Mathematical Modelling in the Natural Sciences SS21 Solutions to Exercises on Sheet 7 Exercises und Lecture Notes Contents • Exercise 1: Cerclage Model 1 ◦ Task .............................................. 1 ◦ Solution ............................................ 2 • Exercise 2: Bungee Cord 10 ◦ Task .............................................. 10 ◦ Solution ............................................ 10 • Exercise 3: Pastry Sheet 14 ◦ Task .............................................. 14 ◦ Solution ............................................ 14 • Exercise 4: Meditation Model 23 ◦ Task .............................................. 23 ◦ Solution ............................................ 23 • Exercise 5: Comparison of Membrane Models 38 ◦ Task .............................................. 38 ◦ Solution ............................................ 38 • Exercise 1: Cerclage Model ◦ Task Derive the necessary stationarity condition on page 190 of the lecture notes, - T (u) uu φ q u 2 + u 2 φ sin φ φ + T (u) 2u 2 + u 2 φ q u 2 + u 2 φ sin φ = f (u sin φ, u cos φ) (u cos φ) φ q u 2 + u 2 φ +(p 1 + λ) u 2 sin φ, 0 < φ < π, u φ =0, φ =0,π, Z π 0 u 3 sin φdφ =2r 3 1 Write a Matlab code to solve this system as indicated on pages 191 - 192, i.e., by iterating a cell centered finite difference approximation of A(u) K(u) K * (u) 0 v λ = F (u, p) G(u) u = u + αv p = p + αλ 1
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Mathematical Modelling in the Natural Sciences SS21Solutions to Exercises on Sheet 7
In the lecture notes the model of the band force f(u sin(φ), u cos(φ)) is given on page 187 and themodel of tension T (u) is given on pages 188-189. The Lagrangian functional on page 186,
L(u, λ) = 12π [Ji(u) + Je(u)− λJc(u)]
has the variational derivatives seen on page 184,
δJi
δu(u; v) = 2π
∫ π
0T (u)
u2φv + 2u2v + uuφvφ√
u2 + u2φ
sin(φ)dφ
and on page 186
δJe
δu(u; v) = −2π
∫ π
0
f(u cos(φ), u sin(φ))(u cos(φ))φ√u2 + u2
φ
+ p
vu2 sin(φ)dφ
and the variational derivative of the contraint functional Jc on page 186 is given by
δJc
δu(u; v) = 2π
∫ π
0u2v sin(φ)dφ.
Setting δL(u; v)/δu = 0 gives the condition∫ π
0T (u)
u2φv + 2u2v + uuφvφ√
u2 + u2φ
sin(φ)dφ
−∫ π
0
f(u cos(φ), u sin(φ))(u cos(φ))φ√u2 + u2
φ
+ p
vu2 sin(φ)dφ
−λ2π
∫ π
0u2v sin(φ)dφ = 0, ∀v ∈ C([0, π])
The only term among these which contains a derivative of a test function v is∫ π
0T (u)
uuφvφ√u2 + u2
φ
sin(φ)dφ = −∫ π
0
T (u)sin(φ)uuφ√u2 + u2
φ
φ
vdφ+
T (u)sin(φ)uuφ√u2 + u2
φ
φ
v
∣∣∣∣∣π
0
2
Setting the test function v to be concentrated in the iterior of the interval (0, π) gives the necessarystationarity condition on u,
−
T (u)uuφ√u2 + u2
φ
sinφ
φ
+ T (u)2u2 + u2
φ√u2 + u2
φ
sinφ
=
f(u sinφ, u cosφ)(u cosφ)φ√u2 + u2
φ
+ (p1 + λ)
u2 sinφ, 0 < φ < π,
,
∫ π
0u3 sinφdφ = 2r3
1
Letting the test function v be concentrated on the boundary of the interval gives the Neumannboundary condition,
uφ = 0, φ = 0, π
Finally setting δL(λ;µ)/δλ = 0 gives the condition∫ π
0u?3 sin(φ)dφ = 2r3
1.
The saddle point problem is solved using the following Matlab code, in which various parametersappear as indicated in the lecture notes.
The results from this code are shown graphically as follows.
