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Mathematics LibreTexts

6.3: Maximum Principle

  • Page ID
    2158
  • [ "article:topic" ]

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    Let \(\Omega\subset \mathbb{R}^n\) be a bounded domain. Set

    \begin{eqnarray*}
    D_T&=&\Omega\times(0,T),\ \ T>0,\\
    S_T&=&\{(x,t):\ (x,t)\in\Omega\times\{0\}\ \mbox{or}\ (x,t)\in\partial\Omega\times[0,T]\},
    \end{eqnarray*}

    see Figure 6.3.1.

    Notations to the maximum principle

    Figure 6.3.1: Notations to the maximum principle

    Theorem 6.2

    Assume \(u\in C(\overline{D_T})\), that \(u_t\), \(u_{x_ix_k}\) exist and are continuous in \(D_T\), and

    $$u_t-\triangle u\le 0\ \ \mbox{in}\ D_T.$$

    Then

    $$\max_{\overline{D_T}} u(x,t)=\max_{S_T} u.$$

    Proof

    Assume initially \(u_t-\triangle u<0\) in \(D_T\). Let \(\varepsilon>0\) be small and \(0<\varepsilon<T\). Since \(u\in C(\overline{D_{T-\varepsilon}})\), there is an \((x_0,t_0) \in \overline{D_{T-\varepsilon}}\) such that

    $$u(x_0,t_0)=\max_{\overline{D_{T-\varepsilon}}} u(x,t).$$

    Case (i)

    Let \((x_0,t_0)\in D_{T-\varepsilon}\). Hence, since \(D_{T-\varepsilon}\) is open,

    \(u_t(x_0,t_0)=0\), \(u_{x_l}(x_0,t_0)=0\), \(l=1,\ldots,n\) and

    $$\sum_{l,k=1}^n u_{x_lx_k}(x_0,t_0)\zeta_l\zeta_k\le0\ \ \mbox{for all}\ \zeta\in\mathbb{R}^n.$$

    The previous inequality implies that \(u_{x_kx_k}(x_0,t_0)\le0\) for each \(k\). Thus we arrived at a contradiction to \(u_t-\triangle u<0\) in \(D_T\).

    Case (ii)

    Assume \((x_0,t_0)\in\Omega\times\{T-\varepsilon\}\). Then it follows as above \(\triangle u\le 0\) in \((x_0,t_0)\), and from \(u(x_0,t_0)\ge u(x_0,t)\), \(t\le t_0\), one concludes that \(u_t(x_0,t_0)\ge0\). We arrived at a contradiction to \(u_t-\triangle u<0\) in \(D_T\) again.

    Summarizing, we have shown that

    $$\max_{\overline{D_{T-\varepsilon}}} u(x,t)=\max_{T-\varepsilon} u(x,t).$$

    Thus there is an \((x_\varepsilon,t_\varepsilon)\in S_{T-\varepsilon}\) such that

    $$u(x_\varepsilon,t_\varepsilon)=\max_{\overline{D_{T-\varepsilon}}} u(x,t).$$

    Since \(u\) is continuous on \(\overline{D}_T\), we have

    $$\lim_{\varepsilon\to 0}\max_{\overline{D_{T-\varepsilon}}} u(x,t)=\max_{\overline{D_T}} u(x,t).$$

    It follows that there is \((\overline{x},\overline{t})\in S_T\) such that

    $$u(\overline{x},\overline{t})=\max_{\overline{D_T}} u(x,t)$$

    since \(S_{T-\varepsilon}\subset S_T\) and \(S_T\) is compact. Thus, theorem is shown under the assumption \(u_t-\triangle u<0\) in \(D_T\). Now assume \(u_t-\triangle u\le 0\) in \(D_T\). Set

    $$v(x,t):=u(x,t)-kt,$$

    where \(k\) is a positive constant. Then

    $$v_t-\triangle v=u_t-\triangle u-k<0.$$

    From above we have

    \begin{eqnarray*}
    \max_{\overline{D_T}} u(x,t)&=&\max_{\overline{D_T}} (v(x,t)+kt)\\
    &\le&\max_{\overline{D_T}} v(x,t)+kT\\
    &=&\max_{S_T} v(x,t)+kT\\
    &\le&\max_{S_T}u(x,t)+kT,
    \end{eqnarray*}

    Letting \(k\to 0\), we obtain

    $$\max_{\overline{D_T}} u(x,t)\le\max_{S_T} u(x,t).$$

    Since \(S_T\subset\overline{D_T}\), the theorem is shown.

