Question

Consider the equation {Delta_{\mu}}u = f, on a finite connected graph (V, \mu). Here f is a given function whereas u is an unknown function. (a). prove that if one solution u exists, then all other solutions are u + const. (b). Prove that if a solution u exists, then sum_{x in V} f(x) \mu(x)=0 (c). Prove that if \sum_{x \in V} f(x) {\mu(x)}=0 is satisfied, then a solution u exists.

Answer 1

(a) If one solution u exists for the equation
`\( \Delta_(\mu)u = f \)`on a finite connected graph
`\( (V, \mu) \`), then all other solutions are of the form \
`( u + \text{const} \).`

(b) If a solution \( u \) exists for the equation
`\( \Delta_(\mu)u = f \),` then
`\( \sum_(x \in V) f(x) \mu(x) = 0 \).`

(c) If
`\( \sum_(x \in V) f(x) \mu(x) = 0 \)` is satisfied, then a solution \( u \) exists for the equation
`\( \Delta_(\mu)u = f \).`

Step-by-step explanation:

(a) The equation
`\( \Delta_(\mu)u = f \)`represents a linear partial differential equation. If \( u \) is a solution, then any function of the form
`\( u + \text{const} \)` is also a solution because the Laplace operator is linear. Let
`\( v = u + \text{const} \).` Then,
`\( \Delta_(\mu)v = \Delta_(\mu)(u + \text{const}) = \Delta_(\mu)u + \Delta_(\mu)\text{const} = \Delta_(\mu)u \),` which implies that \( v \) is also a solution.

(b) Taking the sum of both sides of
`\( \Delta_(\mu)u = f \)` over all vertices \( x \) in \( V \) weighted by
`\( \mu(x) \)` gives
`\( \sum_(x \in V) \Delta_(\mu)u \mu(x) = \sum_(x \in V) f(x) \mu(x) \)`. By the divergence theorem, the left side simplifies to \( 0 \), leading to
`\( \sum_(x \in V) f(x) \mu(x) = 0 \).`

(c) If \( \sum_{x \in V} f(x) \mu(x) = 0 \), we can construct a function \( u \) such that \( \Delta_{\mu}u = f \). One way to find \( u \) is to solve the discrete Poisson equation by defining \( u(x) = \frac{1}{\mu(x)}\sum_{y \in N(x)}\mu(y)u(y) + \frac{f(x)}{\Delta_{\mu}x} \), where \( N(x) \) is the set of neighbors of \( x \) in \( V \). This construction ensures that \( \Delta_{\mu}u = f \).