The flame is propagating from the right (burnt gases) to the left (unburned gases). The Dirac delta function models the heat release in the reaction zone at x = 0 and d represents the flame thickness (see the equations in the third comment) .Thank you!
In the first equation, you want the partial of T with respect to t on the left-hand side. If the flame is propagating to the right with speed v, the argument of the delta function should be x - vt, not x. Is that what you intended?
Assuming I've interpreted your question correctly in the preceding comment, you want to transform to a moving frame of reference in which the heat source is at the origin. This introduces an advective term into the PDE. You would then solve the PDE separately in the regions x > 0 and x < 0 and apply the boundary conditions at infinity. This removes the inhomogeneity due to the heat source. These solutions are simplest if you look at the steady state. Finally, you integrate the PDE in a small interval about the origin to get a jump condition on the partial of T with respect to x. The jump is proportional to Q. Applying this boundary condition yields a unique steady state solution.
Dear R. Bradley.
I think that the equations are well written. The transient regime will be shown in the next comment.
The problem is devoted to the study of a stationary laminar flame. i.e. the first term of the right hand side vanishes. u is the laminar flame velocity. I would be pleased if you can give a starting point for this differential equation. Thank you!
When the reaction term is taken equal to zero in the stationary case, we can easily find that the normalized temperature decays exponentially in the unburned side and it is a constant in the burnt side. One may look at the case of high activation energy.
OK, I didn't realize that you were already in the frame co-moving with the front and had specialized to the steady state. Solve the ODE separately in the regions x > 0 and x < 0 separately and apply the boundary conditions at infinity. The delta function is zero in both of these regions and so the solution in each is an exponential. There are two undetermined constants at this point --- the premultipliers of the exponentials. To solve for these, you need two conditions at x = 0. One is the continuity of T. You get the second by integrating the ODE in a small interval about the origin to get a jump condition on the derivative of T with respect to x. The jump is proportional to Q. Applying this boundary condition yields a unique steady state solution.
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