1.Introduction

2.First-order Differential Equations

Exercise2.1.

Find solutons of the following intial-value problems in $\bbR^2$:

(1)$2u_y-u_x+xu=0$ with $u(x,0)=2xe^{x^2/2}$;

(2)$u_y+(1+x^2)u_x-u=0$ with $u(x,0)=\arctan x$.

Solution:

(1)Since $(-1,2)\cdot (0,1)=2\neq 0$, we see $\sed{y=0}$ is noncharacteristic for the PDE, and consequently the characteristic ODE is $$\bex \ba{ccc} \frac{\rd x}{\rd s}=-1,&\frac{\rd y}{\rd s}=2,&\frac{\rd u}{\rd s}=-xu,\\ x(0)=x_0,&y(0)=0,&u(0)=2x_0e^{x_0^2/2}. \ea \eex$$ Solving it yields $$\bex u=(2x+y)e^{x^2/2}. \eex$$

(2)Since $(1+x^2,1)\cdot (0,1)=1\neq 0$, we see $\sed{y=0}$ is noncharacteristic for the PDE, and consequently the characteristic ODE is $$\bex \ba{ccc} \frac{\rd x}{\rd s}=1+x^2,&\frac{\rd y}{\rd s}=1,&\frac{\rd u}{\rd s}=u,\\ x(0)=x_0,&y(0)=0,&u(0)=\arctan x_0. \ea \eex$$ Solving it yields $$\bex u=(\arctan x-y)e^y. \eex$$

Exercise2.2.

Solve the following initial-value problems:

(1)$u_y+u_x=u^2$ with $u(x,0)=h(x)$;

(2)$u_z+xu_x+yu_y=u$ with $u(x,y,0)=h(x,y)$.

Solution:

(1)Since $(1,1)\cdot (0,1)=1\neq 0$, we see $\sed{y=0}$ is noncharacteristic for the PDE, and consequently the characteristic ODE is $$\bex \ba{ccc} \frac{\rd x}{\rd s}=1,&\frac{\rd y}{\rd s}=1,&\frac{\rd u}{\rd s}=u^2,\\ x(0)=x_0,&y(0)=0,&u(0)=h(x_0). \ea \eex$$ Solving it yields $$\bex u=\frac{h(x-y)}{1-yh(x-y)}. \eex$$

(2)Since $(x,y,1)\cdot (0,0,1)=1\neq 0$, we see $\sed{z=0}$ is noncharacteristic for the PDE, and consequently the characteristic ODE is $$\bex \ba{cccc} \frac{\rd x}{\rd s}=x,&\frac{\rd y}{\rd s}=y,&\frac{\rd z}{\rd s}=1,&\frac{\rd u}{\rd s}=u,\\ x(0)=x_0,&y(0)=y_0,&z(0)=0,&u(0)=h(x_0,y_0). \ea \eex$$ Solving it yields $$\bex u=h\sex{\frac{x}{e^z},\frac{y}{e^z}}e^z. \eex$$

Exercise2.3.

Let $B_1$ be the unit disc in $\bbR^2$ and $a$ and $b$ be continuous functions in $\bar B_1$ with $a(x,y)x+b(x,y)y>0$ on $\p B_1$. Assume $u$ is a $C^1$-solution of $$\bex a(x,y)u_x+b(x,y)u_y=-u\mbox{ in }\bar B_1. \eex$$ Prove that $u$ vanishes identically.

Solution:  The $u$, as a $C^1(\bar B_1)$-function, attains its maximum and minimum on $\bar B_1$.

(1)If $u$ attains its maximum in $B_1$, then $$\bex 0=au_x+bu_y=-u, \eex$$ and hence $u\leq 0$ in $\bar B_1$.

(2)If $u$ attains its maximum on $\p B_1$, then since $ax+by>0$, we find the directional derivative $\dps{\frac{\p}{\p_{(a,b)}}}$ of $u$ satisfies $$\bex 0\leq au_x+bu_y=-u, \eex$$ and hence $u\leq 0$ in $\bar B_1$.

(3)If $u$ attains its minimum in $B_1$, then $$\bex 0=au_x+bu_y=-u, \eex$$ and hence $u\geq 0$ in $\bar B_1$.

(4)If $u$ attains its minimum on $\p B_1$, then since $ax+by>0$, we find the directional derivative $\dps{\frac{\p}{\p_{(a,b)}}}$ of $u$ satisfies $$\bex 0\geq au_x+bu_y=-u, \eex$$ and hence $u\geq 0$ in $\bar B_1$. Consequently, in any case, we find $u\equiv 0$.

Exercise2.4.

Find a smooth solution $a=a(x,y)$ in $\bbR^2$ such that, for the equation of the form $$\bex u_y+a(x,y)u_x=0, \eex$$ there does not exist any solution in the entire $\bbR^2$ for any nonconstant initial value prescribed on $\sed{y=0}$.

Solution:  Clearly, $\sed{y=0}$ is noncharacteristic for the PDE, and consequently the characteristic ODE is $$\bex \ba{cccc} \frac{\rd x}{\rd s}=a,&\frac{\rd y}{\rd s}=1,&\frac{\rd u}{\rd s}=0,\\ x(0)=x_0,&y(0)=0,&u(0)=u_0(x_0). \ea \eex$$ Hence, $$\bex \ba{cccc} \frac{\rd x}{\rd y}=a(x,y),&u(x,y)=u_0(x_0),\\ x(0)=x_0. \ea \eex$$ For the purpose in the exercise, we need only to ensure that the ODE blows up at finite time. Consequently, we may choose $a(x,y)=x^2$.

Exercise2.5.

Let $\alpha$ be a number and $h=h(x)$ be a continuous function in $\bbR$. Consider $$\bex \ba{rcl} yu_x+xu_y&=&\alpha u,\\ u(x,0)&=&h(x). \ea \eex$$

(1)Find all points on $\sed{y=0}$ where $\sed{y=0}$ is characteristic. What is the compatibility condition on $h$ at these points?

(2)Away from the points in $(1)$, find the solution of the initial-value problem. What is the domain of this solution in general?

(3)For the cases $h(x)=x, \alpha=1$ and $h(x)=x, \alpha=3$, check whether this solution can be extended over the points in $(1)$.

(4)For each point in $(1)$, find all characteristic curves containing it. What is the relation of these curves and the domain in $(2)$?

Solution:

(1)From $(y,x)\cdot (0,1)=x$, we see the only characteristic point is $(0,0)$. At $(0,0)$, $$\bex 0=yu_x+xu_y=\alpha u=\alpha h(0). \eex$$ Thus the compatibility condition is $\alpha h(0)=0$.

(2)If $x_0\neq 0$, then the characteristic ODE is $$\bex \ba{cccc} \frac{\rd x}{\rd s}=y,&\frac{\rd y}{\rd s}=x,&\frac{\rd u}{\rd s}=\alpha u,\\ x(0)=x_0,&y(0)=0,&u(0)=h(x_0). \ea \eex$$ Hence, $$\bex x=\frac{x_0}{2}(e^{-s}+e^s),\quad y=\frac{x_0}{2}(-e^{-s}+e^s),\quad u=h(x_0)e^{\alpha s}. \eex$$ Consequently, $$\bex u=h(\sqrt{x^2-y^2})\sex{\sqrt{\frac{x+y}{x-y}}}^\alpha, \eex$$ and the domain of this solution is $$\bex \sed{(x,y);\ x^2-y^2\geq 0,\ x\neq y}. \eex$$

(3)For $h(x)=x, \alpha=1$, we have $$\bex u=x+y, \eex$$ which can be extended to $(0,0)$. For $h(x)=x, \alpha=3$, we have $$\bex u=\frac{(x+y)^2}{x-y}, \eex$$ which can not be extended to $(0,0)$.

(4)From the ODE in $(2)$, we have $$\bex \frac{x+y}{x-y}=e^{2s},\quad x_0=\sqrt{x^2-y^2}. \eex$$ Thus the characteristic curve contain $(0,0)$ is $x=\pm y$, which is the boundary of the domain in $(2)$.

Exercise2.6.

Let $\alpha\in\bbR$ be a real number and $h=h(x)$ be continuous in $\bbR$ and $C^1$ in $\bbR\bs \sed{0}$. Consider $$\bex \ba{rcl} xu_x+yu_y&=&\alpha u,\\ u(x,0)&=&h(x). \ea \eex$$

(1)Check that the straight line $\sed{y=0}$ is characteristic at each point.

(2)Find all $h$ satisfying the compatibility condition on $\sed{y=0}$. (Consider three cases, $\alpha>0$, $\alpha=0$ and $\alpha<0$).

(3)For $\alpha>0$, find two solutions with the given initial value on $\sed{y=0}$. (This is easy to do simply by inspecting the equation, especially for $\alpha=2$).

Solution:

(1)$(x,0)\cdot (0,1)=0$.

(2)From the compatibility condition $xh_x=\alpha h$, we see $h=h(1)x^\alpha$.

(3)For $h=x^\alpha$, we have two solutions $$\bex u_1=x^\alpha,\quad u_2=x^\alpha+y^\alpha. \eex$$

Exercise2.7.

In the plane, solve $u_y=4u_x^2$ near the origin with $u(x,0)=x^2$ on $\sed{y=0}$.

Solution:  Denote by $$\bex F(x,y,u,p_1,p_2)=4p_1^2-p_2. \eex$$ Since $F_{p_2}=-1\neq 0$, we may solve $$\bex \ba{rcl} 4u_x^2-u_y&=&0,\\ u(x,0)&=&x^2,\\ u_y(x,0)&=&4\cdot (2x)^2=16 x^2 \ea \eex$$ near the origin. The characteristic ODE is $$\bex \ba{lllll} \frac{\rd x}{\rd s}=8p_1,&\frac{\rd y}{\rd s}=-1,& \frac{\rd u}{\rd s}=8p_1^2-p_2,&\frac{\rd p_1}{\rd s}=0,&\frac{\rd p_2}{\rd s}=0,\\ x(0)=x_0,&y(0)=0,&u(0)=x_0^2,&p_1(0)=2x_0,&p_2(0)=16x_0^2. \ea \eex$$ Solving it yields $$\bex u=\frac{x^2}{1-16y}. \eex$$

Exercise2.8.

In the plane, find two solutions of the initial-value problem $$\bex \ba{rcl} xu_x+yu_y+\frac{1}{2}(u_x^2+u_y^2)&=&u,\\ u(x,0)&=&\frac{1}{2}(1-x^2). \ea \eex$$

Solution:  Denote by $$\bex F(x,y,u,p_1,p_2)=xp_1+yp_2+\frac{1}{2}(p_1^2+p_2^2)-u. \eex$$ On $\sed{y=0}$, $$\bex u=\frac{1}{2}(1-x^2),\quad p_1=-x. \eex$$ Solving $p_2$ as the solution to $$\bex x\cdot (-x)+\frac{1}{2}(x^2+p_2^2)-\frac{1}{2}(1-x^2)=0, \eex$$ we find $p_2=\pm 1$. Thus we may seek the solution to $$\bee\label{2.8.1} \ba{rcl} xu_x+yu_y+\frac{1}{2}(u_x^2+u_y^2)&=&u,\\ u(x,0)&=&\frac{1}{2}(1-x^2),\\ u_y(x,0)&=&1; \ea \eee$$ $$\bee\label{2.8.2} \ba{rcl} xu_x+yu_y+\frac{1}{2}(u_x^2+u_y^2)&=&u,\\ u(x,0)&=&\frac{1}{2}(1-x^2),\\ u_y(x,0)&=&-1. \ea \eee$$ For \eqref{2.8.1}, the characteristic ODE is $$\bex \ba{lllll} \frac{\rd x}{\rd s}=x+p_1, &\frac{\rd y}{\rd s}=y+p_2, &\frac{\rd u}{\rd s}=xp_1+yp_2+p_1^2+p_2^2, &\frac{\rd p_1}{\rd s}=0, &\frac{\rd p_2}{\rd s}=0;\\ x(0)=x_0, &y(0)=0, &u(0)=\frac{1}{2}(1-x_0^2), &p_1(0)=-x_0, &p_2(0)=1. \ea \eex$$ Solving it yields $$\bex u=\frac{1}{2}(1-x^2)+y. \eex$$ For \eqref{2.8.2}, the characteristic ODE is $$\bex \ba{lllll} \frac{\rd x}{\rd s}=x+p_1, &\frac{\rd y}{\rd s}=y+p_2, &\frac{\rd u}{\rd s}=xp_1+yp_2+p_1^2+p_2^2, &\frac{\rd p_1}{\rd s}=0, &\frac{\rd p_2}{\rd s}=0;\\ x(0)=x_0, &y(0)=0, &u(0)=\frac{1}{2}(1-x_0^2), &p_1(0)=-x_0, &p_2(0)=-1. \ea \eex$$ Solving it yields $$\bex u=\frac{1}{2}(1-x^2)-y. \eex$$

Exercise2.9.

In the plane, find two solutions of the initial-value problem $$\bex \ba{rcl} \frac{1}{4}u_x^2+uu_y&=&u,\\ u\sex{x,\frac{1}{2}x^2}&=&-\frac{1}{2}x^2. \ea \eex$$

Solution:  Change variable as $$\bex s=x,\quad t=y-\frac{1}{2}x^2, \eex$$ we get $$\bex u_x=u_s-su_t,\quad u_y=u_t, \eex$$ and hence $$\bex \frac{1}{4}u_s^2-\frac{1}{2}su_su_t+\frac{1}{4}s^2u_t^2 +uu_t=u,,\quad u(s,0)=-\frac{s^2}{2}. \eex$$ Denote by $$\bex F(s,t,u,p_1,p_2)=\frac{1}{4}p_1^2-\frac{1}{2}sp_1p_2+\frac{1}{4}s^2p_2^2 +up_2-u. \eex$$ On $\sed{t=0}$, $$\bex u=-\frac{s^2}{2},\quad p_1=-s, \eex$$ we find $$\bex F(s,t,u,u_s,u_t)=\frac{1}{4}s^2-\frac{1}{2}(-s)p_2+\frac{1}{4}s^2p_2^2 -\frac{s^2}{2}p_2+\frac{s^2}{2} =\frac{s^2}{4}(3+p_2^2)\neq 0,\quad (s\neq 0). \eex$$ Consequently, the exercise is not well posed.

Exercise2.10.

Let $a_i$, $b$ and $f$ be continuous functions satisfying $$\bex (2.3.1)\qquad |a|\leq \frac{1}{\kappa},\quad \mbox{in }\bbR^n\times [0,\infty) \eex$$ and $u$ be a $C^1$-solution of $$\bex (2.3.2)\qquad \ba{rcll} u_t+a(x,t)\n u+b(x,t)u&=&f(x,t),&\mbox{in }\bbR^n\times (0,\infty),\\ u(x,0)&=&u_0(x),&\mbox{in }\bbR^n \ea \eex$$ in $\bbR^n\times [0,\infty)$. Prove that, for any $P=(X,T)\in \bbR^n\times (0,\infty)$, $$\bex \sup_{C_\kappa(P)}|e^{-\alpha t}u| \leq \sup_{\p_-C_\kappa(P)}|u_0| +\frac{1}{\alpha+\inf_{C_\kappa(P)}b}\sup_{C_\kappa(P)}|e^{-\alpha t}f|, \eex$$ where $\alpha$ is a constant such that $$\bex \alpha+\inf_{C_\kappa(P)}b>0. \eex$$

Solution:  Let $$\bex M=\sup_{\p_-C_\kappa(P)}|u_0| +\frac{1}{\alpha+\inf_{C_\kappa(P)}b}\sup_{C_\kappa(P)}|e^{-\alpha t}f|. \eex$$ Then we need only to show that $$\bex |e^{-\alpha t}u|\leq M,\quad \forall\ (x,t)\in C_\kappa(P). \eex$$ For the upper bound, we consider $$\bex w=e^{-\alpha t}u-M. \eex$$ A simple calculation show that $$\beex \bea &\quad w_t+a\cdot\n w+(b+\alpha)w\\ &=-\alpha e^{-\alpha t} u+e^{-\alpha t}u_t +e^{-\alpha t}a\cdot \n u+(b+\alpha)(e^{-\alpha t}u-M)\\ &=e^{-\alpha t}f-(b+\alpha)M\\ &=(b+\alpha)\sez{\frac{1}{b+\alpha}e^{-\alpha t}f-M}\\ &\leq 0. \eea \eeex$$ Arguing as in Theorem 2.3.1 (see Page 34 in the book, which is named as ``maximum principle type argument''), we see $$\bex w\leq 0,\mbox{ in }C_\kappa(P), \eex$$ as desired. The lower bound is proved similarly and is omitted.

