Past exam of the year 1 of the computer science course of the University of Oxford / 2015 / Discrete mathematics, continuous mathematics, and probability / 6

With step size $h=1/N$ and nodes $x_j=j/N$, the composite trapezoidal rule givesThe endpoint weights are half the interior weights, and the endpoint values are $1$ and $1/2$.

Write the sum in the question as the right Riemann sumRelative to the composite trapezoidal rule, the only change is in the two endpoint weights:The integrand $f(x)=1/(1+x^2)$ is continuous on $[0,1]$, so $T_N\to I$. Since an antiderivative is $\arctan x$,Thus

For a nonzero separable solution $u=XT$, separation of variables givesThese Dirichlet boundary conditions follow because $T$ is not identically zero. If $\kappa =0$, the affine function $X=A+Bx$ obeys both conditions only when $A=B=0$. If $\kappa =-{\rho }^2<0$, $X=A\cosh (\rho x)+B\sinh (\rho x)$ also has only the zero solution under both conditions.

For $\kappa ={\omega }^2>0$, $X=A\cos (\omega x)+B\sin (\omega x)$. The first boundary condition forces $A=0$, and the second forces $\omega L=n\pi$ for a nonzero solution. Hence all nonzero spatial modes areThe identically zero function is also a solution. Allowing $B_n=0$ includes it in the displayed family. Here $L>0$ is the interval length and $n$ labels the eigenvalues of the operator $-d^2/d{x}^2$ with these boundary conditions.

For the spatial mode with eigenvalue ${\kappa }_n=(n\pi /L{)}^2$, the temporal ordinary differential equation isIts general solution isThe printed reference to “each solution $X(x)$ ... in part (b)” refers to (b)(i). The zero spatial mode gives $u=0$ irrespective of $T$ and contributes no additional nonzero solution.^[ref]

The heat equation and its homogeneous Dirichlet boundary conditions are linear, so superposition of the separated modes givesThe Fourier sine basis represents the initial profile on the interior of the interval. At $t=0$ its Fourier sine series must equal $x(1-x)$. Multiplying by $\sin (m\pi x/L)$, integrating from $0$ to $L$, and using orthogonality givesTherefore the required Fourier sine coefficients areFor every $t>0$, the exponential factors ensure convergence of the differentiated series, and every term satisfies the partial differential equation and the boundary conditions. The Fourier sine series recovers the smooth initial profile at every $0<x<L$ as $t\downarrow 0$.

The scan explicitly prints $x(1-x)$ on $(0,L)$; it must not silently become $x(L-x)$. For $L\ne 1$, the interior profile tends to $L(1-L)\ne 0$ at the right endpoint, whereas the prescribed boundary condition is zero for $t>0$. The solution is therefore not continuous at the corner $(L,0)$, although it satisfies every condition on the printed open ranges. A solution continuous on the closed space-time domain with these data would require $L=1$.^[ref]