Show that \(f(z) = z^2\) is uniformly continuous in any open disk \(|z| < R\), where \(R>0\) is fixed, but it is not uniformly continuous on \(\mathbb C\).
(1) Show that the function $u=u(x,y)$ given by
$$u(x,y)=\frac{e^{ny}-e^{-ny}}{2n^2}\sin nx\quad \text{for}\ n\in {\mathbf N}$$
is the solution on $D=\{(x,y)\ | x^2+y^2<1\}$ of the Cauchy problem for
the Laplace equation
$$\frac{\partial ^2u}{\partial x^2}+\frac{\partial ^2u}{\partial y^2}=0,\quad
u(x,0)=0,\quad \frac{\partial u}{\partial y}(x,0)=\frac{\sin nx}{n}.$$
(2) Show that there exist points $(x,y)\in D$ such that
$\displaystyle{\limsup_{n\to\infty} |u(x,y)|=\infty}$.Solution
Goal:
Prove that \(f(z) = z^2\) is uniformly continuous on the open disk \(D(0, R) = \{z \in \mathbb{C} : |z| < R\}\) for any fixed \(R > 0\), but is not uniformly continuous on \(\mathbb{C}\).
For the function \(u_n(x,y) = \frac{e^{ny}-e^{-ny}}{2n^2}\sin(nx) = \frac{\sinh(ny)}{n^2}\sin(nx)\) on \(D = \{(x,y) \in \mathbb{R}^2 : x^2+y^2 < 1\}\), verify it solves the Laplace equation with Cauchy initial data \(u_n(x,0) = 0\) and \(\frac{\partial u_n}{\partial y}(x,0) = \frac{\sin(nx)}{n}\).
Show that there exist points \((x,y) \in D\) where \(\limsup_{n\to\infty} |u_n(x,y)| = \infty\).
Part 1: Uniform Continuity of \(f(z) = z^2\)
1. \(f(z) = z^2\) is uniformly continuous on \(D(0, R) = \{z \in \mathbb{C} : |z| < R\}\).
1.1. For any \(z_1, z_2 \in D(0, R)\), \(|f(z_1) - f(z_2)| = |z_1^2 - z_2^2| = |z_1 + z_2| |z_1 - z_2|\).Proof: By the algebraic factorization of difference of squares and multiplicativity of the complex modulus.
1.2. For any \(z_1, z_2 \in D(0, R)\), \(|z_1 + z_2| \leq |z_1| + |z_2| < R + R = 2R\).Proof: By the triangle inequality and the condition \(|z_1| < R, |z_2| < R\).
1.3. Thus, for any \(z_1, z_2 \in D(0, R)\), \(|f(z_1) - f(z_2)| \leq 2R |z_1 - z_2|\), so \(f\) is Lipschitz continuous on \(D(0,R)\) with Lipschitz constant \(2R\).Proof: Follows directly from 1 and
1.2.
1.4. For every \(\varepsilon > 0\), choosing \(\delta = \frac{\varepsilon}{2R} > 0\), if \(|z_1 - z_2| < \delta\) with \(z_1, z_2 \in D(0, R)\), then \(|f(z_1) - f(z_2)| \leq 2R |z_1 - z_2| < 2R \delta = \varepsilon\).Proof: Direct substitution of \(\delta = \frac{\varepsilon}{2R}\) into
1.3.
1.5. Q.E.D.
2. \(f(z) = z^2\) is not uniformly continuous on \(\mathbb{C}\).
2.1. A function \(f: \mathbb{C} \to \mathbb{C}\) is uniformly continuous if and only if \(\forall \varepsilon > 0, \exists \delta > 0, \forall z_1, z_2 \in \mathbb{C}, |z_1 - z_2| < \delta \implies |f(z_1) - f(z_2)| < \varepsilon\).The negation is: \(\exists \varepsilon_0 > 0, \forall \delta > 0, \exists z_1, z_2 \in \mathbb{C}\) such that \(|z_1 - z_2| < \delta\) and \(|f(z_1) - f(z_2)| \geq \varepsilon_0\).Proof: Standard logical negation of uniform continuity.
2.2. Set \(\varepsilon_0 = 1\).For any given \(\delta > 0\), choose \(z_1 = \frac{1}{\delta} \in \mathbb{R} \subset \mathbb{C}\) and \(z_2 = \frac{1}{\delta} + \frac{\delta}{2} \in \mathbb{R} \subset \mathbb{C}\).Proof: Explicit construction of test points.
2.3. \(|z_1 - z_2| = \frac{\delta}{2} < \delta\).Proof: By definition of \(z_1\) and \(z_2\).
2.4. \(|f(z_2) - f(z_1)| = |z_2^2 - z_1^2| = |(z_2 - z_1)(z_2 + z_1)| = \frac{\delta}{2} \cdot \left(\frac{2}{\delta} + \frac{\delta}{2}\right) = 1 + \frac{\delta^2}{4} > 1 = \varepsilon_0\).Proof: By direct calculation since \(z_1, z_2\) are positive real numbers.