Exercise2.11.

Let $a_i$, $b$ and $f$ be $C^1$-functions in $\bbR^n\times [0,\infty)$ satisfying (2.3.1) and $u_0$ be a $C^1$-function in $\bbR^n$. Suppose $u$ is a $C^2$-solution of (2.3.2) in $\bbR^n\times [0,\infty)$. Prove that, for any $P=(X,T)\in \bbR^n\times (0,\infty)$, $$\bee\label{2.11:eq} |u|_{C^1(C_\kappa(P))}\leq C\sez{|u_0|_{C^1(\p_-C_\kappa(P))}+|f|_{C^1(C_\kappa(P))}}, \eee$$ where $C$ is a positive constant depending only on $T$ and the $C^1$-norm of $a_i$ and $b$ in $C_\kappa(P)$.

Solution:  From Theorem 2.3.1 or Exercise 2.10, we see $|u|_{C^0(C_\kappa(P))}$ is already bounded by the RHD of \eqref{2.11:eq}. To dominate $|u|_{C^1(C_\kappa(P))}$, we differentiate $$\bex u_t+a\cdot\n u+bu=f \eex$$ with respect to $x_i$ and $t$ as $$\bex \ba{rcl} u_{x_i,t}+a_{j,x_i}u_{x_j}+a_ju_{x_j,x_i}+b_{x_i}u+bu_{x_i}&=&f_{x_i},\\ u_{tt}+a_{j,t}u_{x_j}+a_ju_{x_j,t}+b_tu+bu_t&=&f_t. \ea \eex$$ Thus, denoting by $v=u_{x_i}$, $w=u_t$, we see $$\bex \ba{rcl} v_t+a\cdot\n v+bv&=&f_{x_i}-a_{j,x_i}u_{x_j}-b_{x_i}u,\\ w_t+a\cdot\n w+bw&=&f_t-a_{j,t}u_{x_j}-b_tu. \ea \eex$$ By Exercise 2.10, we obtain $$\beex \bea &\quad\sup_{C_\kappa(P)}|e^{-\alpha t}(v,w)|\\ & \leq \sup_{\p_-C_\kappa(P)}|(v_0,u_0)| +\frac{1}{\alpha+\inf_{C_\kappa(P)}b} \sup_{C_\kappa(P)}|e^{-\alpha t} (f_{x_i}-a_{j,x_i}u_{x_j}-b_{x_i}u, f_t-a_{j,t}u_{x_j}-b_tu)|. \eea \eeex$$ Thus, $$\beex \bea &\quad\sup_{C_\kappa(P)}|e^{-\alpha t} \n u|\leq \sup_{\p_-C_\kappa(P)}|\n u_0|\\ & +\frac{1}{\alpha+\inf_{C_\kappa(P)}b} \sez{\sup_{C_\kappa(P)}|e^{-\alpha t}\n f| +\sup_{C_\kappa(P)}|\n a|\cdot \sup_{C_\kappa(P)}|e^{-\alpha t}\n u|+\sup_{C_\kappa(P)}|\n b|\cdot \sup_{C_\kappa(P)}|u|}. \eea \eeex$$ Choosing $\alpha$ sufficiently large so that $$\bex \frac{1}{\alpha+\inf_{C_\kappa(P)}b}\sup_{C_\kappa(P)}|\n a|<\frac{1}{2}, \eex$$ we deduce $$\bex \sup_{C_\kappa(P)}|e^{-\alpha t} \n u| \leq \sup_{\p_-C_\kappa(P)}|\n u_0| +\frac{1}{\alpha+\inf_{C_\kappa(P)}b} \sez{\sup_{C_\kappa(P)}|e^{-\alpha t}\n f| +\sup_{C_\kappa(P)}|\n b|\cdot \sup_{C_\kappa(P)}|u|}. \eex$$ Combining this with the estimate for $|u|_{C^0(C_\kappa(P))}$, we get the desired \eqref{2.11:eq}.

Exercise2.12.

Let $a$ be a $C^1$-function in $\bbR\times [0,\infty)$ satisfying $$\bex |a(x,t)|\leq\frac{1}{\kappa}, \eex$$ and let $b_{ij}$ be continuous in $\bbR\times [0,\infty)$, for $i,j=1,2$. Suppose $(u,v)$ is a $C^1$-solution in $\bbR\times (0,\infty)$ of the first-order differential system $$\bee\label{2.12:eq} \ba{rcl} u_t-a(x,t)v_x+b_{11}(x,t)u+b_{12}(x,t)v&=&f_1(x,t),\\ v_t-a(x,t)u_x+b_{21}(x,t)u+b_{22}(x,t)v&=&f_2(x,t), \ea \eee$$ with $$\bex u(x,0)=u_0(x),\quad v(x,0)=v_0(x). \eex$$ Derive an $L^2$-estimate of $(u,v)$ in appropriate cones.

Solution:  Multiplying $\eqref{2.12:eq}_1$ with $u$, $\eqref{2.12:eq}_2$ with $v$, we find $$\bex \frac{1}{2}(u^2+v^2)_x-a(uv)_x +b_{11}u^2+(b_{12}+b_{21})uv +b_{22}v^2=f_1u+f_2v, \eex$$ $$\bex \frac{1}{2}(u^2+v^2)_x-(auv)_x +b_{11}u^2+(a_x+b_{12}+b_{21})uv +b_{22}v^2=f_1u+f_2v, \eex$$ $$\beex \bea &\quad\frac{1}{2}[e^{-\alpha t}(u^2+v^2)]_x-(e^{-\alpha t}auv)_x\\ &\quad +e^{-\alpha t}\sez{\sex{\frac{\alpha }{2}+b_{11}}u^2+(a_x+b_{12}+b_{21})uv +\sex{\frac{\alpha }{2}+b_{22}}v^2}\\ &=e^{-\alpha t}\sex{f_1u+f_2v}. \eea \eeex$$ Integrating the above equation in $C_\kappa(P)$, we see $$\beex \bea &\quad\int_{\p_sC_\kappa(P)}e^{-\alpha t}\sez{-a\nu_xuv+\nu_t\frac{u^2+v^2}{2}}\rd S +\int_{C_\kappa(P)}e^{-\alpha t}\sez{\frac{u^2+v^2}{2}-auv}\rd x\rd t\\ &\quad+\int_{C_\kappa(P)} e^{-\alpha t}\sez{\sex{\frac{\alpha }{2}+b_{11}}u^2+(a_x+b_{12}+b_{21})uv +\sex{\frac{\alpha }{2}+b_{22}}v^2}\rd x\rd t\\ &=\int_{\p_-C_\kappa(P)} e^{-\alpha t}\frac{u_0^2+v_0^2}{2}\rd x +\int_{C_\kappa(P)}e^{-\alpha t}(f_1u+f_2v)\rd x\rd t. \eea \eeex$$ Since $$\bex \nu_t\geq |a\nu_x|,\quad \frac{u^2+v^2}{2}, \eex$$ we may omit the first term to get $$\beex \bea &\quad \int_{C_\kappa(P)} e^{-\alpha t}\sez{\sex{\frac{\alpha+1 }{2}+b_{11}}u^2+(a_x-a+b_{12}+b_{21})uv +\sex{\frac{\alpha+1}{2}+b_{22}}v^2}\rd x\rd t\\ &\leq \int_{\p_-C_\kappa(P)} e^{-\alpha t}\frac{u_0^2+v_0^2}{2}\rd x +\int_{C_\kappa(P)}e^{-\alpha t}(f_1u+f_2v)\rd x\rd t. \eea \eeex$$ Now, choose $\alpha$ sufficiently large so that $$\beex \bea &\quad\sex{\frac{\alpha+1 }{2}+b_{11}}u^2+(a_x-a+b_{12}+b_{21})uv +\sex{\frac{\alpha+1}{2}+b_{22}}v^2\\ &\geq \frac{\alpha}{4}u^2-\sez{|a|_{C^1(C_\kappa(P))}+|b_{ij}|_{C(C_\kappa(P))}} \frac{u^2+v^2}{2} +\frac{\alpha}{4}v^2\\ &\geq \frac{\alpha}{8}(u^2+v^2), \eea \eeex$$ we deduce by applying Cauchy inequality, $$\beex \bea &\quad \frac{\alpha}{8}\int_{C_\kappa(P)}e^{-\alpha t}(u^2+v^2)\rd x\rd t\leq \int_{\p_-C_\kappa(P)} e^{-\alpha t}\frac{u_0^2+v_0^2}{2}\rd x\\ & +C\int_{C_\kappa(P)}e^{-\alpha t}(f_1^2+f_2^2)\rd x\rd t +\frac{\alpha}{16}\int_{C_\kappa(P)}e^{-\alpha t}(u^2+v^2)\rd x\rd t. \eea \eeex$$ Consequently, $$\bex \int_{C_\kappa(P)}e^{-\alpha t}(u^2+v^2)\rd x\rd t \leq C\sez{\int_{\p_-C_\kappa(P)} e^{-\alpha t}\frac{u_0^2+v_0^2}{2}\rd x +\int_{C_\kappa(P)}e^{-\alpha t}(f_1^2+f_2^2)\rd x\rd t}, \eex$$ where $C$ depends only on $\kappa$, $T$ and $|a|_{C^1(C_\kappa(P))},|b_{ij}|_{C(C_\kappa(P))}$.

3.An Overview of Second-Order PDEs

Exercise3.1.

Classify the following second-order PDEs:

(1)$\dps{\sum_{i=1}^n u_{x_ix_i}+\sum_{1\leq i<j\leq n}u_{x_ix_j}=0}$.

(2)$\dps{\sum_{1\leq i<j\leq n}u_{x_ix_j}=0}$.

Solution:

(1)For any $k$, $$\beex \bea \sev{\ba{cccc} 1&\frac{1}{2}&\cdots&\frac{1}{2}\\ \frac{1}{2}&1&\cdots&\frac{1}{2}\\ \vdots&\vdots&\ddots&\vdots\\ \frac{1}{2}&\frac{1}{2}&\cdots&1 \ea}_{k\times k} &=\sex{\frac{k-1}{2}+1} \sev{\ba{cccc} 1&\frac{1}{2}&\cdots&\frac{1}{2}\\ 1&1&\cdots&\frac{1}{2}\\ \vdots&\vdots&\ddots&\vdots\\ 1&\frac{1}{2}&\cdots&1 \ea}\\ &=\frac{k+1}{2}\sev{\ba{cccc} 1&\frac{1}{2}&\cdots&\frac{1}{2}\\ 0&\frac{1}{2}&\cdots&0\\ \vdots&\vdots&\ddots&\vdots\\ 0&0&\cdots&\frac{1}{2} \ea}=\frac{k+1}{2}\cdot \frac{1}{2^{k-1}}. \eea \eeex$$ Thus the matrix is positive definite, and the PDE is elliptic.

(2)For any $k$, $$\beex \bea \sev{\ba{cccc} 0&\frac{1}{2}&\cdots&\frac{1}{2}\\ \frac{1}{2}&0&\cdots&\frac{1}{2}\\ \vdots&\vdots&\ddots&\vdots\\ \frac{1}{2}&\frac{1}{2}&\cdots&0 \ea}_{k\times k} &=\frac{k-1}{2} \sev{\ba{cccc} 1&\frac{1}{2}&\cdots&\frac{1}{2}\\ 1&0&\cdots&\frac{1}{2}\\ \vdots&\vdots&\ddots&\vdots\\ 1&\frac{1}{2}&\cdots&0 \ea}=\frac{k-1}{2}\sev{\ba{cccc} 1&\frac{1}{2}&\cdots&\frac{1}{2}\\ 0&-\frac{1}{2}&\cdots&0\\ \vdots&\vdots&\ddots&\vdots\\ 0&0&\cdots&-\frac{1}{2} \ea}\\ &=\frac{k-1}{2}\cdot \sex{-\frac{1}{2}}^{k-1} =(-1)^{k-1}\frac{k-1}{2^k}. \eea \eeex$$ Thus the PDE is hyperbolic\footnote{In fact, the matrix has eigenvalues $\dps{\lambda_1=\frac{n-1}{2}<0}$, $\dps{\lambda_{2,\cdots,n}=-\frac{1}{2}}$.}.

Exercise3.2.

