Past exam of the mathematics course of the year 1 of the natural sciences course of the University of Cambridge / 2019 / Paper 2 / 20

Use separation of variables with $u=X(x)Y(y)$ on a region where the factors do not vanish. The equation becomes $X^{\prime }Y-xX{Y}^{\prime }=0$, soThe separated equations are $X^{\prime }=\lambda xX$ and $Y^{\prime }=\lambda Y$, giving $u=C\exp [\lambda (y+x^2/2)]$. Choosing $C=\lambda =1$ gives the nonconstant real solutionAlthough the division used $x\ne 0$, the resulting solution extends smoothly across $x=0$ and directly satisfies $u_x=x{u}_y$ everywhere.

With the same separation of variables ansatz, $x{X}^{\prime }Y-2yX{Y}^{\prime }=0$ givesOn $x,y>0$, integration yields $X=A{x}^{\lambda }$ and $Y=B{y}^{\lambda /2}$. The choice $\lambda =2$, $AB=1$ gives a polynomial that extends smoothly to all real $x,y$:It is nonconstant and real everywhere, and $x{u}_x-2y{u}_y=2x^{2}y-2x^{2}y=0$, including on the axes.

For the Gaussian ansatz, partial differentiation givesCancel the positive exponential and compare the constant and quadratic powers of $x$ in the diffusion equation. The two coupled ordinary differential equations areThe initial profile requires $F(0)=T_0$ and $H(0)=L^{-2}$. The second equation is a separable differential equation:Substitute this into the first equation and integrate its logarithmic derivative:Thusand the solution isThis is Gaussian diffusion of a smooth initial profile: its width grows while its peak decreases. The Gaussian integral gives ${\int }_{-\infty }^{\infty }T(x,t)dx=\sqrt{\pi }{T}_{0}L$, independent of time, which checks conservation of the total amount under the diffusion equation.