Upload
others
View
0
Download
0
Embed Size (px)
Math 53: Fall 2020, UC Berkeley Lecture 015 Copyright: J.A. Sethian
All rights reserved. You may not distribute/reproduce/display/post/upload any course materials in any way, regardless of whether or not a fee is charged, without my express written consent. You also may not allow anyone else to do so. If you do so, you will be prosecuted under UC Berkeley student proceedings Secs. 102.23 and 102.25 [email protected]
Section 15.10: Change of Variable This lecture is one of those key moments in the course when all of sudden you really understand what is going on!
Let’s go back to 1D calculus
g(u) u x
Then we can rewrite by finding the ``magic factor’’
(1) Let x = g(u) (2) Then dx = g’(u) du
So we can write
Finally, we need to get the limits of integration correct, so if g(c)= a and g(d)=b, then
Example: Let f(x)=(2x)3, and suppose we want We can let x = g(u)= (1/2) u. Then dx = g’(u) du = ½ du, so
You have been doing this all along
Math 53: Fall 2020, UC Berkeley Lecture 015 Copyright: J.A. Sethian
All rights reserved. You may not distribute/reproduce/display/post/upload any course materials in any way, regardless of whether or not a fee is charged, without my express written consent. You also may not allow anyone else to do so. If you do so, you will be prosecuted under UC Berkeley student proceedings Secs. 102.23 and 102.25 [email protected]
Section 15.10: Change of Variable
The key to changing variables was the calculation of the magic factor: dx = g’(u) du
g(u) u x Which allowed us to go from
Okay, what if we had a function of two variables?
u
v
x
y
f(u,v)
g(u,v)
Could we find a magic factor?
that allows us to transform variables?
Example: We have already done this for polar coordinates!!!
r
q
x
y
So, for polar coordinate transformations, the magic factor is r
How do we find it in general?
Math 53: Fall 2020, UC Berkeley Lecture 015 Copyright: J.A. Sethian
All rights reserved. You may not distribute/reproduce/display/post/upload any course materials in any way, regardless of whether or not a fee is charged, without my express written consent. You also may not allow anyone else to do so. If you do so, you will be prosecuted under UC Berkeley student proceedings Secs. 102.23 and 102.25 [email protected]
Section 15.10: Change of Variable
Finding the magic factor general: (by the way, there’s nothing magic about it-it’s just really cool)
Step 1: Start with an arbitrary two-dimensional transformation
u
v
x
y S
(u1,v1) R
(x1,y1)
u
v
x=g(u,v)
y=h(u,v)
g(u,v)
h(u,v)
Step 2: Suppose we have We can write this in general as
T(u,v)= (x,y)
Math 53: Fall 2020, UC Berkeley Lecture 015 Copyright: J.A. Sethian
All rights reserved. You may not distribute/reproduce/display/post/upload any course materials in any way, regardless of whether or not a fee is charged, without my express written consent. You also may not allow anyone else to do so. If you do so, you will be prosecuted under UC Berkeley student proceedings Secs. 102.23 and 102.25 [email protected]
Section 15.10: Change of Variable
Example: Consider the transformation: x = u2 – v2, y = 2uv
What happens to the unit square 0≤u≤𝟏, 𝟎≤𝒗≤𝟏, under this transformation?
u
v
x
y R
(x1,y1) T (u1,v1)
What does the output region R look like?
Part 1: I am now going to do a funky example to show you that nice regions can be
transformed intro strange ones.
Math 53: Fall 2020, UC Berkeley Lecture 015 Copyright: J.A. Sethian
All rights reserved. You may not distribute/reproduce/display/post/upload any course materials in any way, regardless of whether or not a fee is charged, without my express written consent. You also may not allow anyone else to do so. If you do so, you will be prosecuted under UC Berkeley student proceedings Secs. 102.23 and 102.25 [email protected]
Example: x = u2 – v2, y = 2uv
u
v
x
y R
(x1,y1) T (u1,v1)
What does the output region R look like?
