MATHEMATICS

GU LIAN-KUN

ON THE BEHAVIOR OF THE SOLUTION OF THE STEFAN PROBLEM UNDER UNBOUNDED INCREASE OF TIME

(Presented by Academician I. G. Petrovskii, 2 I 1961)

In the present note the behavior of the solution of the Stefan problem with a Cauchy condition for a parabolic equation as \(t \to \infty\) is studied. For the heat-conduction equation, the asymptotic behavior as \(t \to \infty\) of the solution of the Stefan problem with boundary conditions was considered in \((^1)\).

We formulate the Stefan problem with a Cauchy condition as follows: find functions \(s(t)\) and \(u_i(x,t)\) such that \(u_i(x,t)\) \((i=1,2)\) are bounded and continuous in the domains \(\overline{D_i}\) \((D_1=\{x<s(t),\, t>0\}\) and \(D_2=\{x>s(t),\, t>0\})\) everywhere, except possibly at the point \((0,0)\), have derivatives \(\partial u_i/\partial x\), \(\partial^2u_i/\partial x^2\), \(\partial u_i/\partial t\) in \(D_i\), and satisfy the equation

\[ L_i(u_i)\equiv \frac{\partial^2 u_i}{\partial x^2} +B_i(x,t)\frac{\partial u_i}{\partial x} +C_i(x,t)u_i -A_i(x,t)\frac{\partial u_i}{\partial t}=0 \quad \text{in } D_i \tag{1} \]

and the conditions

\[ u_1(x,0)=\varphi_1(x)\leqslant 0 \quad \text{for } x<0; \qquad u_2(x,0)=\varphi_2(x)\geqslant 0 \quad \text{for } x>0; \tag{2} \]

\[ u_i(s(t),t)=0; \qquad \left(\frac{\partial u_1}{\partial x}-\frac{\partial u_2}{\partial x}\right)_{x=s(t)} =\frac{ds(t)}{dt}, \tag{3} \]

where \(A_i(x,t)\geqslant a>0\); \(C_i(x,t)\leqslant 0\); \(A_i,B_i,C_i,\varphi_i\) are sufficiently smooth and bounded functions; \(s(t)\) is a differentiable function, \(s(0)=0\).

Using the maximum principle, it is easy to prove the lemma (see \((^1)\)):

Lemma. Let \(y(t)\) be a differentiable function, and let \(V_i(x,t)\) \((i=1,2)\) be bounded and continuous in the domains \(\overline{\Omega_i}\) \((\Omega_1=\{x<y(t),\, t>0\}\), \(\Omega_2=\{x>y(t),\, t>0\})\) everywhere except possibly at the point \((y(0),0)\), and suppose that \(V_i(x,t)\) have derivatives \(\partial V_i/\partial x\), \(\partial^2V_i/\partial x^2\), \(\partial V_i/\partial t\) in \(\Omega_i\). Let, further, \(u_i(x,t)\), \(s(t)\) be a solution of problem (1)—(3), and let \(L_i(V_i)\leqslant 0\) in \(\Omega_i\) \((i=1,2)\),

\[ u_1(x,0)\leqslant V_1(x,0) \quad \text{for } x<y(0); \qquad u_2(x,0)\leqslant V_2(x,0) \quad \text{for } x>s(0); \]

\[ V_i(y(t),t)=0; \qquad \frac{dy(t)}{dt} < \left(\frac{\partial V_1}{\partial x}-\frac{\partial V_2}{\partial x}\right)_{x=y(t)} \quad \text{for } 0\leqslant t\leqslant T, \]

and let there exist \(t_0>0\) such that \(y(t)\leqslant s(t)\) for \(0<t\leqslant t_0\). Then for all \(t\in(0,T)\) the inequality \(y(t)\leqslant s(t)\) holds.

It is clear that if one changes the sign of the inequalities in all conditions of the lemma, then one obtains the relation \(y(t)\geqslant s(t)\) for \(t\in(0,T)\).

