3
$\begingroup$

Heat and Burgers equations for example are both conservation laws

$du/dt+dq/dx=0$, where $q=-u_x$

$du/dt+df/dx=0$, where $f=u^2$

The former is usually solved by common finite differences and finite elements while the latter usually by those specially designed for conservative PDEs like DG and finite volumes. Though heat equations are actually dissipative, why can't they be treated as conservative PDEs like the form above and solved numerically by methods designed for hyperbolic PDEs? To rephrase my question, I'm actually asking why heat equations can't be treated as conservative PDEs and hyperbolic equations when writing $q=-u_x$ and why Burgers can when $q=u^2$.

$\endgroup$
9
  • 1
    $\begingroup$ Because the heat equation is not actually a conservation equation? $\endgroup$ Feb 25 at 12:28
  • 1
    $\begingroup$ It looks conservative when written with a flux $\endgroup$
    – feynman
    Feb 26 at 3:24
  • 1
    $\begingroup$ @DanDoe Heat is a form of energy. Why, then, is it not conserved? $\endgroup$
    – NNN
    Feb 26 at 14:47
  • 1
    $\begingroup$ Sorry, I meant temperature. I deleted the original answer. $\endgroup$
    – Dan Doe
    Feb 26 at 15:51
  • 2
    $\begingroup$ @feynman $q=-u_x$ s indeed a flux in the physical meaning of the word, but it is not a flux in the language of hyperbolic equations. $\endgroup$ Feb 27 at 11:25

2 Answers 2

8
$\begingroup$

The core difference between the two PDEs you presented is that the heat equation is parabolic while Burgers equation is hyperbolic.

This means that for Burgers equation changes in the solution travel at finite speed through the domain, i.e., for Burgers equation you e.g. do not "see" the upstream shock until it is passing you.

In contrast, for parabolic PDEs the whole domain "knows" what is going on at any point in the domain, i.e., information spread infinitely fast. I elaborated on this difference in this Math SE post.

To give a more physical example consider incompressible vs compressible fluid flow modeling. When using an incompressible model you are simulating the case with infinite speed of sound. In contrast when using a compressible model you have actually finite speed of sound which is absolutely crucial for physical meaningful simulations of e.g. supersonic aerodynamics.


A consequence of this is that if you discretize (with a stable scheme) the PDE in space in a method of lines fashion to obtain an ODE system like

\begin{align} \boldsymbol U(t_0) &= \boldsymbol U_0 \\ \boldsymbol U(t)' &= \boldsymbol F\,(t, \boldsymbol U) \end{align} and take a look at the Jacobian of the right-hand-side $$ J = \frac{\partial \boldsymbol F}{\partial \boldsymbol U}$$ and compute its eigenvalues $\lambda \in \sigma(J)$ (i.e. the spectrum of the discrete operator) you will have purely real, negative eigenvalues $\lambda$ for the heat equation while for Burgers Equation you will obtain truly complex (conjugated) eigenvalues (also with negative real part).

This reflects the dissipative nature of the heat equation (recall from dynamic system theory that negative real eigenvalues correspond to damping). On the other hand, the truly complex-conjugated eigenvalues for Burgers' equation correspond in the dynamical systems context to oscillations which reflect the transport behaviour of the PDE, i.e., the PDE transports "mass" between the discretization points.


Finally, the implementation of boundary conditions is for hyperbolic PDEs more complex than for parabolic PDEs as one can, in general, not prescribe Dirichlet boundary conditions on each boundary. The reason for this is that the characteristics (which carry the solution information at finite speed through the domain) will eventually hit the boundaries. If the prescribed Dirichlet value and the value due to the characteristic information disagree you have an ill-posed problem and your simulation will crash. Consequently, boundary conditions for hyperbolic PDEs need to be prescribed much more carefully through solving the Riemann problem at the boundary either analytically or numerically.


Having said all of this, it is just natural to use different techniques for different problems.

$\endgroup$
8
  • 1
    $\begingroup$ OTOH it is quite common to have an equation that has both parts. Be it a viscous Burgers equation, viscous shallow water equations or real fluid flow. $\endgroup$ Feb 22 at 13:38
  • $\begingroup$ Sure, but the original question was on a purely parabolic PDE, not a mixed one (where the answer would most likely be: It depends). $\endgroup$
    – Dan Doe
    Feb 22 at 14:52
  • 3
    $\begingroup$ Just as an additional point, nonlinear hyperbolic equations require far more conditions to be satisfied to ensure existence and uniqueness of solutions, and numerical methods have to be designed specifically to satisfy these condtions, e.g., integral conservation law, entropy conditions, etc. Parabolic equations tend to be more robust to approximation as solutions are often smooth $\endgroup$
    – whpowell96
    Feb 22 at 16:22
  • 1
    $\begingroup$ @feynman You might find the following interesting en.wikipedia.org/wiki/… $\endgroup$
    – NNN
    Feb 29 at 3:10
  • 1
    $\begingroup$ @feynman Since it makes the heat equation hyperbolic, I thought you might be interested. $\endgroup$
    – NNN
    Mar 5 at 8:25
6
$\begingroup$

Actually, the heat equation can be, and often is, solved by hyperbolic methods. Instead of writing $q=-u_x$, write $q_t=-(u_x+q)/\tau$. Instead of the heat flux becoming instantaneously equal to the temperature gradient, it relaxes toward it. The relaxation time $\tau$ can be chosen to match the dispersion relationship at low frequency. Advantages are that this is probably closer to how heat really behaves and the domain of dependence is now finite, bounded by characteristic speeds $\pm\tau^{-1/2}$. This form is preferred for applications such as thermal shock, and for numerical work by some researchers because the methods can now be explicit rather than implicit. The drawback is that we now have a stiff source term. For more detail, Google Hyperbolic Navier-Stokes.

$\endgroup$
2
  • $\begingroup$ thank you very much for the answer! Can a heat equation be solved by hyperbolic methods when $q=-u_x$? $\endgroup$
    – feynman
    Feb 25 at 4:11
  • 2
    $\begingroup$ No, if you take $q=-u_x$ the problem is parabolic and implies action at a distance with an infinite domain of dependence. The classification of a pde describes how the highest frequencies behave and that depends on the highest derivatives along each coordinate. $\endgroup$
    – Philip Roe
    Feb 27 at 22:47

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.