A
B
C
D
(1,0)
(0,1)
(0,0)
(1,1)
Let’s see what happens to the edges of the unit square
Edge A: For edge A, v=0 and 0≤u≤1, so the transformation is x=u2-v2 which = u2; y=2uv which = 0
So, x=u2 and y= 0 is a line from x=0 to x=1 u x
y
A (1,0) (0,0) (1,0) (0,0) A’
Edge B: For edge B, u=1 and 0≤v≤1, so the transformation is x=u2-v2 which = 1-v2; y=2uv which = 2v
So (y/2)=v, so x = 1-v2=1-(y/2)2=1-y2/4 which is a parabola
x
B
(1,0)
(1,1)
(1,0)
(0,2)
A’
B’
v
u
Edge C: For edge C, v=1 and 1≥u≥0, so the transformation is x=u2-v2 which = u2 -1; y=2uv which = 2u
So (y/2)=u, so x = u2 -1=(y/2)2-1=y2/4 -1 which is a parabola u x
C (1,1) (0,1)
(1,0) (0,0) A’
C’
x
Edge D: For edge D, u=0 and 1≥v≥0, so the transformation is x=u2-v2 which = -v2; y=2uv which = 0
So, x=-v2 and y= 0 is a line from x=-1 to x=0 u x
y
(0,0) (-1,0)
D’
v
D
Math 53: Fall 2020, UC Berkeley Lecture 015 Copyright: J.A. Sethian
All rights reserved. You may not distribute/reproduce/display/post/upload any course materials in any way, regardless of whether or not a fee is charged, without my express written consent. You also may not allow anyone else to do so. If you do so, you will be prosecuted under UC Berkeley student proceedings Secs. 102.23 and 102.25 [email protected]
Example: x = u2 – v2, y = 2uv So we just found that the uv unit square maps into:
u
v
x
y
(x1,y1)
T (u1,v1)
A
B
C
D
(1,0)
(0,1)
(0,0)
(1,1)
A’
B’ C’
D’
So, how do we find the magic factor such that ?
Here goes: Let’s agree that (x0,y0) = T(u0,v0) Standing at input (u0,v0), edges of the box DuDv get sent to different vectors in xy space
T(u,v)
u
v
The output (x,y) comes from an input (u,v). Then any point (x,y) in output space can be written as vector whose components depend on u and v
Our strategy will be to take the two vectors in the u,v plane making up a DuDv box, and see what they get transformed into. We will then take their cross-product to find the area of the transformed box—which is the magic stretch factor…
T(u,v)
u
v
T(bottom side)
T(left side) Area
[x coordinate of vector = g(u,v) y coordinate = h(u,v)]
Math 53: Fall 2020, UC Berkeley Lecture 015 Copyright: J.A. Sethian
All rights reserved. You may not distribute/reproduce/display/post/upload any course materials in any way, regardless of whether or not a fee is charged, without my express written consent. You also may not allow anyone else to do so. If you do so, you will be prosecuted under UC Berkeley student proceedings Secs. 102.23 and 102.25 [email protected]
So, how do we find the magic factor such that ?
Here goes: Let’s agree that (x0,y0) = T(u0,v0) Standing at input (u0,v0), edges of the box DuDv get sent to different vectors in xy space
T(u,v)
u
v
The output (x,y) comes from an input (u,v). Then any point (x,y) in output space can be written as vector whose components depend on u and v
Our strategy will be to take the two vectors in the u,v plane making up a DuDv box, and see what they get transformed into. We will then take their cross-product to find the area of the transformed box—which is the magic stretch factor…
T(u,v)
u
v
T(bottom side)
T(left side) Area
[x coordinate of vector = g(u,v) y coordinate = h(u,v)]
Part 2: I am now going to go back to the main question: and find magic factors in general!
