User:Egm6322.s09.three.liu/HW reports

=General Nonlinear PDE=

Variable
If $$  \left \{ {x}_{i} \right \}=   \left \{ {x}_{1},...,{x}_{n}   \right \} $$,

then $$\left \{ {x}_{i} \right \}$$has n independent variables.

Example 1: 1-D Case With Time Variable (i.e. 1-D time dependent problem)

If $$(x, t)=({x}_{1},{x}_{2})$$,then


 * $${x}_{1}=x$$
 * $$ {x}_{2}=t$$

$${x}_{1}$$ is the spatial variable; $${x}_{2}$$ is the temporal variable.

Example 2: 3-D Time Dependent Problems

$$ (x,y,z,t)=({x}_{1},{x}_{2},{x}_{3},{x}_{4})$$


 * $$ x=x_{1}$$
 * $$ y=x_{2}$$
 * $$ z=x_{3}$$
 * $$ t=x_{4}$$

Components of PDE
The unknown functions are expressed as $${u}_{1}$$,$${u}_{2}$$,...

e.g.

Navier-Stokes Equation in 3D is a typical PDE where (u1,u2,u3) represents the velocity field along (x,y,z) coordinates and (x1,x2,x3,x4) represent the independent variables (x,y,z,t).

$${u}_{i}(\left \{ {x}_{j} \right \});$$ $$i=1,2,3$$ $$j=1,2,3,4$$

In this case, the PDE has one unknown function u ;

n independent variable $$\left \{ {x}_{i} \right \}$$, i=1,2,...,n;

m th partial derivative: $$\frac{\partial^m u}{\partial x_{i_1}\,...,\partial x_{i_m}}$$

i1,...,im=1,...,n

Nonlinear PDEs
$$F( \left \{ {x}_{i} \right \}, \left \{ \frac{\partial u}{\partial x_i} \right \},\frac{\partial^2 u}{\partial x_i\,\partial x_j},...)=0$$, i=1,2,...,n

This general form of PDE has n independent variables: $$\left \{ {x}_{i} \right \}$$ and 1 unknown function u.

$$\left \{ {x}_{i} \right \}$$ has n arguments

$$\left \{ \frac{\partial u}{\partial x_i} \right \}$$ is computation of gradient of u, and has n arguments

$$\frac{\partial^2 u}{\partial x_i\,\partial x_j}$$ is the Hessian derivative, and has $$\frac{n(n+1)}{2}$$ arguments

$${H}_{ij}$$ is the Hessian Matrix, and $${H}_{ij}:=\frac{\partial^2 u}{\partial x_i\,\partial x_j}$$, (":=" means "equal by definition")

$$H^T=H$$



The PDE F[(x,y),u,(ux,uy),(uxx,uxy,uyy)]=0 has 2 independent variables and 1 unknown function.

e.g. $$2{u}_{xx}+3{u}_{yy}=ax^2+bx$$

Laplace Equation: $${u}_{xx}+{u}_{yy}=0$$

$$5({u}_{xx})^2+7({u}_{yy})^2=0$$ is a nonlinear PDE since the second derivatives uxx and uyy are squared.

Linearity of PDE
 u  is an unknown function; and  L  is an operator.

 L(u)  is linear with respect to  u  if

$$L(\alpha\,\!u+\beta\,\!u)=\alpha\,\!L(u)+\beta\,\!L(v)$$

$$F((x,y),u,({u}_{x},{u}_{y}),({u}_{xx}{u}_{xy}{u}_{yy}))=0$$

$$2{u}_{xx}+3{u}_{yy}=ax^2+bx$$

$${u}_{xx}+{u}_{yy}=0$$

If  L  is a operator, and $$L(u)=2{u}_{xx}+3{u}_{yy}-7x^2-x$$

$$ L(\alpha u+ \beta v)= \alpha \left (2 {u}_{xx}+3{u}_{yy} \right ) +\beta \left ( 2 {v}_{xx} +3 {v}_{yy} \right ) +f(x) $$.

$$ \alpha L(u)+ \beta L(v)= \alpha \left [2 {u}_{xx} +3 {u}_{yy} +f(x) \right ] +\beta \left [2 {v}_{xx}+3 {v}_{yy} + f(x) \right ] $$

$$ L(\alpha u+ \beta v) \ne \alpha L(u) +\beta L(v) $$

and then, the operator  L  is nonlinear.

