Past exam of the mathematics course of the University of Cambridge 2007 ii Paper 2 13B Solution Created 2026-09-24 Updated 2026-09-24
Past exam of the mathematics course of the University of Cambridge 2010 ii Paper 3 6A Solution Created 2026-09-24 Updated 2026-09-24
Past exam of the mathematics course of the University of Cambridge 2022 ii Paper 3 13C Solution Created 2026-09-23 Updated 2026-09-25
For , conservation of the deposited amount requiresThe dimensions satisfy and . Thus has dimension , so the similarity length is . Requiring the concentration scale to have dimension givesHence
In polar coordinates, substitution into
givesMultiplying by and integrating once, with regularity and zero radial flux at the origin, yieldsWhere , this reduces to , soNonnegativity and zero flux at the moving front give the compactly supported profileThe normalization givesand hence and . This is the Two-dimensional Barenblatt solution with diffusivity proportional to concentration, and its support radius isso .
givesMultiplying by and integrating once, with regularity and zero radial flux at the origin, yieldsWhere , this reduces to , soNonnegativity and zero flux at the moving front give the compactly supported profileThe normalization givesand hence and . This is the Two-dimensional Barenblatt solution with diffusivity proportional to concentration, and its support radius isso .
Now add the linear reaction term and writeSubstitution cancels the terms and leavesChoosingproduces , as in the linear-reaction time change for quadratic nonlinear diffusion. Therefore
Past exam of the mathematics course of the University of Cambridge 2026 iii Paper 332 2 c Solution Created 2026-09-24 Updated 2026-09-25
At times long compared with the encounter time at depth , the fixed depth is negligible relative to the growing diffusion lengths. Salt obeys the diffusion equation with diffusivity . Withthe self-similar concentration profile in the liquid isIt has and tends to in the far field.
Because the solid contains no salt, conservation of solute at the moving boundary requires the rejected solute flux from Fick's first law to equal the rate at which the interface sweeps up salt:Substituting the similarity profile givesThis relation determines if is already known. More generally it must be solved together with the thermal Stefan condition and the interfacial phase-equilibrium relation from the liquidus. For , salt occupies a much thinner boundary layer than heat and can be much larger than .