Algorithms for Numerical Solution of the Goursat
Problem on a Triangular Domain with Mixed Boundary
Conditions
Kun Gou
Michigan State University
Great Lakes SIAM 2013 Conference
April 20, 2013
Motivation
1
2
My current research: new mathematical models for the large strain
swelling response of biological tissues supported by Qatar National
Research Fund (QNRF)
In bio-medical engineering, spectral techniques are used to recover the
material parameters like shear modulus, residual stress etc. This may
generate a Sturn-liouville problem with known eigenvalues
u 00 (x) + v (x)u(x) = ωu(x),
3
from which we need to recover the half-known coefficient function v (x).
The algorithm of such inverse problem produces the so-called Goursat
problem.
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Content of the presentation
Problem description and related analysis
Construction of numerical algorithms
Examples and summary
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Description of the Problem
1
Goursat problem (hyperbolic equation):
(x, y ) ∈ D
uxy = f (x, y , u, ux , uy ),
for
u(0, y ) = s(y )
for y ∈ [0, 1],
H1 ux (x, x) + H2 uy (x, x) + u(x, x) = r (x)
for x ∈ [0, 1],
2
f (x, y , X , Y , Z ), s(y ) and r (x) are known.
3
H1 and H2 are two given real constants called impedance numbers.
4
(Remark: They can also be infinite. When H1 = ∞ and H2 is finite,
it is ux (x, x) = 0, the Neumann boundary condition.)
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The Triangular Domain
D : a triangular domain defined by D = {(x, y ) : 0 ≤ x ≤ 1, x ≤ y ≤ 1}
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Examples of Goursat problem
Example 1:
uxy = e 2u ,
y
u(0, y ) = − ln(1 + e y )
2
77ux (x, x) + 5uy (x, x) + u(x, x) = x − ln(2e x )
for y ∈ [0, 1],
for x ∈ [0, 1].
Example 2:
uxy = u + sin(uy ) + (ux )2 − y ,
u(0, y ) = y + e y
for y ∈ [0, 1],
100ux (x, x) + 2uy (x, x) + u(x, x) = 2 + x + 103e 2x
for x ∈ [0, 1].
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Assumptions
1
2
The real valued function f is continuous for all arguments of
x, y , X , Y , Z as (x, y ) ∈ D;
For some D ∗ ⊂ R 3 , f (x, y , X , Y , Z ) is Lipschitz continuous in
X , Y , Z , i.e. for any (X 1 , Y 1 , Z 1 ), (X 2 , Y 2 , Z 2 ) ∈ D ∗ ,
|f (x, y , X 1 , Y 1 , Z 1 ) − f (x, y , X 2 , Y 2 , Z 2 )|
≤ L(|X 1 − X 2 | + |Y 1 − Y 2 | + |Z 1 − Z 2 |),
where L is the Lipschitz constant;
3
s(y ), r (x) ∈ C 1 [0, 1], i.e. both of them are continuously differentiable
on [0, 1].
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Existence and Uniqueness of Solution
Proof for a special case as H1 = H2 is considered. The proofs for other
cases could be devised in a similar manner
Theorem: The exact solution for the Goursat problem exists and is unique
as H1 = H2 .
The proof is given by approach of generalized Picard iteration
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Content of the presentation
Problem description and related analysis
Construction of numerical algorithms
Examples and summary
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Preparation for the Algorithms
A uniform mesh with mesh size h =
x0 = 0, x1 = h . . . xi = ih . . . xn = 1
y0 = 0, y1 = h . . . yi = ih . . . yn = 1.
1
n
Denotation:
p(x, y ) = ux (x, y ), q(x, y ) = uy (x, y ),
uij = u(xi , yj ), pij = p(xi , yj ), qij = q(xi , yj ),
fij = f (xi , yj , uij , pij , qij ), sj = s(yj ), ri = r (xi ).
