second integral approaches zero as t → 0+, because Φ → 0 uniformly, and exponentially, away from y = x as t → 0+. 64 6. THE HEAT EQUATION

Theorem: Let g ∈ C(R) be bounded, and let u(x, t) be given be the formula (2.6). Then

1. u is C∞ in (x, t) for t > 0;

2. u satisfies the heat equation

ut = kuxx, x ∈ R, t > 0;

3. lim u(x, t) = g(x0) for all x0 ∈ R. (x, t) → (x0, 0) t > 0

Proof:

Property 1 follows because Φ is C∞ for t > 0, and of Φ all decay exponentially as |x| → ∞, so the integrals converge.

Property 2 follows from Φt = kΦxx, t > 0.

To prove property 3, we look at the difference |u(x, t) − g(x0)|, estimate and show that the pieces we get behave as we expect, i.e., as in the rough argument preceeding the proof. R Let  > 0. (This measures |u(x, t)−g(x0)|.) Then, since Φ dx = 1, we have Z ∞

(2.7) |u(x, t) − g(x0)| = Φ(x − y, t)(g(y) − g(x0)) dy , −∞

Let δ > 0 (we choose δ below), and break up the integrals in (2.7): (2.8) Z Z

|u(x, t)−g(x0)| ≤ Φ(x−y, t)(g(y)−g(x0))dy − Φ(x−y, t)(g(y)−g(y0))dy |x0−y|<δ |x0−y|≥δ Now we use δ two ways to show the two integrals are small:

1st: g(y) ∼ g(x0). For y near x, Φ(x − y, t) blows up as t → 0, but R Φ(x − y, t) dy is bounded uniformly by 1.

2nd: R Φ(x − y, t) dy → 0 as t → 0, provided |x − x | < δ , while |x0−y|≥δ 0 2 g(y) is bounded.

We write the right hand side of (2.8) as

Iδ + Jδ

Choose δ > 0 so that |g(y) − g(x0)| <  for |y − x0| < δ (by continuity of g at x0). 6.2. THE CAUCHY PROBLEM FOR THE HEAT EQUATION 65

Then Z Iδ ≤ Φ(x − y, t)|g(y) − g(x0)|dy |x0−y|<δ Z ≤  Φ(x − y, t)dy ≤ . |x0−y|<δ The second integral is somewhat trickier: Since g is bounded, there is K > 0 such that |g(y)| ≤ K, for all y. Thus, Z Jδ ≤ Φ(x − y, t)|g(y) − g(x0)|dy |x0−y|>δ Z 2 1 − (x−y) ≤ 2K √ e 4kt dy |x0−y|>δ 4πkt

δ Here is the second use of δ : Consider x satisfying |x − x0| < 2 . Then δ |x − y| > 2 in the range of integration |x0 − y| > δ. But this is not a good enough estimate of the exponential, because we would still be left with an integral over an infinite interval of a small but positive quantity. So, we 1 need to observe that in the region of integration, |x − y| ≥ 2 |y − x0|. Then 2 K R 1 − (x0−y) Jδ ≤ √ √ e 16kt dy πk |x0−y|>δ t R −z2 ≤ C |z|≥ √δ e dz < , 4 kt for t > 0 sufficiently small.

δ Thus, |u(x, t) − g(x0)| < 2 for |x − x0| < 2 , t > 0 sufficiently small. This proves property 3.

Remarks

1. limt→0+ Φ(x, t) is not a function in the usual sense, but is a distribution or generalized function called the δ(x): Z δ(x − y)g(y) dy = g(x), R where the integral is a notational convenience. One way to interpret the delta function and the integral is to recognize the delta function as a measure that places unit at x = 0 and zero mass at each x 6= 0. R ∞ 2. −∞ Φ(x − y, t) g(y) dy is the of the function Φ(·, t) with g More generally, for integrable functions φ, ψ on R the convolution product of φ and ψ is also a function φ ∗ ψ defined by Z ∞ φ ∗ ψ(x) = φ(x − y)ψ(y) dy. −∞ 66 6. THE HEAT EQUATION

6.2.1. Using the Fundamental Solution to Solve Quarter-plane Problems. :

Consider the so-called quarter-plane problem:

1) ut = kuxx, x > 0, t > 0

2) u(0, t) = 0, t > 0. (Homogeneous Dirichlet boundary condition.)

3) u(x, 0) = g(x), x > 0

As for the wave equation, we reflect the initial data so the solution satisfies the boundary conditions:

Letg ˜(x) be the odd extension of g(x).

 g(x), x > 0 g˜(x) = −g(−x), x < 0 and define Z ∞ (2.9) u(x, t) = Φ(x − y, t)˜g(y)dy −∞ Then

ut = kuxx,

u(x, 0) =g ˜(x) = g(x), x > 0 Z ∞ u(0, t) = Φ(−y, t)˜g(y)dy = 0, t > 0, −∞ since Φ(y, t) is an even function of y, andg ˜ is an odd function.

