Steady-State temperature in heat equation over a wedge In preparation of a upcoming exam I have came across the following problem and I am not quite sure how to complete this problem and am looking for some help finding this solution, thanks!
I need to find the steady state temperature in a wedge shaped flat plate occupying the region $0\leq r \leq1, 0\leq \theta \leq \frac{\pi}{4} $ (which are polar coordinates). The edges $\theta = 0 $ and $\theta = \frac{\pi}{4}$ are kept at zero temprature and the arc $ r = 1 (0 < \theta < \frac{\pi}{4})$ is kept at constant temperature, 
Do you need to use the laplace transform to solve this?
$$\Delta u = \frac{\partial^2 u}{\partial x^2 } + \frac{\partial^2 u }{\partial y^2} = 0 $$
I know that the steady state solution is $$k^2 \frac{\partial^2 u }{\partial x^2} = \frac{\partial u}{\partial t } = 0 $$
 so then $$\frac{\partial u^2}{\partial x^2 } =0 $$ and then $$u(x,t)=Ax+b $$ but then how do i use my conditions to solve this problem?
 A: You should start by posing the heat-conduction problem. Under some assumptions, the temperature distribution in the solid is dictated by the heat equation
$$ \rho c_p T_t = \nabla \cdot (k \nabla T),$$
where $k$ is the thermal conductivity, $\rho$ is the density and $c_p$ is the specific heat of the material, $t$ is time and $T$ is temperature. 
Let $\alpha = k/(\rho c_p)$ be thermal diffusivity and define the nondimensional time $\tau = t/t_c$, with $t_c = r_o^2/\alpha$, and the nondimensional radius $ r=r^*/r_o$ where $r_o$ is the radius of your plate and ${}^*$ denotes a dimensional quantity. On defining 
$$ u = \frac{T-T_e}{T_i - T_e},$$ 
where $T_e$ is the temperature at the edges and $T_i$ is an initial temperature distribution, your problem reduces to integrate
$$u_\tau = \nabla^2 u,$$
to be integrated with initial condition $u = 1$ and boundary conditions $u = 0$ at the edges and $u = u_a$, say, at the arc.
Now let's have a look at the specific system of reference you have. Since the domain is a sector of a circle, it makes sense using cylindrical polar coordinates $(r,\theta)$. Since there's no addition of heat, the problem reaches a steady state and you don't have to care about initial conditions. Thus, the heat equation reduces to integrate:
$$ 0 = \frac{1}{r} \partial_r (r \, \partial_r u) + \frac{1}{r^2} \partial_{\theta\theta}u, \quad u = 0 \text{ at } \theta = 0,\pi/4, \quad u = u_a \text{ at } r=1 $$
This second-order PDE can be solved using, for instance, separation of variables. Let $u = G(r)\phi(\theta),$ where $G$ and $\phi$ are nonzero functions of their arguments. Introduction of the ansatz yields:
$$ 0 = \underbrace{\frac{r}{G} (r G')'}_{\lambda^2} + \underbrace{\frac{\phi''}{\phi}}_{-\lambda^2}  $$
Introducing the suitably signed separation constant $\lambda^2$ the problem in $\theta$ becomes $\phi''+\lambda^2 \phi = 0$ with $\phi(0) = \phi(\pi/4) = 0$. Nontrivial solutions are
$$\phi_n = \sin(4n\theta), \quad n = 1,2,\ldots$$ 
for which one may define $\lambda_n = 4n > 0$. On the other hand, for $G$, the solution to 
$$  (r \, G')' - \frac{\lambda^2}{r} G = 0 \implies r^2 G'' + r G' - \lambda^2 G = 0$$
can be sought of in the form $G \propto r^\sigma$ giving 
$$ \sigma (\sigma -1)+\sigma - \lambda^2 = \sigma^2 - \lambda^2 = 0$$
which provides $\sigma = \pm \lambda$ and produces solutions of the form $G\propto r^\lambda$ and $G \propto r^{-\lambda}$. On eliminating solutions that blow up at $r \to 0$, we expand $u = \sum_{n=1}^\infty B_n \phi_n(\theta)G_n(r)$, where $B_n$ are real constants and $G_n(r) = r^{\lambda_n}$. The constant(s) of integration are found with the help of the boundary condition at $r = 1$ which yields:
$$u_a = \sum_{n=1}^\infty B_n G_n(1) \phi_n(\theta)$$
Note that $G_n(1) = 1$. Multiply both sides by $\phi_m$ and integrate over $0<\theta<\pi/4$ to find:
$$ B_n =  \frac{u_a \int^{\pi/4}_0 \phi_n(\theta) \, \mathrm{d}\theta}{\int^{\pi/4}_0 \phi^2_n(\theta) \, \mathrm{d}\theta} $$
The problem is solved upon evaluation of the integrals. To that end, it's convenient to note that $\sin^2(4n\theta) = (1-\cos(8n\theta))/2$. Therefore, the integral in the denominator is just $ [ \theta/2 - \sin{(8 n\theta)}/(16n)]^{\pi/4}_0 = \pi/8$ because the sines evaluate to $0$ in $0$ and in $2 n \pi$. For the one upstairs, simply use 
$$ \int^{\pi/4}_0 \sin{4n\theta} \, \mathrm{d}\theta = \left.\frac{\cos{4n\theta}}{4n}\right|^0_{\pi/4} = \frac{1 - \cos{n\pi}}{4n} = \frac{1 - (-1)^n}{4n} $$
Note that this is $0$ for $n = 2,4,\ldots$. Thus, on defining $n$ to be an odd number, $n = 2k+1$, $k = 0, 1, \ldots$, we can simplify the result to
$$B_k = \frac{2}{4(2k+1)} = \frac{1}{2(2k+1)} $$
Can you take it from here?
