I have a question about the uniqueness argument. Usually if I write the heat equation on $[a,b]$ I can see how maximum principle is used to prove its solution uniqueness, however, when the domain is changed to $[a,\infty]$, how to apply the same max principle? The domain is now unbounded on one side, so I can't see that as the same case as with bounded boundary $b$.

  • $\begingroup$ my domain is $(t,x)=[0,T]*[a,\infty)$ $\endgroup$ – Medan Aug 4 '16 at 17:19

Uniqueness for the heat equation fails on unbounded domains, without some additional growth restrictions on the solution.

Here is an example:

$$u(x,t) = \sum_{n=0}^\infty \frac{g^{(n)}(t)}{(2n)!} x^{2n}$$


$$g(t) = \begin{cases} e^{-\frac{1}{t^2}},&\text{if } t>0\\ 0,&\text{if } t\leq 0.\end{cases}$$

The function $u$ satisfies the heat equation $u_t = u_{xx}$ on the entire real line $\mathbb{R}\times(0,\infty)$, and $u(x,0)=0$ (the zero function is also a solution). There are in fact infinitely many solutions constructed this way, and the construction can easily be adapted to a half line.

Notice the function $u$ grows exponentially fast as $x \to \pm \infty$ (compare it to the Taylor series for $\cosh(x)$ for example). This corresponds to an infinite amount of heat propagating in from $x=\pm \infty$ in finite time, and for this reason is considered a "non-physical" solution of the heat equation. It is possible to use the maximum principle to prove that there is a unique solution satisfying the growth estimate

$$u(x,t) \leq A e^{ax^2}$$

where $A,a>0$ are arbitrary constants. You can find the proof of this in Chapter 2 of L.C. Evans PDE textbook (Section 2.3.3).

The proof is a fair bit easier if we assume the solutions are bounded. I'll sketch the proof here. Let $u(x,t)$ be a bounded solution of the heat equation $u_t = u_{xx}$ on $\mathbb{R}\times (0,\infty)$ satisfying $u(x,0)=0$. By bounded I mean that $|u(x,t)|\leq C$ for all $(x,t)$, where $C>0$ is some constant.

We want to show that $u(x,t)=0$, as this is the physically reasonable solution with zero initial conditions. The difficulty with using the maximum principle directly is that $u$ may not attain its maximum value on the unbounded domain. So we have to find some way of replacing the entire real line with a bounded interval.

Along these lines, consider the following: Let $\varepsilon>0$, $a \in \mathbb{R}$ and define

$$v(x,t) = u(x,t) - (2t + (x-a)^2)\varepsilon$$

Check that $v(x,t)$ is again a solution of the heat equation $v_t=v_{xx}$. Now consider a very big rectangle $R=[a-N,a+N] \times [0,T]$ where $N>0$ is large. By the usual maximum principle, $v$ must attain its maximum value over $R$ on the sides or base of the rectangle. On the sides we have

$$v(x,t) \leq C-(N-a)^2\varepsilon \leq 0$$

provided we choose $N$ sufficiently large. On the base we have $u(x,0)=0$ so

$$v(x,t) \leq -(x-a)^2\varepsilon \leq 0.$$

Therefore $v\leq 0$ throughout the rectangle $R$. In particular,

$$v(a,t) = u(a,t) - 2t\varepsilon \leq 0$$

and so $u(a,t) \leq 2t\varepsilon$ for all $t \in [0,T]$. We can send $\varepsilon \to 0^+$ and $T\to \infty$ to find that $u(a,t)\leq 0$ for all $t\geq 0$. Since $a\in \mathbb{R}$ was an arbitrary real number, we have that $u\leq 0$ everywhere. Since $-u$ satisfies the same heat equation, we also have $-u \leq 0$, so $u=0$.

Thus, the only bounded solution of the heat equation on the entire real line with zero initial data is the zero function $u=0$. From this, you can easily show that bounded solutions with the same initial data are unique (use linearity of the PDE).

The proof from Evans PDE book using the exponential growth estimate is similar, except that the definition of $v$ is more clever. All of these arguments can be easily adapted to half lines as well.

  • $\begingroup$ thanks for your answer, however, assume I have this growth restriction, then could I use the max principle at all? I just don't see how it can possibly be applied since the domain is not bounded from one end. That is, if I do impose the condition, how would I go with proving uniqueness? $\endgroup$ – Medan Aug 4 '16 at 19:16
  • $\begingroup$ The proof for the more general growth condition is the same as what I wrote, except you will need to define $$v(x,t) = u(x,t) - \frac{\varepsilon}{(T+\mu -t)^{1/2}}e^{\frac{(x-a)^2}{4(T+\mu-t)}}.$$ You can check again that $v$ also satisfies the heat equation. $\endgroup$ – Jeff Aug 4 '16 at 19:21
  • $\begingroup$ Great counterexample! $\endgroup$ – Shalop May 5 '18 at 10:48

If $u : [0,T] \times [a, \infty) \to \Bbb R$ is the solution of

$$\partial_t u - \Delta u = 0 \\ u(0,x) = f(x) \quad \forall x \in [a, \infty) \\ u(t,a) = g(t) \quad \forall t \in [0,T] ,$$

then a weak maximum principle and an associated uniqueness theorem may still be given, provided that one adds some growth condition on $u(t, \cdot)$. Namely, we require that there exist $M, c > 0$ such that $|u(t,x)| \le M \Bbb e ^{c x^2} \ \forall t,x$.

Let $D$ be the "partial boundary" $\big( \{0\} \times [a, \infty) \big) \cup \big( [0,T] \times \{a\} \big)$. The weak maximum principle in this case is: if $\partial_t u - \Delta u \le 0$ with $u(0,x) = f(x) \ \forall x \in [a, \infty)$ and $u(t,a) = g(t) \ \forall t \in [0,T]$, and $u(t,x) \le M \Bbb e ^{c x^2} \ \forall t,x$, then

$$\sup \limits _{(t,x) \in [0,T] \times [a, \infty)} u = \sup \limits _{(t,x) \in D} u = \max (\sup f, \sup g) .$$

Armed with this maximum principle, assume $v$ and $w$ to satisfy the equation given above, with those boundary conditions and the growth conditions $v(t,x) \le M \Bbb e ^{c x^2}$ and $w(t,x) \le N \Bbb e ^{d x^2}$. Then $v - w$ satisfies the problem

$$\partial_t u - \Delta u = 0 \\ u(0,x) = 0 \quad \forall x \in [a, \infty) \\ u(t,a) = 0 \quad \forall t \in [0,T] $$


$$(v-w) (t,x) \le |(v-w) (t,x)| \le |v(t,x)| + |w(t,x)| \le M \Bbb e ^{c x^2} + N \Bbb e ^{d x^2} \le \max (M,N) \ \Bbb e ^{\max (c,d) x^2} .$$

This means that $v-w$ satisfies the same type of growth condition as required by the maximum principle, so you may apply the maximum principle to $v-w$ to obtain that

$$\sup \limits _{(t,x) \in [0,T] \times [a, \infty)} (v-w) = \sup \limits _{(t,x) \in D} (v-w) = \max (\sup 0, \sup 0) = 0 ,$$

so $v-w \le 0$.

Do the same thing for $w-v$ to obtain that $w-v \le 0$. Therefore, $v=w$, so the solution is unique.

This is adapted from theorems 5 and 12 found in Xinwei Yu's lecture notes at the University of Alberta (in them, Yu works with $\Bbb R$ instead of $[a, \infty)$, but with a bit of dilligence and intelligence they may be adapted into what I have written above).


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