变分原理
[习题1.1] 若函数\(u\in L^2(a,b)\) 满足
\[\int_a^bu(x)v(x)\mathrm{~d}x=0,\quad v\in C_0^2[a,b],
\]
证明 \(u\xlongequal{\text{几乎处处}}0,~x\in(a,b)\).
证明：利用反证法, 假设在\((a,b)\) 内 \(u\) 不几乎处处为 \(0\), 则一定存在一个 \(E\subset(a,b)\), 满足 \(m(E)>0\), 使得在 \(E\) 上 \(u\neq0\), 从而必有 \(\Vert u\Vert_2>0\). 又 \(u\in L^2(a,b)\), 于是存在 \(0<M<\infty\), s.t. \(\Vert u\Vert<M\).
注意到\(C_0^{\infty}(a,b)\) 在 \(L^2(a,b)\) 中稠密, 而 \(C_0^{\infty}(a,b)\subset C_0^2[a,b]\), 因此一定能找到一列 \(\{v_n(x)\}_{n=0}^{\infty}\subset C_0^{\infty}(a,b)\subset C_0^2[a,b]\), 使得 \(v_n(x)\overset{L^2}{\rightarrow}u(x)\), 即 \(\Vert v_n-u\Vert_2\to0~(n\to\infty)\). 且有 \((u,v_n)=0\), 从而
\[\begin{aligned}
0<\Vert u\Vert_2^2&=|(u,u-v_n+v_n)|=|(u,u-v_n)+(u,v_n)|\\
&=|(u,u-v_n)|\le\Vert u\Vert_2\Vert u-v_n\Vert_2<M\Vert u-v_n\Vert_2
\end{aligned}
\]
令\(n\to\infty\), 有 \(\Vert u\Vert_2=0\), 矛盾! #
[习题1.2] 考察如下变分问题:
\[\min_{y \in K} J(y) = \int_{-1}^{1} x^2 (y'(x))^2\, \mathrm{~d}x,
\]
其中\(K = \{ y \in C^1[-1, 1] : y(-1) = -1, y(1) = 1 \}\).
(a) 设( a ) 为任一正数, \(y_a = \frac{\arctan(x/a)}{\arctan(1/a)}\). 求证
\[J(y_a) < \frac{2a}{\arctan(1/a)}.
\]
(b) 证明以上变分问题无解.
证明：(a) \(y_a(x)=\frac{\arctan(x/a)}{\arctan(1/a)}\), \(a>0\). 于是
\[y_a'(x)=\frac{a}{(a^2+x^2)\arctan(1/a)}.
\]
从而
\[\begin{aligned}
J(y_a)&=\int_{-1}^1x^2\left(\frac{a}{(a^2+x^2)\arctan(1/a)}\right)^2\mathrm{~d}x\\
&=\frac{a^2}{\arctan^2(1/a)}\int_{-1}^1\frac{x^2}{(a^2+x^2)^2}\mathrm{~d}x\\
&=\frac{2a^2}{\arctan^2(1/a)}\int_{0}^1\frac{x^2}{(a^2+x^2)^2}\mathrm{~d}x\\
&<\frac{2a^2}{\arctan^2(1/a)}\int_{0}^1\frac{x^2+a^2}{(a^2+x^2)^2}\mathrm{~d}x\\
&=\frac{2a^2}{\arctan^2(1/a)}\int_{0}^1\frac{1}{a^2+x^2}\mathrm{~d}x\\
&= \frac{2a}{\arctan(1/a)}.\quad \#
\end{aligned}
\]
(b) 用反证法,假设 \(y\) 是上述变分问题的解, 则它也是 Euler-Lagrange 方程
\[\frac{\partial F}{\partial y}-\frac{\mathrm d}{\mathrm dx}\left(\frac{\partial F}{\partial y'}\right)=0 \tag{1}
\]
的解, 其中\(F(x,y,y')=x^2(y')^2\). 注意到
\[\frac{\partial F}{\partial y}=0,\quad \frac{\partial F}{\partial y'}=2x^2y'.
