Q:

Nonclassical solution to $u_t-\Delta u=f$ in one space dimension?

I know that one may find continuous $f$ (say in a ball $\bar B$ centered at $0$) such that there exists no classical ($\mathcal C^2$) solution $u$ to the Poisson equation $\Delta u =f$ in $B$, and obviously this requires to be in dimension 2 or more. (But a classical solution exists as soon as $f$ is assumed Hölder continuous).
This counterexample may be found for instance in Qi Han's book "A Basic Course in Partial Differential Equations" (link) pp.136-137.
Now it is easy to see that for such an $f$ there exists no classical solution to the inhomogeneous heat equation
$$u_t - \Delta u = - f(x) \;\;\;\mbox{  on } (0,T) \times B.$$
But the case of one space dimension seems to require a new argument.
So my question is this :
Is there a continuous function $f$ on $[0,T] \times [-a,a] $ s.t. the equation $u_t - u_{xx}=f(t,x)$ has no classical solution ?
(I know that if $f$ is Hölder continuous in $x$, uniformly in $t$, then one may prove by explicit formulas that there is a classical solution.)

A:

An example can be constructed analogically to the elliptic case.
Let $u(t,x)=\left(2 t+x^2\right) \sqrt{\log \left(\frac{1}{t^2+x^4}\right)}\ $ in some neighbourhood of the origin for $(t,x)\ne0$ and $u(0,0)=0\,$. Denote
$$
f(t,x)=u_t(t,x)-u_{xx}(t,x)=
$$
$$
\frac{\left(-2 t^4+11 t^3 x^2+12 t^2 x^4-5 t x^6+6 x^8\right) \log
   \left(\frac{1}{t^2+x^4}\right)+4 x^6 \left(2 t+x^2\right)}{\left(t^2+x^4\right)^2
   \log ^{\frac{3}{2}}\left(\frac{1}{t^2+x^4}\right)}.
$$
Using Young's inequality 
$$
|ab|\le \frac{|a|^p}p+\frac{|b|^q}q,
$$
there $q=p/(p-1)\,$, it is straightforward to estimate monomials in the numerator by the denominator polynomial $(t^2+x^4)^2$. For example, for $t^3 x^2$ we can put $p=4/3$ to obtain $t^4$ from $t^3$. Then  $q=4$ and
$$
|t^3 x^2|\le \frac{t^4}{4/3}+\frac{(x^2)^4}{4}\le C(t^2+x^4)^2.
$$
So, defining $f(0,0)=0$ we have a function, continuous in some neighbourhood of the origin. 
From the other hand
$$
u_t(t,x)=
2 \sqrt{\log \left(\frac{1}{t^2+x^4}\right)}-\frac{t \left(2
   t+x^2\right)}{\left(t^2+x^4\right) \sqrt{\log \left(\frac{1}{t^2+x^4}\right)}}
$$
and
$$
\lim_{(x,t)\to0}u_t(x,t)=+\infty
$$
due to the first summand.

