For a second order PDE, for example heat conduction equation $\frac{\partial T}{\partial t} = \frac{\alpha}{C_p} \nabla^2 T$, is it possible to determine the steady-state (or even transient) solution with two Dirichlet conditions? I have two different questions regarding this

  1. From my understanding, the solution is non unique for all equal valued ratios of $\alpha$ and $C_p$. so two Dirichlet conditions say nothing about how fast the disturbance propagates with a temporal change of one boundary condition. So only the knowledge of $T$ and $\nabla T$ together can fix the solution curve for specific values of $\alpha$ and $C_p$ instead of the ratio.

  2. Integrating the 1-D second order (steady-state) equation gives $T=C_1x+C_2$ where $C_1 = \frac{\partial T}{\partial x}$. So, two Dirichlet conditions are two values of $C_2$ and therefore still don't give us the value of $C_1$ which is required to fix the solution curve. So in this case, how is it possible to assume we know the solution with two just Dirichlet conditions?


1 Answer 1


Before I begin answering your question, I just need to clarify one key point.

On the notion of "Uniqueness"

The way you use the word "unique" in your question is not correct. "Uniqueness" has a very precise meaning in a mathematical context and is very different from the way you are using it. When $\alpha$ and $C_p$ are of equal ratio, they produce the same solution curve. However, this does not imply that the solution of the PDE is not unique. Uniqueness refers to whether two completely different functions solve the same PDE with the same data (i.e. data = coefficients and initial/boundary conditions). It is more appropriate to say that the problems with the same ratio $\frac{\alpha}{C_p}$ are equivalent problems and must have the same solution, if one exists.

Please keep this in mind as I answer your questions more directly below:

  1. Both the steady state and transient PDE's are well posed (under sufficient assumptions on the coefficients & initial/boundary conditions) for both pure dirichlet and mixed boundary conditions. Thus, a solution exists, is unique, and depends continuously on the data. Therefore, you can uniquely determine the steady state solution using only dirichlet boundary conditions. The transient solution can also be uniquely determined from a combination of the dirichlet boundary conditions and the initial condition.

  2. You can determine the coefficients of the analytical solution (1D case) by setting up a system of equations. Suppose the boundary conditions of the PDE at $x_1$ and $x_2$ are given as $g(x_1)=g_1$ and $g(x_2)=g_2$, respectively. Then, the coefficients $C_1$ and $C_2$ are uniquely determined by the system of equations

\begin{align}C_1x_1 + C_2 = g_1\\ C_1x_2+C_2=g_2\end{align}

  • $\begingroup$ As an extension to my question, might i ask, considering a transient equation, if there is a change in the boundary at one end and the system readjusts the gradient to the new values, we still lose the information of the dynamics, won't we? By that i mean, we still can only say the gradient will adjust itself to the new boundary value based on the ratio of heat capacity to conductivity, and this ratio can be the same for several values of the two. where as having a flux condition at one end will determine the conductivity and therefore the solution is suitable to the problem at hand. thoughts? $\endgroup$
    – vkumar
    Commented Jan 8, 2014 at 13:00
  • $\begingroup$ Also for inverse theories $\endgroup$
    – vkumar
    Commented Jan 8, 2014 at 13:14
  • $\begingroup$ @vkumar: it is difficult to say without more details. Please formulate a new question and we, as a community, will try our best to answer it. In your new question, be specific about your transient boundary conditions (i.e. Give them a specific equation of interest to you). $\endgroup$
    – Paul
    Commented Jan 8, 2014 at 17:40

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.