\]
代入公式 \((1)\) 中, 有
\[(2x^2y')'=0\Rightarrow2x^2y'=c\Rightarrow y'=\frac{c}{2x^2}\Rightarrow y=-\frac{c}{2x}+d,
\]
其中\(c,d\) 为任意常数. 注意到 \(y\in C^1[-1,1]\), 故必有 \(c=0\), 从而 \(y=d\) 为常值函数, 这与 \(y(-1)=-1,~y(1)=1\) 矛盾! 故变分问题无解. #
[习题1.3] 记\(K = \{v \in C^1[a, b] : v(a) = u_a, v(b) = u_b\}\). 考察如下变分问题: 找 \(u \in K\) 满足
\[J(u) = \min_{v \in K} J(v), \quad \text{其中 } J(v) = \int_a^b F(x, v, v')\, dx.
\]
分两步研究其解\(u\) 满足的条件:
(a) 设\(f(x) \in C[a, b]\), 如果
\[\int_a^b f(x) \phi'(x)\, dx = 0, \quad \phi\in C_0^\infty(a, b),
\]
则\(f(x) = \text{常数}, \, x \in [a, b]\).
(b) 解\(u\) 满足
\[\int_a^x F_{u'}(x, u(x), u'(x))\, dx - F_u(x, u(x), u'(x)) = \text{常数}, \quad x \in [a, b].
\]
证明： (a)\(f(x)\in C[a,b]\subset L^1[a,b]\), 令 \(f'\) 为 \(f\) 的弱导数, 则 \(L^1\in L^1[a,b]\), 且对 \(\forall\phi(x)\in C_0^{\infty}(a,b)\), 有
\[0=\int_a^bf(x)\phi'(x)\mathrm{~d}x=-\int_a^bf'(x)\phi(x)\mathrm{~d}x.
\]
从而由变分法基本引理可知 \(f'\xlongequal{a.e}0\), 又 \(f\in C[a,b]\), 故 \(f(x)=\)常数, \(x\in[a,b]\). #
(b) 对\(\forall\phi\in C_0^{\infty}(a,b)\), 有 \(u+\alpha\phi\in K~(\alpha\in\mathbb R)\). 令
\[\varphi(\alpha)\triangleq J(u+\alpha\phi)=\int_a^bF(x,u+\alpha\phi,u'+\alpha\phi')\mathrm{~d}x.
\]
由于\(u\) 是变分问题的解, 故 \(\alpha=0\) 必是 \(\varphi(\alpha)\) 的极小值点, 从而 \(\varphi'(\alpha)\vert_{\alpha=0}=0\), 即
\[\int_a^b\left(F_u\phi+F_{u'}\phi'\right)\mathrm{~d}x=0
\]
令
\[f(x)\triangleq\int_a^xF_u(x,u,u')\mathrm{~d}x,
\]
则
\[\int_a^b\phi(x)\mathrm{~d}f(x)+\int_a^bF_{u'}(x,u,u')\phi'(x)\mathrm{~d}x=0
\]
于是
\[\int_a^b\left[F_{u'}(x,u,u')-f(x)\right]\phi'\mathrm{~d}x=0
\]
由 (a) 可知
\[F_{u'}(x,u,u')-\int_a^xF_u(x,u,u')\mathrm{~d}x=\text{常数},\quad x\in[a,b].\quad\#
\]
[习题1.4] 证明非齐次两点边值问题
\[\begin{cases}
Lu = -(pu')' + qu = f, & x \in (a, b),\\
u(a) = \alpha,\, u'(b) = \beta
\end{cases}
\]
与下列变分问题等价: 求\(u \in H^1(a, b), u(a) = \alpha\), 使
\[J(u) = \min_{v \in H^1(a, b), v(a) = \alpha} J(v),
\]
其中
\[J(v) = \frac{1}{2} a(v, v) - (f, v) - p(b) \beta v(b), \quad a(u, v) = \int_a^b (pu'v' + quv) \mathrm{~d}x.
\]
进一步给出该问题的虚功原理.
证明：注意到\(a(u,v)=a(v,u),~a(u,v_1+v_2)=a(u,v_1)+a(u,v_2),~a(u_1+u_2,v)=a(u_1,v)+a(u_2,v)\), 因此 \(a(u,v)\) 是对称双线性泛函. 且
\[\begin{aligned}
J(u+v) &=\frac12a(u+v,u+v)-(f,u+v)-p(b)\beta(u+v)(b)\\
&=\frac12a(u,u)-(f,u)-p(b)\beta u(b)+\frac12a(v,v)-(f,v)-p(b)\beta v(b)+a(u,v)\\
&=J(u)+\frac12a(v,v)+a(u,v)-p(b)\beta v(b)-(f,v).
