Areg Hayrapetian (#53)
Problem Set #6
18.085
3.3: 6, 8, 10, 16, 27
6) v1 = (y 2 , x2 ) is not a gradient of any function because the curl of v1 can be shown to be 0. ∇ × v1 =
∂(y 2 )
∂y
∇×
∂(x2 )
∂x = 2y − 2x 6= 0 for all x, y. However, v2
2
)
∂(2xy)
v2 = ∂(y
= 2y − 2y = 0 for all x, y.
∂y −
∂x
−
= (y 2 , 2xy) is a gradient of the function u(x, y) because
∂u
= y2
∂x
u(x, y) = xy 2 + f1 (y)
∂u
= 2xy
∂y
u(x, y) = xy 2 + f2 (x)
So, f1 (y) = f2 (x) = C, but the constant C can be arbitrary set. By letting C = 0, we find u(x, y) = xy 2 .
8) w = ∇u with u(x, y) = x2 in the square S = −1 < x, y < 1. w = ∇u = (2x, 0), and ∇ · w = 2.
ZZ
Z
Divergence Theorem:
∇ · (∇u) dx dy =
n · ∇u ds
S
C
ZZ
Z
1
Z
1
∇ · (∇u) dx dy =
Z
Z
1
n · ∇u ds =
Z
1
ex · (2, 0) dy +
−1
Z 1
C
=
2 dx dy = 8
−1
S
−1
Z
ey · (2x, 0) dx +
−1
Z
Z
1
(−ex ) · (−2, 0) dy +
−1
(−ey ) · (2x, 0) dx
−1
1
2 dy + 0 +
−1
1
2 dy + 0 = 4 + 4 = 8
−1
10) v = (2xy, x2 − y 2 ) = ∇u.
∂u
∂s
=
= 2xy
∂y
∂x
u(x, y) = x2 y + f1 (y)
s(x, y) = xy 2 + f3 (x)
−
∂s
∂u
=
= x2 − y 2
∂x
∂y
1
u(x, y) = x2 y − y 3 + f2 (x)
3
1
s(x, y) = xy 2 − x3 + f4 (y)
3
Equating the two equations for u(x, y) and s(x, y), and letting arbitrary constants be 0, we find f1 (y) = − 31 y 3 ,
1
1
f2 (x) = 0, f3 (x) = − 13 x3 , and f4 (x) = 0. So, u(x, y) = x2 y − y 3 and s(x, y) = xy 2 − x3 . The
3
3
equipotentials (u(x, y) = c) and streamlines (s(x, y) = c) are shown in the figure below for flow into a 30◦
√
wedge. To verify that flow does not pass through the upper boundary, we can show that v ·n = 0 for y = x/ 3.
√
√
√ 2
◦
◦
2
On the upper boundary, n = (− sin(30 ), cos(30 )) = (−1/2, 3/2), and v = (2x(x/ 3), x − (x/ 3) ) =
√
x2 (2/ 3, 2/3).
√
1
2
3 22
v · n = − x2 √ +
x
2
2
3
3
2
2
x
x
= −√ + √ = 0 X
3
3
11/3/10
Page 1 of 10
Areg Hayrapetian (#53)
Problem Set #6
18.085
°
Equipotentials and Streamlines for Flow into a 30 Wedge
3
Streamline Curves
Equipotential Curves
Boundary
2.5
y
2
1.5
1
0.5
0
0
0.5
1
1.5
2
2.5
x
3
3.5
4
4.5
5
16) v(x, y, z) is a rotation field given by the cross product of axis A = (a1 , a2 , a3 ) with R = (x, y, z),
v = A × R = (v1 , v2 , v3 ) = (a2 z − a3 y, a3 x − a1 z, a1 y − a2 x).
