Discover MakerZone

MATLAB and Simulink resources for Arduino, LEGO, and Raspberry Pi

Learn more

Discover what MATLAB® can do for your career.

Opportunities for recent engineering grads.

Apply Today

Thread Subject:
solve differential equation

Subject: solve differential equation

From: Peter Schreiber

Date: 3 Apr, 2012 03:42:11

Message: 1 of 3

Hello,
If someone could guide me in the right direction that would be great. I'm trying to solve follwing partial differential equation in matlab. Is there any chance to analytically integrate the following equation?

( (x-x1)*dx + (y-y1)*dy + (z-z1)*dz ) /sqrt( (x-x1)^2 + (y-y1)^2 + (z-z1)^2 ) = c*dx - sqrt(1-c^2)*dz

x1,y1,z1 and c are constants

Thanks,
Peter

Subject: solve differential equation

From: Roger Stafford

Date: 3 Apr, 2012 05:46:10

Message: 2 of 3

"Peter Schreiber" <schreiber.peter15@gmail.com> wrote in message <jldrij$ntc$1@newscl01ah.mathworks.com>...
> Hello,
> If someone could guide me in the right direction that would be great. I'm trying to solve follwing partial differential equation in matlab. Is there any chance to analytically integrate the following equation?
>
> ( (x-x1)*dx + (y-y1)*dy + (z-z1)*dz ) /sqrt( (x-x1)^2 + (y-y1)^2 + (z-z1)^2 ) = c*dx - sqrt(1-c^2)*dz
>
> x1,y1,z1 and c are constants
>
> Thanks,
> Peter
- - - - - - - - - -
  This equation implies that

 f(x,y,z) = sqrt((x-x1)^2 + (y-y1)^2 + (z-z1)^2) - c(x-x1) + s*(z-z1)

must be some constant which we call K and where we define s = sqrt(1-c^2). It defines a surface f(x,y,z) = K which we can write as

 sqrt((x-x1)^2 + (y-y1)^2 + (z-z1)^2) = c*(x-x1) - s*(z-z1) + K

Squaring both sides and simplifying gives

 (c*(x-x1)-s*(z-z1)+K/2 = ((s*(x-x1)+c*(z-z1))^2+(y-y1)^2)/(2*K)

If we define u = (c*(x-x1)-s*(z-z1), v = y-y1, and w = s*(x-x1)+c*(z-z1), this is just

 u+K/2 = (v^2+w^2)/(2*K)

where u, v, w are also cartesian coordinates translated to (x1,y1,z1) and rotated about a line parallel to the v-axis by an angle given by c and s. This equation clearly shows that this is a family of paraboloids of revolution about u-axis. When K is zero it degenerates to the single point (x1,y1,z1).

  Does this help you?

Roger Stafford

Subject: solve differential equation

From: Roger Stafford

Date: 3 Apr, 2012 06:05:11

Message: 3 of 3

"Roger Stafford" wrote in message <jle2r2$eqj$1@newscl01ah.mathworks.com>...
> "Peter Schreiber" <schreiber.peter15@gmail.com> wrote in message <jldrij$ntc$1@newscl01ah.mathworks.com>...
> > Hello,
> > If someone could guide me in the right direction that would be great. I'm trying to solve follwing partial differential equation in matlab. Is there any chance to analytically integrate the following equation?
> >
> > ( (x-x1)*dx + (y-y1)*dy + (z-z1)*dz ) /sqrt( (x-x1)^2 + (y-y1)^2 + (z-z1)^2 ) = c*dx - sqrt(1-c^2)*dz
> >
> > x1,y1,z1 and c are constants
> >
> > Thanks,
> > Peter
> - - - - - - - - - -
> This equation implies that
>
> f(x,y,z) = sqrt((x-x1)^2 + (y-y1)^2 + (z-z1)^2) - c(x-x1) + s*(z-z1)
>
> must be some constant which we call K and where we define s = sqrt(1-c^2). It defines a surface f(x,y,z) = K which we can write as
>
> sqrt((x-x1)^2 + (y-y1)^2 + (z-z1)^2) = c*(x-x1) - s*(z-z1) + K
>
> Squaring both sides and simplifying gives
>
> (c*(x-x1)-s*(z-z1)+K/2 = ((s*(x-x1)+c*(z-z1))^2+(y-y1)^2)/(2*K)
>
> If we define u = (c*(x-x1)-s*(z-z1), v = y-y1, and w = s*(x-x1)+c*(z-z1), this is just
>
> u+K/2 = (v^2+w^2)/(2*K)
>
> where u, v, w are also cartesian coordinates translated to (x1,y1,z1) and rotated about a line parallel to the v-axis by an angle given by c and s. This equation clearly shows that this is a family of paraboloids of revolution about u-axis. When K is zero it degenerates to the single point (x1,y1,z1).
>
> Does this help you?
>
> Roger Stafford
- - - - - - - - -
  I should have said, when K is zero it becomes a degenerate half infinite line along the positive u-axis and ending at the point (x1,y1,z1).

Roger Stafford

Tags for this Thread

No tags are associated with this thread.

What are tags?

A tag is like a keyword or category label associated with each thread. Tags make it easier for you to find threads of interest.

Anyone can tag a thread. Tags are public and visible to everyone.

Contact us