\end{aligned}
\tag{2}
\]
\((\Longrightarrow):\) 若 \(u\) 是边值问题的解, 对 \(\forall v\in H^1(a,b),~v(a)=\alpha\), 由公式 \((2)\) 可得
\[\begin{aligned}
J(v) &=J(u+v-u)\xlongequal{w\triangleq v-u} J(u+w)\\
&=J(u)+\frac12a(w,w)+a(u,w)-p(b)\beta w(b)-(f,w)
\end{aligned}
\tag{3}
\]
注意到\(w\in H^1(a,b),~w(a)=0\), 有
\[\begin{aligned}
(f,w)&=\int_a^b\left(-(pu')'+qu\right)w\mathrm{~d}x\\
&=a(u,w)-p(b)\beta v(b)
\end{aligned}
\]
于是\(a(u,w)-p(b)\beta v(b)-(f,w)=0\), 代入公式 \((3)\) 中可得
\[J(v)=J(u)+\frac12a(w,w)\ge J(u).
\]
从而\(u\) 也是变分问题的解.
\((\Longleftarrow):\) 若 \(u\) 是变分问题的解. 对 \(\forall v\in H^1(a,b),~v(a)=0\), 有 \(u+tv\in H^1(a,b),~t\in\mathbb R\), 且 \((u+tv)(a)=\alpha\). 于是由公式 \((2)\) 可知
\[\begin{aligned}
J(u+tv)&=J(u)+\frac12a(tv,tv)+a(u,tv)-p(b)\beta tv(b)-(f,tv)\\
&=J(u)+\frac12 t^2a(v,v)+t\left[a(u,v)-(f,v)-p(b)\beta v(b)\right]
\end{aligned}
\]
由于\(u\) 是变分问题的解, 故 \(\frac{\mathrm dJ(u+tv)}{\mathrm dt}\vert_{t=0}=0\), 即
\[a(u,v)-(f,v)-p(b)\beta v(b)=0.
\]
写成积分的形式, 运用分部积分, 可得
\[\int_a^b\left(-(pu')'+qu-f\right)v\mathrm{~d}x+p(b)v(b)(u'(b)-\beta)=0.
\]
因此当\(u'(b)=\beta\) 时,
\[\int_a^b\left(-(pu')'+qu-f\right)v\mathrm{~d}x=0
\]
再由变分法基本引理可知此时\(-(pu')' + qu = f\), 因此 \(u\) 是边值问题的解. #
[习题1.5] 试建立与边值问题等价的变分问题.
\[\begin{cases}
Lu = -\frac{\mathrm{d}^4u}{\mathrm{d}x^4} + u = f, & x \in (a, b),\\
u(a) = u'(a) = 0,\, u(b) = u'(b) = 0
\end{cases}
\]
证明：设函数集合
\[V=\{v\in H^2(a,b)\vert v(a)=v'(a)=v(b)=v'(b)=0\}.
\]
对\(\forall v\in V\), 则
\[\int_a^b(-u^{(4)}+u)v\mathrm{~d}x=\int_a^bfv\mathrm{~d}x.
\]
作两次分部积分可得
\[\int_a^b\left(uv-u''v''\right)\mathrm{~d}x=\int_a^bfv\mathrm{~d}x.
\]
令
\[D(u,v)\triangleq\int_a^b\left(uv-u''v''\right)\mathrm{~d}x,\quad
F(v)=\int_a^bfv\mathrm{~d}x.
\]
则不难看出\(F(v)\) 是线性泛函, \(D(u,v)\) 是对称双线性泛函. 于是边值问题转化为了变分问题: 求 \(u\in V\), 使得
\[D(u,v)=F(v),\quad v\in V.\quad \#
\]
[习题1.6] 试导出与二维 Poisson 方程 Robin 边界条件问题
\[\begin{cases}
-\Delta u = f(x, y), & (x, y) \in\Omega,\\
(\partial_n u + \alpha u)|_{\partial\Omega} = \beta, &\alpha \ge 0
\end{cases}
\]
等价的变分问题.
