Let $h$ be such that $\hat{h}(\xi)=e^{-\xi^{6}}$. We note that $u_{t}=u_{x x x x x x}$ implies that
$$
\frac{d}{d t} \hat{u}(\xi, t)=(i \xi)^{6} \hat{u}(\xi, t)=-\xi^{6} \hat{u}(\xi, t)
$$
That means
$$
\hat{u}(\xi, t)=e^{-t \xi^{6}} \hat{u}(\xi, 0)
$$
Then we first note that (with $\left.z=t^{1 / 6} y d z=t^{1 / 6} d y\right)$
$$
\begin{aligned}
e^{-t \xi^{6}} &=\frac{1}{\sqrt{2 \pi}} \int e^{-i t^{1 / 6} \xi y} h(y) d y \
&=t^{-1 / 6} \frac{1}{\sqrt{2 \pi}} \int e^{-i \xi z} h\left(t^{-1 / 6} z\right) d z
\end{aligned}
$$
This implies with Fourier inversion formula that
$$
\begin{aligned}
u(x, t) &=\int e^{i \xi x} \hat{u}(\xi, t) \frac{d x}{\sqrt{2 \pi}} \
&=\int e^{i \xi x} e^{-t \xi^{6}} \hat{u}(\xi, 0) \frac{d x}{\sqrt{2 \pi}} \
&=t^{-1 / 6} \iint e^{i \xi x} e^{-i \xi z} \hat{u}(\xi, 0) \frac{d x}{\sqrt{2 \pi}} h\left(t^{-1 / 6} z\right) d z \
&=t^{-1 / 6} \int u(x-z) h\left(t^{-1 / 6} z\right) \frac{d z}{\sqrt{2 \pi}}
\end{aligned}
$$
Now it suffices to plug in $u(x)=\sin (x)+\cos (x)$. For the second solution, we just use periodicity and deduce that
$$
u(x, t)=\sum_{k \in \mathbb{Z}} e^{-t k^{6}} \hat{u}(k, 0) e^{-i k x} .
$$
Note that $\sin (x)=\frac{e^{i x}-e^{-i x}}{2 i}, \cos (x)=\frac{e^{i x}+e^{i x}}{2}$ and hence
$$
\hat{u}(1,0)=\frac{1}{2}(1-i), \hat{u}(-1,0)=\frac{1}{2}(1+i)
$$
Therefore
$$
2 u(x, t)=(1-i) e^{-t(1)^{6}} e_{1}+(1+i) e^{-t(-1)^{6}} e_{-1}=e^{-t} 2 u(x, 0)
$$

implies $u(x, t)=e^{-t}(\cos (x)+\sin (x))$

Solution: Uniqueness follows from the maximum principle. The first possibility is ‘Ansatz’ (meaning try and error)
$$
u(x, t)=T(t) X(x)
$$
Then the pde implies
$$
T^{\prime} X=T X^{\prime \prime}
$$
in other words $T^{\prime}=e^{-\lambda t} T(0)$ and
$$
X^{\prime \prime}=-\lambda X
$$
Our initial condition $X(x)=\sin (\pi x)$ satisfies $X^{\prime \prime}(x)=-\pi^{2} X(x)$ and hence $\lambda=\pi^{+2}$ and we are done (because we were lucky).

The systematic solution is given by extending $\phi(x)=\sin (\pi x)$ to an odd solution to the write of 0, meaning $\phi_{e x t}=\sin (\pi x)$ and then extend to a 2-periodic function. Let $f_{k}$ be an orthonormal basis for all 2 periodic solutions, indeed $f_{k}(x)=e^{i \pi k}$ will do. Then
$$
u_{t}=u_{x x}
$$
means
$$
\frac{d}{d t} \hat{u}(k, t)=e^{-t \pi^{2} k^{2}} \hat{u}(k, 0)
$$
In our case $\sin (\pi x)=\frac{f_{1}+f_{-1}}{2 i}$ and hence
$$
u(x, t)=\frac{1}{2 i} e^{-t \pi^{2}} f_{1}-\frac{1}{2 i} e^{-t \pi^{2}} f_{-1}=e^{-t \pi^{2}} \sin (\pi x)
$$

The solution formula for the heat equation and the $N$ –
part of the solution for the wave equation are both given by integrals and
hence ‘smoothen out’ singularities by integrating them. However, the term $\frac{\phi(x+t)+\phi(x-t)}{2}$ keeps the singularity, and the only cancellation occurs from points which are $t$ opposite with respect to $x$. However, this cancelation
can only occur for specific $x$ at a given time, and hence in a neighbourhood we still detect the singularity (think of $(x-1)^{-1}-(x-3)^{-1}$ then at $t=1=x$ we will see some cancelation).

Let $u_{k j}(x, y)=e_{k}(x) e_{j}(y) .$ Then
$$
L(u)=-\left(k^{2}+j^{2}\right) u_{k j}
$$
is an eigenvalue and hence
$$
u_{t}=L(u)
$$
is satisfied for $u(x, y, t)=e^{-t\left(k^{2}+j^{2}\right)} u_{k j} .$ This means
$$
S(t)(\phi)=\sum_{k j}\left(e_{k j}, \phi\right) e^{-t\left(k^{2}+j^{2}\right)} e_{k j}
$$
is the desired solution operator.

Solution: Assume $u(x, t)=\sum_{k} \hat{u}(k, t) e_{k} .$ Then we need
$$
\frac{d^{2}}{d t^{2}} \hat{u}(k, t)+r \frac{d \hat{u}}{d t}=k^{2} \hat{u}(k, t)
$$
or
$$
f^{\prime \prime}(t)+r f^{\prime}(t)=k^{2} f(t)
$$
This ODE is solved trying $f(t)=e^{\lambda t}$ first. Then we get
$$
\lambda^{2}+r \lambda+\frac{r^{2}}{4}=k^{2}+\frac{r^{2}}{4}
$$
This means
$$
\lambda_{p m}=\frac{r}{2} \pm \sqrt{k^{2}+\frac{r^{2}}{4}}
$$
We have exactly the two solutions we are looking for and hence
$$
u(x, t)=\sum_{k \in \mathbb{Z}} a_{k} e^{\lambda_{+} t} e_{k}+\sum_{k \in \mathbb{Z}} b_{k} e^{\lambda_{-} t} e_{k}
$$
is the general form of the solution. With the help of $a_{k}$ and $b_{k}$ we can now adjust the initial conditions. Indeed,
$$
u(x, 0)=\sum_{k}\left(a_{k}+b_{k}\right) e_{k}
$$
and
$$
u_{t}(x, 0)=\sum_{k}\left(\lambda_{+}(k) a_{k}+\lambda_{-}(k) b_{k}\right) e_{k}
$$
uniquely determines $a_{k}$ and $b_{k}$. Meaning
$$
\hat{u}(k, 0)=a_{k}+b_{k}, \hat{u}{t}(k, 0)=\left(\lambda{+}(k) a_{k}+\lambda_{-}(k) b_{k}\right)
$$
has a unique solution.