The heat equation on the domain [0, 1] × [0, ∞) with boundary condition u = 0 on ({0} ∪ {1}) × [0, ∞) is given by:

∂u/∂t = ∂²u/∂x²

To solve this equation, we assume that the solution has a separable form:

u(x, t) = X(x)T(t)

Substituting this into the heat equation, we get:

X(x)T'(t) = X”(x)T(t)/

Dividing both sides by X(x)T(t) gives:

T'(t)/T(t) = X”(x)/X(x)

Since the left-hand side only depends on t and the right-hand side only depends on x, both sides must be equal to a constant, say -λ²:

T'(t)/T(t) = -λ² = X”(x)/X(x)

Solving for T(t), we get:

T(t) = e^(-λ²t)

Solving for X(x), we get:

X”(x) + λ²X(x) = 0

This is a standard second-order linear homogeneous differential equation, and its solutions are:

X(x) = Acos(λx) + Bsin(λx)

where A and B are constants that depend on the initial conditions.

Using the boundary condition that u = 0 on ({0} ∪ {1}) × [0, ∞), we get:

X(0) = A = 0 X(1) = B*sin(λ) = 0

Since sin(λ) = 0, we must have λ = nπ for some integer n > 0. Therefore, the solutions are:

u_n(x, t) = sin(nπx)e^(-n²π²t)

where n = 1, 2, 3, …

The general solution is a linear combination of these solutions:

u(x, t) = ∑[n=1 to ∞] c_n*u_n(x, t)

where the coefficients c_n are determined by the initial condition u(x, 0).

Therefore, the solutions of the heat equation on [0, 1] × [0, ∞) with the boundary condition u = 0 on ({0} ∪ {1}) × [0, ∞) are given by:

u(x, t) = ∑[n=1 to ∞] c_n*sin(nπx)e^(-n²π²t)

where the coefficients c_n are determined by the initial condition u(x, 0).

We can start by plugging the given form of the solution, $u=f(t)g(x)$, into the heat equation:

\begin{align*} \frac{\partial^2 u}{\partial x^2} &= \frac{\partial}{\partial x} \frac{\partial u}{\partial x} \ &= \frac{\partial}{\partial x} \left(\frac{\partial}{\partial x} (f(t) g(x))\right) \ &= \frac{\partial}{\partial x} \left(f(t) \frac{\partial g(x)}{\partial x}\right) \ &= f(t) \frac{\partial^2 g(x)}{\partial x^2} \end{align*}

Similarly, we have

\begin{align*} \frac{\partial u}{\partial t} &= \frac{\partial}{\partial t} (f(t) g(x)) \ &= \frac{\partial f(t)}{\partial t} g(x) \end{align*}

Plugging these expressions into the heat equation, we get

\begin{align*} f(t) \frac{\partial^2 g(x)}{\partial x^2} &= \frac{\partial f(t)}{\partial t} g(x) \ \frac{1}{g(x)} \frac{\partial^2 g(x)}{\partial x^2} &= \frac{1}{f(t)} \frac{\partial f(t)}{\partial t} \ \end{align*}

The left-hand side only depends on $x$, and the right-hand side only depends on $t$. Therefore, both sides must be constant, say $-k^2$, where $k$ is a constant. This gives us two separate ordinary differential equations:

\begin{align*} \frac{\partial^2 g(x)}{\partial x^2} + k^2 g(x) &= 0 \ \frac{\partial f(t)}{\partial t} + k^2 f(t) &= 0 \ \end{align*}

The first equation is the well-known differential equation for harmonic oscillation, and its solutions are of the form $g(x) = A\cos(kx) + B\sin(kx)$, where $A$ and $B$ are constants determined by the initial/boundary conditions.

The second equation is a first-order linear differential equation, and its solutions are of the form $f(t) = Ce^{-k^2t}$, where $C$ is a constant determined by the initial/boundary conditions.

Combining the solutions for $g(x)$ and $f(t)$, we get

\begin{align*} u(x,t) &= f(t) g(x) \ &= Ce^{-k^2t} (A\cos(kx) + B\sin(kx)) \end{align*}

where $A$, $B$, $C$, and $k$ are constants determined by the initial/boundary conditions.

Therefore, all solutions of the form $u=f(t)g(x)$ to the heat equation $\frac{\partial^2 u}{\partial x^2}=\frac{\partial u}{\partial t}$ are linear combinations of functions of the form $Ce^{-k^2t} (A\cos(kx) + B\sin(kx))$.

Let $G(x)$ be a Green’s function for $\mathbb{R}^n$. Then, by definition, $G(x)$ is harmonic on $\mathbb{R}^n\backslash{0}$ and depends only on the radius $r=|x|$.

For $n=1$, there is no nontrivial Green’s function, since any non-constant harmonic function on $\mathbb{R}\backslash{0}$ is periodic and hence cannot depend only on the radius.

For $n=2$, we can take $G(x)=\log |x|$ as a Green’s function. To see that $G$ is harmonic, note that for $r=|x|>0$, $$\frac{\partial^2}{\partial x_1^2}G(x) + \frac{\partial^2}{\partial x_2^2}G(x) = \frac{\partial^2}{\partial r^2}\log r + \frac{1}{r}\frac{\partial}{\partial r}\left(r\frac{\partial}{\partial r}\log r\right) = 0,$$ where we have used the fact that $\frac{\partial}{\partial r}\log r = \frac{1}{r}$ and $\frac{\partial^2}{\partial r^2}\log r = -\frac{1}{r^2}$. It is clear that $G(x)$ depends only on $r=|x|$.