(1)Let $(r,\theta)$ be polar coordinates in $\bbR^2$, i.e., $$\bex x=r\cos\theta,\quad y=r\sin \theta. \eex$$ Prove that the Laplace operator $\lap$ can be expressed by $$\bex \lap u=u_{rr}+\frac{1}{r}u_r+\frac{1}{r^2}u_{\theta\theta}. \eex$$

(2)Let $(r,\theta,\phi)$ be spherical coordinates in $\bbR^3$, i.e., $$\bex x=r\sin \theta\cos \phi,\quad y=r\sin \theta\sin \phi,\quad z=r\cos \theta. \eex$$ Prove that the Laplace operator $\lap$ can be expressed as $$\bex \lap u=\frac{1}{r^2}\frac{\p}{\p r}\sex{r^2\frac{\p u}{\p r}} +\frac{1}{r^2\sin \theta} \frac{\p}{\p \theta}\sex{\sin \theta\frac{\p u}{\p \theta}} +\frac{1}{r^2\sin^2\theta}\frac{\p^2u}{\p \phi^2}. \eex$$

Solution:

(1)Since $$\bex x_r=\cos \theta,\quad x_\theta=-r\sin \theta=-y,\quad y_r=\sin \theta,\quad y_\theta=r\cos \theta=x, \eex$$ we have $$\beex \bea u_r&=u_x\cos \theta+u_y\sin \theta,\\ u_{rr}&=u_{xx}\cos^2\theta +2u_{xy}\sin \theta\cos \theta +u_{yy}\sin^2\theta,\\ u_\theta&=-yu_x+xu_y,\\ u_{\theta\theta}&= -xu_x-y(-yu_{xx}+xu_{xy})\\ &\quad-yu_y+x(-yu_{xy}+xu_{yy})\\ &=y^2u_{xx}-2xyu_{xy} +x^2u_{yy}-(xu_x+yu_y). \eea \eeex$$ Hence, $$\bex u_{rr}+\frac{1}{r}u_r+\frac{1}{r^2}u_{\theta\theta} =u_{xx}+u_{yy}=\lap u. \eex$$

(2)Let $\rho=r\sin \theta$, then $$\bex \ba{ll} x=\rho \cos \phi,&y=\rho\sin \phi,\\ z=r\cos \theta,&\rho=r\sin \theta. \ea \eex$$ Thus by (1), $$\beex \bea u_{xx}+u_{yy}&=u_{\rho\rho} +\frac{1}{\rho}u_\rho+\frac{1}{\rho^2}u_{\phi\phi},\\ u_{zz}+u_{\rho\rho} &=u_{rr}+\frac{1}{r}u_r+\frac{1}{r^2}u_{\theta\theta}. \eea \eeex$$ Adding these together yields $$\bee\label{3.2:1} \lap u=u_{rr}+\frac{1}{r}u_r+\frac{1}{\rho}u_\rho +\frac{1}{r^2}u_{\theta\theta}+\frac{1}{\rho^2}u_{\phi\phi}. \eee$$ We now calculate $u_\rho$ as $$\bee\label{3.2:2} \bea u_\rho&=u_rr_\rho+u_\theta\theta_\rho\quad\sex{z=r\cos\theta,\ \rho=r\sin \theta,\quad u(z,\rho)=u(r,\theta)}\\ &=\frac{\rho}{r}u_r+\frac{z}{r^2}u_\theta\\ &=u_r\sin \theta+u_\theta\frac{\cos\theta}{r}. \eea \eee$$ Plugging \eqref{3.2:2} into \eqref{3.2:1}, we obtain $$\beex \bea \lap u&=u_{rr}+\frac{1}{r}u_r+\frac{1}{r\sin\theta} \sex{u_r\sin \theta+u_\theta\frac{\cos \theta}{r}} +\frac{1}{r^2}u_{\theta\theta} +\frac{1}{r^2\sin^2\theta}u_{\phi\phi}\\ &=u_{rr}+\frac{2}{r}u_r+ \frac{1}{r^2\sin\theta}\cdot u_\theta\cos \theta +\frac{1}{r^2}u_{\theta\theta} +\frac{1}{r^2\sin^2\theta}u_{\phi\phi}\\ &=\frac{1}{r^2}(r^2u_{rr}+2ru_r) +\frac{1}{r^2\sin\theta} (u_\theta\cos \theta+u_{\theta\theta}\sin \theta) +\frac{1}{r^2\sin^2\theta}u_{\phi\phi}\\ &=\frac{1}{r^2}(r^2u_r)_r +\frac{1}{r^2\sin\theta}(u_\theta\sin \theta)_\theta +\frac{1}{r^2\sin^2\theta}u_{\phi\phi}. \eea \eeex$$

Exercise3.3.

Discuss the uniqueness of the following problems using energy methods:

(1)$\dps{\left\{\ba{ll} \lap u-u^3=f,&\mbox{in }\Omega,\\ u=\varphi,&\mbox{on }\p \Omega; \ea\right.}$

(2)$\dps{\left\{\ba{ll} \lap u-u\int_\Omega u^2(y)\rd y=f,&\mbox{in }\Omega,\\ u=\varphi,&\mbox{on }\p\Omega. \ea\right.}$

Solution:

(1)Let $u_1, u_2$ be two solutions of the PDE, then $v=u_1-u_2$ verifies $$\bee\label{3.3:1} \left\{\ba{ll} \lap v-v(u_1^2+u_1u_2+u_2^2)=0,&\mbox{in }\Omega,\\ u=0,&\mbox{on }\p \Omega. \ea\right. \eee$$ Multiplying $\eqref{3.3:1}_1$ with $-v$, and integrating over $\Omega$, we obtain $$\bex \int_\Omega|\n v|^2\rd x +\int_\Omega v^2\sez{\sex{u_1+\frac{u_2}{2}}^2+\sex{\frac{\sqrt{3}}{2}u_2}^2}\rd x =0. \eex$$ Thus, $v$ is constant, and this constant is zero since $v$ is zero on the boundary.

(2)Let $u_1, u_2$ be two solutions of the PDE, then $v=u_1-u_2$ verifies $$\bee\label{3.3:2} \left\{\ba{ll} \lap v-\sex{u_1\int_\Omega u_1^2\rd y -u_2\int_\Omega u_2^2\rd y}=0,&\mbox{in }\Omega,\\ u=0,&\mbox{on }\p \Omega. \ea\right. \eee$$ Multiplying $\eqref{3.3:1}_1$ with $-v$, and integrating over $\Omega$, we obtain $$\beex \bea &\quad\int_\Omega|\n v|^2\rd x +\int_\Omega (u_1-u_2)\sex{u_1\int_\Omega u_1^2\rd y -u_2\int_\Omega u_2^2\rd y}\rd x\\ &=\int_\Omega |\n v|^2\rd x +\sex{\int_\Omega u_1^2\rd x}^2 -\int_\Omega u_1u_2\rd x\cdot \int_\Omega (u_1^2+u_2^2)\rd x +\sex{\int_\Omega u_2^2\rd x}^2\\ &\geq \int_\Omega |\n v|^2\rd x +\sex{\int_\Omega u_1^2\rd x}^2 -\int_\Omega \frac{u_1^2+u_2^2}{2}\rd x \cdot \int_\Omega (u_1^2+u_2^2)\rd x +\sex{\int_\Omega u_2^2\rd x}^2\\ &\geq \int_\Omega |\n v|^2\rd x\\ &\quad\sex{a=\int_\Omega u_1^2\rd x, b=\int_\Omega u_2^2\rd x\ra a^2+b^2\geq \frac{1}{2}(a+b)^2}. \eea \eeex$$ Consequently, $v$ is constant, and this constant is zero since $v$ is zero on the boundary.

Exercise3.4.

Let $\Omega$ be a bounded $C^1$-domain in $\bbR^n$ and $u$ be a $C^2$-function in $\bar \Omega\times [0,T]$ satisfying $$\bee\label{3.4:1} \ba{rcl} u_t-\lap u=f,&\mbox{in }\Omega\times (0,\infty),\\ u(\cdot,0)=u_0,&\mbox{in }\Omega,\\ u=0,&\mbox{on }\p\Omega \times (0,\infty). \ea \eee$$ Prove that $$\bex \sup_{0\leq t\leq T}\int_\Omega |\n u(\cdot,t)|^2\rd x +\int_0^T \int_\Omega u_t^2\rd x\rd t \leq C\sez{\int_\Omega |\n u_0|^2\rd x +\int_0^T \int_\Omega f^2\rd x\rd t}, \eex$$ where $C$ is a positive constant depending only on $\Omega$.

Solution:  Multiplying $\eqref{3.4:1}_1$ with $u_t$, and integrating over $\Omega$, we obtain $$\beex \bea \int_\Omega u_t^2\rd x +\frac{1}{2}\frac{\rd }{\rd t}\int_\Omega |\n u|^2\rd x =\int_\Omega fu_t\rd x \leq \int_\Omega \frac{f^2+u_t^2}{2}\rd x. \eea \eeex$$ And thus $$\bex \int_\Omega u_t^2\rd x+\frac{\rd }{\rd t}\int_\Omega |\n u|^2\rd x\leq \int_\Omega f^2\rd x. \eex$$ Integrating with respect to $t$, we get the desired result with $C=1$.

Exercise3.5.

Prove that the Poisson kernel in $$\bex (3.3.6)\qquad K(r,\theta,\eta)=\frac{1}{2\pi}+\frac{1}{\pi}\sum_{k=1}^\infty r^k \cos k(\theta -\eta) \eex$$ is given by $$\bex (3.3.7)\qquad K(r,\theta,\eta)=\frac{1}{2\pi}\cdot\frac{1-r^2}{1-2r\cos(\theta -\eta)+r^2}. \eex$$

Solution:  $$\beex \bea K(r,\theta,\eta) &=\frac{1}{2\pi}+\frac{1}{\pi}\sum_{k=1}^\infty r^k\cos k(\theta -\eta)\\ &=\frac{1}{2\pi}+\frac{1}{\pi}\sum_{k=1}^\infty \Re \sez{r^ke^{ik(\theta-\eta)}}\\ &=\Re\sez{\frac{1}{2\pi}+\frac{1}{\pi}\cdot \frac{re^{i(\theta-\eta)}}{1-re^{i(\theta-\eta)}}}\\ &=\frac{1}{2\pi} +\frac{1}{\pi} \frac{re^{i(\theta-\eta)} \sex{1-re^{-i(\theta-\eta)}}}{|1-re^{i(\theta-\eta)}|^2}\\ &=\frac{1}{2\pi}+\frac{1}{\pi} \frac{r\cos(\theta-\eta)-r^2}{1-2r\cos(\theta-\eta)+r^2}\\ &=\frac{1}{2\pi}\cdot \frac{1-r^2}{1-2r\cos(\theta -\eta)+r^2}. \eea \eeex$$

Exercise3.6.

For any $u_0\in L^2(0,\pi)$, let $u$ be given by $$\bex (3.3.11)\qquad u(x,t)=\sqrt{\frac{2}{\pi}}\sum_{k=1}^\infty a_ke^{-k^2t}\sin kx. \eex$$ For any nonnegative integers $i$ and $j$, prove $$\bex \sup_{[0,\pi]}|\p^i_x\p^j_t u(\cdot,t)|\to 0,\mbox{ as }t\to\infty. \eex$$

Solution:  $$\beex \bea \sup_{[0,\pi]}|\p^i_x\p^j_t u(\cdot,t)| &\leq \sqrt{\frac{2}{\pi}} \sum_{k=1}^\infty k^{i+2j}|a_k|e^{-k^2t}\\ &\leq \sqrt{\frac{2}{\pi}} e^{-\frac{t}{2}} \sex{\sum_{k=1}^\infty k^{2(i+2j)}e^{-k^2t}}^\frac{1}{2} \sex{\sum_{k=1}^\infty a_k^2}^\frac{1}{2}\\ &\to 0,\quad(t\to\infty). \eea \eeex$$

Exercise3.7.

Let $G$ be defined as in (3.3.16), $$\bex G(x,y;t)=\frac{2}{\pi}\sum_{k=1}^\infty e^{-k^2t}\sin kx\sin ky. \eex$$ Prove $$\bex |G(x,y;t)|\leq\frac{1}{\sqrt{\pi t}},\mbox{ for all }x,y\in [0,\pi]\mbox{ and }t>0. \eex$$

Solution:  $$\beex \bea |G(x,y;t)| &\leq \frac{2}{\pi}\sum_{k=1}^\infty e^{-k^2t}\\ &\leq \frac{2}{\pi}\sum_{k=1}^\infty \int_{k-1}^k e^{-x^2t}\rd x\\ &=\frac{2}{\pi}\int_0^\infty e^{-x^2t}\rd x\\ &=\frac{2}{\pi \sqrt{t}}\int_0^\infty e^{-s^2}\rd s\\ &=\frac{2}{\pi \sqrt{t}}\cdot \frac{\sqrt{\pi}}{2}\\ &\quad\sex{\sex{\int_0^\infty e^{-s^2}\rd s}^2 =\iint_{\bbR^2}e^{-(s^2+t^2)}\rd s\rd t =\int_0^\infty 2\pi re^{-r^2}\rd r=\pi}\\ &=\frac{1}{\sqrt{\pi t}}. \eea \eeex$$

Exercise3.8.

For any $u_0\in L^2(0,\pi)$, solve the following problem by separation of variables: $$\bee\label{3.8.1} \ba{ll} u_t-u_{xx}=0,&\mbox{in }(0,\pi)\times (0,\infty),\\ u(x,0)=u_0(x),&\mbox{for any }x\in (0,\pi),\\ u_x(0,t)=u_x(\pi,t)=0,&\mbox{for any }t\in (0,\infty). \ea \eee$$

Solution:

(1)We first find solutions of $$\bex \ba{ll} u_t-u_{xx}=0,&\mbox{in }(0,\pi)\times (0,\infty),\\ u_x(0,t)=u_x(\pi,t)=0,&\mbox{for any }t\in (0,\infty), \ea \eex$$ which has the form $$\bex u(x,t)=a(t)w(x). \eex$$ Then $$\bex a'(t)w(x)-a(t)w''(x)=0\ra \frac{a'(t)}{a(t)}=-\lambda=\frac{w''(x)}{w(x)}. \eex$$ Consequently, $$\bee\label{3.8.2} a'(t)+\lambda a(t)=0, \eee$$ and $$\bee\label{3.8.3} w''(x)+\lambda w(x)=0,\quad w'(0)=w'(\pi)=0. \eee$$ To ensure \eqref{3.8.3} has non-trivial solution, we need $$\bex \lambda=k^2,\quad k=0,1,2,\cdots, \eex$$ and the corresponding solution is $$\bex \sqrt{\frac{2}{\pi}}\cos kx. \eex$$ Now that $\lambda=k^2$, we see by \eqref{3.8.2} that $$\bex a(t)=e^{-k^2t}, \eex$$ and $$\bex u(x,t)=e^{-k^2t}\cos kx. \eex$$

(2)Now, we suppose that the solution to \eqref{3.8.1} is $$\bex u(x,t)=\sqrt{\frac{2}{\pi}} \frac{a_0}{2} +\sqrt{\frac{2}{\pi}}\sum_{k=1}^\infty a_ke^{-k^2t}\cos kx. \eex$$ Then $$\bex u_0(x)=u(x,0)=\sqrt{\frac{2}{\pi}} \frac{a_0}{2} +\sqrt{\frac{2}{\pi}}\sum_{k=1}^\infty a_k\cos kx. \eex$$ Consequently, $a_k$ is the Fourier coefficients of $u_0$, $$\bex a_k=\sqrt{\frac{2}{\pi}}\int_0^\pi u_0(x)\cos kx\rd x. \eex$$

Exercise3.9.

For any $u_0\in L^2(0,\pi)$ and $f\in L^2((0,\pi)\times (0,\infty))$, find a formal explicit expression of a solution of the problem $$\bee\label{3.9.1} \ba{ll} u_t-u_{xx}=f,&\mbox{in }(0,\pi)\times (0,\infty),\\ u(x,0)=u_0(x),&\mbox{for any }x\in (0,\pi),\\ u(0,t)=u(\pi,t)=0,&\mbox{for any }t\in (0,\infty). \ea \eee$$

Solution:

(1)As is well-known, the problem $$\bex w''(x)+\lambda w(x)=0,\quad w(0)=w(\pi)=0 \eex$$ has non-trivial solution iff $$\bex \lambda=k^2,\quad k=1,2,\cdots, \eex$$ and the corresponding solution is $$\bex w=\sqrt{\frac{2}{\pi}}\sin kx. \eex$$

(2)We now suppose the solution of \eqref{3.9.1} is $$\bex u(x,t)=\sqrt{\frac{2}{\pi}}\sum_{k=1}^\infty u_k(t) \sin kx. \eex$$ Then $$\bex \ba{rcl} \sqrt{\frac{2}{\pi}}\sum_{k=1}^\infty [u_k'(t)+k^2u_k(t)]\sin kx&=&f(x,t),\\ u(x,0)=\sqrt{\frac{2}{\pi}}\sum_{k=1}^\infty u_k(0)\sin kx&=&u_0(x). \ea \eex$$ Hence, $$\bee\label{3.9.2} \ba{rcl} u_k'(t)+k^2u_k(t)&=&a_k(t),\\ u_k(0)&=&b_k, \ea \eee$$ where $a_k,b_k$ are the Fourier coefficients of $f$, $u_0$ respectively, $$\bex \ba{rcl} a_k(t)&=&\sqrt{\frac{2}{\pi}}\int_0^\pi f(x,t)\sin kx\rd x,\\ b_k&=&\sqrt{\frac{2}{\pi}}\int_0^\pi u_0(x)\sin kx\rd x. \ea \eex$$ Solving \eqref{3.9.2} yields $$\bex u_k(t)=b_ke^{-k^2t}+\int_0^t a_k(s)e^{k^2(s-t)}\rd s, \eex$$ and thus $$\bex u(x,t)=\sqrt{\frac{2}{\pi}}\sum_{k=1}^\infty \sez{b_ke^{-k^2t}+\int_0^t a_k(s)e^{k^2(s-t)}\rd s}\sin kx. \eex$$

Exercise3.10.