=Linearity for N-Dimensions=

Example of Linear map for $$R^n$$ (n-dimensional case)
 R n $$\rightarrow$$ $$\mathbb{R}$$n is a set-to-set map.

If x $$\in$$  R  n; y $$\in$$ $$\mathbb{R}$$n,

then x $$\rightarrow$$ y=Ax is a point-to-point map.

A $$\in$$ $$\mathbb{R}$$n×n, and is a n×m matrix.



D (domain) =  R n

R (range) = $$\mathbb{R}$$n

General Case
When $$\mathbb{R}$$m $$\rightarrow$$ $$\mathbb{R}$$n; m $$\neq$$ n

If x $$\in$$ $$\mathbb{R}$$n; y $$\in$$ $$\mathbb{R}$$n; and y=Ax,

then A is a n×m matrix, A $$\in$$ $$\mathbb{R}$$n×m

Now consider: y=Ax+b;

y is a n×1 matrix; A is a n×m matrix; x is a m×1 matrix; b is a n×1 matrix.

M($$\cdot$$): $$\mathbb{R}$$m $$\rightarrow$$ $$\mathbb{R}$$n

If x $$\in$$ $$\mathbb{R}$$m; y $$\in$$ $$\mathbb{R}$$n;

then x $$\longmapsto$$ y=Ax+b

So clearly,M( 0 )= b $$\neq$$ 0 $$\Rightarrow$$ M($$\cdot$$) is not a linear map, and M($$\cdot$$) is not homogeneous, either.

 Example of Rotation 

In the case y=Rx+b

$$\begin{Bmatrix} y_1 \\ y_2 \end{Bmatrix}$$ = $$\begin{bmatrix} R_{11} &R_{12}    \\ R_{21} &R_{22} \end{bmatrix}$$$$\begin{Bmatrix} x_1 \\ x_2 \end{Bmatrix}$$ + $$\begin{Bmatrix} b_1 \\ b_2 \end{Bmatrix}$$ For more: Link title

 Note:  If A $$\Leftrightarrow$$ B, then A is equivalent to B.

If A $$\Rightarrow$$ B, then A is sufficient condition for B;

If A $$\Leftarrow$$ B, then B is sufficient condition for A, or A is necessary condition for B.

[A $$\Rightarrow$$ B]$$\Leftrightarrow$$ [negation of B $$\Rightarrow$$ negation of A]

=Transformation of Coordinates=

Assume $$R^m$$ and $$R^n$$ are expressions of vectors (tensors). Divergent maps of vector field (vector valued function) into a scalar function, in other words, domain and range of div(·) are functions spaces.

 Note:  In P9.1, it can't be transformed to coordinates of Eqn(5) by 2nd order linear PDEs (e.g. Eqn(2))

In coordinates:

$$x$$=$$\Phi$$($$\bar{x}$$,$$\bar{y}$$)

$$y$$=$$\psi$$($$\bar{x}$$,$$\bar{y}$$)

in coordinate transformation of Eqn(5) in P9-1

$$u$$(x,y)=$$u$$($$\Phi$$($$\bar{x}$$,$$\bar{y}$$),$$\psi$$($$\bar{x}$$,$$\bar{y}$$))

=$$u$$($$\bar{x}$$,$$\bar{y}$$)           (abuse of notation by using "u")

=$$\bar{u}$$($$\bar{x}$$,$$\bar{y}$$)      (more rigorous notation)

 Example： 

Let $$u(x)=ax+b$$

$$ x=x(\bar{x})

