04. Second Order Semi-Linear PDEs

4.1 Classification

Second Order Semi-Linear PDEs

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

PDE is linear if $f$ is linear in $u,u_x,u_y$ otherwise is semi-linear
PDE is quasi-linear if $a,b,c$ also depend on $u,u_x,u_y$

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4.1.1 The Idea

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

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

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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

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

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4.1.2 The Classification

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

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4.2 Characteristics

Characteristics of Second Order Semi-Linear PDEs

Analogous Properties to the Characteristic Projections of First Order Semi-Linear PDEs

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Discontinuities in Second Derivatives

Discontinuities in Second Derivatives of a solution across a given curve $\Gamma$ means

$$\text{curve }\Gamma \text{ is characteristic curve}$$

Proof

Suppose curve $\Gamma$ defined parametrically by

$$(x(s),y(s))$$

is a curve across which there are discontinuities in the second derivative of solution

Let $u_{xx}^{+}$ denote values on one side of $\Gamma$ and $u_{xx}^{-}$ denoting values on the other side of $\Gamma$

Differentiating $u_x(x(s),y(s))$ and $u_y(x(s),y(s))$ along $\Gamma$
with $u,u_x,u_y$ being continuous across the curve then

\begin{alignat*}{3} \frac{du_{x}}{ds} &= \quad && \frac{dx}{ds}u_{x x}^\pm + \quad && \frac{dy}{ds} u_{xy}^\pm \\ \frac{du_{y}}{ds} &= && && \frac{dx}{ds}u_{y x}^\pm + \quad && \frac{dy}{ds} u_{y y}^\pm \\ \end{alignat*}

with

$$f(x,y,u,u_x,u_y)=a(x,y)u_{xx}^{\pm }+2b(x,y)u_{xy}^{\pm }+c(x,y)u_{yy}^{\pm }$$

Subtracting the minus equation from the plus equations we get

\begin{alignat*}{3} 0 &= \quad && \frac{dx}{ds}[u_{x x}]^+_{-} + \quad && \frac{dy}{ds} [u_{xy}]^+_{-} \\ 0 &= && && \frac{dx}{ds}[u_{y x}]^+_{-} + \quad && \frac{dy}{ds} [u_{y y}]^+_{-} \\ 0 &= &&a(x, y)[u_{x x}]^+_{-} + &&2b(x, y)[u_{x y}]^+_{-} + &&c(x, y) [u_{yy}]^+_{-} \end{alignat*}

where $[u_{xx}{]}_{-}^{+}=u_{xx}^{+}-u_{xx}^{-}$ denotes the jump in $u_{xx}$ across $\gamma$, etc

If there are discontinuities in the second derivatives then
the set of equations in the jumps must have a non-zero solution hence

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

Hence $\Gamma$ is a characteristic

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4.3 Type and Data: Well Posed Problems

Well Posed PDE

Problem consisting of a PDE with data is well posed if solution

1. exists
2. is unique
3. depends continuously on data

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Solution Depends Continuously on Data

Let $u(x,y)$ be a solution of a PDE in a bounded subset of plane $D$ with $u$ given on curve $\Gamma$

Solution $u(x,y)$ depends continuously on data if

$\forall ϵ>0\exists \delta >0$ such that if $u_i,i=1,2$ are solutions with $u_i=f_i$ on $\Gamma$ then

$$\underset{\Gamma }{\sup }|f_1-f_2|<\delta ⟹\underset{D}{\sup }|u_1-u_2|<ϵ$$

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Models from Prelims

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4.4 The Maximum Principle

4.4.1 Poisson’s Equation

Normal form for Second Order Elliptic PDE

$$u_{xx}+u_{yy}=f(x,y,u,u_x,u_y)$$

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07 - The Maximum Principle for the Laplacian

The Maximum Principle for the Laplacian

Suppose $u$ satisfies

$$\Delta u:=u_{xx}+u_{yy}\ge 0\text{ for }(x,y)\in D$$

everywhere within a bounded domain $D$ then

$$u\text{ attains maximum value on }\partial D$$

If $\Delta u\le 0$ on $D$ then $u$ attain minimum value on $\partial D$ (substitute $u$ with $-u$)
If $\Delta u=0$ then $u$ attains both maximum and minimum values on $\partial D$

Proof

Denote boundary of $D$ as $\partial D$ so $D\cup \partial D$ is a closed boundary set so compact
Hence $u$ must attain the maximum either on $D$ or on the boundary $\partial D$