For any $u_0,u_1\in L^2(0,\pi)$ and $f\in L^2((0,\pi)\times(0,\infty))$, find a formal explicit expression of a solution of the problem $$\bee\label{3.10.1} \ba{ll} u_{tt}-u_{xx}=f,&\mbox{in }(0,\pi)\times (0,\infty),\\ u(x,0)=u_0(x),\ u_t(x,0)=u_1(x),&\mbox{for any }x\in (0,\pi),\\ u(0,t)=u(\pi,t)=0,&\mbox{for any }t\in (0,\infty). \ea \eee$$

Solution:

(1)As is well-known, the problem $$\bex w''(x)+\lambda w(x)=0,\quad w(0)=w(\pi)=0 \eex$$ has non-trivial solution iff $$\bex \lambda=k^2,\quad k=1,2,\cdots, \eex$$ and the corresponding solution is $$\bex w=\sqrt{\frac{2}{\pi}}\sin kx. \eex$$

(2)We now suppose the solution of \eqref{3.10.1} is $$\bex u(x,t)=\sqrt{\frac{2}{\pi}}\sum_{k=1}^\infty u_k(t) \sin kx. \eex$$ Then $$\bex \ba{rcl} \sqrt{\frac{2}{\pi}}\sum_{k=1}^\infty [u_k'(t)+k^2u_k(t)]\sin kx&=&f(x,t),\\ u(x,0)=\sqrt{\frac{2}{\pi}}\sum_{k=1}^\infty u_k(0)\sin kx&=&u_0(x),\\ u_t(x,0)=\sqrt{\frac{2}{\pi}}\sum_{k=1}^\infty u_k'(0)\sin kx&=&u_1(x). \ea \eex$$ Hence, $$\bee\label{3.10.2} \ba{rcl} u_k''(t)+k^2u_k(t)&=&a_k(t),\\ u_k(0)&=&b_k,\\ u_k'(0)&=&c_k, \ea \eee$$ where $a_k,b_k,c_k$ are the Fourier coefficients of $f$, $u_0$ and $u_1$ respectively, $$\bex \ba{rcl} a_k(t)&=&\sqrt{\frac{2}{\pi}}\int_0^\pi f(x,t)\sin kx\rd x,\\ b_k&=&\sqrt{\frac{2}{\pi}}\int_0^\pi u_0(x)\sin kx\rd x,\\ c_k&=&\sqrt{\frac{2}{\pi}}\int_0^\pi u_1(x)\sin kx\rd x. \ea \eex$$ Solving \eqref{3.10.2} yields $$\bex u_k(t)=b_k\cos kt+\frac{c_k}{k}\sin kt+\tilde u_k(t), \eex$$ where $\tilde u_k(t)$ is a particular solution of \eqref{3.10.2}, which could be calculated only if $k$ and $a_k(t)$ are known. In summary, we have $$\bex u(x,t)=\sqrt{\frac{2}{\pi}}\sum_{k=1}^\infty \sez{b_k\cos kt+\frac{c_k}{k}\sin kt+\tilde u_k(t)}\sin kx. \eex$$

Exercise3.11.

Let $T$ be a positive constant, $\Omega$ be a bounded $C^1$-domain in $\bbR^n$ and $u$ be $C^2$ in $x$ and $C^1$ in $t$ in $\bar \Omega \times [0,T]$. Suppose $u$ satisfies $$\bee\label{3.11.1} \ba{ll} u_t-u_{xx}=0,&\mbox{in }\Omega\times (0,T),\\ u(\cdot,T)=0,&\mbox{in }\Omega,\\ u=0,&\mbox{on }\p \Omega\times (0,T). \ea \eee$$ Prove that $u=0$ in $\Omega \times (0,T)$. {\sl Hint:} The function $\dps{J(t)=\ln \int_\Omega u^2(x,t)\rd x}$ is a decreasing convex function.

Solution:  Denote by $$\bex J(t)=\ln \int_\Omega u^2\rd x. \eex$$ Then\footnote{The last inequality follows from the fact $$\bex \sez{\int_\Omega |\n u|^2\rd x}^2 =\sez{\int_\Omega \n u\cdot\n u\rd x}^2 =\sez{\int_\Omega u\cdot \lap u\rd x}^2 \leq \int_\Omega |\lap u|^2\rd x\cdot \int_\Omega u^2\rd x. \eex$$} $$\beex \bea J'(t)&=\frac{ \int_\Omega2uu_t\rd x}{ \int_\Omega u^2\rd x}=\frac{\int_\Omega 2u\lap u\rd x}{\int_\Omega u^2\rd x}\leq 0,\\ J''(t)&=-2\frac{\int_\Omega 2\n u\cdot \n u_t\rd x \cdot \int_\Omega u^2\rd x-\int_\Omega |\n u|^2\rd x\cdot \int_\Omega 2uu_t\rd x}{\sez{\int_\Omega u^2\rd x}^2}\\ &=4\frac{ -\int_\Omega \n u\cdot\n\lap u\rd x\cdot \int_\Omega u^2\rd x+\int_\Omega |\n u^2|\rd x\cdot \int_\Omega u\lap u\rd x }{ \sez{\int_\Omega u^2\rd x}^2 }\\ &=4\frac{\int_\Omega |\n^2u|^2\rd x\cdot \int_\Omega u^2\rd x-\sez{\int_\Omega |\n u|^2\rd x}^2}{\sez{\int_\Omega u^2\rd x}^2} \geq 0. \eea \eeex$$ Thus $J$ is decreasing\footnote{This ensures the $J(t)$ is well-defined on $(0,T)$.} and convex. Consequently, for $t=(1-\theta)t_1+\theta t_2$, $$\bex J(t)\leq (1-\theta)J(t_1)+\theta J(t_2), \eex$$ $$\bex \int_\Omega u^2(x,t)\rd x \leq \sez{\int_\Omega u^2(x,t_1)\rd x}^{1-\theta} \cdot \sez{\int_\Omega u^2(x,t_2)\rd x}^\theta. \eex$$ Taking $t_0=0$, $t_2=T$, $\theta=t/T$, we deduce $$\bex \int_\Omega u^2(x,t)\rd x \leq \sez{\int_\Omega u^2(x,0)\rd x}^{1-t/T} \cdot \sez{\int_\Omega u^2(x,T)\rd x}^{t/T}=0. \eex$$ The proves the statement.

Exercise3.12.

Classify homogeneous harmonic polynomials in $\bbR^3$ by following the steps outlined below. let $(r,\theta,\phi)$ be spherical coordinates in $\bbR^3$. (Refer to Exercse 3.2.) Suppose $u$ is a homogenous harmonic polynomial of degree $m$ in $\bbR^3$ and set $u=r^mQ_m(\theta,\phi)$ for some function $Q_m$ defined in $\mathbb{S}^2$.

(1)Prove that $Q_m$ satisfies $$\bex m(m+1)Q_m+\frac{1}{\sin \theta}\frac{\p}{\p \theta}\sex{\sin \theta\frac{\p Q_m}{\p \theta}} +\frac{1}{\sin^2\theta}\frac{\p^2Q_m}{\p \phi^2}=0. \eex$$

(2)Prove that, if $Q_m$ is of the form $f(\theta)g(\phi)$, then $$\bex Q_m(\theta,\phi)=(A\cos k\phi+B\sin k\phi)f_{m,k}(\cos\theta), \eex$$ where $$\bex f_{m,k}(\mu)=(1-\mu^2)^\frac{k}{2} \frac{\rd^{m+k}}{\rd \mu^{m+k}}(1-\mu^2)^m,\quad \mbox{for }\mu\in [-1,1], \eex$$ for $k=0,1,\cdots,m$.

(3)Sketch the zero set of $Q_m$ on $\mathbb{S}^2$ according to $k=0$, $1\leq k\leq m-1$, $k=m$.

Solution:

(1)Now that $u=r^mQ_m(\theta,\phi)$, and by Exercise 3.2, $$\bex \lap u=\frac{1}{r^2}\frac{\p}{\p r}\sex{r^2\frac{\p u}{\p r}} +\frac{1}{r^2\sin \theta} \frac{\p}{\p \theta}\sex{\sin \theta\frac{\p u}{\p \theta}} +\frac{1}{r^2\sin^2\theta}\frac{\p^2u}{\p \phi^2}, \eex$$ we have $$\beex \bea \frac{1}{r^2}\frac{\p}{\p r}\sex{r^2\frac{\p u}{\p r}} &=\frac{1}{r^2}(r^m mr^{m-1})_rQ_m(\theta,\phi) =m(m+1)r^{m-2}Q_m,\\ \frac{1}{r^2\sin \theta} \frac{\p}{\p \theta}\sex{\sin \theta\frac{\p u}{\p \theta}} &=\frac{1}{r^2\sin \theta}\sex{r^m\frac{\p Q_m}{\p \theta}}_\theta=r^{m-2}\frac{1}{\sin \theta}\frac{\p}{\p \theta}\sex{\sin \theta\frac{\p Q_m}{\p \theta}},\\ \frac{1}{r^2\sin^2\theta}\frac{\p^2u}{\p \phi^2} &=r^{m-2}\frac{1}{\sin^2\theta}\frac{\p^2Q_m}{\p \phi^2}. \eea \eeex$$ Consequently, we have $$\bex m(m+1)Q_m+\frac{1}{\sin \theta}\frac{\p}{\p \theta}\sex{\sin \theta\frac{\p Q_m}{\p \theta}} +\frac{1}{\sin^2\theta}\frac{\p^2Q_m}{\p \phi^2}=0. \eex$$

(2)If $Q_m=f(\theta)g(\phi)$, then $$\bex m(m+1)fg +\frac{1}{\sin \theta} (f'\sin \theta)'g+\frac{f}{\sin^2\theta}g''=0. \eex$$ Dividing by $fg,$ we obtain $$\bex m(m+1)+\frac{1}{\sin\theta}\frac{(f'\sin \theta)'}{f} +\frac{1}{\sin^2\theta}\frac{g''}{g}=0. \eex$$ Thus $$\bex \sin^2\theta\sez{m(m+1)+\frac{1}{\sin\theta}\frac{(f'\sin\theta)'}{f}} =\lambda=-\frac{g''}{g}. \eex$$ Consequently, $$\bee\label{3.12.1} g''+\lambda g(0)=g(2\pi),\quad g'(0)=g'(2\pi); \eee$$ $$\bee\label{3.12.2} m(m+1)+\frac{1}{\sin\theta}\frac{(f'\sin \theta)'}{f}=\frac{\lambda}{\sin^2\theta}. \eee$$ Solving \eqref{3.12.1} yields $$\bex \ba{ll} \lambda_0=0,&g_0=1;\\ \lambda_k=k^2(k\in \bbZ_+),&g_k=A\cos k\phi+B\sin k\phi. \ea \eex$$ Plugging $\lambda_k=k^2(k\in\bbN)$ into \eqref{3.12.2}, we find $$\bex \frac{1}{\sin\theta}(f'\sin\theta)' +\sez{m(m+1)-\frac{k^2}{\sin^2\theta}}f=0, \eex$$ $$\bee\label{3.12.3} f''+f'\cdot \cot \theta+\sez{m(m+1)-\frac{k^2}{\sin^2\theta}}f=0. \eee$$ Setting $$\bex f(\theta)=y(\mu),\quad \mu=\cos\theta, \eex$$ we see $$\bex f'=-y'\sin\theta,\quad f''=y''\sin^2\theta-y'\cos\theta, \eex$$ and \eqref{3.12.3} becomes $$\bee\label{3.12.4} \bea 0&=(y''\sin^2\theta-y'\cos\theta)-y'\cos\theta +\sez{m(m+1)-\frac{k^2}{\sin^2\theta}}y\\ &=y''\sin^2\theta -2y'\cos\theta +\sez{m(m+1)-\frac{k^2}{\sin^2\theta}}y\\ &=(1-\mu^2)y''-2\mu y' +\sez{m(m+1)-\frac{k^2}{1-\mu^2}}y. \eea \eee$$ Now defining $$\bex y(\mu)=(1-\mu^2)^\frac{k}{2}z(\mu), \eex$$ we have $$\beex \bea y'&=-k\mu(1-\mu^2)^{\frac{k}{2}-1}z +(1-\mu^2)^{\frac{k}{2}}z',\\ y''&=-k(1-\mu^2)^{\frac{k}{2}-1}z +k(k-2)\mu^2(1-\mu^2)^{\frac{k}{2}-2}z -2k\mu(1-\mu^2)^{\frac{k}{2}-1}z' +(1-\mu^2)^{\frac{k}{2}}z'', \eea \eeex$$ and \eqref{3.12.4} reduces to $$\beex \bea 0&=-k(1-\mu^2)^{\frac{k}{2}}z+k(k-2)\mu^2(1-\mu^2)^{\frac{k}{2}-1}z -2k\mu(1-\mu^2)^{\frac{k}{2}}z' +(1-\mu^2)^{\frac{k}{2}+1}z''\\ &\quad+2k\mu^2(1-\mu^2)^{\frac{k}{2}-1}z-2\mu(1-\mu^2)^{\frac{k}{2}}z' +m(m+1)(1-\mu^2)^{\frac{k}{2}}z -k^2(1-\mu^2)^{\frac{k}{2}-1}z\\ &=(1-\mu^2)^{\frac{k}{2}-1}z \sez{-k(1-\mu^2)+k(k-2)\mu^2+2k\mu^2+m(m+1)(1-\mu)^2-k^2}\\ &\quad-2(k+1)\mu(1-\mu^2)^{\frac{k}{2}}z'+(1-\mu^2)^{\frac{k}{2}+1}z''\\ &=(1-\mu^2)^{\frac{k}{2}}[m(m+1)-k(k+1)]z -2(k+1)\mu(1-\mu^2)^{\frac{k}{2}}z'+(1-\mu^2)^{\frac{k}{2}+1}z''. \eea \eeex$$ Hence, $$\bee\label{3.12.5} (1-\mu^2)z'' -2(k+1)\mu z' +[m(m+1)-k(k+1)]z=0. \eee$$ Observing that if $k=0$, \eqref{3.12.5} is the Legendre equation $$\bex (1-\mu^2)z''-2\mu z'+m(m+1)z=0, \eex$$ which has the solution\footnote{参见丁同仁、李承志编《常微分方程教程(第二版)》第 7.3 节; 要删除在 $\pm 1$ 处不收敛的幂级数解.} $$\bex z_{m,0}=\frac{\rd^m}{\rd \mu^m}(1-\mu^2)^m,\quad \mu\in [-1,1]. \eex$$ Now, for general $k$, we differentiate \eqref{3.12.5} with respect to $\mu$, $$\bee\label{3.12.6} (1-\mu^2)(z')'' -2(k+2)\mu(z')' +[m(m+1)-(k+1)(k+2)]z'=0. \eee$$ Comparing \eqref{3.12.6} with \eqref{3.12.5}, we find, by denoting the solution of \eqref{3.12.5} as $z_{m,k}$, that $$\bex z_{m,k}=z_{m,k-1}'=\cdots =z_{m,0}^{(k)} =\frac{\rd^{m+k}}{\rd \mu^{m+k}}(1-\mu^2)^m. \eex$$ Now that $z_{m,0}$ is a polynomial of degree $m$, the non-trivial solution exists iff $k=0,1,\cdots,m$. Returning back to the $\theta$ variable, we get $$\beex \bea f_{m,k}(\theta)&=y_{m,k}(\mu)\quad(\mu=\cos\theta)\\ &=(1-\mu^2)^\frac{k}{2}z_{m,k}(\mu)\\ &=(1-\mu^2)^\frac{k}{2}\frac{\rd^{m+k}}{\rd \mu^{m+k}}(1-\mu^2)^m|_{\mu=\cos\theta}. \eea \eeex$$

(3)In case $k=0$, $$\bex Q_m=A\frac{\rd^m}{\rd \mu^m}(1-\mu^2)^m|_{\mu=\cos\theta}, \eex$$ which is a polynomial of $\cos\theta$, with degree $m$. By the fundamental theorem of algebra, there exists at most $m$ zeros of $Q_m$\footnote{Note that $0\leq \theta\leq \pi$.}. In case $1\leq k\leq m-1$, $$\bex Q_m=\sqrt{A^2+B^2}\cos(k\phi-\phi_0) (1-\mu^2)^\frac{k}{2}\frac{\rd^{m+k}}{\rd \mu^{m+k}}(1-\mu^2)^m|_{\mu=\cos\theta}, \eex$$ with appropriate $\phi_0$. Hence, there exists at most $$\bex 1+2+(m-k)=m-k+3 \eex$$ zeros. In case $k=m$, similar arguments can be applied to deduce that there exists at most $$\bex 1+2=3 \eex$$ zeros.

