 # intuitive PDE to ODE (characteristics)

Here is a fun way to interpret PDE.

\begin{aligned} x&=(x_1,...,x_n)\\ p&=(p_1,...,p_n)\\ u=u(x)&=u(x_1,...,x_n) \qquad u:U→ℝ \qquad U⊂ℝ^n \text{ an open set}\\ ∇u&=(u_{x_1},...,u_{x_n}) \qquad \text{often, }p=∇u \\ %u_{x_i}&=∂_{x_i}u=(∂u/∂x_i)(x_1,...,x_n) \\ (x,y,p)&=(x_1,...,x_n,y,p_1,...,p_n) \\ T_x(U)&=\text{the tangent space of surface U at point x} \end{aligned}

A $1^{st}$ order PDE for $u$ can be formulated as $$\begin{cases} F(x,y,p)=0\qquad F:U×ℝ×ℝ^n→ℝ \\ y=u \\ p_i=\frac{∂u}{∂x_i} \end{cases}$$

Indeed by replacing $y$ and $p_i$, we get a relationship between the solution $u$, it's inputs $x$ and partial derivatives $∂u/∂x_i$

The space J^1(U) := \left\{(x,y,p)\;\middle|\; \begin{aligned} &x∈U \\ &y∈ℝ \\ &p∈T_x(U) \end{aligned} \right\} is called the 1-Jet space

They are the points $(x,y,p)$ such that $x$ is an element of the space $U$, $p$ is an element of the tangent space of $U$ at $x$.

For $U⊂ℝ^n$ a subspace of dim $n$ we get J^1(U) := \left\{(x,y,p)\;\middle|\; \begin{aligned} &x∈U \\ &y∈ℝ \\ &p∈T^∗_x(U)≈ℝ^n \end{aligned} \right\}≈ℝ^{2n+1}

it has a contact structure $K$ that will be explaned later.

Given $u:U→ℝ$, $G^1(u):=\{(x,u(x),∇u(x))\mid x∈U\}$ $⊂ J^1(U)$ is the 1-graph of $u$

Now our PDE has a new interpretation : $V := \{(x,y,p) \mid F(x,y,p)=0\}$ defines some hypersurface in $J^1(U)$

This implies for a solution $u$, $G^1(u) ⊂ V$, must share their tangent space $$∀z∈G^1(u), \quad T_zG^1(u)⊂T_zV$$ $$T_zV=\{(\dot{x},\dot{y},\dot{p}) \mid \dot{x}F_x+\dot{y}F_y+\dot{p}F_p = 0\}$$

Any smooth $u:ℝ^n→ℝ$ verifies the chain rule $du(x) = ∑_i u_{x_i}(x)dx_i$

So the hypersurface $G^1(u) ⊂ J^1(U)$ must verify $dy = ∑_i p_idx_i$, the contact structure $K$ of $J^1(U)$

This means that $G^1(u)$ must be tangent to $K$ $$∀z∈G^1(u), \quad T_zG^1(u) ⊂ K_z$$

Here is a visualization of our plane distribution $K$ in the 1 dimentional case : the red axis corresponds to $x$, the green axis corresponds to $y$, the blue axis corresponds to $p$.

For each point $z=(x,0,p)$ I drew a plane representing $K_z$. This represents the only allowed directions.

Note that this plane distribution exists for any point $(x,y,p)$, it's just that it doesn't depend on $y$, and having a volume filled with planes makes it unreadable. (it doesn't depend on $x$ either, but I drew it anyway)

a direction where $p=0=∂_xu$ but $u$ grows with $x$ is impossible : you can see how our planes are "flat" relative to $x$ when $p=0$

When $p$ grows, the plane slope change to allow $u$ to grow with $x$ with the according partial derivative.

I can only make an illustration for the 1d case as $\dim(J^1(U))=2n+1$ which is already $3$ for $n=1$... for $n=2$ we would need to draw in a $5$ dimensional space !

We could try to build a surface back from those tangents : rising each planes so that their neighbor stick so that we get a "mountain shaped" thing.

This works if the tangent planes are of an actual surface. Sometimes, there is no way the plane distribution can build up a surface... and this plane distribution is one of those problematic case.

Indeed : TODO : replace all of this with an animation if we try to stick planes from the origin along axis $p$, we see a twist, but the planes don't get higher. So no increase in $y$ direction. Let's make $p$ increase to reach a value $p=p_0$.
Then from there, if we continue to stitch those planes along $x$, given that $p=p_0≠0$, $y$ will increase accordingly, the planes must be higher and higher as we stitch them together.
Now if we try to stick planes from the origin along axis $x$, we have $p=0$ so there is no increase in $y$. Then from there, if we continue to stitch those planes along $p$, then again, on increase for $y$. These two constructions are contradicting : $y$ has to both increase and stay at $0$. There is no way we can build a surface.