Using the similarity method (see \((^2)\)), one can prove the following proposition:

Theorem 1. The solution of the equation \(\partial^2u_i/\partial x^2=a_i^2\partial u_i/\partial t\) with the conditions \(u_1(x,0)=u_-<0\) for \(x<0\), \(u_2(x,0)=u_+>0\) for \(x>0\) and the conditions (3) has the form \(s(t)=\alpha_0\sqrt{t}\) and

\[ u_1(x,t)=u_-\,[\operatorname{erf}(a_1\alpha_0/2)-\operatorname{erf}(a_1x/2\sqrt{t})]/[1+\operatorname{erf}(a_1\alpha_0/2)] \quad \text{for } x<s(t), \]

\[ u_2(x,t)=u_+\,[\operatorname{erf}(a_2x/2\sqrt{t})-\operatorname{erf}(a_2\alpha_0/2)]/[1-\operatorname{erf}(a_2\alpha_0/2)] \quad \text{for } x>s(t), \]

where

\[ \operatorname{erf}(x)=\frac{2}{\sqrt{\pi}}\int_0^x e^{-\xi^2}\,d\xi; \]
\(\alpha_0\) is the root of the equation

\[ F(\alpha;u_-,u_+;a_1,a_2)\equiv \frac{a_1u_-e^{-a_1^2\alpha^2/4}}{1+\operatorname{erf}(a_1\alpha/2)} + \frac{a_2u_+e^{-a_2^2\alpha^2/4}}{1-\operatorname{erf}(a_2\alpha/2)} + \frac{\alpha\sqrt{\pi}}{2}=0. \]

It is easy to see that \(\dot F_\alpha>0\) and \(\alpha_0<0\) when \(a_1u_-+a_2u_+>0\); \(\alpha_0>0\) when \(a_1u_-+a_2u_+<0\).

Theorem 2. Let \(u_i(x,t)\) be a solution of the equation

\[ L_i(u_i)\equiv \partial^2u_i/\partial x^2+B_i(x,t)\partial u_i/\partial x-A_i(x,t)\partial u_i/\partial t=0 \]

with conditions (2) and (3), and let the coefficients \(A_i(x,t)\to a_i^2\), \(B_i(x,t)=o(1/\sqrt t)\) as \(t\to\infty\), uniformly in \(x\). Suppose \(xB_i(x,t)\ge 0\) for \(|x|\ge x_0\), where \(x_0>0\) is a certain constant. Then, if \(\varphi_1(x)\to u_-<0\) as \(x\to-\infty\) and \(\varphi_2(x)\to u_+>0\) as \(x\to\infty\), with \(a_1u_-+a_2u_+>0\), then \(\lim_{t\to\infty}s(t)=-\infty\) and

\[ \lim_{t\to\infty}u_2(x,t)=u_+\operatorname{erf}(a_2|\alpha_0|/2)/[1+\operatorname{erf}(a_2|\alpha_0|/2)]\equiv U_+. \]

Proof. Since \(a_1u_-+a_2u_+>0\), we have \(\alpha_0<0\) and \(F(\alpha_1;u_-,u_+;a_1,a_2)<0\), where \(\alpha_1=\alpha_0-\varepsilon\) and \(\varepsilon>0\). Construct the functions \(y(t)=\alpha_1\sqrt t-N\) and

\[ V_1(x,t)=(u_-+3\varepsilon_1) \frac{\operatorname{erf}(e_1\alpha_1/2)-\operatorname{erf}\bigl(e_1(x+N)/2\sqrt t\bigr)} {1+\operatorname{erf}(e_1\alpha_1/2)} +\varepsilon_2 v(x,t) \]

for \(x<y(t)\),

\[ V_2(x,t)=(u_++3\varepsilon_1) \frac{\operatorname{erf}\bigl(e_2(x+N)/2\sqrt t\bigr)-\operatorname{erf}(e_2\alpha_1/2)} {1-\operatorname{erf}(e_2\alpha_1/2)} +\varepsilon_2 v(x,t) \]