• Exercise 2: Bungee Cord
Task
Recall the constructions in the lecture notes on pages 204-207 for modelling the dynamic state of abungee cord. Develop a Matlab code to simulate the cord displacement. Hint: Consider this code,which contains yet some riddles.
d = 0.6; axis([0.5-d,0.5+d,0.0-d,0.0+d,0.1-2*d,0.1]);
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else
d = 1.0; axis([0.5-d,0.5+d,0.0-d,0.0+d,0.1-2*d,0.1]);
end
pbaspect([1 1 1]);
drawnow;
if (norm(U-Ua) <= tol*norm(U)) % stop when iterates
break; % do not differ
end
end
The result at time t = 1 for a cord falling while fastened at both ends is shown graphically as follows.
The result at time t = 2 for a cord falling while fastened at only one end is shown graphically as
13
follows.
• Exercise 3: Pastry Sheet
Task
Recall the definitions on page 210 of the lecture notes. Based upon the principle of least action, derivethe stationary state for the Lagrangian functional
with gravity f = (0, 0,−φ/ρ) and ρ, κ, φ > 0. Develop a Matlab code to simulate the membranedisplacement. Hint: Consider this code, which contains yet some riddles.
Solution
Let Ω = (0, 1)2 and Q = Ω×(0, T ) and suppose a test function v is sufficiently smooth and vanishes atthe spatial boundary η = 0 and at initial and final times t = 0, T . For the computation of variational
Since v = 0 at t = 0, T , for the first integral vanishes and implicitly requires initial and final timeconditions on u, but instead the initial position and velocity are imposed,
u(ξ, η, 0) = (ξ, η, 0), ut(ξ, η, 0) = (0, 0, 0)
and under the condition of well-posedness of the resulting intial- and boundary-value problem, thefinal time values u(ξ, η, T ) may be regarded as having been imposed as a condition on the Lagrangianfunctional L. The condition that v = 0 hold at η = 0 corresponds to the given Dirichlet boundarycondition
u(ξ, 0, t) = (ξ, 0, 0), ξ ∈ [0, 1], t ∈ [0, T ].
Letting v be concentrated on ∂Q at η = 1 gives the natural boundary condition
B(u)uη(ξ, 1, t) = (0, 0, 0), ξ ∈ [0, 1], t ∈ [0, T ].
Letting v be concentrated on ∂Q at ξ = 0, 1 gives the natural boundary conditions
where the term−(1, 0, 0) is required for compatibility with the Dirichlet boundary condition u(ξ, 0, t) =(ξ, 0, 0) at η = 0. The necessary optimality condition for a stationary u is
if (norm(U-Ua) <= tol*norm(U)) % stop when iterates
break; % do not differ
end
end
The result at time t = 5 for a membrane falling while fastened on all sides is shown graphically as
21
follows.
The result at time t = 4 for a membrane falling while fastened only along one edge is shown graphically
22
as follows.
• Exercise 4: Meditation Model
Task
Develop a wave equation model for the phenomena illustrated in this video of a simulation, andimplement your model in Matlab.
Solution
The following model is proposed for the simulation of known neurophysiological phenomena associatedwith waking and dream or meditative states. First, the spatial domain Ω = (0, 1)2 is divided intohalves,
where Ωl models the left hemisphere of the brain and Ωr the right. Let n denote a unit vector which isoutwardly directed with respect to ∂Ωl and ∂Ωr. Let A(u) = diag(α, ω) be the diagonal matrix withwave speeds
√α and
√ω in the x- and y-directions respectively. Let f be a forcing function yet to be
specified. Then the wave displacement u is modelled according to:utt = ∇ · (A∇u) + f(u), (x, y) ∈ (0, 1)2, t ∈ (0, T )
u(0, y, t) = u(1, y, t), y ∈ [0, 1], t ∈ [0, T ]ut(x, y, t) = n>A∇u(x, y, t), x ∈ [0, 1], y ∈ 0, 1/2±, 1, t ∈ [0, T ]
u = 0, (x, y) ∈ (0, 1)2, t = 0ut = 0, (x, y) ∈ (0, 1)2, t = 0.
Since the left hemisphere has been found to carry out serial processing, this characteristic is modelledby allowing waves to propagate in only one direction according to
α = 0, ω > 0, Ωl.
On the other hand, the right hemisphere has been found to carry out parallel processing, and thischaracteristic is modelled by letting waves travel unhindered in any direction according to
α, ω > 0, Ωr.
At the boundary between the hemispheres transmission is accomplished through f(u) discussed below,but otherwise
ω = 0, ∂Ωr ∩ ∂Ωl.