It is straightforward to check directly that u(x, t) is an odd function of x ∈ R. That is, the in the initial data is carried through to the same symmetry in the solution.

Now replaceg ˜(y) with g(y) usingg ˜(y) = −g(−y), y < 0. Then

Z ∞ Z 0 u(x, t) = Φ(x − y, t)g(y)dy + Φ(x − y, t)(−g(−y))dy 0 −∞ Z ∞ Z ∞ = Φ(x − y, t)g(y)dy − Φ(x + y, t)g(y)dy 0 0 Z ∞ = (Φ(x − y, t) − Φ(x + y, t))g(y)dy 0 6.3. THE ENERGY METHOD. 67

With a homogeneous Neumann boundary condition instead, we extend the initial data to be even:

ut = kuxx, x > 0, t > 0

ux(0, t) = 0 (Homogeneous Neumann Condition) u(x, 0) = g(x), x > 0

Solution: Extend g using the even extension, so that the first is zero at x = 0. Z ∞ Then u(x, t) = (Φ(x − y, t) + Φ(x + y, t)) g(y) dy. 0 To check the boundary condition: R ∞ ux(0, t) = 0 (Φx(−y, t) + Φx(y, t) g(y) dy.

Φ(y, t) is even in y so Φx(y, t) is odd. Thus, ux(0, t) = 0.

6.3. The Energy Method.

Consider

ut = k uxx, a ≤ x ≤ b, t > 0. multiplying by u and integrating over [a, b], Z b Z b uut dx = k uuxxdx. a a Therefore, Z b Z b d 1 2 b 2 u dx = kuux|a − k uxdx. dt a 2 a R b 1 2 Thus, the energy integral (not the heat energy) E(t) = a 2 u (x, t)dx is b decreasing in time if uux|a ≤ 0.

For example, if either u or ux is zero at each end point x = a, x = b, then the energy integral decreases in t.

In this case, we have the important comparison to the initial data: E(t) ≤ E(0), t > 0.

I.e., Z b Z b (3.10) u2(x, t)dx ≤ u2(x, 0)dx, t > 0. a a 68 6. THE HEAT EQUATION

6.3.1. Using the Energy Method to prove Uniqueness. Consider the initial   ut = kuxx + f(x, t), 0 < x < l, t > 0    (IBVP) u(0, t) = g(t), t > 0  u(L, t) = h(t), t > 0    u(x, 0) = φ(x), 0 < x < L where f, g, h, φ are given functions.

The energy inequality works in any number of dimensions (will involve di- vergence theorem and Green’s identities instead of integration by parts).

2 Theorem. If u1, u2 solve (IBVP) and are C functions, then u1 = u2 everywhere.

Proof: Let u = u1 − u2

Then u satisfies (IBVP) with zero data: f ≡ g ≡ h ≡ φ ≡ 0. But we know that E(t) is decreasing, from the energy inequality

Z L Z L 0 ≤ (u(x, t))2dx ≤ u(x, 0)2 dx = 0 0 0

Therefore, u ≡ 0, so that u1 ≡ u2.

Remark. In general, the existence of solutions is harder to establish than uniqueness.

6.3.2. The Energy Principle in Higher Dimensions: Consider a n bounded open subset U of R , and the initial boundary value problem

ut = k∆u, x ∈ U t > 0

u(x, t) = 0, x ∈ ∂U, t > 0

u(x, 0) = φ(x), x ∈ U

We begin by defining the energy integral as in one : 1 Z (3.11) E(t) = u(x, t)2 dx. 2 U Then 6.4. THE MAXIMUM PRINCIPLE 69 Z Z 0 E (t) = uut dx = k u∆u dx U U Z Z = k u∇u · n dS − k ∇u · ∇u dx ∂u U R 2 = −k U |∇u| dx ≤ 0.

Exercise: Prove the energy inequality for ut = ∇.(k(x, u)∇u) where n k(x, u) ∈ R is a given positive function. The only difference is that k gets moved under the integral sign. Does this enable us to prove uniqueness of solutions for the quasilinear heat equation?

6.4. The Maximum Principle

n For T > 0, and U ⊂ R , we use the notation UT = U × (0,T ]. Note that UT includes the top U × {t = T }. The following theorem, the Maximum Principle, states that the maximum of any (smooth) solution of the heat equation occurs either initially (at t = 0), or on the boundary of the domain. These parts of the boundary of UT are known as the parabolic boundary ΓT :

ΓT = U T − UT = (∂U × [0,T ]) ∪ (U × {t = 0}). Solutions of the heat equation should have two spatial derivatives and one time derivative, so we define the appropriate space of functions on UT : 2 2 C1 (UT ) = {u = u(x, t): u, ut,Dxu, Dxu ∈ C(UT )}.