    \(\Box\)

    If we replace in the above theorem the bounded domain \(\Omega\) by \(\mathbb{R}^n\), then the result remains true provided we assume an {\it additional} growth assumption for \(u\). More precisely, we have the following result which is a corollary of the previous theorem. Set for a fixed \(T\), \(0<T<\infty\),

    $$D_T=\{ (x,t):\ x\in\mathbb{R}^n,\ 0<t<T\}.$$

    Proposition 6.2

    Assume \(u\in C(\overline{D_T})\), that \(u_t\), \(u_{x_ix_k}\) exist and are continuous in \(D_T\),

    $$u_t-\triangle u\le 0\ \ \mbox{in}\ D_T,$$

    and additionally that \(u\) satisfies the growth condition

    $$u(x,t)\le Me^{a|x|^2},$$

    where \(M\) and \(a\) are positive constants.

    Then

    $$\max_{\overline{D_T}} u(x,t)=\max_{S_T} u.$$

    It follows immediately the 

    Corollary

    The initial value problem \(u_t-\triangle u=0\) in \(D_T\), \(u(x,0)=f(x)\), \(x\in\mathbb{R}^n\), has a unique solution in the class defined by \(u\in C(\overline{D_T})\), \(u_t\), \(u_{x_ix_k}\) exist and are continuous in \(D_T\) and \(|u(x,t)|\le Me^{a|x|^2}\).

    Proof of Proposition 6.2.

    See [10], pp. 217. We can assume that \(4aT<1\), since the finite interval can be divided into finite many intervals of equal length \(\tau\) with
    \(4a\tau<1\). Then we conclude successively for \(k\) that

    $$u(x,t)\le\sup_{y\in\mathbb{R}^n}u(y,k\tau)\le\sup_{y\in\mathbb{R}^n}u(y,0)$$

    for \(k\tau\le t\le (k+1)\tau\), \(k=0,\ldots, N-1\), where \(N=T/\tau\).

    There is an \(\epsilon>0\) such that \(4a(T+\epsilon)<1\). Consider the comparison function
    \begin{eqnarray*}
    v_\mu(x,t):&=&u(x,t)-\mu\left(4\pi(T+\epsilon-t)\right)^{-n/2}e^{|x-y|^2/(4(T+\epsilon-t))}\\
    &=&u(x,t)-\mu K(ix,iy,T+\epsilon-t)
    \end{eqnarray*}

    for fixed \(y\in\mathbb{R}^n\) and for a constant \(\mu>0\). Since the heat kernel \(K(ix,iy,t)\) satisfies \(K_t=\triangle K_x\), we obtain

    $$\frac{\partial}{\partial t}v_\mu-\triangle v_\mu=u_t-\triangle u\le0.$$

    Set for a constant \(\rho>0\)

    $$D_{T,\rho}=\{(x,t):\ |x-y|<\rho,\ 0<t<T\}.$$

    Then we obtain from Theorem 6.2 that

    $$v_\mu(y,t)\le\max_{S_{T,\rho}}v_\mu,$$

    where \(S_{T,\rho}\equiv S_T\) of Theorem 6.2 with \(\Omega=B_\rho(y)\), see Figure 6.3.1.

