Question

Find the nature of the partial differential equation
4uxx + 4uxy + Uyy+2ux -Uy = 0
​

Answer 1

Given:

A partial differential equation -

$\[4Uxx{\text{ }} + {\text{ }}4Uxy{\text{ }} + {\text{ }}Uyy + 2Ux{\text{ }} - Uy{\text{ }} = {\text{ }}0\]$

To Find:

Nature of given partial differential equation.

Solution:

Since,

$\[4uxx{\text{ }} + {\text{ }}4uxy{\text{ }} + {\text{ }}Uyy + 2ux{\text{ }} - Uy{\text{ }} = {\text{ }}0\]$ is given.

On comparing above equation with Canonical form of partial differential equation, we get

A= 4 , B= 4 and C= 1

Therefore,

$B^{2} - 4AC$ =

$=4^{2} - 4(4)(1)\\=16 - 16\\=0$

Here, $B^{2} - 4AC$ = 0

Thus, nature of given partial differential equation is parabolic.

Answer 2

TO DETERMINE

The nature of the partial differential equation

$\sf{4u_{xx} + 4u_{xy} + u_{yy} + 2u_x - u_y = 0}$

FORMULA TO BE IMPLEMENTED

A general partial differential equation is of the form

$\sf{A \: u_{xx} +B \: u_{xy} + C \: u_{yy} + D u_x - E u_y + F u = G}$

Where A, B, C, D, E, F, G are functions of x & y or constants

$\sf{Consider \: the \: term \: \: \: {B}^{2} - 4AC}$

$\sf{ 1. \: \:{B}^{2} - 4AC < 0 \: \: then \: equation \: is \: Elliptic }$

$\sf{ 2. \: \:{B}^{2} - 4AC > 0 \: \: then \: equation \: is \: Hyperbolic }$

$\sf{ 3. \: \:{B}^{2} - 4AC = 0 \: \: then \: equation \: is \:Parabolic }$

EVALUATION

The given partial differential equation is

$\sf{4u_{xx} + 4u_{xy} + u_{yy} + 2u_x - u_y = 0}$

Comparing with the general equation

$\sf{A \: u_{xx} +B \: u_{xy} + C \: u_{yy} + D u_x - E u_y + F u = G}$

We get

$\sf{A=4, B=4,C=1,D=2,E=-1,F=0,G=0}$

Now

$\sf{{B}^{2} - 4AC} =$

$\sf = {4}^{2} - (4 \times 4 \times 1)$

$= 16 - 16$

$= 0$

Hence the given differential equation represents Parabolic

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