- Find the heat kernel for the half-line, $x \geq 0$, and use it to solve the IVP,
$$
\begin{array}{rlrl}
u_t & =\frac{1}{2} u_{x x}, & x \in \mathbb{R}^{+}, t \in \mathbb{R}^{+} \
\text {IC } & \
u(x, 0) & =f(x), & x \in \mathbb{R}^{+} \
\mathrm{BC} & \
u_x(0, t) & =0, \quad t \in \mathbb{R}^{+}
\end{array}
$$
This is an example of Kelvin’s method of images.
To find the heat kernel for the half-line $x\geq 0$, we need to solve the heat equation with the boundary condition $u(0,t)=0$ and initial condition $u(x,0)=\delta(x)$, where $\delta(x)$ is the Dirac delta function. We can use the method of images, also known as Kelvin’s method, to solve this problem.
Let $G(x,t;y)$ be the heat kernel for the half-line. We want to find a function $G(x,t;y)$ that satisfies the heat equation
$\frac{\partial G}{\partial t}=\frac{1}{2} \frac{\partial^2 G}{\partial x^2}$
subject to the boundary condition $G(0,t;y)=0$ and initial condition $G(x,0;y)=\delta(x-y)$.
We will use the method of images to construct a solution to this problem. Specifically, we will introduce an auxiliary problem on the whole line and use the reflection principle to relate the solutions of the two problems.
Let $G(x,t;y)$ be the solution to the problem above. Consider the function
$H(x, t ; y)=G(|x|, t ; y)$
where $|x|$ denotes the absolute value of $x$. Since $|x|$ is an even function, it satisfies the symmetry condition $H(-x,t;y)=H(x,t;y)$. Note that $H(x,t;y)$ is a solution to the heat equation on the whole line with the initial condition $H(x,0;y)=\delta(x-y)$.
To apply the reflection principle, we introduce the reflected function
$R(x, t ; y)=H(x, t ;-y)=G(|x|, t ;-y)$
Prove or disprove that the quotient of two harmonic functions is harmonic.
The quotient of two harmonic functions is not necessarily harmonic. Here is a counterexample:
Consider the functions $u(x,y) = \sin(x+y)$ and $v(x,y) = \sin(x-y)$. Both $u$ and $v$ are harmonic functions, since they satisfy Laplace’s equation $\nabla^2 f = 0$: $$\frac{\partial^2 u}{\partial x^2} + \frac{\partial^2 u}{\partial y^2} = -\sin(x+y) – \sin(x+y) = 0,$$ $$\frac{\partial^2 v}{\partial x^2} + \frac{\partial^2 v}{\partial y^2} = -\sin(x-y) – \sin(x-y) = 0.$$ However, their quotient $f(x,y) = \frac{u(x,y)}{v(x,y)} = \frac{\sin(x+y)}{\sin(x-y)}$ is not harmonic. To see this, we compute the Laplacian of $f$: $$\nabla^2 f = \frac{\partial^2 f}{\partial x^2} + \frac{\partial^2 f}{\partial y^2} = \frac{2\sin(x+y)\cos(x-y)}{\sin^3(x-y)} – \frac{2\cos(x+y)\sin(x-y)}{\sin^3(x-y)}.$$ This is not zero in general, so $f$ is not harmonic.
Therefore, the quotient of two harmonic functions is not necessarily harmonic.
Solve the Neumann BVP $$ \Delta u=0, \quad x \in \mathbb{R}, y \in \mathbb{R}^{+} $$ BC $$ \frac{\partial u}{\partial n}(x, 0)=x e^{-x^2} $$
To solve this problem, we use separation of variables. Let $u(x,y) = X(x)Y(y)$, then the Laplace equation becomes $$X”(x)Y(y) + X(x)Y”(y) = 0.$$ Dividing both sides by $XY$, we obtain $$\frac{X”(x)}{X(x)} = -\frac{Y”(y)}{Y(y)} = \lambda,$$ where $\lambda$ is a constant. This gives us two ordinary differential equations to solve: $$X”(x) – \lambda X(x) = 0,$$