证明：设\(H^1(\Omega)=\{v\in L^2(\Omega)~:~\partial_iv\in L^2(\Omega)\}\). 则由 Poisson 方程可知
\[-\int_{\Omega}\Delta uv\mathrm{~d}x=\int_{\Omega}fv\mathrm{~d}x.
\]
由 Green 公式可得,
\[\int_{\Omega}\nabla u\cdot\nabla v\mathrm{~d}x-\int_{\partial\Omega}\partial_n u v\mathrm{~d}s=\int_{\Omega}fv\mathrm{~d}x.
\]
于是
\[\int_{\Omega}\nabla u\cdot\nabla v\mathrm{~d}x+\int_{\partial\Omega}\alpha uv\mathrm{~d}s=\int_{\Omega}fv\mathrm{~d}x+\beta\int_{\partial\Omega}v\mathrm{~d}s.
\]
于是对应的 Garlekin 变分问题为找\(u\in H^1(\Omega)\), 使得
\[\int_{\Omega}\nabla u\cdot\nabla v\mathrm{~d}x+\int_{\partial\Omega}\alpha uv\mathrm{~d}s=\int_{\Omega}fv\mathrm{~d}x+\beta\int_{\partial\Omega}v\mathrm{~d}s,\quad v\in H^1(\Omega).\quad\#
\]
[习题1.7] 用 Galerkin 方法或 Ritz 方法求边值问题
\[\begin{cases}
-u'' + u = x^2, & x \in (0, 1), \\
u(0) = 0, \, u(1) = 1
\end{cases}
\]
的第 \(n\) 次近似 \(u_n(x)\), 基函数 \(\varphi_i(x) = \sin(i \pi x), i = 1, 2, \ldots, n\).
解：设 \(w=u-x\), 则 \(w(0)=w(1)=0\), 边值问题等价于
\[\begin{cases}
-w''+w=x^2-x, & x\in(0,1),\\
w(0)=w(1)=0.
\end{cases}
\]
令 \(H_0^1(0,1)=\{v\in H^1(0,1)~:~v(0)=v(1)=0\}\), 则对 \(\forall v\in H_0^1(0,1)\), 有
\[\int_0^1w'v'\mathrm{~d}x+\int_0^1wv\mathrm{~d}x=\int_0^1(x^2-x)v\mathrm{~d}x.
\]
则相应的 Galerkin 变分问题为: 求 \(w_n\in H_0^1(0,1)\), 使得
\[a(w,v)=f(v),\quad v\in H_0^1(0,1),
\]
其中 \(a(w,v)=\int_0^1w'v'\mathrm{~d}x+\int_0^1wv\mathrm{~d}x\), \(f(v)=\int_0^1(x^2-x)v\mathrm{~d}x\). Galerkin 变分问题的 Ritz 变分问题为: 求 \(w_n\in H_0^1(0,1)\), 使得
\[J(w)=\min\limits_{\nu\in H_0^1(0,1)}J(\nu),
\]
其中 \(J(\nu)=\frac12a(\nu,\nu)-f(\nu)\).
构造试探函数空间 \(U_n=\text{span}\{\varphi_1,\varphi_2,\ldots,\varphi_n\}\), 则 Ritz 变分问题的近似问题为: 求 \(w_n\in U_h\), 使得
\[J(w_n)=\min\limits_{\nu\in U_n}J(\nu).
\]
其中 \(\nu_n=\sum\limits_{i=1}^{n}c_i\varphi_i\), 代入可得
\[J(\nu_n) = \frac12\sum_{i=1}^{n}\sum_{j=1}^{n}c_ic_ja(\varphi_i,\varphi_j)-\sum_{i=1}^{n}c_if(\varphi_i).
\]
令 \(\boldsymbol{A}=(a(\varphi_i,\varphi_j))_{n\times n}\), \(\boldsymbol{F}=(f(\varphi_i))_{n\times1},~ \boldsymbol{c} = (c_i)_{n\times1}\), 则
\[J(\nu_n)=\frac12\boldsymbol{c}^T\boldsymbol{A}\boldsymbol{c}-\boldsymbol{F}^T\boldsymbol{c}.
