There is a function \(u \in C^\infty(\mathbb{R} \times \mathbb{R}_+) \cap C^0(\mathbb{R} \times \overline{\mathbb{R}}_+)\) solving
\[ \begin{cases} \partial_t u - \Delta u = 0 & \text{in } \mathbb{R} \times \mathbb{R}_+ \\ u|_{t = 0} = 0 & \text{in } \mathbb{R} \end{cases} \]and not being \(0\).
Proof
Consider
\[ u(x,t) = \sum_{n \geq 0} \frac{a_n(t)}{n!} x^n. \]The problem in (0x683fe8f3) implies \( a_n'(t) = a_{n+1}(t) \). Set \( a_{2n} = g^{(n)} \) and \( a_{2n+1} = 0 \) with
\[ g(t) = \begin{cases} e^{-1/t^2}, & t > 0, \\ 0, & t \leq 0. \end{cases} \]It remains to show that the power series of \( u \) and \( \partial_t u \) converges locally uniformly. Especially, we have for some \( \varepsilon > 0 \)
\[ |u(x,t)| \leq \exp\left( \frac{|x|^2}{\theta t} - \frac{1}{2t^2} \right), \]This implies
\[ |u(x,t)| \to 0 \quad \text{as } t \to 0. \]
Remarks
- A proof can be found in the script of Chris.
- This result is also found by Tychonoff.