**Example 2:** Find the steady-state temperature \( u(r, \theta) \) in a semicircular plate. The boundary value problem is:\[\frac{\partial^2 u}{\partial r^2} + \frac{1}{r} \frac{\partial u}{\partial r} + \frac{1}{r^2} \frac{\partial^2 u}{\partial \theta^2} = 0, \quad 0 < \theta < \pi, \quad 0 < r < c\]Boundary conditions:\[u(c, \theta) = u_0, \quad 0 < \theta < \pi\]\[u(r, 0) = 0, \quad u(r, \pi) = 0, \quad 0 < r < c\]This problem involves solving a partial differential equation to find the temperature distribution in a semicircular region, considering specific boundary conditions at the edge of the plate and along its diameter.

We need to solve the following boundary problem. ∂2u∂r2+1r∂u∂r+1r2∂2u∂θ2=0, 0<θ<π, 0<r<c u(c, θ)=u0, 0<θ<π u(r, 0)=0, u(r, π)=0, 0<r<c----------------(1) We will be solving the above by the method of variation of parameters. Let u(r, θ)=R(r) Θ(θ). Then after using u(r, θ) in the above equation then we get, R''Θ+1rR'Θ+1r2RΘ''=0=>Θ R''+1rR'+1r2RΘ''=0=>r2 R''+1rR'R=-Θ''Θ=λ (Say) Thus, we have, r2R''+rR'-λR=0 -----(2) and Θ''+λΘ=0, Θ(0)=0, Θ(π)=0. ----------(3)Now, we will be solving the above equations. Let us first solve (3). The general solution of (3) is Θ=A cos(λθ)+B sin(λθ). By relation, Θ(0)=0=>A=0. And we have, Θ(π)=0=>B sin(λπ)=0 =>sin(λπ)=0 (as B≠0) =>λπ=nπ =>λ=n2 The eigenvalue of equation (3) is λn=n2. Thus, the general solution becomes, Θ=Bn sin(nθ). Now, we will be solving equation (2).By using the values of λ, the equation (2) becomes, r2R''+rR'-n2R=0. The indicial polynomial of this equation is, m(m-1)+m-n2=0=>m2-n2=0=>m=±n. Thus, the general solution is Rn=c1rn+c2r-n. To obtain a solution that remains bounded as r→0+, we take c2=0. Because of the Dirichlet condition at r=c, it is convenient to have r(c)=1, therefore we take c1=c-n. Thus, the general solution of equation (2) becomes, Rn=c-nrn=>Rn=rcn.Now, un(r, θ)=Rn(r)Θ(θ)=rcnsin(nθ). Thus, the general solution of equation (1) is u(r, θ)=∑n=1∞Dnrcnsin(nθ). We now need to find the value of Dn. We are given that, u(c, θ)=u0=>∑n=1∞Dnccnsin(nθ) =u0=>∑n=1∞Dn sin(nθ) =u0=>Dn=2π∫0πu0 sin(nθ)dθ Answer: Thus, the general solution to the given system is u(r, θ)=∑n=1∞Dnrcnsin(nθ) where Dn=2π∫0πu0 sin(nθ)dθ. Note: We can find the exact values of Dn if we know the expression of u0.