\]
于是 Ritz 变分问题等价于以 \(\boldsymbol{c}\) 为未知数的极值问题
\[\min_{\boldsymbol{c}\in\mathbb R^n}\left\{\frac12\boldsymbol{c}^T\boldsymbol{A}\boldsymbol{c}-\boldsymbol{F}^T\boldsymbol{c}\right\}.
\]
这等价于求解线性方程组 \(\boldsymbol{A}\boldsymbol{c}=\boldsymbol{F}\).
由基函数的正交性可得 \(a_{\varphi_i,\varphi_j}=0~(i\neq j)\), 而
\[a(\varphi_i,\varphi_i)=\int_0^1(\varphi_i')^2\mathrm{~d}x+\int_0^1\varphi_i^2\mathrm{~d}x=\frac{1+i^2\pi^2}{2}.
\]
于是 \(A=\text{diag}\left\{\frac{1+\pi^2}{2},\frac{1+4\pi^2}{2},\ldots,\frac{1+n^2\pi^2}{2}\right\}\). 又
\[f(\varphi_i)=\int_0^1(x^2-x)\varphi_i\mathrm{~d}x=\frac{2}{(i\pi)^3}\left[(-1)^i-1\right],
\]
求解 \(\boldsymbol{A}\boldsymbol{c}=\boldsymbol{F}\) 即可得到
\[c_i^*=\frac{4\left[(-1)^i-1\right]}{(i\pi)^3(i^2\pi^2+1)}.
\]
于是
\[u_n(x)=x+\sum\limits_{i=1}^{n}c_i^*\sin(i\pi x)=x+\sum\limits_{i=1}^n\frac{4\left[(-1)^i-1\right]}{(i\pi)^3(i^2\pi^2+1)}\sin(i\pi x)
\]
[习题1.8] 用 Galerkin 方法或 Ritz 方法求解二维边值问题
\[\begin{cases}
-\Delta u = xy, & (x, y) \in \Omega, \\
u|_{\partial \Omega} = 0,
\end{cases}
\]
其中 \(\Omega = \{(x, y) : 0 < x < 1, 0 < y < 1\}\). 试探函数空间的基选为
\[\begin{aligned}
\varphi_1 &= \sin(\pi x) \sin(\pi y), & \varphi_2 &= \sin(\pi x) \sin(2\pi y), \\
\varphi_3 &= \sin(2\pi x) \sin(\pi y), & \varphi_4 &= \sin(2\pi x) \sin(2\pi y).
\end{aligned}
\]
解：设 \(H_0^1(\Omega)=\{v\in H^1(\Omega)~:~v|_{\partial\Omega}=0\}\), 则对 \(\forall v\in H_0^1(\Omega)\), 有
\[(-\Delta u, v) = \langle xy,v\rangle:=\langle f,v\rangle.
\]
由 Green 公式可得
\[\int_{\Omega}\nabla u\cdot\nabla v\mathrm{~d}x\mathrm{~d}y=\int_{\Omega}xyv\mathrm{~d}x\mathrm{~d}y.
\]
记为 \(D(y,v)=F(v)\). 于是 \(u\) 满足 Galerkin 变分问题: 求 \(u\in H_0^1(\Omega)\), 使得
\[D(u,v)=F(v),\quad v\in H_0^1(\Omega).
\]
对应的 Ritz 变分问题为: 求 \(u\in H_0^1(\Omega)\), 使得
\[J(u)=\min\limits_{w\in H_0^1(\Omega)}J(w),
\]
其中 \(J(w)=\frac12D(w,w)-F(w)\).
构造试探函数空间 \(U_h=\text{span}\{\varphi_1,\varphi_2,\varphi_3,\varphi_4\}\), 则 Ritz 变分问题的近似问题为: 求 \(u_h\in U_h\), 使得
\[J(u_h)=\min\limits_{w_h\in U_h}J(w).
\]
其中 \(w_h=\sum\limits_{i=1}^{4}c_i\varphi_i\), 代入可得
\[J(w_h)=\frac12\sum_{i=1}^{4}\sum_{j=1}^{4}c_ic_jD(\varphi_i,\varphi_j)-\sum_{i=1}^{4}c_iF(\varphi_i).