4.Laplace Equations

Exercise4.1.

Suppose $u(x)$ is harmonic in some domain in $\bbR^n$. Prove that $$\bex v(x)=|x|^{2-n}u\sex{\frac{x}{|x|^2}} \eex$$ is also harmonic in a suitable domain.

Solution:  $$\beex \bea v_{x_i}&=(2-n)\frac{x_i}{|x|^n}u+|x|^{2-n}\sum_j u_{y_j}\sex{\frac{\delta_{ij}}{|x|^2}-2\frac{x_ix_j}{|x|^4}},\\ v_{x_ix_i} &=(2-n)\sex{\frac{1}{|x|^n}-\frac{nx_i^2}{|x|^{n+2}}}u +(2-n)\sum_j \frac{x_i}{|x|^n} u_{y_j}\sex{\frac{\delta_{ij}}{|x|^2}-2\frac{x_ix_j}{|x|^4}}\\ &\quad+(2-n)\frac{x_i}{|x|^n} \sum_ju_j\sex{\frac{\delta_{ij}}{|x|^2}-2\frac{x_ix_j}{|x|^4}}\\ &\quad +|x|^{2-n}\sum_{j,k}u_{y_jy_k} \sex{\frac{\delta_{ij}}{|x|^2}-2\frac{x_ix_j}{|x|^4}} \sex{\frac{\delta_{ik}}{|x|^2}-2\frac{x_ix_k}{|x|^4}}\\ &\quad +|x|^{2-n}\sum_j u_{y_j} \sex{\frac{-2\delta_{ij}x_i}{|x|^4} -2\frac{x_j+x_i\delta_{ij}}{|x|^4} +8\frac{x_i^2x_j}{|x|^6}}\quad\sex{(x_j)_{x_i}=\delta_{ij}},\\ \lap v &=2(2-n)\sum_{i,j} \frac{x_i}{|x|^n} u_{y_j}\sex{\frac{\delta_{ij}}{|x|^2}-2\frac{x_ix_j}{|x|^4}}\\ &\quad +|x|^{2-n}\sum_{j,k}u_{y_jy_k} \sex{\frac{\delta_{ij}}{|x|^2}-2\frac{x_ix_j}{|x|^4}} \sex{\frac{\delta_{ik}}{|x|^2}-2\frac{x_ix_k}{|x|^4}}\\ &\quad +|x|^{2-n}\sum_j u_{y_j} \sex{\frac{-2\delta_{ij}x_i}{|x|^4} -2\frac{x_j+x_i\delta_{ij}}{|x|^4} +8\frac{x_i^2x_j}{|x|^6}}\\ &=\frac{2(2-n)}{|x|^{n+2}}\sum_i x_iu_{y_i} -\frac{4(2-n)}{|x|^{n+2}}\sum_j x_ju_{y_j}\\ &\quad +\frac{1}{|x|^{n+2}}\lap_yu -\frac{2}{|x|^{n+4}}\sum_{i,k} x_ix_ku_{y_iy_k} -\frac{2}{|x|^{n+4}}\sum_{i,j} x_ix_ju_{y_jy_i} +\frac{4}{|x|^{n+4}}\sum_{j,k} x_jx_ku_{y_jy_k}\\ &\quad -\frac{2}{|x|^{n+2}}\sum_i x_iu_{y_i} -\sex{\frac{2n}{|x|^{n+2}}\sum_j x_ju_{y_j} +\frac{2}{|x|^{n+2}}\sum_i x_iu_{y_i}} +\frac{8}{|x|^{n+2}}\sum_j x_ju_{y_j}\\ &=\sez{-2(2-n)-2-2n-2+8}\frac{1}{|x|^{n+2}}\sum_i x_iu_{y_i}\\ &=0. \eea \eeex$$

Exercise4.2.

For $n=2$, find the Green's function for the Laplace operator on the first quadrant.

Solution:  We use complex variable in $\bbR^2$. Then the Green's function for the first quadrant $\Omega$ is $$\bex G(z,w)=\vGa(|w-z|)+\vGa(|w+z|)-\vGa(|w-\bar z|)-\vGa(|w+ \bar z|) =\vGa\sex{\frac{|w-z|\cdot |w^2-z^2|}{\cdot |w^2-\bar z^2|}}, \eex$$ where $\dps{\vGa(x)=\frac{1}{2\pi}\ln |x|}$ is the fundamental solution of the two-dimensional Laplacian. Indeed, for $z\in \omega$, we have $-z,\bar z, -\bar z\not\in \Omega$, and hence $G(z,\cdot)$ satisfies $$\bex \lap_wG(z,w)=0,\quad w\neq z. \eex$$ For the boundary behavior, we have $$\beex \bea \Re w=0&\ra |w-z|=|w+\bar z|,\ |w+\bar z|=|w-\bar z|,\\ \Im w=0&\ra |w-z|=|w-\bar z|,\ |w+\bar z|=|w+z|. \eea \eeex$$ Consequently, $\vGa(z,\cdot)|_{\p \Omega}=0$.  \bem In rectangular coordinates, we have $$\bex G((x_1,x_2),(y_1,y_2))=\frac{1}{2\pi}\ln\sqrt{\frac{[(y_1-x_1)^2+(y_2-x_2)^2]\cdot [(y_1+x_1)^2+(y_2+x_2)^2]}{[(y_1-x_1)^2+(y_2+x_2)^2]\cdot [(y_1+x_1)^2+(y_2-x_2)^2]}}. \eex$$ \eem

Exercise4.3.

Find the Green's function for the Laplace operator in the upper half-space $\sed{x_n>0}$ and then derive a formal integral representation for a solution of the Dirichlet problem $$\bex \ba{ll} \lap u=0,&\mbox{in }\sed{x_n>0};\\ u=\varphi,&\mbox{on }\sed{x_n=0}. \ea \eex$$

Solution:  The Green's function for the half-plane $\Omega =\sed{x_n>0}$ is $$\bex G(x,y)=\vGa(x-y)-\vGa(x-\bar y), \eex$$ where $\bar y=(y',-y_n)$ if $y=(y',y_n)$. Thus $$\beex \bea u(x) &=\int_{\p \Omega}\frac{\p G(x,y)}{\p \nu_y}\varphi(y)\rd S_y\\ &=\int_{\bbR^{n-1}} -\frac{\p}{\p y_n} \sez{\vGa(x'-y',x_n-y_n)-\vGa(x'-y',x_n+y_n)}\varphi(y')\rd y'\\ &=\int_{\bbR^{n-1}} \sez{\frac{x_n-y_n}{\omega_n|x-y|^n} +\frac{x_n+y_n}{\omega_n|x-y|^n}}\varphi(y')\rd y'\\ &=\frac{2x_n}{\omega_n}\int_{\bbR^{n-1}} \frac{\varphi(y')}{(|x'-y'|^2+x_n^2)^{n/2}}\rd y'. \eea \eeex$$

Exercise4.4.

(1)Suppose $u$ is a nonnegative harmonic funciton in $B_R(x_0)\subset \bbR^n$. Prove by the Poisson integral formula the following Harnack inequality: $$\bex \sex{\frac{R}{R+r}}^{n-2}\frac{R-r}{R+r}u(x_0) \leq u(x)\leq \sex{\frac{R}{R-r}}^{n-2}\frac{R+r}{R-r}u(x_0), \eex$$ where $r=|x-x_0|<R$.

(2)Prove by (1) the Liouville theorem: If $u$ is a harmonic function in $\bbR^n$ and bounded above or below, then $u$ is constant.

Solution:

(1)Up to a translation, we may assume $x_0=0$. Then by the Poisson' integral formula, $$\bex u(x)=\int_{\p B_R(0)}\frac{R^2-|x|^2}{\omega_n R|x-y|^n}u(y)\rd S_y. \eex$$ Since $$\bex R-r=|y|-|x|\leq |x-y|\leq |y|+|x| =R+r, \eex$$ we have $$\beex \bea u(x)&\leq \frac{R^2-r^2}{\omega_n R(R-r)^n}\int_{\p B_R(0)}u(y)\rd S_y\\ &=\frac{(R+r)R^{n-2}}{(R-r)^{n-1}}\cdot \frac{1}{\omega_nR^{n-1}}\int_{\p B_R(0)}u(y)\rd S_y\\ &=\sex{\frac{R}{R-r}}^{n-2}\frac{R+r}{R-r}u(x_0);\\ u(x)&\geq \frac{R^2-r^2}{\omega_nR(R+r)^n} \int_{\p B_R(0)}u(y)\rd S_y\\ &=\frac{(R-r)R^{n-2}}{(R+r)^{n-1}}\cdot \frac{1}{\omega_nR^{n-1}}\int_{\p B_R(0)}u(y)\rd S_y\\ &=\sex{\frac{R}{R+r}}^{n-2}\frac{R-r}{R+r}u(x_0). \eea \eeex$$

(2)Taking $R\to \infty$ in (1), we find $u(x)=u(x_0)$, $\forall\ x\in\bbR^n$.

Exercise4.5.

Let $u$ be a harmonic function in $\bbR^n$ with $\dps{\int_{\bbR^n} |u|^p\rd x<\infty}$ for some $p\in (1,\infty)$. Prove that $u\equiv 0$.

Solution:  By H\"older inequality, $$\beex \bea |u(x)| &=\sev{\frac{n}{\omega_n R^n}\int_{B_R(x)}u(y)\rd y}\\ &\leq \frac{n}{\omega_n R^n} \sex{\int_{B_R(x)}|u(y)|^p\rd y}^\frac{1}{p} \sex{\int_{B_R(x)}1^\frac{p}{p-1}\rd y} ^\frac{p-1}{p}\\ &\leq \frac{n}{\omega_n R^\frac{n}{p}} \sex{\int_{\bbR^n}|u(y)|^p\rd y}^\frac{1}{p}\\ &\to 0\quad(R\to\infty). \eea \eeex$$ Thus, $u\equiv 0$.

Exercise4.6.

Let $m$ be a positive integer and $u$ be a harmonic function in $\bbR^n$ with $u(x)=O(|x|^m)$ as $|x|\to \infty$. Prove that $u$ is a polynomial of degree at most $m$.

Solution:  For any multi-index $\alpha$ with $|\alpha|=m+1$, $$\beex \bea |\p^\alpha u(x)| &\leq \frac{C^{m+1}e^m(m+1)!}{R^{m+1}} \max_{\bar B_R(x)}|u|\quad\sex{Theorem\ 4.1.12}\\ &\leq \frac{C^{m+1}e^m(m+1)!}{R^{m+1}}\max_{\p B_R(x)}|u| \quad\sex{Theorem\ 4.2.3}\\ &\leq \frac{C^{m+1}e^m(m+1)!}{R^{m+1}}\cdot \max_{y\in \p B_R(x)}\frac{|u(y)}{|y|^m}\cdot (R+|x|)^m\\ &\to 0\quad(R\to\infty). \eea \eeex$$ Thus, $\p^\alpha u\equiv 0$, $\forall\ \alpha$ with $|\alpha|=m+1$. This verifies the statement.

Exercise4.7.

Suppose $u\in C(\bar B^+_1)$ is harmonic in $B_1^+=\sed{x\in B_1;\ x_n>0}$ with $u=0$ on $\sed{x_n=0}\cap B_1$. Prove that the odd extension of $u$ to $B_1$ is harmonic in $B_1$.

Solution:  The odd extension of $u$ to $B_1$ is $$\bex \tilde u(x',x_n)=\left\{\ba{ll} u(x',x_n),&x\in B_1^+,\\ 0,&x\in \sed{x_n=0}\cap B_1,\\ -u(x',-x_n),&x\in B_1^-. \ea\right. \eex$$ Then $$\bex \lap \tilde u=0,\quad x\in B_1^+\cup B_1^-, \eex$$ and $u\in C(\bar B_1)$. Now, by Poisson's integral formula, we can find a $v$ satisfying $$\bex \ba{ll} \lap v=0,&\mbox{in }B_1,\\ v=\tilde u,&\mbox{on }\p B_1. \ea \eex$$ Moreover, on $\sed{x_n=0}$, $$\beex \bea v(x',0)&=\int_{\p B_1}\frac{R^2-|x|^2}{\omega_n R(|x'-y'|^2+y_n^2)^\frac{n}{2}}\tilde u(y',y_n)\rd S_{y',y_n}\\ &=\int_{\p B_1\cap \sed{y_n>0}}\frac{R^2-|x|^2}{\omega_n R(|x'-y'|^2+y_n^2)^\frac{n}{2}}u(y',y_n)\rd S_{y',y_n}\\ &\quad +\int_{\p B_1\cap \sed{y_n<0}}\frac{R^2-|x|^2}{\omega_n R(|x'-y'|^2+y_n^2)^\frac{n}{2}}[-u(y',-y_n)]\rd S_{y',y_n}\\ &=\int_{\p B_1\cap \sed{y_n>0}}\frac{R^2-|x|^2}{\omega_n R(|x'-y'|^2+y_n^2)^\frac{n}{2}}u(y',y_n)\rd S_{y',y_n}\\ &\quad +\int_{\p B_1\cap \sed{z_n>0}}\frac{R^2-|x|^2}{\omega_n R(|x'-y'|^2+z_n^2)^\frac{n}{2}}[-u(y',z_n)]\rd S_{y',z_n}\\ &\quad\sex{\mbox{In the second integral,} y_n=-z_n}\\ &=0. \eea \eeex$$ Consequently, $$\bex \ba{ll} \lap \tilde u=\lap v=0,&\mbox{in }B_1^+,\\ \tilde u=v,&\mbox{on }\p B_1^+;\\ \lap \tilde u=\lap v=0,&\mbox{in }B_1^-,\\ \tilde u=v,&\mbox{on }\p B_1^-. \ea \eex$$ By the maximum principle, $$\bex \tilde u=v,\mbox{in }B_1^+;\quad \tilde u=v,\mbox{in }B_1^-. \eex$$ And hence $\tilde u=v$ is smooth, and harmonic in $B_1$.