But this is actually good news because we will use that : the hypersurface given by $F=0$ is an actual surface, so there is no way the intersection between the tangents of the surface defined by $F$ always matches exactly $K$, it can do that only on certain points !

Combining the two previous observations : $∀z∈G^1(u), \quad T_zG^1(u) ⊂ K_z ∩ T_zV$

In dark is shown $T_zV$ for an arbitraty $F$

TODO : customizable function F

In the previous illustration, we observe that the intersection of our 2 dimensional spaces are a 1 dimensional line. This is great as we could just follow those lines to find how a point evolves. If we start with a set of points $Γ$ and that $dim(Γ)=n-1+1=n$ : a subspace of $x$ of dimension $n-1$ with the chosen initial condition $u$ (dim $1$)

However that works only for $n=1$ : $K_z$ and $T_zV$ have dimension $2n+1-1=2n$, thus their intersection $K_z∩T_zV$ has dimension $2n-1$ (if those two space do not coincide, otherwise we have $dim=2n$). This means we need more tools to use the same trick for $ℝ^n$.

$ω$ is a linear symplectic structure if $ω$ is a non-degenrate linear 2-form (billinear + antisym) on $V⊂ℝ^n$, ie

antisym : $∀a,b,c∈ℝ^n, \quad ω(a,b)=-ω(b,a)$
bilinear : $∀a,b,c∈ℝ^n, ∀λ,μ∈ℝ, \quad ω(a,λb+μc)=λω(a,b) + μω(a,c)$
non-degenerate : $∀a∈ℝ^n, ∃b∈ℝ^n$ such that $ω(a,b)≠0$

the standard symplectic form is $ω=dx∧dp=∑_idx_i∧dp_i$

the standard symplectic form verifies $ω(a,b)=∑_i \underset{\text{area of parallelepiped}}{\underbrace{(a_{x_i}b_{p_i}-b_{x_i}a_{p_i})}}$

the orthogonal space of $V$ is given by $V^\perp:=\{x∈ℝ^n\mid ∀a∈V, ω(a,x)=0\}$

$(V^\perp)^\perp=V$

$a∈ℝ^{2n},$ $⟨a⟩⊂⟨a⟩^\perp$ because $ω(a,a) = -ω(a,a) ⇒ ω(a,a) = 0$

an other observation is $\dim(⟨a⟩^\perp)=2n-1$

because if there were $2$ or more non-colinear vectors $u,v$ such that $ω(a,u)≠0$ and $ω(a,v)≠0$ we would get $ω(a,λu+μv)≠0$ but $ω(a,λu+μv)=λω(a,u)+μω(a,v)$ so we can find $λ,μ$ such that this is $0$, which should not be possible : $u,v$ cannot co-exist.

This is generalized in the next remark

$V⊂ℝ^{2n},\dim(V)=k ⇒ \dim(V^\perp)=2n-k$

$V$ is dimension $k$, so it has a basis $\{v_1,...,v_k\}$, by linearity of $ω$, we get

\begin{aligned} V^\perp &= \{x \mid ∀i, ω(v_i,x)=0\} \\&= ∩_i \; \{x \mid ω(v_i,x)=0\} \\&= ∩_i \; ⟨v_i⟩^\perp \qquad \text{(the intersection ofk$spaces of dimension$2n-1)} \end{aligned}

If $⟨v_i⟩$ are all different, then $\dim(V^\perp) = \dim(∩_i \; ⟨v_i⟩^\perp) = 2n-k$ and we are done.

Let's suppose $∃i,j⟨v_i⟩=⟨v_j⟩$, thus define $v$ the direction such that $\begin{cases}ω(v,v_i)=α_i≠0\\ω(v,v_j)=α_j≠0\end{cases}$ TODO TODO TODO

Using $\dim(K_z ∩ T_zV)=2n+1-2=2n-1$ we can conclude

$\dim((K_z ∩ T_zV)^\perp) = 2n-(2n-1) = 1$ for orthogonality in $ℝ^{2n}$ (ie ignore $y$ in $(x,y,p)$)

Given that this space is 1 dimensional, from the previous remark, we also have that $$(K_z ∩ T_zV)^\perp ⊂ ((K_z ∩ T_zV)^\perp)^\perp = K_z ∩ T_zV$$

TODO : illustration with numerical solver (customizable function F) 14 juin 2021 15:48:42 9 juin 2021