I have seen Separation of Variables before. 1. Yes, and I'm comfortable with it. 2. Yes, but I don't quite remember. 3. Nope 4. I'm triggered. PS. Hi from San Antonio -DC
Our example problem has the following boundary conditions: * $V(0,y>0) = 0; V(a,y>0) = 0$ * $V(x_{0\rightarrow a},y=0) = V_0; V(x,y\rightarrow \infty) = 0$ If $X''= c_1 X$ and $Y'' = c_2Y$ with $c_1 + c_2 = 0$, which is constant is positive? 1. $c_1$ 2. $c_2$ 3. It doesn't matter either can be Note: * CORRECT ANSWER: B * Exponential solutions are expected in the y direction
Given the two diff. eq's : $$\dfrac{1}{X}\dfrac{d^2X}{dx^2} = C_1 \qquad \dfrac{1}{Y}\dfrac{d^2Y}{dy^2} = C_2$$ where $C_1+C_2 = 0$. Given the boundary conditions in the figure, which coordinate should be assigned to the negative constant (and thus the sinusoidal solutions)? <img src="./images/cq_cartesian_bc_1.png" align="right" style="width: 400px";/> 1. x 2. y 3. $C_1 = C_2 = 0$ here 4. It doesn't matter. Note: * CORRECT ANSWER: B * Because it has to vanish on both ends
Given the two diff. eq's : $$\dfrac{1}{X}\dfrac{d^2X}{dx^2} = C_1 \qquad \dfrac{1}{Y}\dfrac{d^2Y}{dy^2} = C_2$$ where $C_1+C_2 = 0$. Given the boundary conditions in the figure, which coordinate should be assigned to the negative constant (and thus the sinusoidal solutions)? <img src="./images/cq_cartesian_bc_2.png" align="right" style="width: 400px";/> 1. x 2. y 3. $C_1 = C_2 = 0$ here 4. It doesn't matter. Note: * CORRECT ANSWER: C * It's constant throughout!
When does $\sin(ka)e^{-ky}$ vanish? 1. $k = 0$ 2. $k = \pi/(2a)$ 3. $k = \pi/a$ 4. A and C 5. A, B, C Note: * CORRECT ANSWER: D * It's actuall n*pi/a
Suppose $V_1(r)$ and $V_2(r)$ are linearly independent functions which both solve Laplace's equation, $\nabla^2 V = 0$. Does $aV_1(r)+bV_2(r)$ also solve it (with $a$ and $b$ constants)? 1. Yes. The Laplacian is a linear operator 2. No. The uniqueness theorem says this scenario is impossible, there are never two independent solutions! 3. It is a definite yes or no, but the reasons given above just aren't right! 4. It depends... Note: * CORRECT ANSWER: A