Radial Heat Equation with Nonzero Boundary Condition
To solve \( \dot{T} = \alpha \nabla^2 T \) with boundary condition \( T_{\text{boundary}} = T_0, \) we break it into two problems.
- \( \nabla^2 T = 0 \) with \( T_{\text{boundary}} = T_0 \) and \( T \) independent of time (steady-state problem)
- \( \dot{T} = \alpha \nabla^2 T \) with \( T_{\text{boundary}} = 0 \)
Then add the two solutions. The steady-state solution captures the long-time equilibrium behavior, while the transient solution describes how the system approaches equilibrium. Consider the heat equation with boundary conditions
$$ T(R,\theta,z,t) = T_0 $$ $$ \partial_{\theta} \, T(r,0,z,t) = \partial_{\theta} \, T(r,\pi,z,t) = 0 $$ $$ \partial_z \, T(r,\theta,0,t) = \partial_z \, T(r,\theta,H,t) = 0 $$ $$ T(r,\theta,z,0) = T_i $$Because the boundary and initial conditions are independent of \( \theta \) and \( z \), we look for solutions depending only on \( r \). Therefore, we can solve for \( T(r,t) \)
$$ \dot{T} = \alpha \nabla^2 T $$Assume \( T(r,t) = f(r) \, h(t) \)
$$ \frac{h'(t)}{h(t)} = \alpha \frac{\nabla^2 f(r)}{f(r)} = \alpha\lambda $$Since the left-hand side depends only on \( t \) and the right-hand side only on \( r \), both sides must equal a constant.
$$ h(t) = e^{\alpha\lambda t} $$ $$ \nabla^2 f(r) = \lambda f(r) $$ $$ \frac{1}{r} \frac{\partial}{\partial r} \left( r\frac{\partial f}{\partial r} \right) = \lambda f(r) $$Let \( \lambda = - k^2 \)
$$ \frac{1}{r}\frac{\partial}{\partial r} \left( r\frac{\partial f}{\partial r} \right) = - k^2 f(r) $$Introduce the dimensionless variable \( z \) where
$$ z = kr, \quad f(r(z)) = Z(z) $$ $$ \frac{\partial}{\partial z} = \frac{1}{k} \frac{\partial}{\partial r} $$ $$ \frac{1}{z} \frac{\partial}{\partial z} \left( z\frac{\partial Z}{\partial z} \right) = -Z(z) $$ $$ Z''(z) + \frac{1}{z}Z'(z) + Z(z) = 0 $$This is the Bessel equation with \( n=0 \), since there is no \( \theta \)-dependence. Solutions finite at \( r=0 \) are of the form \( J_0(z) \) where \( J_n \) are Bessel functions of the first kind. The second independent solution diverges at the origin and must therefore be discarded. Therefore,
$$ f(r) = J_0(kr) $$To deal with the inhomogeneous boundary condition on \( r \), divide and conquer:
I. \( \nabla^2 T_{ss} = 0, \quad T_{ss}(R)=T_0, \) where \( T_{ss} \) is independent of time. This is the steady-state problem.
II. \( \dot{T}=\alpha\nabla^2T, \quad T(R,t)=0. \)
I. Steady-State Problem
$$ \frac{1}{r} \frac{\partial}{\partial r} \left( r\frac{\partial T_{ss}}{\partial r} \right) = 0 $$ $$ \frac{\partial}{\partial r} \left( r\frac{\partial T_{ss}}{\partial r} \right) = 0 $$ $$ r\frac{\partial T_{ss}}{\partial r} = c $$ $$ \frac{\partial T_{ss}}{\partial r} = \frac{c}{r} $$ $$ T_{ss} = c\ln r + b $$We discard the \( \ln r \) term to keep the temperature finite at \( r=0 \).
$$ T_{ss}(r) = b $$ $$ T_{ss}(R) = b = T_0 $$ $$ T_{ss}(r)=T_0 $$II. Transient Problem
$$ T(R,t)=0 $$ $$ f(R)=J_0(kR)=0 $$We must solve
$$ J_0(kR)=0 $$With vanishing boundary conditions, we know the eigenvalues of \( \nabla^2 \) are all negative, so \( \lambda < 0 \).
The solutions are labelled by a positive integer \( m \), \( k_m \), where \( \lambda_m = - k_m^2 \). The boundary condition therefore quantizes the allowed radial modes.
\( r^{-1}\partial_r(r\partial_r) \) is symmetric with the inner product
$$ \langle u,v\rangle = \int_0^R u(r)v(r)r\,dr, \quad u(R)=v(R)=0 $$Therefore the eigenvectors
$$ f_m(r) = J_0(k_m r) $$form an orthogonal basis:
$$ \langle f_m,f_k\rangle = \delta_{mk}\|f_m\|^2 $$Hence the transient solution is
$$ T(r,t) = \sum_{m=1}^{\infty} c_m f_m(r) e^{\alpha\lambda_m t} $$Overall General Solution
$$ T(r,t) = T_0 + \sum_{m=1}^{\infty} c_m f_m(r) e^{\alpha\lambda_m t} $$Apply the initial condition:
$$ T(r,0) = T_0 + \sum_{m=1}^{\infty} c_m f_m(r) = T_i $$ $$ \sum_{m=1}^{\infty} c_m f_m(r) = T_i - T_0 $$Therefore,
$$ c_m = (T_i-T_0) \frac{\langle f_m,1\rangle}{\|f_m\|^2} $$ $$ = (T_i-T_0) \frac{\langle f_m,1\rangle}{\langle f_m,f_m\rangle} $$Therefore,
$$ T(r,\theta,z,t) = T_0 + (T_i-T_0) \sum_{m=1}^{\infty} \frac{ \int_0^R J_0(k_m r)r\,dr }{ \int_0^R J_0^2(k_m r)r\,dr } J_0(k_m r) e^{\alpha\lambda_m t} $$where \( J_0 \) is a Bessel function and \( k_m \) is the \( m \)-th positive solution of \( J_0(k_m R)=0 \)
As \( t \to \infty \), all transient modes decay exponentially, leaving only the steady-state temperature \( T_0 \).