for \(x>y(t)\), where \(e_i^2=a_i^2-\varepsilon_3\) and \(v(x,t)=e^{-\theta(x+N)^2/t}-e^{-\theta\alpha_1^2}\). Choose \(\theta\) so small that

\[ L_i(v)\le -[\gamma+2\theta(x+N)B_i]\exp(-\theta(x+N)^2/t)/t, \]

where \(\gamma>0\) is a constant and \(\theta<\min(a_1^2/8,a_2^2/8)\). In this case

\[ \left(\frac{\partial V_1}{\partial x}-\frac{\partial V_2}{\partial x}\right)_{x=y(t)} = \frac{-1}{\sqrt{\pi t}} \left[ \frac{e_1(u_-+3\varepsilon_1)e^{-e_1^2\alpha_1^2/4}}{1+\operatorname{erf}(e_1\alpha_1/2)} + \frac{e_2(u_++3\varepsilon_1)e^{-e_1^2\alpha_1^2/4}}{1-\operatorname{erf}(e_2\alpha_1/2)} \right], \]

\[ L_1(V_1)= \]

\[ = \frac{-e_1(u_-+3\varepsilon_1)} {\sqrt{\pi}[1+\operatorname{erf}(e_1\alpha_1/2)]} \left[ (A_1(x,t)-e_1^2)\frac{x+N}{2t\sqrt t} +\frac{B_1(x,t)}{\sqrt t} \right] e^{-e_1^2(x+N)^2/4t} +\varepsilon_2L_1(v), \]

\[ L_2(V_2)= \]

\[ = \frac{e_2(u_++3\varepsilon_1)} {\sqrt{\pi}[1-\operatorname{erf}(e_2\alpha_1/2)]} \left[ (A_2(x,t)-e_2^2)\frac{x+N}{2t\sqrt t} +\frac{B_2(x,t)}{\sqrt t} \right] e^{-e_2^2(x+N)^2/4t} +\varepsilon_2L_2(v). \]

First choose \(\varepsilon_1>0\) and \(\varepsilon_3>0\) such that \(u_-+3\varepsilon_1<0\), \(F(\alpha_1;u_-+3\varepsilon_1,u_++3\varepsilon_1;e_1,e_2)<0\), and \(\varepsilon_3<\min(a_1^2/2,a_2^2/2)\). Then choose \(\varepsilon_2>0\) sufficiently small and \(k>0\) large so that \(2\varepsilon_1-\varepsilon_2e^{-\theta\alpha_1^2}>0\) and

\[ (u_++3\varepsilon_1)\, \frac{\operatorname{erf}(e_2k/2)-\operatorname{erf}(e_2\alpha_1/2)} {1-\operatorname{erf}(e_2\alpha_1/2)} +\varepsilon_2(e^{-\theta k^2}-e^{-\theta\alpha_1^2}) \ge u_++\varepsilon_1. \]

For such \(\varepsilon,\varepsilon_1,\varepsilon_2,\theta\), and \(k\), decrease \(\varepsilon_3\) so that

\[ \frac{e_2(u_++3\varepsilon_1)(1+k)\varepsilon_3} {\sqrt{\pi}[1-\operatorname{erf}(e_2\alpha_1/2)]} - \frac{e_1(u_-+3\varepsilon_1)\varepsilon_3} {\sqrt{\pi}[1+\operatorname{erf}(e_1\alpha_1/2)]} + 2\varepsilon_2\theta k\varepsilon_3 \le \varepsilon_2\gamma. \]

Moreover, take all \(\varepsilon_i\) so that \(\varepsilon_i\to0\) as \(\varepsilon\to0\). For any …