=sin\bar{x} =\Phi(\bar{x}$$)

$$u(x)=u(\Phi(\bar{x}))

=asin(\bar{x})+b

=u(\bar{x})

=\bar{u}(\bar{x})$$

$$u(\bar{x})$$is an abuse of notation; $$\bar{u}(\bar{x})$$ is more rigorous.

$$u_x(x,y)=\frac{\partial u}{\partial x}(x,y)=u_x(\Phi(\bar{x},\bar{y}),\Psi(\bar{x},\bar{y})) =\frac{\partial}{\partial x}\bar{u}(\bar{x},\bar{y}) =\frac{\partial \bar{u}}{\partial \bar{x}}\frac{\partial \bar{x}}{\partial x}+\frac{\partial \bar{u}}{\partial \bar{y}}\frac{\partial \bar{y}}{\partial x}$$

 Define:  $$\bar{x}=\bar{x}(x,y)$$=$$\bar{\Phi}(x,y)$$; $$\bar{y}=\bar{y}(x,y)$$=$$\bar{\Psi}(x,y)$$

then $$u_y(x,y)=\frac{\partial}{\partial y}\bar{u}(\bar{x},\bar{y})=\bar{u}_{\bar{x}}\frac{\partial \bar{x}}{\partial y}+\bar{u}_{\bar{y}}\frac{\partial \bar{y}}{\partial y}$$

 Matrix form: 

$$\partial_xu=[\frac{\partial \bar{x}}{\partial x},\frac{\partial \bar{y}}{\partial x}]\begin{Bmatrix} \partial_\bar{x} \\ \partial_\bar{y} \end{Bmatrix}(\bar{u})$$

$$ \begin{Bmatrix} \partial_x \\ \partial_y \end{Bmatrix}=\begin{bmatrix} \frac{\partial \bar{x}}{\partial x} & \frac{\partial \bar{y}}{\partial x}     \\ \frac{\partial \bar{x}}{\partial y} & \frac{\partial \bar{y}}{\partial y} \end{bmatrix} \begin{Bmatrix} \partial_\bar{x} \\ \partial_\bar{y} \end{Bmatrix}$$

 Jacobian Matrix:  $$\begin{bmatrix} \frac{\partial \bar{x}}{\partial x} & \frac{\partial \bar{y}}{\partial x}     \\ \frac{\partial \bar{x}}{\partial y} & \frac{\partial \bar{y}}{\partial y} \end{bmatrix}$$; (Sometimes it is defined as transformation of Jacobian matrix)

Note: An easier and more general expression:

$$(x_1,...,x_n)\longmapsto(\bar{x}_1,...,\bar{x}_n)$$

Indicial notation:

$$\bar{x}_i=\bar{x}_i(x_1,...,x_n)$$

$$\underline{J}$$=$${[\frac{\partial \bar{x}_i}{\partial x_j}]}$$n×n

i: row index; j: column index.

Example of Nonlinear Coordinate Transformation
One example of nonlinear coordinate transformation is from Cartesian Coordinate to polar coordinate:

e.g.

(x,y): Cartesian coordinate;

$$(r,\theta)$$: polar coordinate;

The transformation between the two coordinates is:

$$x=r\cos\theta=\bar{x}\cos\bar{y}$$

$$y=r\sin\theta=\bar{x}\sin\bar{y}$$

Homework: Transform the Laplace Equation into polar coordinates, and compare to published results

Find the Jacobian Matrix $$\bar J$$ and transform from Cartesian Coordinate (x,y) to polar coordinate (r,$$\theta$$)

i.e. a particular case for Eq(1) in P13-1.

Further, particular Eq(2) (P8) to Laplace Equation in Cartesian Coordinate.

Recall Laplace Equation in tensor (coordinate free)

notation: $$ div(grad\ u)=0$$

Transform Laplace Equation into polar coordinate, and compute result to published result.