∂v3
∂v2 ∂v1
∂v3 ∂v2
∂v1
∇×v =
−
,
−
,
−
∂y
∂z ∂z
∂x ∂x
∂y
= (a1 + a1 , a2 + a2 , a3 + a3 ) = 2(a1 , a2 , a3 ) = 2A
∂v1
∂v2
∂v3
+
+
=0+0+0=0
∂x
∂y
∂z
v cannot be a gradient field, because if it were, it would contradict with the curl calculated above: ∇ × v =
∇ × (∇u) = 0 6= 2A, using the vector identity that the curl of a gradient is 0. However, v is a vorticity field
because the flow is solenoidal (∇ · v = 0). The stream function S = (s1 (x, y, z), s2 (x, y, z), s3 (x, y, z)) can be
calculated. We can try the following equations and check to see if they satisfy ∇ × S = v:
∇·v =
s1 = a2 xy + a3 xz
s2 = a1 xy + a3 yz
s3 = a1 xz + a2 yz
∂s3
∂s2
−
= a2 z − a3 y
∂y
∂z
∂s1
∂s3
−
= a3 x − a1 z
∂z
∂x
∂s2
∂s1
−
= a1 y − a2 x
∂x
∂y
X
X
X
So, S = (a2 xy + a3 xz, a1 xy + a3 yz, a1 xz + a2 yz) . If a1 and a2 were selected to be 0, the rotation axis A
would equal (0, 0, a3 ), and the rotation would only occur in the x − y plane. Substituting a1 = a2 = 0 into
the equation for v, it can be seen that the z-component of the velocity would disappear, and so v would be
a plane field.
27) We would like to show that M 2 is a diagonal matrix equal to −∇2 I, where ∇2 = ∂ 2/∂x2 + ∂ 2/∂y2 + ∂ 2/∂z2
and M is given by


0
−∂/∂z ∂/∂y ∂/∂x
−1
 ∂/∂z
C
A
curl grad
0
−∂/∂x ∂/∂y 

M=
=
=
∂/∂z 
−∂/∂y ∂/∂x
−div
0
0
AT
0
−∂/∂x −∂/∂y −∂/∂z
0
11/3/10
Page 2 of 10
Areg Hayrapetian (#53)
Problem Set #6
−1
−1 −1
C
A C −1 A
(C C + AAT ) C −1 A
M2 =
=
AT
0
AT
0
AT C −1
0



∂
∂
∂
∂
0
− /∂z ∂/∂y ∂/∂x
0
− /∂z
/∂y
/∂x
 ∂
 ∂/∂z
0
−∂/∂x ∂/∂y 
0
−∂/∂x ∂/∂y 

  /∂z
=
∂
∂/∂z 
∂
∂


−∂/∂y ∂/∂x
− /∂y
/∂x
0
0
/∂z
−∂/∂x −∂/∂y −∂/∂z
0
−∂/∂x −∂/∂y −∂/∂z
0
2

2

2
2
∂
∂
∂
∂
∂2
∂2
−∇2
− ∂x∂y
− ∂x∂z
− ∂y∂z
+ ∂y∂z
∂x∂y
∂x∂z
 2
2
2



∂
∂2
∂
∂2
∂
∂2
− ∂x∂y
−∇2
−
−
 ∂x∂y
∂y∂z
∂y∂z
∂x∂z
∂x∂z 
 = −∇2 I
2
=


∂2
∂2
∂2
∂2
∂2
∂
2
−∇
− ∂x∂y + ∂x∂y 
 ∂x∂z − ∂x∂z
∂y∂z − ∂y∂z 

∂2
∂2
∂2
∂2
∂2
∂2
2
− ∂y∂z
+ ∂y∂z
−
−
+
−∇
∂x∂z
∂x∂z
∂x∂y
∂x∂y
18.085
From the above calculation, we can see three useful identities. First, the 3 × 1 vector C −1 A equals 0, which
means that the curl of a gradient is 0. Second, the 1 × 3 vector AT C −1 equals 0, which means that the
divergence of a curl is 0. Finally, the 3 × 3 matrix (C −1 C −1 + AAT ), which is the curl of a curl added with
the gradient of a divergence, is equal to −∇2 I.