Exercise4.8.

Let $u$ be a $C^2$-solution of $$\bex \ba{ll} \lap u=0,&\mbox{in }\bbR^n\bs B_R,\\ u=0,&\mbox{on }\p B_R. \ea \eex$$ Prove that $u\equiv 0$ if $$\bex \ba{ll} \lim_{|x|\to \infty}\frac{u(x)}{\ln |x|}=0,&\mbox{for }n=2,\\ \lim_{|x|\to \infty}u(x)=0,&\mbox{for }n=3. \ea \eex$$

Solution:  We first consider the case $n=3$. Since $\dps{\lim_{|x|\to \infty}u(x)=0}$, we have $$\bex \forall\ \ve>0,\ \exists\ \tilde R>R,\ \forall\ x:\ |x|=\tilde R,\ |u(x)|<\ve. \eex$$ On the circular ring $\sed{R\leq |x|\leq \tilde R}$, we may apply the maximum principle to deduce $$\bex |u(x)|\leq \ve,\quad R\leq |x|\leq \tilde R. \eex$$ Taking $\tilde R\to \infty$, we see $u\equiv 0$. Then we treat the case $n=2$. By Exercise 4.1, $$\bex v(x)=u\sex{\frac{x}{|x|^2}} \eex$$ verifies $$\bex \ba{ll} \lap v=0,&\mbox{in }B_{1/R}\bs \sed{0},\\ v=0,&\mbox{on }\p B_{1/R}; \ea \eex$$ and $$\bex \lim_{|x|\to 0}\frac{v(x)}{\ln |x|} =-\lim_{|x|\to 0}\frac{u\sex{\frac{x}{|x|^2}}}{\ln \sev{\frac{x}{|x|^2}}}=0. \eex$$ By Theorem 4.3.16, $v$ may be defined at $0$ so that it is harmonic in $B_{1/R}$. Applying the maximum principle, we obtain $v\equiv 0$, on $\bar B_{1/R}$. Consequently, $u\equiv 0$, on $\bbR^n\bs B_R$.

Exercise4.9.

Let $\Omega$ be bounded $C^1$-domain in $\bbR^n$ satisfying the exterior sphere condition at every boundary point and $f$ be a bounded continuous function in $\Omega$. Suppose $u\in C^2(\Omega)\cap C^1(\bar \Omega)$ is a solution of $$\bex \ba{ll} \lap u=0,&\mbox{in }\Omega,\\ u=0,&\mbox{on }\p \Omega. \ea \eex$$ Prove that $$\bex \sup_{\p \Omega}\sev{\frac{\p u}{\p \nu}}\leq C\sup_\Omega|f|, \eex$$ where $C$ is a positive constant depending only on $n$ and $\Omega$.

Solution:  For any fixed $x_0\in \p\Omega$, let $B_R(y_0)\cap \Omega=\sed{y_0}$, and $$\bex w^\pm (x)=\pm A\sex{\frac{1}{R^n}-\frac{1}{|x-y_0|^n}}, \eex$$ then $$\bex \lap w^\pm =\mp \frac{2nA}{|x-y_0|^{n+2}}. \eex$$ For any $\ve>0$, choose $A$ so that $$\bex \frac{2nA}{|x-y_0|^{n+2}}=\max_\Omega |f|+\ve. \eex$$ Then $$\bex \ba{lll} \lap w^+\leq \lap u\leq \lap w^-,&\mbox{in }\Omega,\\ w^+\geq u\geq w^-,&\mbox{on }\p\Omega. \ea \eex$$ By the maximum principle, we have $$\bex w^-\leq u\leq w^+,\quad x\in \Omega. \eex$$ Thus, $$\bex \frac{w^-(x)-w^-(x_0)}{|x-x_0|} \leq \frac{u(x)-w^-u(x_0)}{|x-x_0|} \leq \frac{w^+(x)-w^+(x_0)}{|x-x_0|},\quad x\in \Omega, \eex$$ $$\bex \sev{\frac{\p u}{\p \nu}(x_0)} \leq \min\sed{\sev{\frac{\p w^-}{\p \nu}(x_0)},\sev{\frac{\p w^+}{\p \nu}(x_0)}} =\frac{nA}{R^{n+1}} \leq C\sez{\max_{\Omega}|f|+\ve}. \eex$$ Taking $\ve\to 0^+$, we get the desired result.  \bem We use a uniform exterior sphere condition in the proof, which is not assumed in the problem. \eem

Exercise4.10.

Let $\Omega$ be smooth bounded domain in $\bbR^n$, $c$ be a continuous function in $\Omega$ with $c\leq 0$ and $\alpha$ be a continuous function on $\p\Omega$ with $\alpha\geq 0.$ Discuss the uniqueness of the problem $$\bex \ba{ll} \lap u+cu=f,&\mbox{in }\Omega,\\ \frac{\p u}{\p \nu}+\alpha u=\varphi,&\mbox{on }\p\Omega. \ea \eex$$

Solution:  Let $v$ be the difference of two solution of the problem, then $v$ verifies $$\bee\label{4.10:1} \ba{ll} \lap v+cv=0,&\mbox{in }\Omega,\\ \frac{\p v}{\p \nu}+\alpha v=0,&\mbox{on }\p \Omega. \ea \eee$$ Taking the inner product of \eqref{4.10:1} with $-u$ in $L^2(\Omega)$, we find $$\beex \bea 0&=\int_\Omega (-v\lap v-cv^2)\rd x\\ &=\int_\Omega [-\Div(v\n v)+|\n v|^2-cv^2]\rd x\\ &=-\int_{\p \Omega}v\frac{\p v}{\p \nu}\rd S +\int_\Omega (|\n v|^2-c^2)\rd x\\ &=\int_{\p \Omega}\alpha v^2\rd S +\int_\Omega (|\n v|^2-c^2)\rd x\\ &\geq \int_\Omega |\n v|^2\rd x. \eea \eeex$$ Thus, $v$ is constant. Moreover, if $c\not \equiv 0$ or $\alpha\not\equiv 0$, then $v\equiv 0$.

Exercise4.11.

Let $\Omega$ be a bounded $C^1$-domain and let $\varphi$ and $\alpha$ be continuous functions on $\p\Omega$ with $\alpha\geq \alpha_0$ for a positive constant $\alpha_0$. Suppose $u\in C^2(\Omega)\cap C^1(\bar \Omega)$ satisfies $$\bex \ba{ll} -\lap u+u^3=0,&\mbox{in }\Omega,\\ \frac{\p u}{\p \nu}+\alpha u=\varphi,&\mbox{on }\p\Omega. \ea \eex$$ Prove that $$\bex \max_{\Omega}|u|\leq \frac{1}{\alpha_0}\max_{\p\Omega}|\varphi|. \eex$$

Solution:

(1)If $u$ attains its maximum at $x_0\in\Omega$, then $$\bex \lap u(x_0)\leq 0\ra 0=-\lap u(x_0)+u^3(x_0)\geq u^3(x_0)\ra u(x_0)\leq0, \eex$$ and $u\leq 0$ in $\Omega$.

(2)If $u$ attains its maximum at $x_0\in \p\Omega$, then $$\bex \frac{\p u}{\p \nu}(x_0)\geq 0\ra \varphi(x_0)\geq \alpha(x_0)u(x_0)\geq \alpha(x_0)u(x),\quad \forall\ x\in\Omega, \eex$$ and $$\bex u(x)\leq \frac{1}{\alpha(x_0)}\varphi(x_0) \leq\frac{1}{\alpha_0}\max_{\p\Omega}|\varphi|. \eex$$

(3)If $u$ attains its minumum at $x_0\in\Omega$, then $$\bex \lap u(x_0)\geq 0\ra 0=-\lap u(x_0)+u^3(x_0)\leq u^3(x_0)\ra u(x_0)\geq0, \eex$$ and $u\geq 0$ in $\Omega$.

(4)If $u$ attains its minumum at $x_0\in \p\Omega$, then $$\bex \frac{\p u}{\p \nu}(x_0)\leq 0\ra \varphi(x_0)\leq \alpha(x_0)u(x_0)\leq \alpha(x_0)u(x),\quad \forall\ x\in\Omega, \eex$$ and $$\bex u(x)\geq \frac{1}{\alpha(x_0)}\varphi(x_0) \leq -\frac{1}{\alpha_0}\max_{\p\Omega}|\varphi|. \eex$$ \bigskip In conclusion, we have $$\bex -\frac{1}{\alpha_0}\max_{\p\Omega}|\varphi| \leq u(x)\leq \frac{1}{\alpha_0}\max_{\p\Omega}|\varphi|. \eex$$

Exercise4.12.

Let $f$ be a continuous function in $\bar B_R$. Suppose $u\in C^2(B_R)\cap C(\bar B_R)$ satisfies $$\bex \lap u=f,\mbox{in }B_R. \eex$$ Prove that $$\bex |\n u(0)|\leq \frac{n}{R}\max_{\p B_R(0)}|u|+\frac{R}{2}\max_{B_R}|f|. \eex$$ Hint: In $B_R^+$, set $$\bex v(x',x_n)=\frac{1}{2}[u(x',x_n)-u(x',-x_n)]. \eex$$ Consider an auxiliary function of the form $$\bex w(x',x_n)=A|x'|^2+Bx_n+Cx_n^2. \eex$$ Use the comparison principle to estimate $v$ in $B_R^+$ and then derive a bound for $v_{x_n}(0)$.

Solution:  Up to a scaling, we may assume $R=1$. Denote by $$\bex M=\max_{\p B_1}|u|,\quad F=\max_{B_1}|f|, \eex$$ and let $$\bex A=M,\quad B=nM+\frac{F}{2},\quad C=(1-n)M-\frac{F}{2}. \eex$$ In $B_1^+$, set $$\bex v(x',x_n)=\frac{1}{2}[u(x',x_n)-u(x',-x_n)],\quad w(x',x_n)=A|x'|^2+Bx_n+Cx_n^2. \eex$$ Then $$\beex \ba{lll} |\lap v|\leq F,&\mbox{in }B_1^+,\\ v=0,&\mbox{on }\sed{x_n=0}\cap B_1^+,\\ |v|\leq M,&\mbox{on }\sed{x_n>0}\cap \p B_1^+; \ea \eeex$$ $$\beex \ba{lll} \lap w=2(n-1)A+2C=-F,&\mbox{in }B_1^+,\\ w=A|x'|^2,&\mbox{on }\sed{x_n=0}\cap B_1^+,\\ w=A(1-x_n)^2+Bx_n+Cx_n^2,&\mbox{on }\sed{x_n>0}\cap \p B_1^+. \ea \eeex$$ We claim $$\bee\label{4.12:1} w\geq M,\quad \mbox{on }\sed{x_n>0}\cap \p B_1^+. \eee$$ Then $$\bex \ba{ll} -\lap w\geq \lap v\geq \lap w,&\mbox{in }B_1^+,\\ -w\leq v\leq w,&\mbox{on }\p B_1^+. \ea \eex$$ The maximum principle then yields $$\bex -w\leq v\leq w,\mbox{in }\bar B_1^+. \eex$$ And hence $$\bex \sev{\frac{v(0,x_n)-v(0,0)}{x_n}} \leq \sev{\frac{w(0,x_n)-w(0,0)}{x_n}}, \eex$$ $$\beex \bea |u_{x_n}(0)|&=|v_{x_n}(0)|\\ &\leq |w_{x_n}(0)|\\ &=B\\ &=nM+\frac{F}{2}. \eea \eeex$$ We now prove the claim \eqref{4.12:1}. Indeed, we need only to prove $$\beex \bea f(t)&=A(1-t^2)+Bt+Ct^2\\ &=-\sex{nM+\frac{F}{2}}t^2+\sex{nM+\frac{F}{2}}t+M\\ &=\sex{nM+\frac{F}{2}}t(1-t)+M\\ &\geq 0\quad\sex{0\leq t\leq 1}, \eea \eeex$$ which is indeed true.

Exercise4.13.

Let $u$ be a nonzero harmonic function in $B_1\subset \bbR^n$ and set $$\bex N(r)=\frac{r\int_{B_r}|\n u|^2\rd x}{\int_{\p B_r}u^2\rd S},\mbox{ for any }r\in (0,1). \eex$$

(1)Prove that $N(r)$ is a nondecreasing function in $r\in (0,1)$ and identify $$\bex \lim_{r\to 0^+}N(r). \eex$$

(2)Prove that, for any $0<r<R<1$, $$\bex \frac{1}{R^{n-1}}\int_{\p B_R}u^2\rd S \leq \sex{\frac{R}{r}}^{2N(R)}\frac{1}{r^{n-1}}\int_{\p B_r}u^2\rd S. \eex$$ \bem The quantity $N(r)$ is called the frequency. The estimate in (2) for $R=2r$ is referred to as the doubling condition. \eem

Solution:

(1)Denote by $$\bex A(r)=\int_{B_r}|\n u|^2\rd x,\quad B(r)=\int_{\p B_r}u^2\rd S. \eex$$ Then $$\beex \bea A(r)&=\int_0^r\rd s\int_{\p B_s}|\n u|^2\rd S,\\ A'(r)&=\int_{\p B_r}|\n u|^2\rd S\\ &=\sum_i \int_{\p B_r}\frac{x_i}{r}|\n u|^2 \nu_i\rd S\\ &=\frac{1}{r}\sum_i \int_{B_r}\frac{\p}{\p x_i}\sex{x_i|\n u|^2}\rd x\\ &=\frac{1}{r}\sum_i \int_{B_r}|\n u|^2\rd x +\frac{2}{r}\sum_{i,j} \int_{B_r}x_iu_{x_j}u_{x_ix_j}\rd x\\ &=\frac{n}{r}\int_{B_r}|\n u|^2\rd x +\frac{2}{r}\sum_{i,j} \sez{\int_{\p B_r}x_iu_{x_j}u_{x_i}\nu_j\rd S-\int_{B_r}(x_iu_{x_j})_{x_j}u_{x_i}\rd x}\\ &=\frac{n}{r}\int_{B_r}|\n u|^2\rd x +2\sum_{i,j}\int_{\p B_r}\nu_iu_{x_i}\nu_ju_{x_j}\rd S -\frac{2}{r}\sum_i \int_{B_r}u_{x_i}u_{x_i}\rd x\\ &=\frac{n-2}{r}\int_{B_r}|\n u|^2\rd x +2\int_{\p B_r}u_\nu^2\rd S\quad\sex{u_\nu=\frac{\p u}{\p \nu}}\\ &=\frac{n-2}{r}A(r)+2\int_{\p B_r}u_\nu^2\rd S;\\ B(r)&=\int_{\p B_r}u^2(x)\rd S_x=r^{n-1}\int_{\p B_1}u^2(ry)\rd S_y,\\ B'(r)&=(n-1)r^{n-2}\int_{\p B_1}u^2(ry)\rd S_y +r^{n-1}\int_{\p B_1}2u(ry)\frac{\p}{\p \nu}u(ry)\rd S_y\\ &=\frac{n-1}{r}B(r)+2\int_{\p B_r} uu_\nu\rd S. \eea \eeex$$ Consequently, $$\beex \bea N(r)&=r\frac{A(r)}{B(r)},\\ N'(r)&=\frac{A(r)}{B(r)}+r\frac{A'(r)}{B(r)} -r\frac{A(r)B'(r)}{B^2(r)}\\ &=\frac{rA(r)}{B(r)} \sez{\frac{1}{r}+\frac{A'(r)}{A(r)}-\frac{B'(r)}{B(r)}}\\ &=N(r)\sez{\frac{1}{r}+\frac{n-2}{r}+\frac{2\int_{\p B_r}u_\nu^2\rd S}{\int_{B_r}|\n u|^2\rd x}-\frac{n-1}{r}-\frac{2\int_{\p B_r}uu_\nu \rd S}{\int_{\p B_r}u^2\rd S}}\\ &=2N(r)\sez{\frac{\int_{\p B_r}u_\nu^2\rd S}{\int_{\p B_r}uu_\nu\rd S}-\frac{\int_{\p B_r}uu_\nu \rd S}{\int_{\p B_r}u^2\rd S}}\\ &\quad\sex{\int_{B_r}|\n u|^2\rd x= \int_{B_r}|\n u|^2+u\lap u\rd x =\int_{\p B_r}uu_\nu\rd S}\\ &\geq 0\quad\sex{\mbox{Cauchy inequality}}. \eea \eeex$$

(2)Let $$\bex D(r)=\frac{1}{r^{n-1}}\int_{\p B_r}u^2\rd S=\frac{1}{r^{n-1}}B(r). \eex$$ Then $$\beex \bea D'(r)&=\frac{1-n}{r^n}B(r)+\frac{1}{r^{n-1}}B'(r)\\ &=\frac{1-n}{r^n}B(r)+\sez{\frac{n-1}{r^n}B(r)+\frac{2}{r^{n-1}}\int_{\p B_r}uu_\nu \rd S}\\ &=\frac{2}{r^{n-1}}\int_{\p B_r}uu_\nu \rd S\\ &=\frac{2}{r^{n-1}}\int_{B_r}|\n u|^2\rd x\quad\sex{\mbox{as in the proof of }(1)}\\ &=\frac{2}{r^{n-1}}A(r)\\ &=\frac{2}{r^n}N(r)B(r)\\ &=\frac{2N(r)}{r}D(r). \eea \eeex$$ Thus $$\bex \frac{\rd \ln D(r)}{\rd r}=\frac{2N(r)}{r}\leq \frac{2N(R)}{r},\quad 0<r<R<1. \eex$$ Integrating with respect to $r$, $$\bex \ln D(R)-\ln D(r)\leq 2N(R)\ln\frac{R}{r}\ra D(R)\leq D(r)\sex{\frac{R}{r}}^{2N(R)}. \eex$$

Exercise4.14.