Homework: Compare and translate notations between those in LP and ours Compare the expansion of Equation 1 on page 12-3 to LP (Lapiduce & Binder, 1982), found on page 14 of their book. Translate theirs into our notation.

1st partial derivative under transform of coordinates: Eq(1) in P12-3. (using chain rule)

In LP, P14: 2nd partial derivative (using chain rule), translate it into new coordinate (ours).

i.e. ours: $$(\bar x_1, \bar x_2)=(\bar x, \bar y)$$

$$\bar {x_i}=\bar{x_i} (x_1,x_2) $$    (i=1,2,...)

LP: $$\xi=\phi(x,y)$$

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

LP (1.2.5 ab): 1st partial derivative

LP (1.2.6 $$\theta$$): 2nd partial derivative

Obtain the complete form:

$$u_{xx}=u_{\xi\xi}(\phi_{x})^2+2 u_{\xi \eta} \phi_{x} \psi_{x}+u_{\eta \eta}+(\psi_{x})^2+...$$

Recall Jacobian Matrix    (P12-3)

$$\bar x_{i}=\bar x_{i}(x_1,...,x_n)$$

$$\bar J=\begin{bmatrix} \partial {\bar x_i}     \\ \partial {\bar x_j} \end{bmatrix} $$

2-Dimensions

$$\bar J=\begin{bmatrix} J_{11} & J_{12} \\ J_{21} & J_{22} \end{bmatrix}=\begin{bmatrix} \phi_{x} & \phi_{y} \\ \psi_{x} & \psi_{y} \end{bmatrix}$$

(1.2.6) in LP is:

$$u_{xx}=u_{\bar x \bar x}(J_{11})^2+2 u_{\bar x \bar y} J_{11} J_{21}+u_{\bar y \bar y}^2$$

Derive above equations: (1.2.6~7~8) in LP at once by matrix operator form: Eq(1) P12-3

$$\begin{Bmatrix} \partial x \\ \partial y \end{Bmatrix} \left \lfloor \partial_x \; \partial_y \right \rfloor =\begin{bmatrix} \partial {xx}     & \partial {xy}      \\ \partial {yx}     & \partial {yy} \end{bmatrix}=\bar J^T \begin{Bmatrix} \partial {\bar x} \\ \partial {\bar y} \end{Bmatrix} \left \lfloor \partial_x \; \partial_y \right \rfloor \bar J $$

Note:

$$\left \lfloor \partial_x \; \partial_y \right \rfloor= \left \lfloor \partial_{\bar x} \; \partial_{\bar y} \right \rfloor \bar J$$

$$(\bar A \bar B)^T={\bar B}^T{\bar A}^T$$

$$\alpha={\bar J}^T \begin{bmatrix} \partial_{\bar x \bar x}     & \partial_{\bar x \bar y}      \\ \partial_{\bar y \bar x}     & \partial_{\bar y \bar y}     \end{bmatrix} \bar J $$

Find $$\partial_{xx}$$

$$\begin{bmatrix} J_{11}     & J_{21}      \\ J_{12}     & J_{22} \end{bmatrix} \begin{bmatrix} \partial_{\bar x \bar x}     & \partial_{\bar x \bar y}      \\ \partial_{\bar y \bar x}     & \partial_{\bar y \bar y}  \end{bmatrix} \begin{bmatrix} J_{11}     & J_{12}      \\ J_{21}     & J_{22} \end{bmatrix} = \begin{bmatrix} a    & ? \\ ?      & ? \end{bmatrix} $$

$$a=(\partial_{\bar x \bar x}(J_{11})^2)+\partial {\bar x \bar y}J_{11}J_{21}+\partial {\bar y \bar x}J_{11}J{21}+\partial {\bar y \bar x}J_{11}J_{21}+\partial {\bar y \bar y}(J_{21})^2 $$

