ON THE OSCILLATION BEHAVIOR OF SOLUTIONS TO THE ONE-DIMENSIONAL HEAT EQUATION

. We study the oscillation behavior of solutions to the one-dimensional heat equation and give some interesting examples. We also demonstrate a sim- ple ODE method to ﬁnd explicit solutions of the heat equation with certain particular initial conditions.


1.
Introduction. Consider the one-dimensional heat equation with initial condition u t (x, t) = u xx (x, t) , x ∈ R, t > 0 u (x, 0) = ϕ (x) , x ∈ R. (1) It is known that if ϕ (x) is a bounded continuous function defined on R, then the function given by convolution integral is a smooth solution of the heat equation on R×(0, ∞) with lim (x,t)→(x0,0) u (x, t) = ϕ (x 0 ) for any x 0 ∈ R. We all know that the solution of the heat equation is not unique even if the initial data ϕ (x) is given (unless we impose the growth condition of u (x, t) as |x| → ∞). From now on, when we say u (x, t) is "the solution" of the heat equation with u (x, 0) = ϕ (x) , x ∈ R, we always mean that it is given by the convolution integral (2). For bounded continuous initial data ϕ (x), there is a nice property of determining, for fixed x 0 , whether we have the convergence of u (x 0 , t) to 0 as t → ∞. Without loss of generality, it suffices to look at the case x 0 = 0 and we have the well-known result: [2,3,6].) Assume ϕ ∈ C 0 (R) L ∞ (R) and u (x, t) is a solution of the heat equation (1) given by formula (2). Then Remark 2. Note that, in order for ϕ (x) to satisfy the limit in (3), it does not have to decay to zero as |x| → ∞. Moreover, one can check that if lim |x|→∞ ϕ (x) = 0, then it also satisfies (5) and so we have (4).
where a, b are any two numbers.

Remark 4.
As ϕ is bounded, if it satisfies the limit in (3), we also have lim R→∞
Remark 5. It is easy to find a bounded function ϕ (x) satisfying the limit in (3), but not (5). For example, take ϕ (x) = tan −1 x, x ∈ R.
As we shall demonstrate, the above theorem fails if the initial data ϕ (θ) , θ ∈ R, is not bounded (see Remark 16 below). The purpose of this paper is to give some interesting examples (for both bounded and unbounded ϕ) of non-convergence of u (0, t) as t → ∞. When ϕ is unbounded but with bounded H (see (14)), the convergence of lim R→∞
ON THE OSCILLATION BEHAVIOR OF SOLUTIONS 4075 2. The case when ϕ is bounded. In this section, we assume ϕ ∈ C 0 (R) L ∞ (R) . We first note the following relation between the value u (0, t) and the average integral 1 2R R −R ϕ (θ) dθ : Lemma 7. Assume ϕ ∈ C 0 (R) L ∞ (R) and u (x, t) is given by (2). Then we have the identity Proof. The first identity is clear. For the second identity, let M > 0 be a constant with and lim R→0 = ϕ (0) . By integration by parts, we obtain for fixed t > 0 the following: The proof is done. By Lemma 7, we can prove the following: Corollary 8. Assume ϕ ∈ C 0 (R) L ∞ (R) and the limit lim R→∞

2R
R −R ϕ (θ) dθ does not exist. Let u (x, t) be given by (2). Then we have Remark 9. Since ϕ (x) is a bounded function and so is u (x, t) , the above four limit values all exist and are all finite. From now on, we shall use p, α, β, q to denote the four limit values in (12) respectively.
Proof. This is an easy consequence of representation formula (9). For each fixed z > 0, by (10) and Fatou's lemma, we have where we note that lim inf Hence the first inequality in (12) follows if we note that 4 √ π ∞ 0 z 2 e −z 2 dz = 1. The second inequality in (12) is due to Theorem 1. The proof of the third inequality in (12) is also due to Fatou's lemma.