Math 53: Fall 2020, UC Berkeley Lecture 015 Copyright: J.A. Sethian
All rights reserved. You may not distribute/reproduce/display/post/upload any course materials in any way, regardless of whether or not a fee is charged, without my express written consent. You also may not allow anyone else to do so. If you do so, you will be prosecuted under UC Berkeley student proceedings Secs. 102.23 and 102.25 [email protected]
Step 1: Tangent vector holding v fixed, at input (u,v) is
T(u,v)
u
v
T(bottom side)
T(left side) Area
Step 2: Remembering that
u
v
x=g(u,v)
y=h(u,v)
g(u,v)
h(u,v)
Step 3: So, now can find the vectors that make up the bottom and left sides:
bottom side vector =
left side vector =
Step 4: So we can take the cross-product to find the area
Called the Jacobian
Tangent vector holding u fixed, at input (u,v) is
Math 53: Fall 2020, UC Berkeley Lecture 015 Copyright: J.A. Sethian
All rights reserved. You may not distribute/reproduce/display/post/upload any course materials in any way, regardless of whether or not a fee is charged, without my express written consent. You also may not allow anyone else to do so. If you do so, you will be prosecuted under UC Berkeley student proceedings Secs. 102.23 and 102.25 [email protected]
u
v
x
y
f(u,v)
g(u,v)
So now have the result:
u
v
x
y S
(u1,v1) R
(x1,y1)
Let’s use it right away: Prove that the Jacobian for the polar transformation x=r cos q, y=r sin q is rdrdq
r
q
x=r cos q
y=r sin q
Jacobian is
[R plays the role of u, and q plays the role of v]
Do you remember that for polar coordinates, we had
Math 53: Fall 2020, UC Berkeley Lecture 015 Copyright: J.A. Sethian
All rights reserved. You may not distribute/reproduce/display/post/upload any course materials in any way, regardless of whether or not a fee is charged, without my express written consent. You also may not allow anyone else to do so. If you do so, you will be prosecuted under UC Berkeley student proceedings Secs. 102.23 and 102.25 [email protected]
u
v
x
y
f(u,v)
g(u,v)
Let’s do another example:
u
v
x
y S
(u1,v1) R
(x1,y1)
Using the change of variables x = u2 – v2, y = 2uv to evaluate where R is the region bounded by the x-axis
and the parabolas y2=4-4x,x≥0 and y2=4-4x, x≤0 x
y
Solution: The Jacobian for this mapping is
..I’ll let you finish it…..
Math 53: Fall 2020, UC Berkeley Lecture 015 Copyright: J.A. Sethian
All rights reserved. You may not distribute/reproduce/display/post/upload any course materials in any way, regardless of whether or not a fee is charged, without my express written consent. You also may not allow anyone else to do so. If you do so, you will be prosecuted under UC Berkeley student proceedings Secs. 102.23 and 102.25 [email protected]
And now you see how to find the magic factor!!!
x (u,v,w)
y (u,v,w)
z (u,v,w)
u
w z
x
y v
dx dy dz= [magic factor] du dv dw
Form the matrix of partials
Then the magic factor = | det A |
Let’s check it for things we know!
Cylindrical coordinates: x (r, q,z)
y (r, q,z)
z (r, q,z)
r
q
z z=z
x=r cos q
y=r sin q
det A= r cos q cos q – ( - r sin q sin q ) = r !!!
dx dy dz = r dr dq dz Cartesian element
Cylindrical element
Geometrically, we got
Math 53: Fall 2020, UC Berkeley Lecture 015 Copyright: J.A. Sethian
All rights reserved. You may not distribute/reproduce/display/post/upload any course materials in any way, regardless of whether or not a fee is charged, without my express written consent. You also may not allow anyone else to do so. If you do so, you will be prosecuted under UC Berkeley student proceedings Secs. 102.23 and 102.25 [email protected]
Spherical coordinates: x (r, q,z)
y (r, q,z)
z (r, q,z)
r
q
z z= r cos f
x=r sin f cos q
y=r sin f sin q
det A= r2 sin f!!!
dx dy dz = r2 sin f dr dq df And geometrically the book got
Here goes!
I leave it to you to check that:
And remembering that if then det A= a (ei-fh) – b (di-fg) + c (dh-eg)