Let $\Omega$ be a bounded domain in $\bbR^n$ and $f$ be a bounded function in $\Omega$. Suppose $w_f$ is the Newtonian potential defined in $$\bex (4.4.2)\qquad w_f(x)=\int_\Omega \vGa(x-y)f(y)\rd y. \eex$$

(1)Prove that $w_f\in C^1(\bbR^n)$ and $$\bex \p_{x_i}w_f(x)=\int_\Omega \p_{x_i}\vGa(x-y)f(y)\rd y, \eex$$ for any $x\in \bbR^n$ and $i=1,\cdots,n$.

(2)Assume, in addition, that $f\in C^\alpha$ in $\Omega$ for some $\alpha\in (0,1)$, i.e., for any $x,y\in\Omega$, $$\bex |f(x)-f(y)|\leq C|x-y|^\alpha. \eex$$ Prove that $w_f\in C^2(\Omega)$, $\lap w_f=f$ in $\Omega$ and the second derivatives of $w_f$ are $C^\alpha$ in $\Omega$.

Solution:

(1)Now that $$\bex w_f(x)=\int_\Omega \p_{x_i}\vGa(x-y)f(y)\rd y, \eex$$ we have for any fixed $x\in \Omega$, $i=1,\cdots,n$, $$\bex \frac{w_f(x+he_i)-w_f(x)}{h} =\int_\Omega \frac{\vGa(x+he_i-y)-\vGa(x-y)}{h}f(y)\rd y. \eex$$ Since $$\bex \sev{\frac{\vGa(x+he_i-y)-\vGa(x-y)}{h}} =\sev{\p_{x_i}\vGa(x+h\theta e_i-y)}\leq M,\quad \forall\ |h|<\frac{\rd(x,\p\Omega)}{2}, \eex$$ for some $M>0$, we may apply Lebesgue's dominated convergence theorem to deduce $$\bex \p_{x_i}w_f(x)=\int_\Omega \p_{x_i}\vGa(x-y)f(y)\rd y. \eex$$

(2)Since $D_{x_ix_j}\vGa$ is not integral around $x$, we need to introduce an auxiliary function $$\bex \zeta\in C^1(\bbR);\quad 0\leq \zeta\leq 1;\quad \zeta(t)=0,\mbox{ for } t\leq 1;\quad \zeta(t)=1,\mbox{ for } t\geq 2. \eex$$ And for $\ve>0$, define $$\bex \zeta_\ve(t)=\zeta\sex{\frac{t}{\ve}}. \eex$$ Then $$\bex u_\ve(x)=\int_\Omega \p_{x_i}\vGa(x-y)\zeta_\ve(x-y)f(y)\rd y \eex$$ verifies $$\beex \bea v_{\ve,x_j} &=\int_\Omega \p_{x_j}\sez{\p_{x_i}\vGa(x-y)\zeta_\ve(x-y)} [f(y)-f(x)]\rd y\\ &\quad +f(x)\int_\Omega \p_{x_j}[\p_{x_i}\vGa(x-y)\zeta_\ve(x-y)]\rd y\\ &=\int_\Omega \p_{x_j}\p_{x_i}\vGa(x-y)\zeta_\ve(x-y)[f(y)-f(x)]\rd y\\ &\quad +\int_\Omega \p_{x_i}\vGa(x-y)\p_{x_j}\zeta_\ve(x-y)[f(y)-f(x)]\rd y\\ &\quad -f(x)\int_\Omega \p_{x_i}\vGa(x-y)\zeta_\ve(x-y)\nu_j\rd S_y. \eea \eeex$$ Taking $\ve \to 0$, we deduce\footnote{All the integrals are well-defined.} $$\bex \p_{x_j}\p_{x_i}w_f(x)=\int_\Omega \p_{x_j}\p_{x_i}\vGa(x-y)[f(y)-f(x)]\rd y -f(x)\int_{\p\Omega}\p_{x_i}\vGa(x-y)\nu_j\rd S_y. \eex$$ Consequently\footnote{See page 93 in the book.}, $$\bex \lap w=f(x)\sum_{i=1}^n \int_{\p\Omega} \p_{y_i}\vGa(x-y)\nu_i\rd S_y=f(x)\int_{\p\Omega}\frac{\p\vGa}{\p \nu}(x-y)\rd S_y=f(x). \eex$$ Finally, for the $C^\alpha$ regularity of $\p_{x_i}\p_{x_j}w_f$, the readers are referred to Lemma 4.4 in [Gilbarg,Trudinger, Page 57]. It is not hard to prove, but long and tedious to write down all the details.

5.Heat Equations

Exercise5.1.

Prove the following statements by straightforward calculations:

(1)$\dps{K(x,t)=t^\frac{n}{2}e^{-\frac{|x|^2}{4t}}}$ satisfies the heat equation for $t>0$;

(2)For any $\alpha>0$, $\dps{G(x,t)=(1-4\alpha t)^{-\frac{n}{2}}e^\frac{\alpha |x|^2}{1-4\alpha t}}$ satisfies the heat equation for $t<1/4\alpha$.

Solution:

(1)$$\beex \bea K_t&=-\frac{n}{2}t^{-\frac{n}{2}-1}e^{-\frac{|x|^2}{4t}} +t^{-\frac{n}{2}}\cdot \frac{|x|^2}{4t^2}e^{-\frac{|x|^2}{4t}};\\ K_{x_i}&=t^{-\frac{n}{2}}\sex{-\frac{x_i}{2t}}e^{-\frac{|x|^2}{4t}};\\ K_{x_ix_i}&=t^{-\frac{n}{2}}\frac{x_i^2}{4t^2} e^{-\frac{|x|^2}{4t}} +t^{-\frac{n}{2}}\sex{-\frac{1}{2t}}e^{-\frac{|x|^2}{4t}};\\ \lap K&=t^{-\frac{n}{2}-2}\frac{|x|^2}{4}e^{-\frac{|x|^2}{4t}}-\frac{n}{2}t^{-\frac{n}{2}-1} e^{-\frac{|x|^2}{4t}};\\ K_t-\lap K&=0. \eea \eeex$$

(2)$$\beex \bea G_t&=-\frac{n}{2}(1-4\alpha t)^{-\frac{n}{2}-1}(-4\alpha)e^\frac{\alpha |x|^2}{1-4\alpha t} +(1-4\alpha t)^{-\frac{n}{2}}\cdot \frac{4\alpha^2|x|^2}{(1-4\alpha t)^2} e^\frac{\alpha |x|^2}{1-4\alpha t};\\ G_{x_i}&=(1-4\alpha t)^{-\frac{n}{2}} e^\frac{\alpha |x|^2}{1-4\alpha t} \frac{2\alpha x_i}{1-4\alpha t},\\ G_{x_ix_i} &=(1-4\alpha t)^{-\frac{n}{2}}e^\frac{\alpha |x|^2}{1-4\alpha t}\frac{4\alpha^2x_i^2}{(1-4\alpha t)^2}+(1-4\alpha t)^{-\frac{n}{2}}e^\frac{\alpha |x|^2}{1-4\alpha t} \frac{2\alpha}{1-4\alpha t},\\ \lap G&=(1-4\alpha t)^{-\frac{n}{2}-2} 4\alpha^2|x|^2e^\frac{\alpha |x|^2}{1-4\alpha t} +(1-4\alpha t)^{-\frac{n}{2}-1}2n\alpha e^\frac{\alpha |x|^2}{1-4\alpha t};\\ G_t-\lap G&=0. \eea \eeex$$

Exercise5.2.

Let $u_0$ e a continuous function in $\bbR^n$ and $u$ be defined in $$\bex (5.2.4)\qquad u(x,t)=\int_{\bbR^n}K(x-y,t)u_0(y)\rd y. \eex$$ Suppose $u_0(x)\to 0$ as $|x|\to\infty$. Prove $$\bex \lim_{t\to\infty}u(x,t)=0,\quad \mbox{uniformly in }x. \eex$$

Solution:  Since $\dps{\lim_{|x|\to\infty}u_0(x)=0}$, we obtain, for fixed $\ve>0$, $$\bex \exists\ Y>0,\ \forall\ |y|>Y,\ |u_0(y)|<\frac{\ve}{2}. \eex$$ Fix this $Y$, $$\bex \exists\ T>0,\ \forall\ t>T,\ \frac{2Y}{(4\pi t)^\frac{n}{2}}\cdot \max|u_0|<\frac{\ve}{2}. \eex$$ Then for $t>T$, $$\beex \bea |u(x,t)|&=\sev{\int_{\bbR^n}K(x-y,t)u_0(y)\rd y}\\ &\leq \int_{|y|\leq Y}K(x-y,t)|u_0(y)|\rd y +\sev{\int_{|y|>Y} K(x-y,t)u_0(y)\rd y}\\ &\leq \frac{2Y}{(4\pi t)^\frac{n}{2}}\cdot \max|u_0| +\max_{|y|>Y}|u_0(y)|\cdot\int_{\bbR^n}K(x-y,t)\rd y\\ &<\frac{\ve}{2}+\frac{\ve}{2}\\ &=\ve. \eea \eeex$$

Exercise5.3.

Prove the convergence in Theorem 5.2.5.

Solution:  Since $\dps{\lim_{y\to x_0}u_0(y)=u_0(x_0)}$, we have, for any fixed $\ve>0$, $$\bex \exists\ \tilde \delta>0,\ \forall\ |y-x_0|<\tilde \delta,\ |u_0(y)-u_0(x_0)|<\frac{\ve}{2}. \eex$$ For this fixed $\tilde \delta$, due to the fact that $$\bex \lim_{t\to 0}\frac{2M}{\pi^\frac{n}{2}}\int_{|\xi|>\frac{\tilde\delta}{4\sqrt{t}}} e^{-\frac{|\xi|^2}{2}}\rd \xi=0, \eex$$ we see $$\bex \exists\ \delta>0,\ \forall\ 0<t<\delta,\ \frac{2M}{\pi^\frac{n}{2}}\int_{|\xi|>\frac{\tilde\delta}{4\sqrt{t}}} e^{-\frac{|\xi|^2}{2}}\rd \xi<\frac{\ve}{2}. \eex$$ Consequently, for any $\dps{0<t<\min\sed{\delta,\frac{\tilde \delta^2}{64A(x_0^2+\delta^2)}}}$, $$\beex \bea &\quad\sev{\int_{\bbR^n}K(x-y,t)u_0(y)\rd y -u_0(x_0)}\\ &=\sev{\int_{\bbR^n}K(x-y,t)[u_0(y)-u_0(x_0)]\rd y}\\ &\leq \int_{|y-x_0|<\tilde \delta} +\int_{|y-x_0|\geq\tilde \delta}\\ &<\frac{\ve}{2}\int_{\bbR^n}K(x-y,t)\rd y +\int_{|y-x_0|\geq\tilde \delta} \frac{1}{(4\pi t)^\frac{n}{2}} e^{-\frac{|x-y|^2}{4t}} [Me^{A(|x_0|+t)^2}+Me^{Ax_0^2}]\rd y\\ &<\frac{\ve}{2}+\frac{2Me^{2A(x_0^2+\delta^2)}}{\pi^\frac{n}{2}} \int_{|y-x_0|\geq\tilde \delta}e^{-\xi^2}\rd \xi\\ &<\frac{\ve}{2}+\frac{2M}{\pi^\frac{n}{2}} \int_{|y-x_0|\geq\tilde \delta}e^{-\frac{\xi^2}{2}} e^{-\frac{\tilde \delta^2}{32t}+2A(x_0^2+\delta^2)}\rd \xi\\ &<\frac{\ve}{2}+\frac{2M}{\pi^\frac{n}{2}} \int_{|y-x_0|\geq\tilde \delta}e^{-\frac{\xi^2}{2}}\rd \xi\\ &<\frac{\ve}{2}+\frac{\ve}{2}\\ &=\ve. \eea \eeex$$

Exercise5.4.

Let $u_0$ be a bounded and continuous function in $[0,\infty)$ with $u_0(0)=0$. Find an integral representation for the solution of the problem $$\bex \ba{ll} u_t-u_{xx}=0,&\mbox{for }x>0,\ t>0,\\ u(x,0)=u_0(x),&\mbox{for }x>0,\\ u(0,t)=0,&\mbox{for }t>0. \ea \eex$$

Solution:  Let $\tilde u_0$ be the odd extension of $u_0$, i.e., $$\bex \tilde u_0(x)=\left\{\ba{ll} -u_0(-x),&x<0,\\ u_0(x),&x\geq0. \ea\right. \eex$$ Then the initial-value problem $$\bex \ba{ll} u_t-u_{xx}=0,&\mbox{for }x\in\bbR,\ t>0,\\ u(x,0)=u_0(x),&\mbox{for }x>0 \ea \eex$$ has a solution $$\bee\label{5.4:1} \bea u(x,t)&=\int_{\bbR}K(x-y,t)\tilde u_0(y)\rd y\\ &=\int_{-\infty}^0\frac{1}{(4\pi t)^\frac{1}{2}} e^{-\frac{|x-y|^2}{4t}}[-u_0(-y)]\rd y+ \int_0^{+\infty}\frac{1}{(4\pi t)^\frac{1}{2}} e^{-\frac{|x-y|^2}{4t}}[u_0(y)]\rd y\\ &=\int_0^{+\infty}\frac{1}{(4\pi t)^\frac{1}{2}} \sez{e^{-\frac{|x-y|^2}{4t}}-e^{-\frac{|x+y|^2}{4t}}}[u_0(y)]\rd y. \eea \eee$$ Notice that $\dps{\lim_{(x,t)\to (x_0,0)}u(x,t)=u_0(x)}$ and $u(0,t)=0$, we see \eqref{5.4:1} with $x>0$ is the desired representation.

