10 - Classifying Second Order Semi-Linear PDEs

Idea of Solving Them

Using Second Order Semi-Linear PDE Form then

Want to have a change of variables

$$(x,y)\to (\phi (x,y),\psi (x,y))$$

with non-vanishing Jacobian (so the map is locally invertible) that is

$$\frac{\partial (\phi ,\psi )}{\partial (x,y)}={\phi }_x{\psi }_y-{\phi }_y{\psi }_x\ne 0$$

Classification Form of Second Order Semi-Linear PDEs

For Second Order Semi-Linear PDE Form

$$\underset{\text{principal part}}{\underbrace{a(x,y)u_{xx}+2b(x,y)u_{xy}+c(x,y)u_{yy}}}=f(x,y,u,u_x,u_y)$$

then we have equation

$$(AC-B^2)=(ac-b^2)({\phi }_x{\psi }_y-{\psi }_x{\phi }_y{)}^2=(ac-b^2){\left(\frac{\partial (\phi ,\psi )}{\partial (x,y)}\right)}^2$$

where

$$\begin{array}{rl}A& =a{\phi }_x^2+2b{\phi }_x{\phi }_y+c{\phi }_y^2\\ B& =a{\phi }_x{\psi }_x+b({\phi }_x{\psi }_y+{\phi }_y{\psi }_x)+c{\phi }_y{\psi }_y\\ C& =a{\psi }_x^2+2b{\psi }_x{\psi }_y+c{\psi }_y^2\end{array}$$

Proof

Using the change of variables

$$(x,y)\to (\phi (x,y),\psi (x,y))$$

then the first partial derviatives are

$$\begin{array}{rl}{u}_x& =u_{\phi }{\varphi }_x+u_{\psi }{\psi }_x\\ u_y& =u_{\phi }{\varphi }_y+u_{\psi }{\psi }_y\end{array}$$

So then the second partial derivatives are

$$\begin{array}{rl}{u}_{xx}& =u_{\phi \phi }{\phi }_x^2+2u_{\phi \psi }{\phi }_x{\psi }_x+u_{\psi \psi }{\psi }_x^2+u_{\phi }{\phi }_{xx}+u_{\psi }{\psi }_{xx}\\ u_{xy}& =u_{\phi \phi }{\phi }_x{\phi }_y+u_{\phi \psi }({\phi }_x{\psi }_y+{\psi }_x{\phi }_y)+u_{\psi \psi }{\psi }_x{\psi }_y+u_{\phi }{\phi }_{xy}+u_{\psi }{\psi }_{xy}\\ u_{yy}& =u_{\phi \phi }{\phi }_y^2+2u_{\phi \psi }{\phi }_y{\psi }_y+u_{\psi \psi }{\psi }_y^2+u_{\phi }{\phi }_{yy}+u_{\psi }{\psi }_{yy}\end{array}$$

Hence the PDE becomes

$$A(\phi ,\psi )u_{\phi \phi }+2B(\phi ,\psi )u_{\phi \psi }+C(\phi ,\psi )u_{\phi \psi }=F(\phi ,\psi ,u,u_{\phi },u_{\psi })$$

where

$$\begin{array}{rl}A& =a{\phi }_x^2+2b{\phi }_x{\phi }_y+c{\phi }_y^2\\ B& =a{\phi }_x{\psi }_x+b({\phi }_x{\psi }_y+{\phi }_y{\psi }_x)+c{\phi }_y{\psi }_y\\ C& =a{\psi }_x^2+2b{\psi }_x{\psi }_y+c{\psi }_y^2\end{array}$$

for some $F$ which will include lower order derivatives from the second partial derivatives

Writing $A,B,C$ in a matrix notation we get

$$\left(\begin{array}{cc}A& B\\ B& C\end{array}\right)=\left(\begin{array}{cc}{\phi }_x& {\phi }_y\\ {\psi }_x& {\psi }_y\end{array}\right)\left(\begin{array}{cc}a& b\\ b& c\end{array}\right)\left(\begin{array}{cc}{\phi }_x& {\psi }_x\\ {\phi }_y& {\psi }_y\end{array}\right)$$

So by taking determinants then

$$(AC-B^2)=(ac-b^2)({\phi }_x{\psi }_y-{\psi }_x{\phi }_y{)}^2=(ac-b^2){\left(\frac{\partial (\phi ,\psi )}{\partial (x,y)}\right)}^2$$

