Problem 1.

Let $u$ be a solution to
$$u_{t}=u_{x x x x x x}, u(x)=\sin (x)+\cos (x)$$
be a $[-\pi, \pi]$ periodic solution. Provide two possibilities to solve this problem. Do you have and energy estimate?

Proof .

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

Problem 2.

Let
$$u_{t}=u_{x x}, u(x, 0)=\sin (\pi x) u(0, t)=u(1, t)=0$$
Provide two possibilities to solve this problem. Do we have uniqueness?

Proof .

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
$$u(x, t)=\frac{1}{2 i} e^{-t \pi^{2}} f_{1}-\frac{1}{2 i} e^{-t \pi^{2}} f_{-1}=e^{-t \pi^{2}} \sin (\pi x)$$

Problem 3.

For diffusion equation (on the real line say) singularities are immediately lost. However for the wave equation they are
kept. Can you explain this on the basis of the solution formulae we have?

Proof .

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).

Problem 4.

Let $L(u)=u_{x x}+u_{y y}$ on $[-\pi, \pi] \times[-\pi, \pi] .$ Find a solution operator of the form
$$S(t)(\phi)(x, y)=\sum_{k, j} e_{k}(x) e_{j}(y)$$
for $u_{t}=u_{x x}+u_{y y}, u(x, y, 0)=\phi(x, y)$.

Proof .

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.

Problem 5.

Wants to solve
$$u_{t t}-u_{x x}=-r u_{t}$$
Assume that we have a periodic solution on $[-\pi, \pi]$ can you find an equation for the Fourier coefficients?

Proof .

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.

## 推荐几本很好的PDE入门教材给大家

Partial Differential Equations, An Introduction, Walter Strauss

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