Suppose that $u_{xx}+u_{yy}>0$ in $D$
If $u$ has an interior maximum at some point $(x_0,y_0)$ inside $D$ then we have the following

$$u_x=u_y=0,u_{xx}\le 0,u_{yy}\le 0$$

As we assumed that $u_{xx}+u_{yy}>0$ in all od $F$ then there is a contradiction
Hence $u$ cannot have an interior maximum within $D$ so it must attain it on boundary $\partial D$

Suppose now we have $u_{xx}+u_{yy}\ge 0$ in $D$
Consider

$$v(x,y)=u(x,y)+\frac{ϵ}{4}(x^2+u^2)$$

for some positive constant $ϵ$

Then

$$v_{xx}+v_{yy}=u_{xx}+u_{yy}+ϵ>0$$

which is in $D$, so $v$ attains maximum value on $\partial D$

So suppose now that

1. maximum value of $u$ on $\partial D$ is $M$
2. maximum value of $(x^2+y^2)$ on $\partial D$ is $R^2$

then maximum value of $v$ on $\partial D$ (and throughout $D$) is

$$M+\left(\frac{ϵ}{4}\right)R^2$$

In other words we have

$$u+\frac{ϵ}{4}(x^2+y^2)=v\le M+\frac{ϵ}{4}{R}^2$$

holding for all $(x,y)\in D$
Then if $ϵ\to 0$ we have that $u\le M$ throughout $D$ hence $u$ attains maximum value on $\partial D$

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Properties of Dirichlet Problem corollary

For Dirichlet Problem for $u$ with boundary of $D$

$$\begin{array}{rl}{u}_{xx}+u_{yy}& =f(x,y)\text{ in }D\\ u& =g(x,y)\text{ on }\partial D\end{array}$$

then

1. If the solution exists it is unique

2. Solution has continuous dependence on data

Proof

1. Suppose that $u_1,u_2$ are two solutions so $u=u_1-u_2$ satisfies

$$\begin{array}{rl}{u}_{xx}+u_{yy}& =0\text{ in }D\\ u& =0\text{ on }\partial D\end{array}$$

So by The Maximum Principle for the Laplacian then
the maximum and minimum of $u$ occur on $\partial D$ hence $u\le 0$ and $u\ge 0$ on $D$ so

$$u=0\text{ in }D$$

2. Need to prove that for all $ϵ>0$ there exists $\delta >0$ such that if

$$u_i,i=1,2\text{ are solutions }\text{ with boundary data }{g}_i$$

then

$$\underset{(x,y)\in \partial D}{\sup }|g_1(x,y)-g_2(x,y)|<\delta ⟹\underset{(x,y)\in D}{\sup }|u_1(x,y)-u_2(x,y)|<ϵ$$

By linearity $u=u_1-u_2$ satisfie

$$\begin{array}{rlrl}{u}_{xx}+u_{yy}& =0& & \text{ in }D\\ u& =g_1-g_2& & \text{ on }\partial D\end{array}$$

Using maximum principle we get that

$$u\le \underset{(x,y)\in \partial D}{\max }(g_1-g_2)$$

So applying same result to $-u$ we get

$$-u\le \underset{(x,y)\in \partial D}{\max }-(g_1-g_2)$$

Hence in $D$

$$|u_1-u_2|\le \underset{(x,y)\in \partial D}{\max }|g_1-g_2|$$

so take $\delta =ϵ$

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4.4.2 The Heat Equation

Normal Form for Second Order Parabolic PD

$$u_{xx}=F(x,t,u,u_t,u_x)$$

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08 - The Maximum Principle for the Heat Equation

The Maximum Principle for the Heat Equation

Suppose $u(x,t)$ satisfies

$$u_t-u_{xx}\le 0$$

in a region $D_{\tau }$ bounded by

1. lines $t=0$ and $t=\tau >0$
2. two non-intersecting smooth curves $C_1$ and $C_2$ which are nowhere parallel to $x-$axis

Suppose also that $f\le 0$ in $D_{\tau }$ then

$u$ takes maximum value either on $t=0$ or on one of the curves $C_1$ or $C_2$

If $u_t-u_{xx}\ge 0$ then $u$ attains minimum value on $\partial D_{\tau }$ but not on $t=tau$

Proof

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Property of Heat Problem

Consider IBVP

where $u(x,t)$ satisfies

$$u_t-u_{xx}\le 0$$

in a region $D_{\tau }$ bounded by

1. lines $t=0$ and $t=\tau >0$
2. two non-intersecting smooth curves $C_1$ and $C_2$ which are nowhere parallel to $x-$axis

with $u$ given $\partial D_{\tau }\{t=\tau \}$ then

$$\text{if solution exists it is unique and depends continuously on initial data}$$

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