=lecture 19=

In the rotation and translation of coordinates,

$$\alpha=\beta+\gamma$$

and in this equation,

$$\alpha = \underline{J}^{T}\begin{Bmatrix} \partial_{\underline{x}}\\ \partial_{\underline{y}} \end{Bmatrix} \left \lfloor \partial_{\overline{x}} \; \partial_{\overline{y}} \right \rfloor \underline{J}+ \begin{Bmatrix} \partial_{x}\\ \partial_{y} \end{Bmatrix} \left \lfloor \partial_{\overline{x}} \; \partial_{\overline{y}} \right \rfloor \underline{J}$$

$$ \ \beta=\underline{J}^{T}\begin{Bmatrix} \partial_{\underline{x}}\\ \partial_{\underline{y}} \end{Bmatrix} \left \lfloor \partial_{\overline{x}} \; \partial_{\overline{y}} \right \rfloor \underline{J}$$

$$ \gamma= \begin{Bmatrix} \partial_{x}\\ \partial_{y} \end{Bmatrix} \left \lfloor \partial_{\overline{x}} \; \partial_{\overline{y}} \right \rfloor \underline{J}$$

$$\displaystyle {\rm div}( {\rm grad} \, u )=u_{xx}+u_{yy}=u_{rr}+\frac {1}{r^2}u_{\theta \theta}+\left \lfloor \frac {1}{r} \; \theta \right \rfloor \begin{Bmatrix} \partial_r u \\ \partial_{\theta} u \end{Bmatrix} $$

In this case,

$$\alpha=\displaystyle {\rm div}( {\rm grad} \, u )$$

$$\beta=u_{rr}+\frac {1}{r^2}u_{\theta \theta}$$

$$\gamma=\left \lfloor \frac {1}{r} \; \theta \right \rfloor \begin{Bmatrix} \partial_r u \\ \partial_{\theta} u \end{Bmatrix}$$

Classifications
In the general form of PDE:

$$au_{xx}+2bu_{xy}+cu_{yy}+du_{x}+eu_{y}+fu+g=0$$

it can be classified by the value of $$ac-b^2$$.

$$ac-b^2=det \underline A$$

$$ac-b^2<0 \Rightarrow$$ PDE is hyperbolic.

$$ac-b^2=0\Rightarrow$$ PDE is parabolic.

$$ac-b^2>0\Rightarrow$$ PDE is elliptic.

For example:

In the equation:

$$u_{xx}+u_{yy}=0 $$...(1)

$$\Rightarrow a=1,b=0,c=1$$

$$\Rightarrow ac-b^2=1>0$$

$$\Rightarrow$$ This PDE is elliptic

Question: what would be the classification of diffusion operator $$\displaystyle {\rm div}( {\rm grad} \, u )$$ in polar coordinate?

Another way to find the type of equation (1):

$$\underline \bar A:=\underline J \underline A \underline J^T =\begin{bmatrix} A     & B      \\ B & C \end{bmatrix} =\begin{bmatrix} \bar a     & \bar b      \\ \bar b & \bar c \end{bmatrix} \to$$ preferred form

$$det \underline \bar A=\bar a \bar c-\bar b^2$$

$$\bar a=1; \bar b=0; \bar c=\frac {1}{r^2} (r\ne0)$$

$$\Rightarrow$$The equation (1) is an elliptic PDE.

Relationship Between Classifications and Transformations
Observations from the verification of PDE classifications:

1. Diffusion operator remains elliptic in polar coordinates.

Question:How about a different transformation of coordinate? Would classification remain the same?

2. Does classification make sense if it changes under transformation of coordinate?

The answer is no, because physics (e.g. distribution of temperature as a result of solution of heat equation) must remain the same regardless of how heat equation was solved (under different coordinate system).

Therefore, classification better remains the same under different coordinate system for it to make sense.

=mtg29= Homework: Plot $$J_n(z)$$ and $$Y_n(z)$$ for $$n=0,1,2 ...$$$$z \in \Re$$ Use Matlab and include code.

