PDE with constant coefficient using orthogonal transformation

[sara_math]

Plz Help :(

Hi

I want 2 know how 2 solve 1st order partial differintial equation (PDE) with constant coefficient using orthogonal transformation

example :
solve: 2Ux + 2Uy + Uz = 0

THnx

 

[quantumdude]

I would tackle this one geometrcially. You can express the left side of your PDE in terms of a directional derivative, as follows:

[tex]D_{\vec{v}}u=0[/tex]
[tex]\vec{v}\cdot\vec{\nabla}u=0[/tex]

Now let me ask you something:

Can you identify [itex]\vec{v}[/itex]? Can you figure out the lines along which [itex]u[/itex] is constant?

 

[sara_math]

Thnx sir

but the dr. who teach me, didn't use this way with us

when we have aUx+bUy+cU=f , where a,b,c,f are constants
he take tan(r)=b/a
then he find r
after that take w=xcosr+ysinr , t=-xsinr+ycosr
then find Ux , Uy
that's it

so if i have aUx+bUy+cUz+dU=f
can i take same w, t and z=z ?

 

[quantumdude]

My approach is not much different from your professor's.

This equation:

aUx+bUy+cU=f , where a,b,c,f are constants

can be written in the following form.

[tex]D_{\vec{v}}u+cu=f[/tex]

where [itex]\vec{v}=<a,b>[/itex].

he take tan(r)=b/a
then he find r
after that take w=xcosr+ysinr , t=-xsinr+ycosr
then find Ux , Uy

What your instructor did was rotate the x-axis so that it coincides with the vector [itex]\vec{v}[/itex]. In that coordinate system there is only one independent variable, and so you effectively have an ODE. The rotation matrix that carries [itex]\{x,y\}[/itex] into [itex]\{w,t\}[/itex] is given by the following.

[tex]R_{z}(r) = \left[\begin{array}{cc} \cos(r) & \sin(r)\\ -\sin(r) & \cos(r) \end{array}\right][/tex]

that's it

No, that's not it. All that does is transform the equation. It still remains to solve it.

so if i have aUx+bUy+cUz+dU=f
can i take same w, t and z=z ?

You will want to do a rotation about some axis so that one coordinate axis is parallel to [itex]\vec{v}=<a,b,c>[/itex].