A heat equation problem with Dirichlet boundary conditions on the domain $[x_1,x_2]$
$$\frac{\delta u}{\delta t} = k \frac{\delta^2 u}{\delta x^2}$$
$$u(x_1,t) = u(x_2,t) = 0$$
Would have eigenfunctions corresponding to sines. Whereas Neumann boundary conditions $$u'(x_1,t) = u'(x_2,t) = 0$$
Results in cosine eigenfunctions.
When evaluating steady-state solutions, with Neumann conditions we can solve for the first coefficients of the Fourier Cosine Series expansion. In the case of Dirichlet conditions, the sine series do not have a $A_0$ coefficient describing steady-state.
Intuitively the steady-state solution for Dirichlet conditions should always decay to zero, as we are allowing heat exchange on the borders and the solution for u has a time-dependent exponential decay. This is constrained with the Neumann conditions (no heat exchange at borders = isolated body), having a non-trivial steady-state.
Is this intuition correct? Is the steady-state solution for the heat equation with Dirichlet B.C always zero? Or is there something I am missing?
Best Answer
Your reasoning is correct. The steady state solutions can be obtained by setting $\partial u/\partial t=0$, leading to $$ u = c_1 x+c_2. $$ For the homogeneous Dirichlet B.C., the only solution is the trivial one (i.e., $u=0$. For the Neumann B.C., a uniform solution $u=c_2$ exists. Also, the steady state solution in this case is the mean temperature in the initial condition. Integrating the heat equation along $x_1<x<x_2$, $$ \frac{\partial}{\partial t} \int_{x_1}^{x_2} u dx =k \left.\frac{\partial u}{\partial x} \right|_{x_1}^{x_2}=0, $$ i.e., the total energy $E=\int_D u dx$ inside the domain is constant. For the steady state solution, $$ \int_{x_1}^{x_2} u(t\to \infty,x) dx=\int_{x_1}^{x_2} u(t=0,x) dx $$ $$ c_2=\frac{1}{x_2-x_1} \int_{x_1}^{x_2} u_0(x) dx, $$ in which $u_0(x)$ is the initial condition for $u$.