2.1.
Some interesting examples for the bounded case. In the following we will give two bounded examples of non-convergence of u (0, t) as t → ∞. One has p = α < β = q and the other has p < α < β < q. We first do some general discussion.
For given initial data ϕ We see that H
Our example is very different from that in [1] and in [4].
Recall the representation formula (9), which is The key observation is the following. If the function H √ 4tz has no oscillation at all (i.e. H (x) ≡ C for some constant C), then u (0, t) is equal to the same constant C for all t ∈ (0, ∞) . On the other hand, if H √ 4tz oscillates too fast as t → ∞, say it oscillates between −1 and +1 rapidly, then similar to the Reimann Lebesgue Lemma, we may have u (0, t) → 0 as t → ∞.
Since we want to have p = α and β = q, we must choose H (x) to be oscillatory (so that p < q), but as slow as possible (so that we can maintain p = α and q = β), and it also has to satisfy the condition H (x) = O 1 x as x → ∞. Therefore, we choose the following function which has a very very slow oscillation as x → ∞ : where and by the identity (15), we can solve ϕ (θ) to get Note that the above function lies in the space ϕ ∈ C 0 (R) L ∞ (R) . By (18), the solution u of the heat equation with initial data (20) satisfies It satisfies lim t→∞ τ (t, z) = 0 for all z ∈ (0, ∞) . Hence by Lebesgue Dominated Convergence Theorem, we have (note that 4

DONG-HO TSAI AND CHIA-HSING NIEN
where we note that the profile of u (0, t) , as t → ∞, is exactly the same as the profile of H (x) = sin [log (log (x + 2))] as x → ∞. For any two numbers α < β, if we pick where ϕ (θ) is from (20), we will have By (23) and (24), we clearly have p = α, q = β, where α < β can be arbitrary values.
2.1.2. Example 2: p < α < β < q for some particular values of p, α, β, q. . The next example is to demonstrate that a strict inequality between p, α can happen, and similarly for β, q. Again, we look for H (x) first and then find the corresponding ϕ (θ) . We still need H (x) to satisfy the condition H (x) = O 1 x as x → ∞ and to oscillate (so that p < q) between two numbers, but not as slow as in the previous example (so that p increases to α and q decreases to β). With this in mind, we now choose where and by the identity (15), we can solve ϕ (θ) to get ϕ (θ) = sin (log (|θ| + 1)) + |θ| |θ| + 1 cos (log (|θ| where ϕ ∈ C 0 (R) L ∞ (R) . By (18), we have (27) Here the dot sign "·" in (27) means the inner product in R 2 . By Lebesgue Dominated Convergence Theorem, we have and Therefore, we conclude the limit and get Combined with the fact that p = −1, q = 1, the example is complete.
Remark 10. Since |ϕ| is bounded by a constant M, by (2) it is easy to derive the gradient estimate which, for each fixed x, implies where U (t) is the function in (30). Thus for each fixed x, u (x, t) also approaches to the same function U (t) as t → ∞. Moreover, due to (31), the convergence is uniform in x ∈ K for any compact set K ⊂ R.
Remark 11. This is to give a comparison. The solution of the heat equation with the initial very slow oscillation profile H (x) in (25) converges to a slightly changed profile in (30). On the other hand, the solution with the initial very very slow oscillation profile H (x) in (19) converges to an unchanged profile in (21).
Remark 12. Is there an initial data ϕ ∈ C 0 (R) L ∞ (R) so that under the heat equation u t = u xx we have lim t→∞ |u (0, t) − P (t)| = 0, where P (t) is a nonconstant 2π-periodic function? The answer is no. Since, by (18), we have where H (x) satisfies Therefore, it is impossible for u (0, t) to converge to a non-constant 2π-periodic function as t → ∞. In view of this, it seems reasonable to see that, instead of converging to A sin t + B cos t, we get convergence to A sin log √ 4t + B cos log √ 4t for some constants A, B. On the other hand, if we allow the initial data ϕ to be unbounded, then the function where λ, A, B are arbitrary constants, is a time-periodic solution of the heat equation with The function u (0, t) is a non-constant 2π-periodic function and so is u (x, t) for each fixed x.
3. The case when ϕ is unbounded. When the initial data ϕ (x) is unbounded (in this paper we always assume ϕ (x) is, at least, a continuous function), Theorem 1 fails. In this section, we assume that the unbounded function ϕ (x) satisfies the following growth condition |ϕ (x)| = o e λ|x| 2 as |x| → ∞, for any constant λ > 0.
Next, in case H (x) is a bounded function on x ∈ (0, ∞) , the classical Fatou's lemma can still be applied and the estimate (13) is still valid. Thus we have proved the first inequality in (34). The proof of the third inequality is similar.