3.4: 2, 4, 5, 6, 17, and Extra Problem
2) We want to solve Poisson’s equation uxx + uyy = 4 with the boundary condition u = 0 on the circle
x2 +y 2 = 1. Trying the function u(x, y) = x2 +y 2 −1, we find that it satisfies Poisson’s equation, uxx +uyy =
2 + 2 = 4, and it satisfies the boundary condition, on the circle u = 1 − 1 = 0. So, u(x, y) = x2 + y 2 − 1 .
4) u = r cos θ + r−1 cos θ solves Laplace’s equation
∂2u
∂r 2
+
1 ∂u
r ∂r
∂2u
2
= 3 cos θ
∂r2
r
1
∂u
= cos θ − 2 cos θ
∂r
r
+
1 ∂2u
r 2 ∂θ 2
= 0, which is verified below.
∂2u
1
= −r cos θ − cos θ
∂θ2
r
∂ 2 u 1 ∂u
1 ∂2u
2
1
1
1
1
+
+
= 3 cos θ + cos θ − 3 cos θ − cos θ − 3 cos θ = 0
2
2
2
∂r
r ∂r
r ∂θ
r
r
r
r
r
u can be expressed in terms of x and y through the change in coordinates r2 = x2 + y 2 and r cos θ = x.
u = r cos θ + r−1 cos θ = r cos θ +
r cos θ
x
= x+ 2
= u(x, y)
r2
x + y2
Then, v = (ux , uy ) is
y 2 − x2
∂u
1(x2 + y 2 ) − x(2x)
=1+
=
1
+
∂x
(x2 + y 2 )2
(x2 + y 2 )2
∂u
0 − x(2y)
2xy
uy =
= 2
=− 2
∂y
(x + y 2 )2
(x + y 2 )2
ux =
v=
2xy
(y 2 − x2 )
1+ 2
,
−
(x + y 2 )2
(x2 + y 2 )2
This is the velocity field for flow past a unit circle x2 + y 2 = 1. Therefore, the flow into or out of the circle
should be 0, or v p
· n = 0 on the circle. This can be verified by first finding the normal vector on the unit
circle, n = (x, y)/ x2 + y 2 , and taking the dot product v · n given that x2 + y 2 = 1 since it is evaluated on
the circle:
1
x(y 2 − x2 )
2xy 2
v·n= p
x+ 2
− 2
= x + x(y 2 − x2 ) − 2xy 2
(x + y 2 )2
(x + y 2 )2
x2 + y 2
= x + xy 2 − x3 − 2xy 2 = x − x3 − xy 2 = x − x(x2 + y 2 ) = x − x = 0
11/3/10
Page 3 of 10
Areg Hayrapetian (#53)
Problem Set #6
18.085
5) u = ln(r) and U = ln(r2 ) both satisfy Laplace’s equation. In fact, U = ln(r2 ) = 2 ln(r) = 2u. So, it is
enough to show that u satisfies Laplace’s equation, because then U automatically will as well. This is shown
below.
∂ 1
1
∂ 2 u 1 ∂u
1 ∂2u
1 1
1
+
=
+
+
+0=− 2 + 2 =0
∂r2
r ∂r
r2 ∂θ2
∂r r
r r
r
r
u and U would have been obtained by selecting the complex functions f (z) = ln(z) = u + is and F (z) =
ln(z 2 ) = U + iS where z = reiθ . f (z) = u + is = (ln(r)) + i(θ) and F (z) = U + iS = (ln(r2 )) + i(2θ) .
The Cauchy-Riemann equations require that ∂U/∂x = ∂S/∂y and ∂U/∂y = −∂S/∂x. This is verified below.
U (x, y) = ln(x2 + y 2 )
S(x, y) = 2 tan−1
y
x
∂U
2x
= 2
∂x
(x + y 2 )
y 2 −1 1 ∂S
=2 1+
∂y
x
x
2x
= 2
(x + y 2 )
∂U
2y
= 2
∂y
(x + y 2 )
y 2 −1 y ∂S
− 2
=2 1+
∂x
x
x
2y
=− 2
(x + y 2 )
6) For each r, the points Z, defined below, lie on an ellipse.