Classifying Second-Order Linear PDEs

From Classification Form of Second Order Semi-Linear PDEs

We have three types

1. Hyperbolic: $ac<b^2$ (e.g. Wave Equation)
2. Elliptic: $ac>b^2$ (e.g. Laplace Equation)
3. Parabolic: $ac=b^2$ (e.g. Heat Equation)

So the class of the equation is invariant under transformations with non-vanishing Jacobian
Note that the type can change depending on value of $x,y,$ etc

Proof

Look at classification in terms of the quadratic polynomial

$$a(x,y){\lambda }^2-2b(x,y)+c(x,y)=0$$

Assuming that $a\ne 0$ in the domain under consideration
If $a=0$ and $c\ne 0$ then swap the roles of $x$ and $y$

Case 1: Hyperbolic Type $ac<b^2$ so quadratic has distinct roots ${\lambda }_1,{\lambda }_2$ so we have characteristic equation

So

$$a(x,y){\left(\frac{dy}{dx}\right)}^2-2b(x,y)\frac{dy}{dx}+c(x,y)=0$$

is equivalent to

$$\frac{dy}{dx}={\lambda }_1(x,y),\frac{dy}{dx}={\lambda }_2(x,y)$$

Suppose respectively the equations have solutions which are characteristic curves

$$\phi (x,y)=\text{constant}\text{ and }\psi (x,y)=\text{constant}$$

Hence we can use those as a change of variables so we get characteristic variables

$$\phi =\phi (x,y),\psi =\psi (x,y)$$

Then on $\phi (x,y)=\text{constant}$ then

$${\phi }_x+{\phi }_y\frac{dy}{dx}=0$$

So

$${\lambda }_1{\phi }_y=-{\phi }_x$$

Thus

$$A(\phi ,\psi )=a(x,y)({\phi }_x{)}^2+2b(x,y){\phi }_x{\phi }_y+c(x,y)({\phi }_y{)}^2=0$$

So similarly $C(\phi ,\psi )=0$
As ${\lambda }_1\ne {\lambda }_2$ so $\frac{{\phi }_x}{{\phi }_y}\ne \frac{{\psi }_x}{{\psi }_y}$ so $B\ne 0$
Hence we get the equation in form

$$u_{\phi \psi }=G(\phi ,\psi ,u,u_{\phi },u_{\psi })$$

Which is the normal / canonical form for a hyperbolic equation

Case 2: Elliptic

We have $ac>b^2$ so characteristic equation

$$a(x,y){\left(\frac{dy}{dx}\right)}^2-2b(x,y)\frac{dy}{dx}+c(x,y)=0$$

has a complex conjugate pair of roots, and the integral curves of

$$y^{\prime }(x)=\lambda (x,y),y^{\prime }(x)=\overline{\lambda }(x,y)$$

are in conjugate complex pairs where

$$\phi (x,y)=\text{const},\psi (x,y)=\overline{\phi }(x,y)=\text{const}$$

These complex coordinates then we get $A=C=0$ and $B\ne 0$ so the equation becomes

$$u_{\phi \overline{\phi }}=G(\phi ,\overline{\phi },u,u_{\phi },u_{\overline{\phi }})$$

By introducing new variables $\xi ,\eta$ given by $\phi =\xi +i\eta$ and $\overline{\phi }=\xi -i\eta$
Get normal form of an elliptic equation

$$u_{\xi \xi }+u_{\eta \eta }=H(\xi ,\eta ,u,u_{\xi },u_{\eta })$$

Note that this closely resembles Laplace’s Equation

Case 3: Parabolic

We have $ac=b^2$ so we have repeated root $\lambda (x,y)$

By solving $y^{\prime }(x)=\lambda (x,y)$ we get coordinate $\phi$ and then pick $\psi$ such that

$${\phi }_x{\psi }_y-{\phi }_y{\psi }_x\ne 0$$

to get the second coordinate

So similarly we get $A=0$ so $B^2=AC=0$ with $C\ne 0$
As $\psi =\text{const}$ is not a characteristic curve then

$$u_{\psi \psi }=G(u,\phi ,\psi ,u_{\phi },u_{\psi })$$

which is the normal form for a parabolic equation