Superposition: 2 separation of variables, constant $$\rho$$, $$\lambda$$

$$\sum_{\lambda}\sum_{\rho}[A(\lambda)exp(-\lambda t)][B(\rho)sin(\sqrt{\rho}\theta)+C(\rho)cos(\sqrt{\rho}\theta)] [D(\rho,\lambda)J_{\sqrt{\rho}}(\sqrt{\lambda} r)+E(\rho,\lambda)Y_{\sqrt{\rho}}(\sqrt{\lambda} r)]$$

Up to now, no restriction on $$\rho$$ and $$\lambda$$, i.e. they can be continuous real numbers (as opposed to discrete integral numbers as in Fourier Series: think of Fourier integral or Fourier Transformation and Laplace Transformation)

$$\Rightarrow$$ Replace discrete sum $$\sum$$ by continuous sum of $$\int$$.

Define:

$$\bar{B}(\rho,\lambda):=A(\lambda)B(\rho)$$

$$\bar{C}(\rho,\lambda):=A(\lambda)C(\rho)$$

General solution:

$$u(r,\theta,t)=\int d\lambda \int d\rho[exp(-\lambda t)][\bar{B}(\rho,\lambda)sin(\sqrt{\rho}\theta)+\bar{C}(\rho,\lambda)cos(\sqrt{\rho}\theta)][D(\rho,\lambda)J_{\sqrt{\rho}}(\sqrt{\lambda} r)+E(\rho,\lambda)Y_{\sqrt{\lambda}r}] $$

The Primitive of Functions
The primitive for function F is:

$$Q(x,\xi):=\int F(x,\xi)d\xi$$

e.g. The primitive for x is:

$$\int xd\xi=\frac {1}{2}x^2+k$$

The definition for the integral of x is:

$$\int_{x=a}^{x=b}x dx=\frac {1}{2}(b^2-a^2)$$

HW: Check the definition of indefinite integral The indefinite integral of a function $$f(x)$$ can be denoted by

$$\int f(x)dx$$

For example, if $$n \ne -1$$, we have

$$\int x^n dx=\frac {x^{n+1}}{n+1}+C$$

In this case, $$C$$ denotes an arbitrary constant. This is because the function $$f(x)$$ has many indefinite integrals, but any two differ by a constant. Consequently, if one antiderivative of a function $$f(x)$$ is found and added by an arbitrary constant, every indefinite integral for the function is found.

Generally, the indefinite integral for $$F(x,\xi)$$ is

$$ \int F(x,\xi)d\xi=G(x,\xi=B(x))-G(x,\xi=A(x))$$

$$\frac {d}{dx}F(x,\xi)

=[\frac {\partial G(x,B(x))}{\partial x}+\frac {\partial G(x,B(x))}{\partial \xi}]-[\frac {\partial G(x,A(x))}{\partial x}+\frac {\partial G(x,A(x))}{\partial \xi}]$$

$$=[\int_{\xi=A(x)}^{\xi=B(x)}\frac {\partial}{\partial x}F(x,\xi) d\xi+F(x,B(x))\frac {dB(x)}{dx}]-[\int_{\xi=\xi_0}^{\xi=A(x)}\frac {\partial}{\partial x}F(x,\xi) d\xi+F(x,A(x))\frac {dA(x)}{dx}]$$

$$\frac {\partial}{\partial x}F(x,\xi)=\int_{\xi=A(x)}^{\xi=B(x)}\frac {\partial}{\partial x}F(x,\xi) d\xi+F(x,B(x))\frac {dB(x)}{dx}-F(x,A(x))\frac {dA(x)}{dx}$$