Z≡
1
1
(z + z −1 ) = (reiθ + r−1 e−iθ ) =
2
2
r + r−1
2
cos θ + i
r − r−1
2
4X 2
4Y 2
+
= cos2 θ + sin2 θ = 1
−1
2
(r + r )
(r − r−1 )2
sin θ = X + iY
(Equation of Ellipse)
We would like to see why the circles |z| = R and |z| = R−1 produce the same ellipse when mapped to Z.
For |z| = R, r = R in the expression z = reiθ , and so we must substitute r = R into the general equation of
the ellipse above to get the equation of the ellipse corresponding to this particular circle:
With r = R:
4X 2
4Y 2
+
=1
−1
2
(r + r )
(r − r−1 )2
−−−−−→
4X 2
4Y 2
+
=1
−1
2
(R + R )
(R − R−1 )2
For |z| = R−1 , r = R−1 in the expression z = reiθ , and so we must substitute r = R−1 into the general
equation of the ellipse above to get the equation of the ellipse corresponding to this particular circle:
With r = R−1 :
4X 2
4Y 2
+
=1
−1
2
(r + r )
(r − r−1 )2
−−−−−→
−−−−−→
4X 2
4Y 2
+ −1
=1
2
+ R)
(R − R)2
4X 2
4Y 2
+
=1
−1
2
(R + R )
(R − R−1 )2
(R−1
The resulting equations for the ellipses for both cases are identical, therefore the circles |z| = R and |z| = R−1
both produce the same ellipse when mapped to Z.
17) We want to verify that uk (x, y) = sin(πkx) sinh(πky)/ sinh(πk) for k = 1, 2, ... solves Laplace’s equation
and satisfies the boundary conditions on a unit square u(x, 1) = sin(πkx) and u(x, 0) = u(0, y) = u(1, y) = 0.
First, we check if uk (x, y) satisfies Laplace’s equation:
∂ 2 uk
sinh(πky)
=
−π 2 k 2 sin(πkx)
2
∂x
sinh(πk)
∂ 2 uk
sin(πkx) 2 2
=
π k sinh(πkx)
2
∂y
sinh(πk)
∂ 2 uk
∂ 2 uk
π2 k2
+
=
[− sin(πkx) sinh(πky) + sin(πkx) sinh(πkx)] = 0
∂x2
∂y 2
sinh(πk)
11/3/10
X
Page 4 of 10
Areg Hayrapetian (#53)
Problem Set #6
18.085
Then, we check if uk (x, y) satisfies the boundary conditions:
sin(0) sinh(πky)
=0X
sinh(πk)
sin(πkx) sinh(0)
uk (x, 0) =
=0X
sinh(πk)
uk (0, y) =
sin(πk) sinh(πky)
=0X
sinh(πk)
sin(πkx)sinh(πk)
uk (x, 1) =
= sin(πkx)
sinh(πk)
uk (1, y) =
X
Extra Problem) Solutions to Laplace’s Equation ∇2 Ψ = 0 come in pairs ui and si . Four of these pairs are
listed below for both 2D Cartesian coordinates and 2D Polar coordinates.