for arbitrary $\varepsilon_1$ and $\varepsilon_3$, in view of the condition $xB_i(x,t)\geqslant 0$ for $|x|\geqslant x_0$, as in (3), one can find such a $T>0$ that
$u_2(x,t)\leqslant u_+ + \varepsilon_1$, $|A_i(x,t)-a_i^2|\leqslant \varepsilon_3$, and $|B_i(x,t)|\leqslant \varepsilon_3/\sqrt t$ for $t\geqslant T$. For such a $T$ we have: $u_1(x,T)\leqslant u_-+\varepsilon_1$ for $x\leqslant -N$, where $N>0$ is sufficiently large and such that $-N<s(T)$ and $N>x_0$ (see (4)). For simplicity we assume that $T=0$. Then it is easy to verify that
$V_1(x,0)\geqslant u_1(x,0)$,
$V_2(x,t)|_{x=k\sqrt t-N}\geqslant u_2(x,t)|_{x=k\sqrt t-N}$,
$V_i(y(t),t)=0$,
$(\partial V_1/\partial x-\partial V_2/\partial x)_{x=y(t)}>dy(t)/dt$, and
$L_1(V_1)\leqslant 0$ for $x<y(t)$, $L_2(V_2)\leqslant 0$ for $y(t)<x\leqslant k\sqrt t-N$. Hence it is easy to see that
$s(t)\geqslant y(t)$. Consequently,
\[ \lim_{t\to\infty} u_2(x,t)\leqslant \lim_{t\to\infty} V_2(x,t) =(u_+ +3\varepsilon_1)\, \]
\[ {}\times \operatorname{erf}(e_2|\alpha_1|/2)/[1+\operatorname{erf}(e_2|\alpha_1|/2)] +\varepsilon_2(1-e^{-\theta a_1^2}). \]
Since $\varepsilon$ is arbitrary and $\varepsilon_i\to 0$ as $\varepsilon\to 0$, it follows that $\lim\limits_{t\to\infty}u_2(x,t)\leqslant U_+$. Analogously one can prove that
$\lim\limits_{t\to\infty}u_2(x,t)\geqslant U_+$ and $\lim\limits_{t\to\infty}s(t)=-\infty$.

If $B_i(x,t)=O(1/\sqrt t)$, then Theorem 2 may turn out to be false. Indeed, take
$A_i\equiv 1$, $B_1\equiv 0$, $B_2=f(x/2\sqrt t)/\sqrt t$, $\varphi_1\equiv 0$ and $\varphi_2\equiv 1$, where $f(z)=0$ for $z\geqslant 1$; $f(z)=2z-2$ for $0\leqslant z\leqslant 1$; $f(z)=-2$ for $z\leqslant 0$. Then
\[ s(t)=\alpha_1\sqrt t,\quad u_1(x,t)=0,\quad u_2(x,t)=[\Phi(x/2\sqrt t)- \]
\[ -\Phi(\alpha_1/2)]/[\Phi(\infty)-\Phi(\alpha_1/2)], \quad \text{where}\quad \Phi(x)=\int_0^x \exp(-z^2-G(z))\,dz,\quad G(z)= \]
\[ =-2\int_0^x f(x)\,dx,\quad \alpha_1\ \text{is the root of the equation}\quad \exp(-\alpha^2/4-G(\alpha/2))/[\Phi(\infty)- \]
\[ -\Phi(\alpha/2)]+\alpha=0. \]
It is easy to verify that $\lim\limits_{t\to\infty}u_2(x,t)\ne U_+$.

Theorem 3. Let $u_i(x,t)$ be the solution of problem (1)—(3), and let the coefficients $A_i(x,t)\to a_i^2$, $B_i(x,t)=o(1/\sqrt t)$ as $t\to\infty$ uniformly in $x$, with $xB_1(x,t)\geqslant 0$ for $|x|\geqslant x_0$, $C_1(x,t)\equiv 0$, $C_2(x,t)\leqslant C<0$. Then, if $\varphi_1(x)\to u_-<0$ as $x\to-\infty$ and $\varphi_2(x)$ is bounded, then
\[ \lim_{t\to\infty}s(t)=\infty \quad\text{and}\quad \lim_{t\to\infty}u_1(x,t)=u_-\,\operatorname{erf}(a_1\alpha_1/2)/[1+\operatorname{erf}(a_1\alpha_1/2)]\equiv U_-, \]
where $\alpha_1$ is the root of the equation $F(\alpha;u_-,0;a_1,a_2)=0$.