Divergence operator in polar coordinate


Basic vector in polar coordinates:

$$\bar{\underline{e_i}}=\frac {\partial P}{\partial \bar{x_i}}; i=1,2,...$$

$$\bar{\underline{e_1}}=\underline{e_r}=\frac {\partial P}{\partial r}=\frac {\partial (\overrightarrow{OP})}{\partial r}; (\bar{x_1}=r)$$

$$\overrightarrow{OP}=x\overrightarrow{i}+y\overrightarrow{j}=(rcos\theta)\underline{i}+(rsin\theta)\underline{j}$$

$$\bar{\underline{e_2}}=\underline{e_\theta}=\frac {\partial P}{\partial \theta}; (\bar{x_x}=\theta)$$

$$ \underline{e_r}=\frac {\partial P}{\partial r}=cos\theta\underline{i}+sin\theta\underline{j}$$

$$\underline{e_\theta}=\frac{\partial P}{\partial \theta}=-rsin \theta \underline{i}+rcos\theta \underline{j}$$

$$\begin{Vmatrix} \underline{e_r} \end{Vmatrix}=1$$

$$\begin{Vmatrix} \underline{e_\theta} \end{Vmatrix}\neq1=r$$

=mtg 39=

Homework: Get the Canonical form for 2nd order linear Elliptic PDE x

Homework: Get the Canonical form for 2nd order linear Hyperbolic PDE For hyperbolic PDE: $$\lambda_1 u_{xx}-\lambda_2 u_{yy}=g$$

$$\begin{Bmatrix} \partial_{x} \\ \partial_{y} \end{Bmatrix} =J_B \begin{Bmatrix} \partial_{\xi} \\ \partial_{\eta} \end{Bmatrix}$$  ... (1)

The Jacobian matrix is:

$$ J_B= \begin{bmatrix} \frac {1}{\sqrt {\lambda_1}} & 0 \\ 0 & \frac {1}{\sqrt {\lambda_2}} \end{bmatrix}$$

so the hyperbolic PDE can be written as:

$$\left \lfloor \partial_x \partial_y \right \rfloor \begin{bmatrix} u \lambda_1 & 0 \\ 0 & -u \lambda_2 \end{bmatrix} \begin{bmatrix} \partial_x \\ \partial_y \end{bmatrix}=g$$

substituting (1) into the above equation gives:

$$\left \lfloor \partial_{\xi} \partial_{\eta} \right \rfloor \begin{bmatrix} \frac {1}{\sqrt {\lambda_1}} & 0 \\ 0 & \frac {1}{\sqrt {\lambda_2}} \end{bmatrix} \begin{bmatrix} u \lambda_1 & 0 \\ 0 & -u \lambda_2 \end{bmatrix} \begin{bmatrix} \frac {1}{\sqrt {\lambda_1}} & 0 \\ 0 & \frac {1}{\sqrt {\lambda_2}} \end{bmatrix} \begin{bmatrix} \partial_{\xi} \\ \partial_{\eta} \end{bmatrix}=g$$

So the canonical form for this hyperbolic PDE is

$$u_{\xi \xi}-u_{\eta \eta}=g$$

Homework: Get the Canonical form for 2nd order linear Parabolic PDE x

Homework: General classification of 2nd order linear PDEs to 2nd order nonlinear PDEs x

Homework: Verify (2) in Solution of (1). Derivation of d'Alembert solution for 1-D wave equation x

Consider: g=0 in Equation $$U_{\xi\xi}-U_{\eta}=0$$

$$U_{\xi\eta}=0$$

$$U{\xi,\eta}=F(\xi)+G(\eta)$$

where $$F(\cdot)$$ and $$G(\cdot)$$ are any function that is differentiable to first order.

$$c^2w_{xx}=w_{tt}$$

$$\tau w_{xx}+P=mw_{tt}$$

Classification of conics and 2nd order linear PDEs.

Transformation of coordinate via eigenvalue problem.

$$\bar a=a(\phi_x)^2+b\phi_x \phi_y+c(\phi_y)^2$$

$$\bar b=...$$

$$\bar c=a(\psi_x)^2+b \psi_x \psi_y+c(\psi_y)^2$$

$$\bar{\underline A}=\underline J \underline A \underline J^T= \begin{vmatrix} \bar a & \bar b \\ \bar b & \bar c \end{vmatrix}$$

To obtain form 2 of canonical form

Eq (3) P 39-1 set $$\bar a=\bar c=0$$