\]
令 \(\boldsymbol{A}=(D(\varphi_i,\varphi_j))_{4\times4}\), \(\boldsymbol{F}=(F(\varphi_i))_{4\times1},~ \boldsymbol{c} = (c_i)_{4\times1}\), 则
\[J(w_h)=\frac12\boldsymbol{c}^T\boldsymbol{A}\boldsymbol{c}-\boldsymbol{F}^T\boldsymbol{c}.
\]
于是 Ritz 变分问题等价于以 \(\boldsymbol{c}\) 为未知数的极值问题
\[\min_{\boldsymbol{c}\in\mathbb R^4}\left\{\frac12\boldsymbol{c}^T\boldsymbol{A}\boldsymbol{c}-\boldsymbol{F}^T\boldsymbol{c}\right\}.
\]
这等价于求解线性方程组 \(\boldsymbol{A}\boldsymbol{c}=\boldsymbol{F}\).
由基函数的正交性可得 \(D(\varphi_i,\varphi_j)=0~(i\neq j)\), 而
\[\begin{aligned}
D(\varphi_1,\varphi_1)&=\int_{\Omega}(\nabla\varphi_1)\cdot(\nabla\varphi_1)\mathrm{~d}x\mathrm{~d}y\\
&=\int_{\Omega} \left[\left(\frac{\partial\varphi_1}{\partial x}\right)^2+\left(\frac{\partial\varphi_1}{\partial y}\right)^2\right]\mathrm{~d}x\mathrm{~d}y\\
&=\int_{\Omega}\pi^2\left(\cos^2(\pi x)\sin^2(\pi y)+\sin^2(\pi x)\cos^2(\pi y)\right)\mathrm{~d}x\mathrm{~d}y\\
&=2\pi^2\int_{\Omega}\cos^2(\pi x)\sin^2(\pi y)\mathrm{~d}x\mathrm{~d}y\\
&=\pi^2\int_0^1\cos^2(\pi x)\mathrm{~d}x\int_0^1\sin^2(\pi y)\mathrm{~d}y\\
&=\frac12\pi^2.
\end{aligned}
\]
同理可求得 \(D(\varphi_2,\varphi_2)=\frac54\pi^2\), \(D(\varphi_3,\varphi_3)=\frac54\pi^2\), \(D(\varphi_4,\varphi_4)=2\pi^2\). 注意到
\[\int_0^1x\sin(k\pi x)\mathrm{~d}x=\frac{1}{k\pi}(-1)^{k+1},
\]
故
\[\begin{aligned}
F(\varphi_1)&=\int_{\Omega}xy\varphi_1\mathrm{~d}x\mathrm{~d}y=\int_0^1x\sin(\pi x)\mathrm{~d}x\int_0^1y\sin(\pi y)\mathrm{~d}y=\frac{1}{\pi^2},\\
F(\varphi_2)&=\int_{\Omega}xy\varphi_2\mathrm{~d}x\mathrm{~d}y=\int_0^1x\sin(\pi x)\mathrm{~d}x\int_0^1y\sin(2\pi y)\mathrm{~d}y=-\frac{1}{2\pi^2},\\
F(\varphi_3)&=\int_{\Omega}xy\varphi_3\mathrm{~d}x\mathrm{~d}y=\int_0^1x\sin(2\pi x)\mathrm{~d}x\int_0^1y\sin(\pi y)\mathrm{~d}y=-\frac{1}{2\pi^2},\\
F(\varphi_4)&=\int_{\Omega}xy\varphi_4\mathrm{~d}x\mathrm{~d}y=\int_0^1x\sin(2\pi x)\mathrm{~d}x\int_0^1y\sin(2\pi y)\mathrm{~d}y=\frac{1}{4\pi^2}.
\end{aligned}
\]
求解 \(\boldsymbol{A}\boldsymbol{c}=\boldsymbol{F}\) 即可得到
\[\boldsymbol{c}^*=\left(\frac{2}{\pi^4},-\frac{2}{5\pi^4},-\frac{2}{5\pi^4},\frac{1}{8\pi^4}\right).
\]
故 \(u_h(x,y)=\sum\limits_{i=1}^{4}c_i^*\varphi_i(x,y)\). #
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