It remains for us to prove that $u_2(x,t)\leqslant \varepsilon_1$ for $x>s(t)$ and $t\geqslant T$, since the rest of the proof can be carried out in the same way as in Theorem 2. For this purpose construct the functions
$w_{\pm}(x,t)=Me^{-\theta t}\pm u_2(x,t)$.
Then
$L_2(w_{\pm})=M(A_2(x,t)\theta+C_2(x,t))e^{-\theta t}\leqslant 0$ for $x>s(t)$,
$w_{\pm}(x,0)=M\pm\varphi_2(x)\geqslant 0$, and
$w_{\pm}(x,t)|_{x=s(t)}\geqslant 0$ for sufficiently small $\theta$ and large $M$. By the maximum principle it follows that
$|u_2(x,t)|\leqslant Me^{-\theta t}$, whence $u_2(x,t)\leqslant \varepsilon_1$ for $x>s(t)$ and $t\geqslant T$.

Theorem 3 is also valid in the case when $A_2(x,t)$ and $B_2(x,t)$ are only bounded.

Theorem 4. Let $u_i(x,t)$ be the solution of problem (1)—(3), and let the coefficients $A_i(x,t)\to a_i^2$, $B_i(x,t)=o(1/\sqrt t)$ as $t\to\infty$ uniformly in $x$, with $B_2(x,t)\leqslant -b<0$, $xB_1(x,t)\geqslant 0$ for $|x|\geqslant x_0$, $C_1(x,t)\equiv 0$, $C_2(x,t)\leqslant 0$. Then, if $\varphi_1(x)\to u_-<0$ as $x\to-\infty$ and $\varphi_2(x)$ is bounded, then
\[ \lim_{t\to\infty}s(t)=+\infty \quad\text{and}\quad \lim_{t\to\infty}u_1(x,t)=U_-. \]

As in Theorem 3, it is enough to prove that for any $\varepsilon_1>0$ the inequality
$u_2(x,t)\leqslant \varepsilon_1$ holds for $s(t)<x<k\sqrt t+N$ and $t\geqslant T$, where $k>0$ is a sufficiently large number and $N>0$. For this purpose construct the functions
$w_{\pm}(x,t)=Me^{(\theta x-\gamma t)}\pm u_2(x,t)$.
Then
\[ L_2(w_{\pm})=M(\theta^2+\gamma A_2+B_2\theta+C_2)\times \]
\[ {}\times e^{(\theta x-\gamma t)}\leqslant 0, \]
$w_{\pm}(x,0)=Me^{\theta x}\pm\varphi_2(x)\geqslant 0$ for $x\geqslant 0$, and
$w_{\pm}(x,t)|_{x=s(t)}\geqslant 0$ for sufficiently small $\theta,\gamma$ and large $M$. Hence we obtain
$|u_2(x,t)|\leqslant Me^{(\theta x-\gamma t)}$. Consequently,
$u_2(x,t)\leqslant \varepsilon_1$ for $s(t)<x<k\sqrt t+N$ and $t\geqslant T$.
This theorem is also valid in the case when $A_2(x,t)$ is only bounded.

Theorem 5. Let \(u_i(x,t)\) be a solution of problem (1)—(3), and let the coefficients \(A_1(x,t)\to a_1^2\), \(B_1(x,t)=o(1/\sqrt{t})\) as \(t\to\infty\), uniformly in \(x\), with \(xB_i(x,t)\geqslant 0\) for \(|x|\geqslant x_0\), \(C_1=0\), \(C_2\leqslant 0\). Then, if \(\varphi_1(x)\to u_-<0\) as \(x\to-\infty\) and \(\varphi_2(x)\to 0\) as \(x\to+\infty\), then \(\lim\limits_{t\to\infty}s(t)=\infty\) and \(\lim\limits_{t\to\infty}u_1(x,t)=U_-\).