2
2
In 2D Cartesian coordinates, ui (x, y) and si (x, y) both satisfying ∂∂xΨ2 + ∂∂yΨ2 = 0:
u1 (x, y) = x
s1 (x, y) = y
2
u2 (x, y) = x − y
3
2
u3 (x, y) = x − 3xy
s2 (x, y) = 2xy
2
s3 (x, y) = 3x2 y − y 3
u4 (x, y) = x4 − 6x2 y 2 + y 4
In 2D Polar coordinates, ui (r, θ) and si (r, θ) both satisfying
11/3/10
s4 (x, y) = 4x3 y − 4xy 3
∂2Ψ
∂r 2
+
1 ∂Ψ
r ∂r
+
1 ∂2Ψ
r 2 ∂θ 2
= 0:
u1 (r, θ) = r cos(θ)
s1 (r, θ) = r sin(θ)
u2 (r, θ) = r2 cos(2θ)
s2 (r, θ) = r2 sin(2θ)
u3 (r, θ) = r3 cos(3θ)
s3 (r, θ) = r3 sin(3θ)
u4 (r, θ) = r4 cos(4θ)
s4 (r, θ) = r4 sin(4θ)
Page 5 of 10
Areg Hayrapetian (#53)
Problem Set #6
18.085
MATLAB Assignment
2
2
We would like to numerically solve Laplace’s equation ∂∂xu2 + ∂∂yu2 = 0 with the boundary conditions u(1, y) = y,
u(x, 1) = x, (w · n)|x=0 = −y, and (w · n)|y=0 = −x. Taking w to be w = ∇u = (∂u/∂x, ∂u/∂y), and n to be
the outward facing normal, the last two boundary conditions can be rewritten in a simpler form. At x = 0,
the normal vector is n = −ex , and the boundary condition can be written as ∂u
∂x x=0 = y. At y = 0, the
normal vector is n = −ey , and the boundary condition can be written as ∂u
∂y y=0 = x.
The MATLAB function laplace was written to numerically solve Laplace’s equation on an N by N 2D
grid, bounded by 0 < x < 1 and 0 < y < 1, using arbitrarily specified boundary conditions. The code is
displayed in Listing 1, and includes comments describing how to use the function. We will use this function
1
, and the boundary conditions
with a grid size of N = 10, meaning a spacing between nodes of h = N1+1 = 11
2
described in the above paragraph. The code generates two matrices K (N × N 2 ) and F (N 2 × 1) which are
used as part of the matrix equation KU = F that solves for U containing the values of u(x, y) at the nodes.
The N 2 × 1 vector U is ordered to go across the columns of a particular row first, before moving on to the
next row. The first component of the vector is the value for u(h, h) and the last component of the vector is
the value for u(N h, N h). Although the right hand side of the Laplace equation is 0, the non-zero vector F
on the right hand side of this matrix equation is a result of the imposed boundary conditions, which change
the matrix K and cause terms to be moved to the right hand side.
To get the matrices K and F we call the laplace function in MATLAB as follows:
>> [K, F] = laplace(10, 1, @(y) y, 0, @(y) y, 1, @(x) x, 0, @(x) x);
We are interested in learning more about the matrix K. First, we can check if it is symmetric. We can
use the MATLAB command max(max(abs(A))) to see what the component in matrix A with the maximum
magnitude is. If the component with maximum magnitude in a matrix is 0, then that means the whole
matrix contains only zeros.
>> max(max(abs(K - K’)))
ans =
0
Since, K − K T = 0, K must be symmetric.
Now, we would like to know whether K is positive definite. A quick way to check this would be to find
the pivots of K and see what their signs are. The command below LU factorizes K and then grabs the
diagonal of the upper triangular matrix U which contains the pivots, and then checks to see if there is any
pivot that is non-negative.
>> [L,U] = lu(K); d = diag(U);
>> find(d >= 0, length(d))
ans =
Empty matrix: 0-by-1
The answer is the empty matrix, which means that there are no non-negative pivots. This means that all
the pivots are negative, and thus the matrix K is actually negative definite.