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Basic Data Formulas
Direct integration for the Goursat equation produces three basic formulas
for algorithms:
Z xi+1 Z yj+1
ui+1,j+1 = ui+1,j + ui,j+1 − ui,j +
f (ξ, η)dξdη,
xi
yj
yj+1
Z
pi,j+1 = pi,j +
f (xi , η)dη,
yj
Z
xi+1
qi+1,j = qi,j +
f (ξ, yj )dξ.
xi
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Data Distribution
uij ,
pij , qij
1
Data on the Left Boundary on y Axis
2
Non-boundary data (include upper boundary data)
3
Data on the slanted boundary
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Finding Data on the Left Boundary on y Axis
1
These data are: p0j , q0j
2
The boundary condition u(0, y ) = s(y ) generates exact q0j .
3
p0j is the solution of ODE from the calculation of Goursat PDE:
∂
p(0, y ) = f (0, y , s(y ), p(0, y ), s 0 (y )),
∂y
1
p(0, 0) =
(r0 − H2 s00 − s0 ).
H1
solved via third Runge-Kutta approach.
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Non-boundary Data
1
These data are: uij , pij and qij as 0 < i < n, i < j ≤ n.
2
In a previous slide
here
, we use trapezoidal rule for them and get
ui+1,j+1 = ui+1,j + ui,j+1 − ui,j +
h2
[fi+1,j+1 + fi+1,j + fi,j+1 + fi,j ].
4
h
pi+1,j+1 = pi+1,j + [fi+1,j+1 + fi+1,j ],
2
h
qi+1,j+1 = qi,j+1 + [fi+1,j+1 + fi,j+1 ].
2
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Non-boundary Data
1
Suppose ui,j , pi,j , qi,j , ui+1,j , pi+1,j , qi+1,j and ui,j+1 , pi,j+1 , qi,j+1
are known
2
An iteration is set up to solve for ui+1,j+1 , pi+1,j+1 , qi+1,j+1 :
(m+1)
ui+1,j+1 = ui+1,j + ui,j+1 − ui,j +
h2 (m)
[f
+ fi+1,j + fi,j+1 + fi,j ],
4 i+1,j+1
h (m)
(m+1)
pi+1,j+1 = pi+1,j + [fi+1,j+1 + fi+1,j ],
2
h (m)
(m+1)
qi+1,j+1 = qi,j+1 + [fi+1,j+1 + fi,j+1 ].
2
3
q
Thoerem: The iteration above is convergent for h < 2( L1 + 1 − 1).
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Slanted-boundary Data
Case 1: H1 = H2 = H
Recalling formulas in a previous slide here , we get the exact expression for u(x, x),
and
Z x
uy (x, x) =s 0 (x) +
f (ξ, x)dξ
0
Rx
−x/H
r (x) − (e
/H) 0 e ξ/H r (ξ)dξ − e −x/H s(0)
ux (x, x) =
H
Z
x
− s 0 (x) −
f (ξ, x)dξ.
0
By the composite trapezoidal rule,
i−1
u(ih, ih) =
X
e −ih/H h
[r (0) + 2
e jh/H r (jh) + e ih/H r (ih)] + e −ih/H s(0),
2H
j=1
i−1
X
h
uy (ih, ih) =s 0 (ih) + [f (0, ih) + 2
f (jh, ih) + f (ih, ih)],
2
j=1
R ih
r (ih) − (e −ih/H /H) 0 e ξ/H r (ξ)dξ − e −ih/H s(0)
ux (ih, ih) =
− s 0 (ih)
H
i−1
X
h
− [f (0, ih) + 2
f (jh, ih) + f (ih, ih)].
2
j=1
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Case 1 Continued
1
2
u(ih, ih) can be obtained explicitly, while the last two equations are
implicit for uy (ih, ih) and ux (ih, ih).
Set up an iteration for uy (ih, ih) and ux (ih, ih):
i−1
X
h
(ih, ih) =s 0 (ih) + [f (0, ih) + 2
f (jh, ih) + f (n) (ih, ih)],
2
j=1
R ih
r (ih) − (e −ih/H /H) 0 e ξ/H r (ξ)dξ − e −ih/H s(0)
(n+1)
− s 0 (ih)
ux
(ih, ih) =
H
i−1
X
h
− [f (0, ih) + 2
f (jh, ih) + f (n) (ih, ih)].