Exercise5.5.

Let $u\in C^{2,1}(\bbR^n\times (-\infty,0))$ be a solution of $$\bex u_t-\lap u=0,\mbox{ in }\bbR^n\times (-\infty,0). \eex$$ Suppose that for some nonnegative integer $m$, $$\bex |u(x,t)|\leq C(1+|x|+\sqrt{t})^m, \eex$$ for any $(x,t)\in\bbR^n\times (-\infty,0)$. Prove that $u$ is a polynomial of degree at most $m$.

Solution:  For any $(x_0,t_0)\in \bbR^n\times (-\infty,0)$, we have by Theorem 5.2.9 that $$\beex \bea |\p^k_t\n^l_xu(x_0,t_0)|&\leq \frac{C^{l+2k}}{R^{l+2k}} n^ke^{l+2k-1} (l+2k)! \sup_{Q(x_0,t_0)}|u|\\ &\leq \frac{C^{l+2k}}{R^{l+2k}} n^ke^{l+2k-1} (l+2k)! [1+(|x_0|+R)+(\sqrt{t_0}+R)]^m. \eea \eeex$$ Thus for $l+2k>m$, we may take $R\to\infty$ to get $$\bex \p^k_t\n^l_xu(x_0,t_0)=0. \eex$$ This implies that $u$ is a polynomial of degree at most $m$.

Exercise5.6.

Prove that $u$ constructed in the proof of Proposition 5.2.6 is smooth in $\bbR\times \bbR$.

Solution:  This is easy, since $$\bex \sum \p_t^k\p_x^l\sez{\frac{1}{(2k)!}a^{(k)}(t)x^{2k}} \eex$$ is uniformly convergent on $(x,t)\in [-R,R]\times [-R,R]$, for each $k,l$.

Exercise5.7.

Let $\Omega$ be a bounded domain in $\bbR^n$ and $u_0\in C(\bar \Omega)$. Suppose $u\in C^{2,1}(\Omega\times(0,\infty))\cap C(\bar \Omega \times [0,\infty))$ is a solution of $$\bex \ba{ll} u_t-\lap u=0,&\mbox{in }\Omega\times (0,\infty),\\ u(\cdot,0)=u_0,&\mbox{on }\Omega,\\ u=0,&\mbox{on }\p\Omega\times (0,\infty). \ea \eex$$ Prove that $$\bex \sup_\Omega |u(\cdot,t)|\leq Ce^{-\mu t} \sup_{\Omega}|u_0|,\mbox{ for any }t>0, \eex$$ where $\mu$ and $C$ are positive constants depending only on $n$ and $\Omega$.

Solution:  Since $\Omega$ is bounded, we can find a ball $B_R\supset \Omega$, and a $y_0\not\in\Omega$ such that $\dps{\dist(y_0,\Omega)\geq 1}$. Set $$\bex M=\sup_\Omega |u_0|,\quad v=e^{-\mu t}|x-y_0|^2M. \eex$$ Then $$\beex \bea v_t-\lap v &=e^{-\mu t}M[2n-\mu(x-y_0)^2]\\ &\geq 2e^{-\mu t}M[n-\mu(R^2+|y_0|^2)]\\ &\geq 0\\ &=(\pm u)_t-\lap(\pm u),\quad\mbox{in }\Omega\times (0,T],\\ v&\geq \pm u,\quad\mbox{on }\p_p(\Omega\times (0,T]). \eea \eeex$$ The comparison theorem (Corollary 5.3.4, Page 177 in the book) then yields $$\bex \pm u\leq v\leq Ce^{-\mu t}\max_{\Omega}|u_0|. \eex$$

Exercise5.8.

Let $\Omega$ be a bounded domain in $\bbR^n$, $c$ is continuous in $\bar \Omega \times [0,T]$ with $c\geq -c_0$ for a nonnegative constant $c_0$, and $u_0$ be continuous in $\Omega$ with $u_0\geq 0$. Suppose $u\in C^{2,1}(\Omega\times (0,T])\times C(\bar \Omega\times [0,T])$ is a solution of $$\bex \ba{ll} u_t-\lap u=-u^2,&\mbox{in }\Omega\times (0,T],\\ u(\cdot,0)=u_0,&\mbox{on }\Omega,\\ u=0,&\mbox{on }\p \Omega\times (0,T). \ea \eex$$ Prove that $$\bex 0\leq u\leq e^{c_0T}\sup_\Omega u_0,\quad\mbox{in }\Omega\times (0,T]. \eex$$

Solution:

(1)Take $$\bex M=\max_{\bar\Omega\times [0,T]}|c+u|+1, \eex$$ and consider $v=e^{-Mt}u$, we have $$\bee\label{5.8:1} \ba{ll} v_t-\lap v+(M+c+u)v =e^{-Mt}(u_t-\lap u+cu+u^2)=0,&\mbox{in }\Omega\times (0,T];\\ v\geq 0,&\mbox{on }\p_p(\Omega\times (0,T]). \ea \eee$$ We prove $v\geq 0$ by contradiction. Suppose $v$ attains its negative minimum at some $(x_0,t_0)\in \bar \Omega\times [0,T]\bs \p_p(\Omega\times (0,T])$, we see $$\bex v_t\leq 0,\quad \lap v\geq 0,\quad v<0,\quad\mbox{at }(x_0,t_0). \eex$$ Thus, $$\bex v_t-\lap v+(M+c+u)v<0,\quad\mbox{at }(x_0,t_0). \eex$$ This contradicts $\eqref{5.8:1}_1$.

(2)Set $$\bex N=\sup_\Omega |u_0|,\quad v=e^{c_0t}v. \eex$$ Then $$\beex \bea v_t-\lap v+cv&=(c+c_0)v\\ &\geq 0\\ &\geq -u^2\\ &= u_t-\lap u+cu,\quad\mbox{in }\Omega\times (0,T];\\ v&\geq u,\mbox{on }\p_p(\Omega\times (0,T]). \eea \eeex$$ The comparison principle (see Corollary 5.3.4, Page 177 in the book) then yields $$\bex v\geq u,\quad\mbox{in }\Omega\times (0,T]. \eex$$

Exercise5.9.

Let $\Omega$ be a bounded domain in $\bbR^n$, $u_0$ and $f$ be continuous in $\bar \Omega$, and $\varphi$ be continuous on $\p\Omega\times [0,T]$. Suppose $u\in C^{2,1}(\Omega\times (0,T])\cap C(\bar \Omega \times [0,T])$ is a solution of $$\bex \ba{ll} u_t-\lap u=e^{-u}-f(x),&\mbox{in }\Omega \times (0,T],\\ u(\cdot,0)=u_0,&\mbox{on }\Omega,\\ u=\varphi,&\mbox{on }\p\Omega\times (0,T). \ea \eex$$ Prove that $$\bex -M\leq M+Te^M+M,\quad\mbox{in }\Omega \times (0,T], \eex$$ where $$\bex M=T\sup_{\Omega}|f|+\sup\sed{\sup_\Omega |u_0|,\sup_{\p\Omega\times (0,T)}|\varphi|}. \eex$$

Solution:

(1)Set $$\bex F=\sup_\Omega|f|,\quad A=\sup\sed{\sup_\Omega |u_0|,\sup_{\p\Omega\times (0,T)}|\varphi|},\quad v=Ft+A. \eex$$ Then $$\beex \bea (-v)_t-\lap (-v)&=-F\\ &\leq -f\\ &\leq e^{-u}-f\\ &=u-\lap u,\quad\mbox{in }\Omega \times (0,T],\\ -v&\leq u,\quad\mbox{on }\p_p(\Omega\times (0,T]). \eea \eeex$$ The comparison theorem then yields $$\bex -M\leq -v\leq u. \eex$$

(2)By (1), $-u\leq M$, and hence by setting $w=e^Mt+Ft+A$, $$\beex \bea u_t-\lap u&=e^{-u}-f\\ &\leq e^M+F\\ &=w_t-\lap w,\quad\mbox{in }\Omega \times (0,T],\\ u\leq w,\quad\mbox{on }\p_p(\Omega\times (0,T]). \eea \eeex$$ The comparison theorem then yields $$\bex u\leq w\leq Te^M+FT+A. \eex$$

Exercise5.10.

Let $Q=(0,l)\times (0,\infty)$ and $u_0\in C^1[0,l]$ with $u_0(0)=u_0(l)=0$. Suppose $u\in C^{2,1}(Q)\cap C(\bar Q)$ is a solution of $$\bee\label{5.10:1} \ba{ll} u_t-u_{xx}=0,&\mbox{in }Q,\\ u(\cdot,0)=u_0,&\mbox{on }(0,l),\\ u(0,\cdot)=u(l,\cdot)=0,&\mbox{on }(0,\infty). \ea \eee$$ Prove that $$\bex \sup_Q|u_x|\leq \sup_{[0,l]}|u_0'|. \eex$$

Solution:  Let $v_0$ be the periodical extension of the odd extension of $u_0$, i.e., $$\bex v_0(x)=v_0(x+2l),\quad v_0(x)=\left\{\ba{ll} u_0(x),&0\leq x\leq l;\\ -u_0(-x),&-1\leq x\leq 0. \ea\right. \eex$$ Then the solution of $$\bex \ba{ll} v_t-v_{xx}=0,&\mbox{in }\bbR\times (0,\infty),\\ v(\cdot,0)=v_0,&\mbox{on }\bbR \ea \eex$$ is given by $$\beex \bea v(x,t)&=\int_{\bbR}K(x-y,t)v_0(y)\rd y\\ &=\sum_{k=-\infty}^\infty \int_{(2k-1)l}^{(2k+1)l}K(x-y,t)v_0(y)\rd y\\ &=\sum_{k=-\infty}^\infty \int_{-l}^l K(x-s-2kl)v_0(y)\rd y\\ &=\sum_{k=-\infty}^\infty \sez{\int_{-l}^0 K(x-y-2kl,t)(-u_0(-y))\rd y +\int_0^lK(x-y-2kl,t)u_0(y)\rd y}\\ &=\sum_{k=-\infty}^\infty \sez{-\int_0^l K(x+y-2kl,t)u_0(y))\rd y +\int_0^lK(x-y-2kl,t)u_0(y)\rd y}\\ &=-\sum_{k=-\infty}^\infty \int_0^lK(x+y-2kl,t)u_0(y)\rd y \sum_{k=-\infty}^\infty \int_0^lK(x-y-2kl,t)u_0(y)\rd y. \eea \eeex$$ Thus, $$\bex v(0,t)=v(l,t)=0. \eex$$ And hence, the restriction of $v$ on $[0,l]$ is the solution of \eqref{5.10:1}. Consequently, for $(x,t)\in Q$, $$\bex |u_x(x,t)| =|v_x(x,t)|=\sev{\int_{\bbR}K(y,t)v_{0,y}(x-y)\rd y} \leq \sup_\bbR |v_0'|=\sup_{[0,l]}|u_0'|. \eex$$

Exercise5.11.

Let $\Omega$ be bounded domain in $\bbR^n$. Suppose $u_1,\cdots,u_m\in C^{2,1}(\Omega\times (0,T])\cap C(\bar\Omega\times [0,T])$ satisfy $$\bex \p_tu_i=\lap u_i,\mbox{ in }\Omega\times (0,T], \eex$$ for $i=1,\cdots,m$. Assume that $f$ is convex function in $\bbR^m$. Prove that $$\bex \sup_{\Omega\times (0,T]}f(u_1,\cdots,u_m)\leq \sup_{\p_p(\Omega\times (0,T])}f(u_1,\cdots,u_m). \eex$$

Solution:  By a simple mollification, we may assume without loss of generality that $f$ is smooth. Direct computations show that $$\bex \p_tf-\lap f=-f_{u_ju_k}u_{j,x_i}u_{k,x_i}\leq 0. \eex$$ Thus the maximum principle (see Page 176 in the book) then yields the desired result.

Exercise5.12.

Let $u_0$ be a bounded continuous function in $\bbR^n$. Suppose $u\in C^{2,1}(\bbR^n\times (0,T])\cap C(\bbR^n\times [0,T])$ satisfies $$\bex \ba{ll} u_t-\lap u=0,&\mbox{in }\bbR^n\times (0,T],\\ u(\cdot,0)=u_0,&\mbox{on }\bbR^n. \ea \eex$$ Assume that $u$ and $\n u$ are bounded in $\bbR^n\times (0,T]$. Prove that $$\bex \sup_{\bbR^n}|\n u(\cdot,t)|\leq \frac{1}{\sqrt{2t}}\sup_{\bbR^n}|u_0|. \eex$$ {\sl Hint:} With $|u_0|\leq M$ in $\bbR^n$, consider $$\bex w=u^2+2t|\n u|^2-M^2. \eex$$

Solution:  Direct computations show $$\beex \bea (\p_t-\lap)|\n u|^2&=-2\sum_{i,j}u_{x_ix_j}^2,\\ (\p_t-\lap)(u^2)&=-2\sum_i u_{x_i}^2,\\ (\p_t-\lap)(u^2+2t|\n u|^2-M^2)=-2\sum_{i,j}u_{x_ix_j}^2&\leq 0,\quad\sex{M=\sup_{\bbR^n}|u_0|}. \eea \eeex$$ Thus, $$\beex \bea &\quad u^2+2t|\n u|^2-M^2 \\ &=\int_{\bbR^n}K(x-y,t)(u_0^2-M^2)\rd y +\int_0^t\int_{\bbR^n}K(x-y,t-s)(-|\n^2u|^2)\rd y\rd s\\ &\leq 0. \eea \eeex$$

Exercise5.13.

Prove Corollary 5.3.15.

Solution:  Taking $\alpha=1/2$ in Theorem 5.3.14, we have for $v=\ln u$, $$\bex v_t-\alpha |\n v|^2+\frac{n}{t}+C\geq 0,\mbox{in }B_{1/2}\times (0,1]. \eex$$ Thus, integrating over the segment $$\bex x(s)=x_1+\frac{x_2-x_1}{t_2-t_1}(s-t_1),\quad t_1\leq s\leq t_2, \eex$$ we obtain $$\beex \bea v(x_2,t_2)-v(x_1,t_1) &=\int_{t_1}^{t_2} \frac{\rd x}{\rd s}\rd s\\ &=\int_{t_1}^{t_2} \sez{v_s+\frac{\rd x}{\rd s}\cdot\n v}\rd s\\ &\geq \int_{t_1}^{t_2} \sez{v_s-\alpha|\n v|^2 -\frac{1}{4\alpha}\sev{\frac{\rd x}{\rd s}}^2}\rd s\\ &\geq -\int_{t_1}^{t_2} \sez{\frac{n}{t}+C+\frac{1}{4\alpha}\sex{\frac{|x_2-x_1|^2}{t_2-t_1}}^2}\rd s\\ &=-\sez{n\ln\frac{t_2}{t_1} +C(t_2-t_1)+\frac{1}{4\alpha}\frac{|x_2-x_1|^2}{t_2-t_1}},\\ u(x_1,t_1)&\leq \sex{\frac{t_2}{t_1}}^n e^{C(t_2-t_1)+\frac{1}{2(t_2-t_1)}}u(x_2,t_2). \eea \eeex$$