    On the bottom of \(S_{T,\rho}\) we have, since \(\mu K>0\),

    $$v_\mu(x,0)\le u(x,0)\le\sup_{z\in\mathbb{R}^n}f(z).$$

    On the cylinder part \(|x-y|=\rho\), \(0\le t\le T\), of \(S_{T,\rho}\) it is

    \begin{eqnarray*}
    v_\mu(x,t)&\le&Me^{a|x|^2}-\mu\left(4\pi(T+\epsilon-t)\right)^{-n/2}e^{\rho^2/(4(T+\epsilon-t))}\\
    &\le&Me^{a(|y|+\rho)^2}-\mu\left(4\pi(T+\epsilon)\right)^{-n/2}e^{\rho^2/(4(T+\epsilon))}\\
    &\le&\sup_{z\in\mathbb{R}^n}f(z)
    \end{eqnarray*}
    for all \(\rho>\rho_0(\mu)\), \(\rho_0\) sufficiently large. We recall that \(4a(T+\epsilon)<1\).
    Summarizing, we have

    $$\max_{S_{T,\rho}}v_\mu(x,t)\le\sup_{z\in\mathbb{R}^n}f(z)$$

    if \(\rho>\rho_0(\mu)\). Thus

    $$v_\mu(y,t)\le\max_{S_{T,\rho}}v_\mu(x,t)\le\sup_{z\in\mathbb{R}^n}f(z)$$

    if \(\rho>\rho_0(\mu)\).

    Since

    $$v_\mu(y,t)=u(y,t)-\mu\left(4\pi(T+\epsilon-t)\right)^{-n/2}$$

    it follows

    $$u(y,t)-\mu\left(4\pi(T+\epsilon-t)\right)^{-n/2}\le\sup_{z\in\mathbb{R}^n}f(z).$$

    Letting \(\mu\to0\), we obtain the assertion of the proposition.

    \(\Box\)

    The above maximum principle of Theorem 6.2 holds for a large class of parabolic differential operators, even for degenerate equations.

    Set

    $$Lu=\sum_{i,j=1}^na^{ij}(x,t)u_{x_ix_j},$$

    where \(a^{ij}\in C(D_T)\) are real, \(a^{ij}=a^{ji}\), and the matrix \((a^{ij})\) is non-negative, that is,

    $$\sum_{i,j=1}^na^{ij}(x,t)\zeta_i\zeta_j\ge 0\ \ \mbox{for all}\ \zeta\in\mathbb{R}^n,$$

    and \((x,t)\in D_T\).

    Theorem 6.3

    Assume \(u\in C(\overline{D_T})\), that \(u_t\), \(u_{x_ix_k}\) exist and are continuous in \(D_T\), and

    $$u_t-L u\le 0\ \ \mbox{in}\ D_T.$$

    Then

    $$\max_{\overline{D_T}} u(x,t)=\max_{S_T} u.$$

    Proof

    (i) One proof is a consequence of the following lemma: let \(A\), \(B\) real, symmetric and non-negative matrices. Non-negative means that all eigenvalues are non-negative. Then trace~\((AB)\equiv\sum_{i,j=1}^na^{ij}b_{ij}\ge0\), see an exercise.

    (ii) Another proof is more directly: let \(U=(z_1,\ldots,z_n)\), where \(z_l\) is an orthonormal system of eigenvectors to the eigenvalues \(\lambda_l\) of the matrix \(A=(a^{i,j}(x_0,t_0))\). Set \(\zeta=U\eta\), \(x=U^T(x-x_0)y\) and \(v(y)=u(x_0+Uy,t_0)\), then

    \begin{eqnarray*}
    0&\le&\sum_{i,j=1}^na^{ij}(x_0,t_0)\zeta_i\zeta_j=\sum_{i=1}^n\lambda_i\eta_i^2\\
    0&\ge&\sum_{i,j=1}^n u_{x_ix_j}\zeta_i\zeta_j=\sum_{i=1}^n v_{y_iy_i}\eta_i^2.
    \end{eqnarray*}

    It follows \(\lambda_i\ge0\) and \(v_{y_iy_i}\le0\) for all \(i\).

    Consequently

    $$\sum_{i,j=1}^na^{ij}(x_0,t_0)u_{x_ix_j}(x_0,t_0)=\sum_{i=1}^n\lambda_iv_{y_iy_i}\le0.$$

     \(\Box\)

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