2
(n+1)
uy
j=1
3
4
5
(0)
Remark: Initial value can be uy = uy ((i − 1)h, ih) and
(0)
ux = ux ((i − 1)h, ih).
We can show the iteration is convergent if hL < 1.
The error for the composite trapezoidal rule is at most O(h2 ), so the
errors for u(ih, ih), ux (ih, ih) and uy (ih, ih) are also at most O(h2 ) given
no error from the iteration.
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Case 2: H1 6= H2
1
The Taylor expansion for u(x, x) at (x − h, x) yields
u(x, x) = u(x − h, x) + ux (x − h, x)h + O(h2 ),
2
The second boundary condition of the Goursat problem
here
produces
ux (x, x) = (r (x) − u(x, x) − H2 uy (x, x))/H1 .
u(x, x)
r (x) H2 − H1
d
u(x, x) +
=
−
uy (x, x).
dx
H1
H1
H1
3
Now applying trapezoidal rule for the integral yields
(
r (x)
1
2
uy (x, x) =
−
[H1 u(x, x) − H1 e −h/H1 u(x − h, x − h)]
H 2 − H 1 H 2 − H1 h
)
− e −h/H1 [r (x − h) − (H2 − H1 )uy (x − h, x − h)]
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+ O(h2 ).
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Case 2 Continued
The formulas for these discretized approximations are
uii =ui−1,i + hpi−1,i ,
ri
1
qii =
−
H 2 − H 1 H 2 − H1
(
2
[H1 uii − H1 e −h/H1 ui−1,i−1 ]
h
)
− e −h/H1 [ri−1 − (H2 − H1 )qi−1,i−1 ] ,
pii =(ri − uii − H2 qii )/H1 .
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The Complete Algorithm
1
Find y -axis boundary data p0j and q0j ;
2
Find u11 , p11 and q11 as slanted boundary data;
3
For fixed j bigger than 1, calculate the non-boundary data ui,j , pi,j ,
qi,j from i = 1 to i = j − 1;
4
For the same j in Step 3, calculate ujj , pjj and qjj as slanted-boundary
data;
5
Then repeat Steps 3 and 4 until we find all the data.
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Content of the presentation
Problem description and related analysis
Construction of numerical algorithms
Examples and summary
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Numerical Examples
Example 1
uxy = u + uy + (ux )2 − y − 1 − e x+y − e 2(x+y ) ,
u(0, y ) = y + e y
100ux (x, x) + 2uy (x, x) + u(x, x) = 2 + x + 103e
for y ∈ [0, 1],
2x
for x ∈ [0, 1].
The exact solution is u(x, y ) = y + e x+y .
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tables
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Discussion and Summary
1
We can also include impedance boundary along y axis. Namely,
I1 ux (0, y ) + I2 uy (0, y ) + u(0, y ) = s(y ) for y ∈ [0, 1]
2
Complexity from the nonlinear function f . Iterative schemes are
designed for it.
3
Algorithms are based on the analysis of the problem
4
Enough information from the boundary conditions is used to design
the algorithms
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Acknowledgement
Special thanks to Qatar National Research Fund (QNRF)
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Selected Literature
K. Gou, B. Sun, Numerical solution of the Goursat problem on a
triangular domain with mixed boundary conditions, Appl. Math.
Comput. 217(2011) 8765-8777.
A. M. Wazwaz, The variational iteration method for a reliable
treatment of the linear and the nonlinear Goursat problem, Appl.
Math. Comput. 193(2007) 455-462.
È. Goursat, A Course in Mathematical Analysis, Vol. 3, Dover, New
York, 2006.
W. Rundell, P. E. Sacks, Reconstruction techniques for classical
inverse Sturm-Liouville problems, Math. Comput. 58(1992) 161-183.
A. M. Wazwaz, On the numerical solution of the Goursat problem,
Appl. Math. Comput. 59(1993) 89-95.
A. I. M. Ismail, M. A. S. Nasir, Numerical solution of the Goursat
problem, Proc. Iasted Inter. Conf. Appl. Simul. Model. June 28-30,
2004, Rhodes, Greece.
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