📂Partial Differential Equations

Linearization of Boundaries in Nonlinear First-Order Differential Equations

Buildup

One method to easily solve the characteristic equation of a nonlinear first-order differential equation is to straighten out a small portion, $\Gamma$, of the boundary, $\partial \Omega$, of the domain, $\Omega$. Since this is always possible, in the vicinity of a point $x^{0}$ on the boundary, one can assume from the start that the boundary is a straight line and approach the problem. This is called straightening the boundary.

Let $\Omega \subset \mathbb{R}^{n}$ be an open set and $\partial \Omega$ be $C^{2}$. Let’s say a partial differential equation $F \in C^{1}(\mathbb{R}^{n} \times \mathbb{R} \times \bar \Omega)$ is given. Also, let’s say the following boundary condition is given.

$$\begin{equation} \left\{ \begin{aligned} F(Du,\ u,\ x)&=0 && \text{in } \Omega \\ u&=g && \text{on } \Gamma \end{aligned} \right. \end{equation}$$

Then $\Gamma \subset \partial \Omega$ and $g : \Gamma \to \mathbb{R}$.

Explanation

There is a fixed point on the boundary, $x^{0} \in \partial \Omega$. And define the transformation $\Phi\ :\ \mathbb{R}^{n} \to \mathbb{R}^{n}$ as follows.

$$\left\{ \begin{align*} \Phi (\Omega) &:= V \\ \Phi (x) &:= \left( \Phi^{1}(x),\ \cdots, \Phi^{n}(x) \right) = (y_{1},\ \cdots,\ y_{n})=y, \quad y \in V \\ y_{i}=\Phi^{i}(x) &:= x_{i}, \quad x \in \mathbb{R}^{n}\ (i=1,\cdots, n-1) \\ y_{n}=\Phi^{n}(x) &:= x_{n}-\gamma (x_{1},\cdots,x_{n-1}), \quad x \in \mathbb{R}^{n} \end{align*} \right.$$

That is, $\Phi$ is a transformation that makes the $n$-th coordinate of some part of the boundary into $0$. This is trivially a bijection by definition. Therefore, an inverse transformation exists, and let’s call it $\Psi$.

$$\left\{ \begin{align*} {l}\Psi (y) &:= \Phi^{{}-1}(y) \\ \Psi^{i}(y) &= x_{i}=y_{i}, \quad y \in \mathbb{R}^{n}\ (i=1,\cdots, n-1) \\ \Psi^{n}(y) &= x_{n}=y_{n}+\gamma (x_{1},\cdots,x_{n-1}), \quad x \in \mathbb{R}^{n} \end{align*} \right.$$

When represented in a diagram, it looks like the following.

Now, let’s say $\Gamma \subset \partial \Omega$ is an open set and $g\in C(\Gamma)$. And let’s say a fixed point $x^{0} \in \Gamma$ is given. And assume that $u \in C^{1} (\Omega)\cap C(\bar \Omega)$ is a solution solving the boundary condition $(1)$. And define $v$ as below.

$$v(y) := u(\Psi (y)) \quad \forall\ y\in V$$

In other words, $v$ is defined to have the same function value in $V$ as in $u$. Then the following holds.

$$u(x)=v(\Phi (x)) \quad \forall\ x\in \Omega$$

Now, let’s see how $Du, u, x$ behaves in $V$. First, if we calculate $u_{x_{i}}$, it looks like the following.

$$u_{x_{i}}(x)=\sum \limits _{k=1}^{n} v_{y_{k}} \left( \Phi (x) \right) \Phi^{k}_{x_{i}}(x)$$

Therefore,

$$\begin{align*} Du(x) &= \left( \sum \limits _{k=1}^{n} v_{y_{k}} \left( \Phi (x) \right) \Phi^{k}_{x_{1}}(x),\ \cdots,\ \sum \limits _{k=1}^{n} v_{y_{k}} \left( \Phi (x) \right) \Phi^{k}_{x_{n}}(x) \right) \\ &= (v_{y_{1}}\Phi^{1}_{x_{1}}+\cdots+v_{y_{n}}\Phi^{n}_{x_{1}},\ \cdots ,\ v_{y_{1}}\Phi^{1}_{x_{n}}+\cdots+v_{y_{n}}\Phi^{n}_{x_{n}} ) \\ &= \begin{pmatrix} v_{y_{1}} & v_{y_{2}} & \cdots & v_{y_{n}} \end{pmatrix} \begin{pmatrix} \Phi^{1}_{x_{1}} & \Phi^{1}_{x_{2}} & \cdots &\Phi^{1}_{x_{n}} \\ \Phi^{2}_{x_{1}} & \Phi^{2}_{x_{2}} & \cdots & \Phi^{2}_{x_{n}} \\ \vdots & \vdots & \ddots & \vdots \\ \Phi^{n}_{x_{1}} & \Phi^{n}_{x_{2}} & \cdots & \Phi^{n}_{x_{n}} \end{pmatrix} \\ &= Dv\left( \Phi (x) \right) D\Phi (x) \end{align*}$$

Or,

$$Du(\Psi (y)) = Dv(y) D\Phi (\Psi (y))$$

Then, the equation of $(1)$ is as follows.

$$F\Big( Du(\Psi (y) ), u( \Psi (y) ), \Psi (y) \Big) = F\Big( Dv(y)D\Phi (\Psi (y)), v(y), \Psi (y) \Big)=0$$

Now, let’s define the following nonlinear first-order partial differential equation.

$$G(q, w, y):=F \Big (q D\Phi (\Psi (y) ), w, \Psi (y) \Big), \quad \forall (q, w, y)\in\mathbb{R}^{n}\times\mathbb{R}\times\bar V$$

Then $G\in C^{1}(\mathbb{R}^{n} \times \mathbb{R} \times \bar V)$, and from the above results, the following equation holds.

$$G \Big( Dv(y),\ v(y),\ y \Big)=0, \quad \forall y\in V$$

And $\Delta:=\Phi (\Gamma)$ and let’s define $h(y):=g(\Phi (y)) \quad y \in \Delta$. Then $\Delta$ is an open set and $\Delta \subset \partial V$. And $\Delta$ is flat in the vicinity of $y^{0}$. In summary, the defined $v \in C^{1}(V) \cap C(\bar V)$ is a solution that satisfies the following boundary condition.

$$\left\{ \begin{align*} G(Dv,\ v,\ y) &= 0 && \text{in } V \\ v &= h && \text{on } \Delta \subset \partial V \end{align*} \right.$$

This is nothing different from $(1)$ and the assumption that some arbitrarily chosen part of the boundary is flat, and everything else is the same. And since the boundary can always be flattened like this, one can assume from the start that this is the problem and solve it.