Past exam of the short options course of the University of Oxford / 2021 / 5 (Integral transforms)

Every $a\in \mathbb{R}$ works. Translation preserves a smooth function, and the support of a function becomes the compact interval $[a-1,a+1]$. Hence $\varphi (\cdot -a)\in C_c^{\infty }(\mathbb{R})$, the space of test functions.

Exactly $a\ne 0$ works. For nonzero $a$, composition with $x\mapsto ax$ preserves smoothness, and the support of a function is $[-1/|a|,1/|a|]$, which is compact. For $a=0$ the result is the strictly positive constant $\varphi (0)$ on all of $\mathbb{R}$, so it lacks compact support and is not a test function.

Exactly $|a|>1$ works. Composition with the hyperbolic tangent is smooth for every $a$. If $|a|>1$, set $R=\mathrm{arctanh}(1/|a|)$. For $|x|\ge R$, $|a\tanh x|\ge 1$, so the result vanishes; it is therefore a test function with support of a function $[-R,R]$.

If $|a|\le 1$, every finite $x$ satisfies $|a\tanh x|<1$. The given positivity makes $\varphi (|a|\tanh x)>0$ for every $x$, so its support is all of $\mathbb{R}$. In particular $|a|=1$ is excluded even though the function tends to zero at infinity: decay is not compact support.

Multiplication of a distribution by a smooth function is defined by moving that function onto the test function. The Dirac delta multiplication identity givesIn fact $x^2\delta$ is the zero distribution, independently of the special positivity assumptions on $\varphi$.

Write $F_j(p)=\mathcal{L}{f}_j(p)$, using the square-root branch positive on real $p>0$. Integrating the differential equation and using boundedness at zero givesThe integration constant is zero since $x{f}_1(x)\to 0$. Continuity of $f_0$ then shows $f_1(x)=O(x)$, so $f_1$ extends continuously with $f_1(0)=0$.

The Laplace transform of a derivative and differentiation under the integral sign giveIntegrating and using $F_1(p)\to 0$ as real $p\to \infty$ fixes the constant:The initial value theorem for Laplace transforms gives $f_0(0)={\lim }_{p\to \infty }p{F}_0(p)=1$. It is tempting to apply the derivative rule immediately to $f_0$, but its stated regularity is only continuity. To justify that step, define $h(x)=1-{\int }_0^x{f}_1(t)dt$. ThenUniqueness of the Laplace transform gives $h=f_0$, and hence $f_0$ is continuously differentiable and

Use the causal convolution $(f_0\ast f_1)(x)={\int }_0^x{f}_0(t)f_1(x-t)dt$. The convolution theorem for Laplace transforms and (i) giveSince $\mathcal{L}(\cos x)=p/(p^2+1)$, uniqueness of the Laplace transform yieldsAt $x=0$ the right side is $1-1=0$, agreeing with the empty convolution integral. Identification of $f_0$ as a Bessel function is unnecessary.

Use the Fourier transform convention $\hat{u}(s,t)={\int }_{\mathbb{R}}{e}^{-isx}u(x,t)dx$. Then $\widehat{u_{xx}}=-s^2\hat{u}$ and $\widehat{xu}=i{\partial }_s\hat{u}$, soThus the transformed partial differential equation and initial condition areThe last equality is the Fourier transform of the Dirac delta distribution. The two terms in ${\partial }_x(xu)=u+x{u}_x$ must be transformed together; dropping $u$ would give the wrong equation.

Substitute $\hat{u}=e^{-b(t)s^2}$ into the transformed partial differential equation. Comparing coefficients gives $b^{\prime }=1-2b$, $b(0)=0$, whenceThe stated Gaussian Fourier transform pair, with its inverse factor $1/(2\pi )$, givesThis is a normal distribution density with mean zero and variance $1-e^{-2t}$. Its total integral is one. As $t\downarrow 0$ it converges to $\delta$ in the sense of distributions, while as $t\to \infty$ it tends to the standard normal distribution density. This is the transition density from zero of the Ornstein-Uhlenbeck process with drift $-x$ and diffusion coefficient $\sqrt{2}$.