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# 3.1: Linear Equations of Second Order

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The general nonlinear partial differential equation of second order is

$$F(x,u,Du,D^2u)=0,$$
where $$n$$, $$u:\ \Omega\subset\mathbb{R}\mapsto\mathbb{R}^1$$, $$Du\equiv\nabla u$$ and $$u$$ stands for all second derivatives. The function $$F$$ is given and sufficiently regular with respect to its $$2n+1+n^2$$ arguments.

In this section we consider the case
\begin{equation}
\label{linsecond}
\sum_{i,k=1}^na^{ik}(x)u_{x_ix_k}+f(x,u,\nabla u)=0.
\end{equation}
The equation is linear if
$$f=\sum_{i=1}^nb^i(x)u_{x_i}+c(x)u+d(x).$$
Concerning the classification the main part
$$\sum_{i,k=1}^n a^{ik}(x)u_{x_ix_k}$$
plays the essential role. Suppose $$u\in C^2$$, then we can assume, without restriction of generality, that $$a^{ik}=a^{ki}$$, since
$$\sum_{i,k=1}^n a^{ik}u_{x_ix_k}=\sum_{i,k=1}^n (a^{ik})^\star u_{x_ix_k},$$
where
$$(a^{ik})^\star=\frac{1}{2}(a^{ik}+a^{ki}).$$
Consider a hypersurface $$\mathcal{S}$$ in $$n$$ defined implicitly by $$\chi(x)=0$$, $$\nabla\chi\not=0$$, see Figure 3.1.1 Figure 3.1.1: Initial Manifold $$\mathcal{S}$$

Assume $$u$$ and $$\nabla u$$ are given on $$\mathcal{S}$$.

Problem: Can we calculate all other derivatives of $$u$$ on $$\mathcal{S}$$ by using differential equation (\ref{linsecond}\) and the given data?

We will find an answer if we map $$\mathcal{S}$$ onto a hyperplane $$\mathcal{S}_0$$ by a mapping
\begin{eqnarray*}
\lambda_n&=&\chi(x_1,\ldots,x_n)\\
\lambda_i&=&\lambda_i(x_1,\ldots,x_n),\ i=1,\ldots,n-1,
\end{eqnarray*}
for functions $$\lambda_i$$ such that
$$\det\frac{\partial(\lambda_1,\ldots,\lambda_n)}{\partial(x_1,\ldots,x_n)}\not=0$$
in $$n$$. It is assumed that $$i$$ and $$i$$ are sufficiently regular. Such a mapping $$\lambda=\lambda(x)$$ exists, see an exercise.

The above transform maps $$\mathcal{S}$$ onto a subset of the hyperplane defined by $$\lambda_n=0$$, see Figure 3.1.2. Figure 3.1.2: Transformed flat manifold $$\mathcal{S}$$

We will write the differential equation in these new coordinates. Here we use Einstein's convention, that is, we add terms with repeating indices. Since
$$u(x)=u(x(\lambda))=:v(\lambda)=v(\lambda(x)),$$
where $$x=(x_1,\ldots,x_n)$$ and $$\lambda=(\lambda_1,\ldots,\lambda_n)$$, we get
\begin{eqnarray}
\label{known}
u_{x_j}&=&v_{\lambda_i}\frac{\partial\lambda_i}{\partial x_j},\\
u_{x_jx_k}&=&v_{\lambda_i\lambda_l}\frac{\partial\lambda_i}{\partial x_j}\frac{\partial\lambda_l}{\partial x_k}+v_{\lambda_i}\frac{\partial^2\lambda_i}{\partial x_j\partial x_k}.\nonumber
\end{eqnarray}
Then the differential equation (\ref{linsecond}) in the new coordinates is given by
$$a^{jk}(x)\frac{\partial\lambda_i}{\partial x_j}\frac{\partial\lambda_l}{\partial x_k}v_{\lambda_i\lambda_l}+\mbox{terms known on}\ \mathcal{S}_0=0.$$
Since $$v_{\lambda_k}(\lambda_1,\ldots,\lambda_{n-1},0)$$, $$n$$, are known, see (\ref{known}), it follows that $$v_{\lambda_k\lambda_l}$$, $$l=1,\ldots,n-1$$, are known on $$\mathcal{S}_0$$. Thus we know all second derivatives $$v_{\lambda_i\lambda_j}$$ on $$\mathcal{S}_0$$ with the only exception of $$v_{\lambda_n\lambda_n}$$.

We recall that, provided $$v$$ is sufficiently regular,
$$v_{\lambda_k\lambda_l}(\lambda_1,\ldots,\lambda_{n-1},0)$$
is the limit of
$$\frac{v_{\lambda_k}(\lambda_1,\ldots,\lambda_l+h,\lambda_{l+1},\ldots,\lambda_{n-1},0)- v_{\lambda_k}(\lambda_1,\ldots,\lambda_l,\lambda_{l+1},\ldots,\lambda_{n-1},0)}{h}$$
as $$h\to0$$.