3.1.
Some interesting examples for the unbounded case. Again, we construct some interesting examples. The examples given below all have the initial condition ϕ (x) satisfying the growth condition (33) and we can evaluate the four limit values (call them p, α, β, q) in (34) directly. They all satisfy the inequality (34).
We first note that if we add an unbounded odd function, say x cos x, to the initial data in the two examples in Section 2.1, it will not affect the values of and u (0, t) . Hence the oscillation behavior of the two examples in Section 2.1 can also occur in the unbounded case. Thus, the following two examples can be attained: Example 1:: p = α < β = q for arbitrary values of α, β. Example 2:: p < α < β < q for some particular values of p, α, β, q.
3.1.1. Example 3: p = α − λ < α = β < β + λ = q for arbitrary values of α, λ, where λ > 0.. For arbitrary numbers α, λ > 0, we take ϕ (x) = α + λx sin x, x ∈ R, which is an even unbounded function. We have which gives p = α − λ, q = α + λ. The solution u (x, t) of the heat equation with initial data ϕ (x) = α + λx sin x is given by which gives lim t→∞ u (0, t) = α. To see the solution formula (36), it suffices to explain how to derive the solution v (x, t) of the heat equation (given by (2)) with initial data x sin x. Instead of using the integral formula (2), we use a little trick here. Note that (v + v xx ) (x, 0) = 2 cos x. As v (x, t) + v xx (x, t) is also a solution of the heat equation, we have v + v xx = w, where w (x, t) is a solution of the heat equation with w (x, 0) = 2 cos x, i.e. w (x, t) = 2e −t cos x. Hence we get and derive v (x, t) = e −t x sin x + 2te −t cos x. The solution (36) is actually defined for all (x, t) ∈ R 2 and it satisfies lim t→∞ u (x, t) = α for all x ∈ R. The convergence is uniform on any compact subset of x. The example is finished.
Remark 17. In the above example, the limit lim t→∞ u (0, t) = α happens to be the average value of p = α − λ and q = α + λ. Here we give a similar example with lim t→∞ u (0, t) = α, but the average value of p and q can be as large as we want. We choose   (20) and choose the initial data as We have and it is not difficult to see that there exists a sequence x k → ∞ so that sin [log (log (x k + 2))] → 1 and cos x k → −1 as k → ∞. Thus we have q = β + λ. Similarly, there exists a sequence x j → ∞ so that sin [log (log (x j + 2))] → −1 and cos x j → 1 as j → ∞. Thus we have p = α − λ. On the other hand, by (24) and (36), we also have Hence we conclude that p = α − λ < α < β < β + λ = q. The example is finished.
3.1.4. Example 6: −∞ = p < α < β < q = ∞ for arbitrary values of α, β.. In this example, we use the idea of time-periodic solutions of the heat equation to help us. Let v (x, t) = e x cos (2t + x) and w (x, t) = e x sin (2t + x) . Both are timeperiodic solutions of the heat equation u t = u xx with initial conditions e x cos x and e x sin x. Hence the function where λ, A, B are constants to be chosen later on, is also a time-periodic solution of the heat equation with The above gives Next, we evaluate lim sup and similarly lim inf The example is completed.
Remark 18. Using Fourier series expansion, it can be shown that, up to parabolic scaling, any 2π time-periodic solution of the heat equation is an infinite superposition of the six solutions: We call them the generating functions for the 2π time-periodic solutions of the heat equation.
3.1.5. Example 7: −∞ = p = α < β = q = ∞.. Similar to the previous example, now we use the idea of space-time periodic solutions of the heat equation to help us. Consider the function One can check that it is a solution of the heat equation satisfying the space-time periodic condition u (x + π, t + π) = u (x, t) for all (x, t) ∈ R 2 . We have and we can easily see that p = α = −∞ and q = β = ∞. The example is finished.
3.1.6. Some remarks about the unbounded case. We have given several interesting examples for values of p, α, β, q. However, there are some situations which we think they probably cannot be attained. The first situation is that for arbitrary three finite numbers p < α = β < q, we probably cannot find an unbounded initial data ϕ (x) so that these three values are attained. We feel that there should be some balance between p and q in order to achieve the limit value α (see Remark 17 for an example of balance). For example, if p = −ε (ε > 0 is small), α = β = 0 and q 0 is sufficiently large, we think it cannot be attained. The second situation is that for arbitrary four finite numbers p < α < β < q, we think it is impossible to find an initial condition ϕ (x) (bounded or unbounded) so that these four values are attained (however, by the Example 4, we can prescribe either p, α, β or α, β, q).