The main diagonal of the matrix K is printed below:
>> diag(K)
ans =
(-2, -3, -3, -3, -3, -3, -3, -3, -3, -3, -3, -4, -4, -4, -4, -4, -4, -4, -4, -4, -3, -4, -4, -4, -4, -4, -4, -4, -4, -4, -3, -4, -4,
-4, -4, -4, -4, -4, -4, -4, -3, -4, -4, -4, -4, -4, -4, -4, -4, -4, -3, -4, -4, -4, -4, -4, -4, -4, -4, -4, -3, -4, -4, -4, -4, -4,
-4, -4, -4, -4, -3, -4, -4, -4, -4, -4, -4, -4, -4, -4, -3, -4, -4, -4, -4, -4, -4, -4, -4, -4, -3, -4, -4, -4, -4, -4, -4, -4, -4, -4 )
11/3/10
Page 6 of 10
Areg Hayrapetian (#53)
Problem Set #6
18.085
K is banded around the main diagonal, and we would like to find the bandwidth w, which is defined as
the minimum w such that Kij = 0 if |i − j| > w. We can find w using the simple script below.
>> Kmod = K - diag(diag(K));
>> for i = 1:50
if max(max(abs(Kmod))) == 0
break;
end
Kmod = Kmod - diag(diag(K,i),i);
Kmod = Kmod - diag(diag(K,-i),-i);
end
>> w = i
w =
11
The script starts from the main diagonal and begins zeroing out the diagonals of the matrix K as it expands
outward from the main diagonal. At each iteration it checks to see if the matrix is equal to the zero matrix
yet. As soon as the matrix is 0, it breaks out of the loop and its index i is equal to the bandwidth w. The
script calculated that the bandwidth of the matrix K is w = 11.
We now want to find the largest and smallest eigenvalues of K. Note that since K is negative definite,
the smallest eigenvalue will be a negative number with the largest magnitude, and the largest eigenvalue will
be a negative number with the smallest magnitude. The eigenvalues can be found by using the MATLAB
command eig to do an eigen decomposition of K, and then extracting the largest and smallest eigenvalues
from the diagonal eigenvalue matrix.
>> [V, E] = eig(K);
>> min(diag(E))
ans =
-7.8223
>> max(diag(E))
ans =
-0.0447
So, the minimum eigenvalue of K is equal to -7.8223, and the maximum eigenvalue of K is -0.0447.
We can now take the inverse of the K matrix and print out the first column v of K −1 .
>> Kinv = inv(K);
>> v = Kinv(:,1)
v =
(-1.75563, -1.25563, -0.95319, -0.74342, -0.58399, -0.45499, -0.34567, -0.24948,
-1.05807, -0.86051, -0.69309, -0.55356, -0.43531, -0.33255, -0.24082, -0.15664,
-0.73769, -0.61487, -0.50185, -0.40013, -0.30840, -0.22461, -0.14661, -0.07238,
-0.52686, -0.43884, -0.35496, -0.27633, -0.20260, -0.13281, -0.06573, -0.58399,
-0.37169, -0.30454, -0.23934, -0.17666, -0.11632, -0.05771, -0.45499, -0.43531,
-0.25218, -0.19982, -0.14838, -0.09810, -0.04878, -0.34567, -0.33255, -0.30840,
-0.15939, -0.11895, -0.07891, -0.03932, -0.24948, -0.24082, -0.22461, -0.20260,
-0.08912, -0.05928, -0.02959, -0.16194, -0.15664, -0.14661, -0.13281, -0.11632,
-0.03951, -0.01974, -0.07971, -0.07718, -0.07238, -0.06573, -0.05771, -0.04878,
-0.00987 )
-0.16194,
-0.07718,
-0.74342,
-0.55356,
-0.40013,
-0.27633,
-0.17666,
-0.09810,
-0.03932,
-0.07971,
-0.95319,
-0.69309,
-0.50185,
-0.35496,
-0.23934,
-0.14838,
-0.07891,
-0.02959,
-1.25563,
-0.86051,
-0.61487,
-0.43884,
-0.30454,
-0.19982,
-0.11895,
-0.05928,
-0.01974,
The ratio of the first component to the last component, which is calculated using v(1)/v(100), is about
177.8556.
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Areg Hayrapetian (#53)
Problem Set #6
18.085
Finally, we can solve the matrix equation KU = F and print out the solution vector U .
