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:

$$D_{\vec{v}}u=0$$
$$\vec{v}\cdot\vec{\nabla}u=0$$

Now let me ask you something:

Can you identify $\vec{v}$? Can you figure out the lines along which $u$ 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.

$$D_{\vec{v}}u+cu=f$$

where $\vec{v}=<a,b>$.

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 $\vec{v}$. In that coordinate system there is only one independent variable, and so you effectively have an ODE. The rotation matrix that carries $\{x,y\}$ into $\{w,t\}$ is given by the following.

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

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 $\vec{v}=<a,b,c>$.