## Examples of PDE

1. PDE Exercises

$f: \mathbb{R}^n \Rightarrow \mathbb{C}$ given ${\tilde{M}f}$ is a solution of the homogeneous Cauchy problem, with ${\tilde{M}f(0,x) = 0}$, ${\frac{\partial \tilde{M}f}{\partial t}(0,x) = f(x)}$ Let ${g:[0,\infty) \times \mathbb{R}^n \Rightarrow \mathbb{C}}$ be sufficiently often continuously differentiable. Show without using any explicit formular for ${\tilde{M}f}$ that the function Qg defined by $Qg(t,x):= \int^{t}_{0} \tilde{M}g_s(t-s,x)ds,$ where ${g_s(x):= g(s,x),}$ is the solution of the Cauchy problem for the inhomogeneous wave equation with right hand side g and zero initial conditions. We want to show that this implies ${\Box Qg = g}$ ${Qg(0,x)= \frac{\partial Qg}{\partial t}(0,x) = 0}$ \newline For n = 1 this gives an alternative solution of Example 1. $\tilde{M}f(0,x) = \frac{1}{2c} \int^{x+t}_{x-t}f(\tau)d\tau$ This implies

$\displaystyle Qg(t,x) = \frac{1}{2c} \int^{t}_{0} \int^{x+c(t-s)}_{x-c(t-s)}g(s,\tau)d\tau ds \ \ \ \ \ (1)$

We can then write Qg(0,x) = ${\int_0^0 \cdots ds = 0}$ ${\frac{\partial Q}{\partial t}(t,x) = \tilde{M} g_s(t-s,x)}$ We let s = t ${+\int^t_0 \frac{\partial}{\partial t} \tilde{M}g_s(t-s,x)ds}$ We know ${\tilde{M}g_t(0,x) =0}$ SO:

$\displaystyle \frac{\partial Qg}{\partial t}(t,x) = \int^{t}_0 \frac{\partial}{\partial t} \tilde{M}g_s(t-s,x)ds \ \ \ \ \ (2)$

In particular if t=0 We get ${\frac{\partial Qg}{\partial t} (0,x) = \int^0_0 \cdots = 0}$ We also have to look at the box operator ${\Box = \frac{\partial^2}{\partial t^2} - \triangle_x}$ we obtain by applying $\frac{\partial}{\partial t}$.

By parameter dependent integrals; $\frac{\partial^2 Qg}{\partial t^2}(t,x) = \int^{t}_0 \frac{\partial^2}{\partial t^2} \tilde{M}g_s(t-s,x) + \frac{\partial}{\partial t} \tilde{M}g_s (t-s,x)|_{s=t}$
$\frac{\partial^2 Qg(t,x)}{\partial t^2} = g(t,x) + \int^t_0 \cdots ds$ By parameter dependent integrals; $\triangle_x Qg(t,x) =\int^t_0 \triangle_x \tilde{M}g_s(t-s,x) ds$ $\Box Qg(t,x) = g(t,x) + \int^t_0 \Box Mg(t-s,x)ds$ The integral above goes to zero by the homogeneous Cauchy problem solution. Therefore we can write

$\displaystyle \Box Qg(t,x) = g(t,x)$

We only need the properties of ${\tilde{M}g_s}$. This should be seen as an exercise in parameter dependent integrals.

2. Examples of distributions and their order

Let us recall the criterion: T:${C^{\infty}_{c}(U) \Rightarrow \mathbb{C}}$ linear T is a distribution ${\iff}$ ${\forall K \subset U}$ compact ${\exists p(K) \in \mathbb{N}_0}$, ${C(K) \in [0,\infty)}$ such that for ${\phi \in C^{\infty}_{c}(U)}$ with supp${\phi \subset K}$ $|T(\phi)| \leq C(K) \sum_{|\alpha|\leq p(K)}\|D^{\alpha}\phi\|_{\infty}$ Where the last norm is the suprenum norm. $\delta_{x_{0}}$ $|\delta_{x_{0}}| = |\phi(x_{0})| \leq \|\phi \|_{\infty}$ (c=1, p = 0, indepedent of K. $f \in L^1_{loc}(U)$ ${T_f(\phi) = \int_U f(x) \phi (x) ds}$ ${|T_f(\phi)| = |\int_U f(x) \phi(x) dx| \leq \int_U|f(x)||\phi(x)|dx}$

Assume ${K \subset U}$ compact ${supp \phi \subset K}$ ${=\int_K|f(x)||\phi(x)|dx}$ ${\leq \|\phi\|_\infty \int_K |f(x)|dx}$ the integral above can be considered as C(K) and it is less than infinity (p=0, independent of K). ${T(\phi) = \int_M \phi|_M . \omega}$ M ${\subset U}$ k-dim oriented subset and we can consider ${\omega}$ a k(top) form There is a trick:

In local coordinates ${Y_1,\cdots Y_k}$ ${V\subset M}$ open ${y_1: V \Rightarrow \mathbb{R}}$ ${\omega = g(y_1,\cdots, y_k)dY_1\wedge \cdots \wedge dY_k}$ We use the pullback in fact, but we can sloppy write $\int_V \phi|_M \omega = \int_V \phi(Y_1 \cdots Y_k)g(Y_1,\cdots,Y_k)dy_1,\cdots, dy_k$ ${\int_M \phi_M \omega =}$ sum over open sets, where we have coordinates using partition of unity. We need a volume form or take the modulus of ${|g(y_1,...,y_k)|}$ If we define, in local coordinates of the correct orientation ${|\omega(x)|=|g(y_1,\cdots,y_k)|dy_1 \wedge \cdots \wedge dy_k}$ then ${|\omega|}$ is well-defined independent of co-ordinates. We can then write the following $|\int_M \phi|_M . \omega| \leq \int_M|\phi|_M . |\omega| = \int_{K \cap M} |\phi|_{|M}.|\omega| \leq \|\phi\|_\infty . \int_{K \cap M} |\omega|$ Because ${supp \phi \subset K}$ we can do this trick. The integral above becomes C(K) and is finite, since ${\omega}$ has a finite number. This is indeed a distribution and clearly of order 0.