In order to compare values of u in UT with values on the boundary, we require in the theorem that u should be continuous on U T .

2 Theorem. The Maximum Principle. Let u ∈ C(UT ) ∩ C1 (UT ) satisfy

ut = k∆u, (x, t) ∈ UT . Then max u(x, t) = max u(x, t). U T ΓT

Remarks:

1. Suppose u has a local maximum at (x, t) ∈ UT . Then ut = 0 = ux = 0, ∆u ≤ 0. If ∆u < 0 at (x, t), then the pde gives us a contradiction:

0 = ut = k∆u < 0 Although this is not a proof, since we have to handle the degenerate case in which ∆u = 0, it has the main idea; the proof merely modifies u to remove a possibly degenerate maximum. 70 6. THE HEAT EQUATION

2. U T is closed, so u achieves it’s maximum somewhere in U T : there is an (x0, t0) such that u(x0, t0) = max u(x, t). U T Proof: Let M = max u(x, t) ΓT

t 6 T

Γ Γ UT

Γ - x 0 L

Goal: Prove that U(x, t) ≤ M for all (x, t) ∈ UT To deal with the possibility ∆u = 0 at a maximum, we perturb u a bit:

Let v(x, t) = u(x, t) + |x|2,  > 0. Then

vt − k∆v = ut − k∆u − 2 k n  < 0 (4.12) | {z } 0

Now suppose v has a local maximum in UT , at P0 = (x0, t0) with t0 < T. Then vt = vx = 0 at P0, and ∆v ≤ 0 at P0. But this contradicts (4.12), so v cannot have a max in the interior of UT .

Now suppose v has a maximum on the line t = T , at P1 = (x1, t = T ). Then 1 vx = 0, ∆vxx ≤ 0 and vt ≥ 0 at P1. Now we have vt − k∆v ≥ 0, again contradicting (4.12).

1 The final inequality is easily proved by contradiction - vt < 0 at P1, then v(x1, t) > v(x1,T ) for t < T close to t − T, contradicting the assumption that v has a maximum at P1. 6.5. DUHAMEL’S PRINCIPLE FOR THE INHOMOGENEOUS HEAT EQUATION 71

Therefore maximum of v on U T occurs on ΓT . I.e., v(x, t) ≤ max v(y, t) (y,t)∈U T for all (x, t) ∈ U T . We have proved 2 2 u(x, t) + |x| ≤ max (u(y, t) + |y| ) ≤ max u + C for all (x, t) ∈ U T , (y,t)∈ΓT ΓT where C = max |y|2. Thus, y∈U u(x, t) ≤ M + (C − |x|2)

≤ M + C

Since  > 0 is arbitrary, we have u(x, t) ≤ M for all (x, t) ∈ U T . Remarks:

1. The weak maximum principle is easy to prove. The related strong maxi- mum principle is somewhat harder to prove. The strong maximum principle states that, provided U is connected, then the maximum of u is achieved only on the parabolic boundary, unless u is constant throughout U T . 2. By applying the maximum principle to −u, which also satisfies the con- ditions of the Theorem, we see that there is a corresponding minimum prin- ciple: min u(x, t) = min u(x, t). U T ΓT

6.5. Duhamel’s Principle for the Inhomogeneous Heat Equation

Consider the example of the heat equation on the whole real line, with a heat source or sink represented by the density function f(x, t): ( ut = kuxx + f(x, t), |x| < ∞, t > 0 (P) u(x, 0) = 0 |x| < ∞

The fundamental solution Φ(x, t) of the homogeneous heat equation satisfies ut = kuxx, for t > 0. But then, for any y, s > 0 the shifted function Φ(x − y, t−s) satisfies the equation for t > s, and we can multiply by an amplitude f(y, s), so that we have a collection of solutions Φ(x − y, t − s)f(y, s) of the heat equation, with y ∈ R and t > s. Summing (i.e., integrating) these solutions over y ∈ R, with s fixed, we have that Z ∞ u˜(x, t; s) = Φ(x − y, t − s)f(y, s) dy is a solution for all t > s, −∞ satisfyingu ˜(x, t = s; s) = f(x, s) 72 6. THE HEAT EQUATION

Now the idea is to integrateu ˜(x, t; s) with respect to s from 0 to t :

Z t Define u(x, t) = u˜(x, t, s)ds. Then 0 Z t ut − kuxx =u ˜(x, t; t) + (˜ut(x, t; s) − ku˜xx(x, t; s))ds 0 = f(x, t)

Note that this calculation is slightly misleading, because we have differen- tiated under the integral sign, when Φ(x − y, t − s) has a singularity at x = y, t = s, on the boundary of the domain of integration. However, this singularity can be handled with the appropriate limit, and the result is the same. (See Evans [PDEs], page 50 for the details.)

Finally, we observe u(x, 0) = 0 − ∞ < x < ∞.