18.085 PROBLEM SET 9 SOLUTIONS
HAOFEI WEI
Problem (3.6.3).
Solution. In 3D, the natural analogue of a triangle is the tetrahedron, which is composed of
4 equilateral triangles. So, for a tetrahedron with standard shape, the four corners must be
at (0, 0, 0), (h, 0, 0), (0, h, 0), and (0, 0, h).
The last step is simply an interpolation. We know that U = a + bx + cy + dz inside the
tetrahedron. We also know from the four corners that
U (0, 0, 0)
U (1, 0, 0)
U (0, 1, 0)
U (0, 0, 1)
=
=
=
=
a = U0
a + bh = U1
a + ch = U2
a + dh = U3
By inspection, we see that a = U0 , b = h1 (U1 − U0 ), c = h1 (U2 − U0 ), d = h1 (U3 − U0 ).
0
0
0
Thus, U = U0 + U1 −U
x + U2 −U
y + U3 −U
z.
h
h
h
Problem (3.6.4).
Solution.
(1) First,we find that

U1 U2
where ci =
U1 U2

(c
+
c
)U
−
c
U
−
c
U
2
3
1
3
2
2
3
U3 Ke = −c3 U1 + (c1 + c3 )U2 − c1 U3 
−c2 U1 − c1 U2 + (c1 + c2 )U3
1
.
2 tan θi
 
U1

Multiplying this with U2 , we get
U3
 
U1

U3 Ke U2  = U12 (c2 + c3 ) + U22 (c1 + c3 ) + U32 (c1 + c2 ) − 2c3 U1 U2 − 2c2 U1 U3 − 2c1 U2 U3
U3
= c1 (U22 + U32 − 2U2 U3 ) + c2 (U12 + U32 − 2U1 U3 ) + c3 (U12 + U22 − U1 U2 )
1 (U2 − U3 )2 (U1 − U3 )2 (U1 − U2 )2
+
+
=
2
tan θ1
tan θ2
tan θ3
as needed.
√
(2) tan 60◦ = 3, so c1 = c2 = c3 =
1
√
.
2 3
Thus,


2 −1 −1
1
Ke = √ −1 2 −1
2 3 −1 −1 2
1
HAOFEI WEI
18.085 PROBLEM SET 9 SOLUTIONS
(3) To construct the matrix, we must add up the elements in each of the six Ke ’s which
share overlapping vertices. For simplicity, we will assume i = 1. The (1, 1) element
of K will be the sum of all the matrix elements of Ke which contribute to U1 Ke U1 .
Since all the ci ’s are identially 2 tan1 60◦ = 2√1 3 , we know that the elements of each Ke
on the diagonal is ci + cj = √13 , and so the sum of this from all six Ke ’s will be √63 .
For all the elements of K not corresponding to U1 Ke Ui6=1 , the summand will be
− 2√1 3 , and there will be only two matrices which contribute elements to U1 Ke Ui6=1 ,
so the element of K will be − √13 .
Combining this, we will have the 1st row of K
1 √ 0 . . . 6 −1 −1 −1 −1 −1 −1 . . .
3
For the ith row of K, the row will have i − 1 zeroes before the 6.
Problem (3.6.7).
Solution. Let U1 be the U of the lower triangle, and U2 be the U of the upper triangle.
Using the condition that U (x, y) = 0 at all nodes besides ( 12 , 12 ), we can find the following
conditions on U1 :
U1 (0, 0)
1
U1 ( , 0)
2
U1 (1, 0)
1
U1 (1, )
2
U1 (1, 1)
= a=0
b d
= a+ + =0
2 4
= a+b+d=0
c e f
= a+b+d+ + + =0
2 2 4
= a+b+c+d+e+f =0
2
HAOFEI WEI
18.085 PROBLEM SET 9 SOLUTIONS
By inspection, we see that a = 0. Since b + d = 2b + d4 , this can only be true for b = d = 0.
Next, we see that c + e + f = 2c + 2e + f4 . This can only be true when f = c + e = 0. Finally,
we know that U1 ( 12 , 12 ) = c( 21 − 41 ) = 1, so c = −e = 4. Thus, U1 = 4y − 4xy.
We see that U2 is simply U1 with x ←→ y, since U2 would be U1 if we interchanged the x
and y axes, so U2 = 4x − 4xy.
3