The solution to Laplace's equation in one Cartesian dimension is always linear. Why? 1. The general solution for $\frac{d^2V(x)}{dx^2} = 0$ is always a line. 2. The only way for a given point $V(x_i)$ (in 1d) to be the arithmetic average of its neighbors is for all point to be on a line. 3. Given boundary conditions, it's a unique solution. 4. None of these. 5. More than one of these.

### Method of Relaxation <img src="./images/convergence_relax.png" align="center" style="width: 700px";/>

Consider a function $f(x)$ that is both continuous and continuously differentiable over some domain. Given a step size of $a$, which could be an approximate derivative of this function somewhere in that domain? $df/dx \approx$ 1. $f(x_i+a) - f(x_i)$ 2. $f(x_i) - f(x_i-a)$ 3. $\frac{f(x_i+a) - f(x_i)}{a}$ 4. $\frac{f(x_i) - f(x_i-a)}{a}$ 5. More than one of these Note: * Correct Answer: E (C and D)

If we choose to use: $$\dfrac{df}{dx} \approx \dfrac{f(x_i+a) - f(x_i)}{a}$$ Where are we computing the approximate derivative? 1. $a$ 2. $x_i$ 3. $x_i + a$ 4. Somewhere else Note: * Correct Answer: D (it's halfway between)

Taking the second derivative of $f(x)$ discretely is as simple as applying the discrete definition of the derivative, $$f''(x_i) \approx \dfrac{f'(x_i + a/2) - f'(x_i - a/2)}{a}$$ Derive the second derivative in terms of $f$.

With the approximate form of Laplace's equation: $\dfrac{V(x_i+a) - 2V(x_i) + V(x_i-a)}{a} \approx 0$ What is a the appropriate estimate of $V(x_i)$? 1. ${1}/{2}(V(x_i+a)-V(x_i-a))$ 2. ${1}/{2}(V(x_i+a)+V(x_i-a))$ 3. ${a}/{2}(V(x_i+a)-V(x_i-a))$ 4. ${a}/{2}(V(x_i+a)+V(x_i-a))$ 5. Something else Note: * Correct answer: B

To investigate the convergence, we must compare the estimate of $V$ before and after each calculation. For our 1D relaxation code, $V$ will be a 1D array. For the kth estimate, we can compare $V_k$ against its previous value by simply taking the difference. Store this in a variable called ``err``. What is the type for ``err``? 1. A single number 2. A 1D array 3. A 2D array 4. ??? Note: * Correct Answer: B

The Method of Relaxation also works for Poisson's equation (i.e., when there is charge!). Given, $\nabla^2 V \approx \dfrac{V(x+a)-2V(x)+V(x-a)}{a^2}$ Which equations describes the appropriate "averaging" that we must do: 1. $V(x) = \dfrac{1}{2}(V(x+a)-V(x-a))$ 2. $V(x) = \dfrac{\rho(x)}{\varepsilon_0}+\dfrac{1}{2}(V(x+a)+V(x-a))$ 3. $V(x) = \dfrac{a^2\rho(x)}{2\varepsilon_0}+\dfrac{1}{2}(V(x+a)+V(x-a))$ Note: * Correct answer: C

### Separation of Variables (Cartesian) <img src="./images/cartesian_sep_variables.png" align="center" style="width: 500px";/>