4.
Using ODE method to solve the heat equation. In the example in Section 3.1.1, we use an ODE method to find explicit solution of the heat equation with initial data x sin x. Here we give a more general discussion on it and look at some interesting examples.
Let ϕ (x) ∈ C 2 (R) be a function such that |ϕ (x)| , |ϕ (x)| and |ϕ (x)| all satisfy the growth condition (33). Let u (x, t) be the solution of the heat equation with initial data ϕ (x) , x ∈ (−∞, ∞). Assume ϕ (x) satisfies an ODE of the form for some constants a = 0, b, c and some function h (x) ∈ C 0 (R) . Let v (x, t) be the solution of the heat equation with initial data v (x, 0) = h (x) , x ∈ (−∞, ∞) . By the growth condition of |ϕ (x)| , |ϕ (x)| and |ϕ (x)| , we see that au xx (x, t) + bu x (x, t)+cu (x, t) is also a solution of the heat equation with initial condition given by Hence, by uniqueness, we have As a consequence, we see that u (x, t) must satisfy the first order PDE (with constant coefficients) (45) can be solved explicitly if we know the function v (x, t) explicitly and know how to integrate the function given below in (46).
More precisely, one can convert (45) into an ODE by change of variables. If b = 0, we let z = x, w = bt − ax, which is a change of variables. Denote the function corresponding to u (x, t) and v (x, t) byũ (z, w) andṽ (z, w) . The equation forũ (z, w) is bũ z (z, w) + cũ (z, w) =ṽ (z, w) and its general solution is where C (w) is an arbitrary function of w. By (46), one can find the general solution u (x, t) of the equation in (45). Finally, we choose suitable C (w) so that the initial condition u (x, 0) = ϕ (x) is satisfied. This solution u (x, t) will be the explicit solution of the heat equation with initial data ϕ (x) . If b = 0, then (45) is already an ODE and we know how to solve it. The above ODE method can be applied to the following initial conditions: We know the solutions with initial condition x and x 2 are given respectively by u (x, t) = x and u (x, t) = x 2 + 2t. For u (x, 0) = x 3 , we see that the function 1 6 u xx (x, t) is a solution with initial condition 1 6 u xx (x, 0) = x. Hence we have u t (x, t) = 6x for all (x, t) and get u (x, t) = 6xt+C (x) for some integration function C (x) . By u (x, 0) = x 3 , we get u (x, t) = 6xt + x 3 .
Keep going and one can obtain the solutions: Each u (x, t) is a polynomial solution with homogeneous terms (view t as x 2 ).