Then the differential equation can be written as
$$\sum_{j,k=1}^na^{jk}(x)\frac{\partial\lambda_n}{\partial x_j}\frac{\partial\lambda_n}{\partial x_k}v_{\lambda_n\lambda_n}=\mbox{terms known on}\ \mathcal{S}_0.$$
It follows that we can calculate $$v_{\lambda_n\lambda_n}$$ if
\begin{equation}
\label{nonchar}
\sum_{i,j=1}^na^{ij}(x)\chi_{x_i}\chi_{x_j}\not=0
\end{equation}

on $$\mathcal{S}$$. This is a condition for the given equation and for the given surface $$\mathcal{S}$$.

Definition. The differential equation
$$\sum_{i,j=1}^na^{ij}(x)\chi_{x_i}\chi_{x_j}=0$$
is called it characteristic differential equation associated to the given differential equation (\ref{linsecond}).

If $$\chi$$, $$\nabla \chi\not=0$$, is a solution of the characteristic differential equation, then the surface defined by $$\chi=0$$ is called characteristic surface.

Remark. The condition (\ref{nonchar}) is satisfied for each $$\chi$$ with $$\nabla\chi\not=0$$ if the quadratic matrix $$(a^{ij}(x))$$ is positive or negative definite for each $$x\in\Omega$$, which is equivalent to the property that all eigenvalues are different from zero and have the same sign. This follows since there is a $$\lambda(x)>0$$ such that, in the case that the matrix $$(a^{ij})$$ is poitive definite,
$$\sum_{i,j=1}^na^{ij}(x)\zeta_i\zeta_j\ge\lambda(x)|\zeta|^2$$
for all $$\zeta\in\mathbb{R}$$. Here and in the following we assume that the matrix $$(a^{ij})$$ is real and symmetric.

The characterization of differential equation (\ref{linsecond}) follows from the signs of the eigenvalues of $$(a^{ij}(x))$$.

Definition. The differential equation (\ref{linsecond}) is said to be of type $$(\alpha,\beta,\gamma)$$ at $$x\in\Omega$$ if $$\alpha$$ eigenvalues of $$(a^{ij})(x)$$ are positive, $$\beta$$ eigenvalues are negative and $$\gamma$$ eigenvalues are zero ($$\alpha+\beta+\gamma=n$$).
In particular, the equation is called

elliptic if it is of type $$(n,0,0)$$ or of type $$(0,n,0)$$, that is, all eigenvalues are different from zero and have the same sign,\\
parabolic if it is of type $$(n-1,0,1)$$ or of type $$(0,n-1,1)$$, that is, one eigenvalue is zero and all the others are different from zero and have the same sign,
hyperbolic if it is of type $$(n-1,1,0)$$ or of type $$(1,n-1,0)$$, that is, all eigenvalues are different from zero and one eigenvalue has another sign than all the others.

## Remarks:

1. According to this definition there are other types aside from elliptic, parabolic or hyperbolic equations.

2. The classification depends in general on $$x\in\Omega$$. An example is the Tricomi equation, which appears in the theory of transsonic flows,
$$yu_{xx}+u_{yy}=0.$$
This equation is elliptic if $$y>0$$, parabolic if $$y=0$$ and hyperbolic for $$y<0$$.

## Examples:

Example 3.1.1:

The Laplace equation in $$\mathbb{R}^3$$ is $$\triangle u=0$$, where
$$\triangle u:=u_{xx}+u_{yy}+u_{zz}.$$
This equation is elliptic since for every manifold $$\mathcal{S}$$ given by $$\{(x,y,z):\ \chi(x,y,z)=0\}$$, where $$\chi$$ is an arbitrary sufficiently regular function such that $$\nabla \chi\not=0$$, all derivatives of $$u$$ are known on $$\mathcal{S}$$, provided $$u$$ and $$\nabla u$$ are known on $$\mathcal{S}$$.

Example 3.1.2:

The wave equation $$u_{tt}=u_{xx}+u_{yy}+u_{zz}$$, where $$u=u(x,y,z,t)$$, is hyperbolic. Such a type describes oscillations of mechanical structures, for example.

Example 3.1.3:

The heat equation $$u_t=u_{xx}+u_{yy}+u_{zz}$$, where $$u=u(x,y,z,t)$$, is
parabolic. It describes, for example, the propagation of heat in a domain.

Example 3.1.4:

Consider the case that the (real) coefficients $$a^{ij}$$ in equation (\ref{linsecond}) are {\it constant}. We recall that the matrix $$A=(a^{ij})$$ is symmetric, that is, $$A^T=A$$. In this case, the transform to principle axis leads to a normal form from which the classification of the equation is obviously. Let $$U$$ be the associated orthogonal matrix, then
$$U^TAU=\left(\begin{array}{llcl} \lambda_1 & 0&\cdots & 0\\ 0 & \lambda_2&\cdots&0\\ ... & ... & ... & ...\\ 0&0&\cdots&\lambda_n \end{array}\right).$$
Here is $$U=(z_1,\ldots,z_n)$$, where $$z_l$$, $$l=1,\ldots,n$$, is an orthonormal system of eigenvectors to the eigenvalues $$\lambda_l$$.

Set $$y=U^Tx$$ and $$v(y)=u(Uy)$$, then
\begin{equation}
\label{hauptachs}
\sum_{i,j=1}^na^{ij}u_{x_ix_j}=\sum_{i=1}^n\lambda_iv_{y_iy_j}.
\end{equation}

## Contributors

• Integrated by Justin Marshall.