In a initial value problem does the initial condition has to satisfy the boundary condition and the governing equation?

For example: If a non-homogeneous Neumann boundary condition for the pressure in a fluid dynamics problem is applied on a boundary, can one initialize a zero pressure field?


1 Answer 1


That depends on the equation you have, and on the situation you want to model. Imagine, for example, that you are considering the advection equation $$ \partial_t u + c \partial_x u = 0, $$ i.e., you are transporting a concentration of a substance along with velocity $c$ from the left to the right. It does make sense to say that at time $t=0$, the concentration everywhere is zero, i.e. $u(x,0)=0$ but that the "reservoir" that lies beyond the left edge has a concentration 1, i.e. $u(0,t)=1$, so that a fluid with a high concentration flows in from the left, replacing the fluid with low concentration that was there before.

Likewise, if you consider the heat equation $$ \partial_t u - k \partial_{xx} u = 0, $$ then you can think of a situation where you model a solid that has initial temperature $u(x,0)=0$ but that starting at time $t=0$, you put a heat bath with temperature 1 in contact with the left or right end of the solid, i.e., $u(0,t)=1$, so that it heats the solid from the left.

On the other hand, if you had the Stokes equation $$ \partial_t u - \eta \Delta u + \nabla p= 0, \\ \nabla\cdot u = 0, $$ then you also have to prescribe initial and boundary values for the velocity. Here, I cannot immediately conjure a situation where the initial velocity is zero (which implies that up until $t=0$, the walls that enclose the fluid are also at rest), but that starting at $t=0$ the walls move at a certain velocity, thereby imparting a nonzero boundary value on the velocity. That is because you would have to accelerate the walls from rest to a nonzero velocity instantaneously -- something that require infinite acceleration and consequently infinite forces.

The considerations above are all from a physical perspective. Mathematically, such non-matching initial/boundary conditions introduce a discontinuity in the solution at the initial time and the boundary. Consequently, you will get a solution that lacks regularity at that location. That is not, a priori, a problem. However, it requires careful thought about the appropriate function spaces for your solution and, in the weak form, the test functions. You will also in general have solutions that only satisfy the weak form, not the strong form of the equation.

  • $\begingroup$ So a non compliant initialization is not necessarily ill posed? $\endgroup$ Feb 19, 2017 at 13:14
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    $\begingroup$ No. It makes physical sense in a lot of situations. It will just lead to a loss of regularity, but if you are careful how you define what is a solution, then the problem is not ill-posed. $\endgroup$ Feb 20, 2017 at 2:19

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