After studding this lecture, the student should be able to: define the initial and boundary conditions to find a solution; find solutions of the diffusion equation in different geometries; find physical interpretation the solutions from physical meaning. In previous chapter we got the one-speed diffusion equation. This equation is true only in the Fick Law approximation frame. This is a general form of the one-speed diffusion equation. Lets consider the simplifications of this equation. The first is the stationary problem. It means that there is no dependence on the temporal variable. The second is the non-multiplying medium. The neutrons are gained by medium only from external sources. The third is the homogeneous medium — it means the macro cross sections do not depend on the spatial variables. The divergence of gradient is the Laplacian of the neutron flux and the ratio of the diffusion coefficient and the macroscopic cross section of absorption is the diffusion length squared. The Laplacian in the simplest case is the second order derivation thus in a solution there are two unknown constants. To determine them the initial and boundary conditions are needed. The first the neutron flux is non negative and limited. The second at an initial time, as usual, the functions are known. The third is the condition on the two media interface. Actually the condition is the equality of the one-way currents at the point of interface. After a simple transformation we can get the equality of neutron flux at a point, but the function of gradient projection (Nabla operator with lower index n) has discontinuity given by the value of ratio of diffusion coefficients ratio. In the case of plane geometry the condition looks as the equality of flux at the point of two media interface and the equality of flux derivatives after taking into account the diffusion coefficients. The fourth is the external boundary condition at the interface of the medium and vacuum. The number of neutrons crossing the outer boundary in the direction from the vacuum to the medium equals zero. The same the negative one way current through the area with outer normal equals zero. After a simple transformation we got the equality of the logarithmic derivative in the left hand side to minus one divided to alpha in the right hand side. The new constant alpha is the length of linear extrapolation of the neutron flux to vacuum. And as follows from the diffusion it equals to two third of the lambda transport. We have to explain the physical meaning of alpha. Let's take the planar geometry and the flux distributed in the medium as depicted on the picture. The question is how the neutron flux in the vacuum is distributed? Actually in the vacuum the neutron flux is distributed as a constant, because in the planar geometry the number of neutrons in vacuum will be the same as the number of neutron fliew out from the medium. In vacuum there is no absorption and the number of neutrons can’t change. Let us denote the real border of the medium as x with the lower index S and linearly extrapolate the neutron flux to zero, the point where the flux equals zero is denoted as X with the lower index 0. Using the expression for the logarithmic derivative it is very simple to ensure that the difference between X0 and XS equals alpha. The point X0 is called the extrapolated outer border. And the boundary condition now looks as the flux at the extrapolated boundary equals zero. This is an important usage of the boundary condition because it is the simplest way compared to the calculation of the logarithmic derivative. From the expression for one way current in the diffusion approximation we got that alpha equals two third of the lambda transport, but for a better description of the flux near the outer border it is recommended to use the improved value — 0.71 of the lambda transport. The fifth is the condition of the localized source. If there is a point source located at the point r with the lower index 0, the source has the intensity of S neutrons per second. The diffusion equation is depicted on the picture, the external source term is described by the Delta function multiplying by intensity S. Let's surround by an imaginary sphere with the radius R capital and square area Г. We can integrate all members of the diffusion equation over the volume of the sphere and tend the radius to zero. In the end we can get the number of neutrons crossing the sphere of an infinitesimal radius equals the intensity of the external localized source. Now we solve several problems of the elementary diffusion as examples. Problem statement. The stationary problem for the homogeneous non-multiplying medium. We need to find the distribution of the neutron flux function. We took the non-multiplying medium because it is simple for studying, the multiplying medium is considered in the next section. The diffusion equation is described there as well. The solution can be found in four steps. The first is to write the equation according to the specific geometry. The second is to find a general solution of the equation containing two unknown constants. The third is to determine the unknown constants. And the fourth is to find and to analyze the solution. The Laplace operator in different geometries is depicted here. In many cases we use the symmetry, that is properties of the medium (homogeneity, isotropic source and so on) allow us to work with the only independent variable in the Laplacian. Thus, for example the Laplacian in the Cartesian geometry becomes a simple second order derivative in the X coordinate.