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?


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 $u$.

  • 1
    $\begingroup$ This is very helpful. How are you getting $u = c_1 x + c_2$ after setting $\delta u / \delta t = 0$? The solution I obtain for $u$, for general $\lambda > or < 0$ is $u(x,t) = \sum_n A_n sin(\lambda_n x) e^{-\lambda_n t}$. I could get your solution only by setting $\lambda = 0$ and getting eigenfunctions for that: $f(x) = c_1 x + c_2$ $\endgroup$
    – hirschme
    Dec 4 '18 at 16:08
  • 1
    $\begingroup$ If $\partial u/\partial t=0$, the 'remaining' equation is $\partial^2 u/\partial x^2=0$. Furthermore, since there is no dependence on $t$ anymore, we can change the partial derivative to total derivative, i.e., $$\frac{d^2 u}{dx^2}=0.$$ Integrating twice, we have $u=c_1 x+c_2$. See that this solution does not correspond to the solution for the original equation for any time $t$, but is the solution of the steady state equation. One can even say that, if $u$ does satisfy $$\frac{\partial u}{\partial t} = k \frac{\partial ^2 u}{\partial x^2},$$ then $$\lim_{t\to \infty} u(x,t) = c_1x+c_2.$$ $\endgroup$
    – rafa11111
    Dec 4 '18 at 16:19

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