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