>> U = K\F
U =
(0.00826, 0.01653, 0.02479, 0.03306, 0.04132, 0.04959, 0.05785, 0.06612, 0.07438, 0.08264,
0.04959, 0.06612, 0.08264, 0.09917, 0.11570, 0.13223, 0.14876, 0.16529, 0.02479, 0.04959,
0.12397, 0.14876, 0.17355, 0.19835, 0.22314, 0.24793, 0.03306, 0.06612, 0.09917, 0.13223,
0.23140, 0.26446, 0.29752, 0.33058, 0.04132, 0.08264, 0.12397, 0.16529, 0.20661, 0.24793,
0.37190, 0.41322, 0.04959, 0.09917, 0.14876, 0.19835, 0.24793, 0.29752, 0.34711, 0.39669,
0.05785, 0.11570, 0.17355, 0.23140, 0.28926, 0.34711, 0.40496, 0.46281, 0.52066, 0.57851,
0.19835, 0.26446, 0.33058, 0.39669, 0.46281, 0.52893, 0.59504, 0.66116, 0.07438, 0.14876,
0.37190, 0.44628, 0.52066, 0.59504, 0.66942, 0.74380, 0.08264, 0.16529, 0.24793, 0.33058,
0.57851, 0.66116, 0.74380, 0.82645 )
0.01653,
0.07438,
0.16529,
0.28926,
0.44628,
0.06612,
0.22314,
0.41322,
0.03306,
0.09917,
0.19835,
0.33058,
0.49587,
0.13223,
0.29752,
0.49587,
The ratio of the last component of U , the value of the numerical solution at u(10h, 10h), to the first component
of U , the value of the numerical solution at u(h, h), can be calculated using U(100)/U(1), and it is equal to
100.
Listing 1: MATLAB function that creates the matrices required to numerically solve 2D Laplace’s Equation.
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function [ K, F ] = l a p l a c e ( N, B x0 type , B x0 func , . . .
B x1 type , B x1 func , . . .
B y0 type , B y0 func , . . .
B y1 type , B y 1 f u n c )
%LAPLACE
G e n e r a t e s m a t r i c e s K and F t h a t w i l l s o l v e Laplace ’ s E q u a t i o n
%
[ K, F ] = l a p l a c e ( N, B x 0 t y p e , B x 0 f u n c , . . . % x=0 boundary
%
B x 1 t y p e , B x 1 f u n c , . . . % x=1 boundary
%
B y 0 t y p e , B y 0 f u nc , . . . % y=0 boundary
%
B y1 type , B y1 func )
% y=1 boundary
%
g e n e r a t e s t h e m a t r i c e s K and F , t h a t s o l v e 2D Laplace ’ s E q u a t i o n w i t h
%
m a t r i x e q u a t i o n KU=F , where U would c o n t a i n t h e v a l u e s o f u a t a l l t h e
%
nodes i n t h e 2D g r i d . K i s an Nˆ2 by Nˆ2 m a t r i x and F and U a r e b o t h
%
Nˆ2 by 1 v e c t o r s . The e l e m e n t s i n U a r e o r d e r e d such t h a t a l l t h e node
%
v a l u e s i n a row y = h i appear i n o r d e r o f columns x = h j b e f o r e t h e node
%
v a l u e s i n t h e n e x t row y = h ( i +1) appear . The u s e r must s p e c i f y t h e
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s i z e o f t h e g r i d N which d e t e r m i n e s t h e s p a c i n g h = 1/(N+1) b e t w e e n each
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node i n t h e 2D g r i d r a n g i n g from 0 < x < 1 and 0 < y < 1 . The u s e r must
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a l s o s p e c i f y t h e boundary c o n d i t i o n s on each o f t h e 4 e d g e s o f t h e g r i d .