Theorem 6. For the solution of problem (1)—(3) the following assertions are valid: 1) \(\lim\limits_{t\to\infty}s(t)=+\infty\), \(\lim\limits_{t\to\infty}u_1(x,t)=u_-\), if \(C_1=0\), \(B_i(x,t)\leqslant -b<0\), \(\varphi_1(x)\to u_-\) as \(x\to-\infty\); 2) \(\lim\limits_{t\to\infty}s(t)=-\infty\), \(\lim\limits_{t\to\infty}u_2(x,t)=u_+\), if \(C_2=0\), \(B_i(x,t)\geqslant b>0\), \(\varphi_2(x)\to u_+\) as \(x\to\infty\); 3) \(\lim\limits_{t\to\infty}u_i(x,t)=0\), if \(C_i(x,t)\leqslant -C<0\) or \(B_1(x,t)\geqslant b_1>0\), \(B_2(x,t)\leqslant -b_2<0\).

The proofs of Theorems 5 and 6 are similar to the proofs of the preceding theorems.

One may consider a generalized solution of problem (1)—(3). To this end we write (1)—(2) in the form
\(u_{xx}+B(x,t,u)u_x+C(x,t,u)=A(x,t,u)u_t\) and \(u(x,0)=u_0(x)\). A continuous bounded function \(u(x,t)\) will be called a generalized solution of problem (1)—(3) if, for every continuously differentiable function \(f(x,t)\) equal to zero outside some finite domain, \(u(x,t)\) satisfies the equality

\[ \iint_{t\geqslant 0}\left[ u\,\frac{\partial^2 f}{\partial x^2} -\overline{B}(x,t,u)\frac{\partial f}{\partial x} +G(x,t,u)f +\Phi(x,t,u)\frac{\partial f}{\partial t} \right]\,dx\,dt - \int_{-\infty}^{\infty}\Phi(x,0,u_0(x))f(x,0)\,dx=0, \]

where

\[ \Phi=\int_0^u A(x,t,u)\,du \quad \text{for } u<0,\qquad \Phi=\int_0^u A\,du+1 \quad \text{for } u>0, \]

\[ \overline{B}=\int_0^u B(x,t,u)\,du \quad\text{and}\quad G(x,t,u)=C(x,t,u)+\int_0^u\frac{\partial A}{\partial t}\,du-\int_0^u\frac{\partial B}{\partial x}\,du. \]

If one assumes that \(u_0(x)\) has a generalized derivative summable with its square, and that \(C_1(x,t)=C_2(x,t)=C(x,t)\), then, as in work (5), one can prove that there exists a unique generalized solution of problem (1)—(3). This solution can be obtained as the limit, as \(h\to 0\), of a uniformly convergent sequence of solutions of the Cauchy problem for the equation
\(u_{xx}+B^h(x,t,u^h)u_x^h+C(x,t)u^h=\Phi_u^h(x,t,u^h)u_t^h\)
with the condition \(u^h(x,0)=u_0^h(x)\), where \(B^h\) and \(\Phi^h\) are the mean functions of \(B\) and \(\Phi\), respectively, and \(u_0^h(x)\) is the mean function of \(u_0(x)\) (see (6)). It can be proved that Theorem 2 for \(A_i(x,t)\equiv a_i^2\), \(B_i\equiv 0\), Theorems 4 and 5 for \(C_2(x,t)\equiv 0\), and Theorem 6 for \(C_1(x,t)=C_2(x,t)\) are valid also for the generalized solution of problem (1)—(3).

I express my sincere gratitude to O. A. Oleinik and A. M. Il’in for the formulation of the problem and for systematic assistance in this work.

Received
28 XII 1960

References

1. L. I. Rubinshtein, DAN, 77, No. 1 (1958).
2. A. N. Tikhonov, A. A. Samarskii, Equations of Mathematical Physics, Moscow, 1953, p. 262.
3. A. M. Il’in, UMN, 16, issue 2 (1961).
4. A. M. Il’in, O. A. Oleinik, DAN, 120, No. 1 (1958).
5. O. A. Oleinik, DAN, 135, No. 5 (1960).
6. S. L. Sobolev, Some Applications of Functional Analysis in Mathematical Physics, Leningrad, 1950, p. 19.