What happens to other eigenvalue? - steady state heat equation A circular plate is bounded by circles of radii $r=2$ and $r=4$, its surface is insulated and temperatures along boundaries are given by $u(2,\theta)=10\cos\theta + 6\sin\theta$ and $u(2,\theta)=17\cos\theta + 15\sin\theta$, how can I find the steady state temperature distribution in the plate?
In this, my professor omitted the solution with eigenvalue $\lambda=0$, why is that? I would also appreciate a full step solution so that I can know I understood the other steps correctly.
as requested i will add image of note of this problem,in that i couldn't understand why solution corresponding to $\lambda$ didn't included in the coefficent calculation of Fourier series,also i didnt understand how did they get coefficent without adding solution corresponding to $\lambda=0!$
 A: The Steady-State Equation
Ok, you are solving the steady-state equation, i.e. finding the temperature distribution at some point far in the future when the system has stabilized. In this case, the temperature isn't changing, so its derivative (in time) is zero. Thus we get
$$
\Delta u = u_{t} = 0
$$
and so actually, we are solving Laplace's equation (if you know what that is). Once we change to polar coordinates, we get the equation
$$
r^2 u_{rr} + r u_{r} + u_{\theta\theta} = 0
$$
as you have written in your notes.
Separation of Variables
Separating variables (i.e. writing $u(r,\theta) = R(r)\Theta(\theta)$), 
$$
r^2 R''\Theta + rR'\Theta + R\Theta'' = 0 \\
\frac{r^2 R'' + rR'}{R} = \mu = -\frac{\Theta''}{\Theta}
$$
so 
$$
\Theta'' + \mu\Theta = 0, \\
r^2 R'' + rR' - \mu R = 0.
$$
Use the periodicity conditions $\Theta(0) = \Theta(2\pi), \Theta'(0) = \Theta'(2\pi)$ to determine $\mu_n = n^2$ and $\Theta_n(\theta)$ can be either $\sin(n\theta)$ or $\cos(n\theta)$. Let $\lambda_n = \sqrt{\mu_n} = n$ (note that $\mu$ is the eigenvalue, though $\lambda$ is sort of more directly useful to us).
Now, solving 
$$
r^2 R'' + rR' - n^2R = 0,
$$
you can check that, if $n$ is not zero, then two linearly independent solutions are $r^n$ and $r^{-n}$ (i.e. $r^\lambda$ and $r^{-\lambda}$, as you have written). So the general solution for $n \neq 0$ is 
$$
R_n = A_n r^n + B_n r^{-n}.
$$
If $n$ is zero then $r^n = r^{-n}$ and we have only found one solution! In this case, another solution is $\log r$, and so
$$
R_0 = E \log r + F
$$
The General Solution
Thus, 
$$
u_n = (A_n r^n + B_n r^{-n})\cos(n\theta) + (C_n r^n + D_n r^{-n})\sin(n\theta)
$$
for $n > 0$ and 
$$
u_0 = (E \log r + F)\cos(0\theta) = (E \log r + F)
$$
(if $n=0$, then $\sin(n\theta) \equiv 0$). (note further that we only have to consider this solution on the annulus; on a disc, the unbounded behavior of $\log r$ near $r=0$ allows us to discard it)
Those $u_n$ satisfy the PDE and periodicity condition for any $n$, and so the general solution is of the form
$$
u = \sum_n u_n = (E \log r + F) + \sum_{n=1}^\infty(A_n r^n + B_n r^{-n})\cos(n\theta) + \sum_{n=1}^\infty(C_n r^n + D_n r^{-n})\sin(n\theta)
$$
as you have written. 
Boundary Conditions and Coefficients
Now, to implement the boundary conditions and determine the coefficients:
$$
10\cos\theta + 6\sin\theta = (E \log 2 + F) + \sum_{n=1}^\infty(A_n 2^n + B_n 2^{-n})\cos(n\theta) + \sum_{n=1}^\infty(C_n 2^n + D_n 2^{-n})\sin(n\theta),\\
17\cos\theta + 15\sin\theta = (E \log 4 + F) + \sum_{n=1}^\infty(A_n 4^n + B_n 4^{-n})\cos(n\theta) + \sum_{n=1}^\infty(C_n 4^n + D_n 4^{-n})\sin(n\theta).
$$
To find the coefficients, we use orthogonality of sines and cosines on $\theta \in [0,2\pi)$; that is:
$$
\int_0^{2\pi} \cos(n\theta)\,d\theta = 0 \\
\int_0^{2\pi} \sin(n\theta)\,d\theta = 0 \\
\int_0^{2\pi} \cos(m\theta)\cos(n\theta)\,d\theta = 0 (m \neq n) \\
\int_0^{2\pi} \sin(m\theta)\sin(n\theta)\,d\theta = 0 (m \neq n) \\
\int_0^{2\pi} \cos(m\theta)\sin(n\theta)\,d\theta = 0
$$
(let me know if this isn't familiar), and so, noting that the boundary conditions only have sine and cosine with $n = 1$, this means all other coefficients, including $n = 0$ are zero. To solve for $n = 1$, write
$$
10 = A_1 2^1 + B_1 2^{-1}\\
17 = A_1 4^1 + B_1 4^{-1}\\
6 = C_1 2^1 + D_1 2^{-1}\\
15 = C_1 4^1 + D_1 4^{-1}
$$
which is a system we can solve for $A_1, B_1$, etc. (and you get $A_1,B_1,C_1 = 4, D_1 = -4$ as you note). 
So now, we know our solution is of the form
$$
u = 4(r + r^{-1})\cos\theta + 4(r - r^{-1})\sin\theta.
$$
Ok, So When Would $E$ and $F$ be Non-Zero?
If you have boundary conditions that looked like 
$$
u(2,\theta) = 9 + 10\cos\theta + 6\sin\theta \\
u(4,\theta) = 3 + 17\cos\theta + 15\sin\theta
$$
you would have non-zero $E,F$. In this case, you would have two additional equations:
$$
9 = E \log 2 + F \\
3 = E \log 4 + F
$$
which you could solve for $E,F$. Basically: if the boundary conditions include constant terms, then you "need" the zero eigenvalue. If not, then you don't.