%
Arguments B ( a ) (m) t y p e s p e c i f y t h e t y p e o f t h e boundary c o n d i t i o n
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on t h e boundary a ( e i t h e r x or y ) = m ( e i t h e r 0 or 1 ) , and t h e v a l u e o f
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t h i s argument must e i t h e r be 0 ( f o r a D i r i c h l e t boundary c o n d i t i o n ) or
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1 ( f o r a Neumann boundary c o n d i t i o n ) . B ( a ) (m) f u n c i s a f u n c t i o n t h a t
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t a k e s one v a r i a b l e , t h e l o c a t i o n i n t h e d i r e c t i o n o f t h e a x i s b t h a t i s
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not a , and s p e c i f i e s t h e r i g h t hand s i d e o f t h e boundary c o n d i t i o n f o r
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t h e c o r r e s p o n d i n g boundary . For D i r i c h l e t boundary c o n d i t i o n s ,
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u ( a t a = m) = B ( a ) (m) f u n c ( b ) . For Neumann boundary c o n d i t i o n s ,
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du/ da ( a t a = m) = B ( a ) (m) f u n c ( b ) .
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Example : S o l v i n g Laplace ’ s e q u a t i o n w i t h t h e boundary c o n d i t i o n s
%
u ( 0 , y ) = u ( 1 , y ) = 0 , u ( x , 0 ) = x , du/ dy ( x , 1 ) = 1−x .
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[K, F ] = l a p l a c e ( 1 0 , 0 , @( y ) 0 , 0 , @( y ) 0 , . . .
%
0 , @( x ) x , 1 , @( x ) 1−x ) ;
%
U = K\F ;
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Areg Hayrapetian (#53)
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Problem Set #6
18.085
% Get i n d e x from row i , column j
indx = @( i , j ) N* ( i −1) + j ;
% S p a c i n g b e t w e e n nodes i n 2D g r i d
h = 1 / (N+1);
% I n i t i a l i z e vectors
F = zeros (N*N, 1 ) ;
F1 = zeros (N*N, 1 ) ;
F2 = zeros (N*N, 1 ) ;
% I n i t i a l l y s e t up K m a t r i x
K = −4* eye (N*N ) ;
f o r i = 1 :N
f o r j = 1 :N
m = indx ( i , j ) ;
if i > 1
K(m, indx ( i −1, j ) ) =
end
if i < N
K(m, indx ( i +1, j ) ) =
end
if j > 1
K(m, indx ( i , j −1)) =
end
if j < N
K(m, indx ( i , j +1)) =
end
end
end
1;
1;
1;
1;
% P r o c e s s x=0 boundary
f o r i = 1 :N
m = indx ( i , 1 ) ;
i f B x 0 t y p e == 1
K(m,m) = K(m,m) + 1 ;
F1 (m) = F1 (m) + B x 0 f u n c ( h * i ) ;
else
F2 (m) = F2 (m) − B x 0 f u n c ( h * i ) ;
end
end
% P r o c e s s x=1 boundary
f o r i = 1 :N
m = indx ( i ,N ) ;
i f B x 1 t y p e == 1
K(m,m) = K(m,m) + 1 ;
F1 (m) = F1 (m) − B x 1 f u n c ( h * i ) ;
else
F2 (m) = F2 (m) − B x 1 f u n c ( h * i ) ;
end
end
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Areg Hayrapetian (#53)
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Problem Set #6
18.085
% P r o c e s s y=0 boundary
f o r j = 1 :N
m = indx ( 1 , j ) ;
i f B y 0 t y p e == 1
K(m,m) = K(m,m) + 1 ;
F1 (m) = F1 (m) + B y 0 f u n c ( h * j ) ;
else
F2 (m) = F2 (m) − B y 0 f u n c ( h * j ) ;
end
end
% P r o c e s s y=1 boundary
f o r j = 1 :N
m = indx (N, j ) ;
i f B y 1 t y p e == 1
K(m,m) = K(m,m) + 1 ;
F1 (m) = F1 (m) − B y 1 f u n c ( h * j ) ;
else
F2 (m) = F2 (m) − B y 1 f u n c ( h * j ) ;
end
end
% F i n a l i z e v e c t o r s and o u t p u t v a r i a b l e s
F1 = F1 * h ;
F2 = F2 ;
F = F1 + F2 ;
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