2.5. Q.E.D.
Part 2: Cauchy Problem for the Laplace Equation
3. \(u(x,y) = \frac{\sinh(ny)}{n^2} \sin(nx)\) satisfies the Laplace equation \(\Delta u = \frac{\partial^2 u}{\partial x^2} + \frac{\partial^2 u}{\partial y^2} = 0\) on \(D\).
3.1. \(\frac{\partial u}{\partial x} = \frac{\sinh(ny)}{n} \cos(nx)\) and \(\frac{\partial^2 u}{\partial x^2} = -\sinh(ny)\sin(nx)\).Proof: Differentiating \(\sin(nx)\) with respect to \(x\) gives \(n\cos(nx)\), and differentiating again gives \(-n^2\sin(nx)\).Dividing by \(n^2\) gives \(-\sinh(ny)\sin(nx)\).
3.2. \(\frac{\partial u}{\partial y} = \frac{\cosh(ny)}{n} \sin(nx)\) and \(\frac{\partial^2 u}{\partial y^2} = \sinh(ny)\sin(nx)\).Proof: Differentiating \(\sinh(ny) = \frac{e^{ny}-e^{-ny}}{2}\) with respect to \(y\) gives \(n\cosh(ny)\), and differentiating again gives \(n^2\sinh(ny)\).Dividing by \(n^2\) gives \(\sinh(ny)\sin(nx)\).
3.3. \(\frac{\partial^2 u}{\partial x^2} + \frac{\partial^2 u}{\partial y^2} = -\sinh(ny)\sin(nx) + \sinh(ny)\sin(nx) = 0\).Proof: Sum of 1 and
3.2.
3.4. Q.E.D.
4. \(u(x,y)\) satisfies the initial conditions \(u(x,0) = 0\) and \(\frac{\partial u}{\partial y}(x,0) = \frac{\sin(nx)}{n}\).
4.1. At \(y=0\), \(\sinh(0) = \frac{e^0 - e^0}{2} = 0\), so \(u(x,0) = \frac{0}{n^2}\sin(nx) = 0\).Proof: Direct evaluation at \(y=0\).
4.2. At \(y=0\), \(\cosh(0) = \frac{e^0 + e^0}{2} = 1\), so from <1>3.<2>2, \(\frac{\partial u}{\partial y}(x,0) = \frac{\cosh(0)}{n}\sin(nx) = \frac{\sin(nx)}{n}\).Proof: Direct evaluation at \(y=0\).
4.3. Q.E.D.
Part 3: Instability / Blowup of the Solution (Hadamard Example)
5. There exist points \((x,y) \in D\) such that \(\limsup_{n\to\infty} |u_n(x,y)| = \infty\).
5.1. Choose \((x_0, y_0) = \left(\frac{1}{2}, \frac{1}{2}\right) \in D\), since \(x_0^2 + y_0^2 = \frac{1}{4} + \frac{1}{4} = \frac{1}{2} < 1\).Proof: Point belongs to \(D\) as its distance to origin is \(\frac{1}{\sqrt{2}} < 1\).
5.2. For any \(y > 0\), \(|u_n(x,y)| = \frac{e^{ny} - e^{-ny}}{2n^2} |\sin(nx)| \geq \frac{e^{ny} - 1}{2n^2} |\sin(nx)|\).Proof: For \(y > 0, n \geq 1\), \(e^{-ny} \leq 1\).
5.3. By Dirichlet’s approximation theorem / equidistribution modulo \(2\pi\), there exists a subsequence \(n_k \to \infty\) such that \(|\sin(n_k x_0)| = |\sin(n_k / 2)| \geq \frac{1}{2}\).Proof: The fractional parts of \(\frac{n}{2\pi}\) are dense in \([0,1]\) (since \(\frac{1}{2\pi}\) is irrational), so \(n/2 \pmod \pi\) visits the interval \([\pi/6, 5\pi/6]\) infinitely often, where \(\sin \geq 1/2\).
5.4. Along this subsequence \(n_k\), \(|u_{n_k}(x_0, y_0)| \geq \frac{e^{n_k / 2} - 1}{2n_k^2} \cdot \frac{1}{2} = \frac{e^{n_k/2} - 1}{4n_k^2}\).Proof: Follows from 2 and
5.3.
5.5. Since \(\lim_{k\to\infty} \frac{e^{n_k/2} - 1}{4n_k^2} = \infty\) (exponential growth dominates polynomial growth), \(\limsup_{n\to\infty} |u_n(x_0, y_0)| = \infty\).Proof: Limit of \(e^{t}/t^2\) as \(t \to \infty\) is \(\infty\).
5.6. Q.E.D.