The squared wave speeds α and ω are otherwise spatially varying,
α = α(x, y), ω = ω(x, y)
where more spatially uniform wave speeds are thought to correspond to the result of meditativepractice as discussed below.
The spatial geometry is torus-like with periodic boundary conditions at x = 0, 1. Also the left andright hemispheres are connected at y = 0, 1 and at y = ±1
2 . For instance, a given wave function udefined on cells Nx×Ny in Ω may be mapped onto a torus-like surface according to
a = kron(ones(Ny,1),linspace(0,2*pi,Nx));
b = kron(linspace(0,2*pi,Ny)’,ones(1,Nx));
y = (2-(1+cos(b)).*cos(b)).*cos(a);
z = (2-(1+cos(b)).*cos(b)).*sin(a);
x = 2*(1+cos(b)).*sin(b);
surf(x,y,z,u);
to illustrate the modelling of waves on and connections between the left and right hemispheres as
24
illustrated in the following graphic.
The boundary conditions at y ∈ 0, 1/2±, 1 are non-reflecting
ut −√ωuy = 0, y = 0
ut +√ωuy = 0, y = 1/2−
ut −√ωuy = 0, y = 1/2+
ut +√ωuy = 0, y = 1.
To illustrate these, note that
a left-travelling (y = −t√ω) wave u = u(y + t
√ω) satisfies (∂t −
√ω∂y)u(y + t
√ω) = 0
and
a right-travelling (y = +t√ω) wave u = u(y − t
√ω) satisfies (∂t +
√ω∂y)u(y −
√ωt) = 0.
Thus, at the boundaries of Ωl or Ωr at y ∈ 0, 1/2±, 1 only out-going waves and no in-coming wavesare supported.
One aspect of the forcing function f(u) is to facillitate transmission of waves between hemispheres.For instance, when a wave from the right hemisphere impinges on (a cell near) its boundary at x = xand y = 1/2− with a certain threshold intensity, then a wave is initiated in the left hemisphere from itsboundary at y = 1/2+. Since in waking states such impulses are networked, this wave is propagatedfrom several locations at (cells near) xj = xj(x) ∈ [0, 1] and y = 1/2+. Since in dream and meditativestates this networking is not active, the wave is propagated only from (a cell at) x = x and y = 1/2+.Because of the non-reflecting boundary conditions at y = 1/2±, the impinging wave from the righthemisphere does not reflect back from where it came, and the wave initiated in the left hemispheremarches only forward into the left hemisphere. This process may be regarded as conscious thoughtsemerging from the subconscious. In a waking states, a wave in the left brain impinges on (a cellnear) its boundary at x = x and y = 0 with a certain intensity, and it then initiates a wave in theright hemisphere from its boundary at (a cell near) x = x and y = 1. Because of the non-reflectingboundary conditions at y = 0, 1, the impinging wave from the left hemisphere does not reflect back
25
from where it came, and the wave initiated in the right hemisphere marches only forward into the righthemisphere. This process may be regarded as a feedback mechanism whereby conscious thoughts affectsubconscious memory. In a dream or meditative state, this feedback mechanism is absent. Finally, theforcing function f(u) initiates impulses in the right brain with a probability which gives advantage toa natural rhythm on the one hand but also allows competing random impulses.
The following Matlab code is used to solve the initial and boundary-value problem.
% set up figure
h1 = figure(1);
close(h1);
h1 = figure(1);
set(h1,’Position’,[10 30 1000 400]);
% true if meditating
med = true;
% max wave speeds and max excitation
al0 = 1;
om0 = 1;
u0 = 2;
% there are Nx x Ny cells, Nt time steps and left-right interface at My
Nx = 51;
Ny = 51;
My = round(Ny/2);
Nt = 1200;
% dimensions of domain
xmin = 0;
xmax = 1;
ymin = 0;
ymax = 1;
T = Nt/500;
% cell sizes
hx = (xmax-xmin)/Nx;
hy = (ymax-ymin)/Ny;
ht = T/Nt;
% cell centers
x = linspace(xmin,xmax,Nx)’;
y = linspace(ymin,ymax,Ny);
xx = kron(x,ones(1,Ny));
yy = kron(ones(Nx,1),y);
% first order differential operators
dx = spdiags(ones(Nx,1), 0,Nx,Nx) ...
- spdiags(ones(Nx,1),-1,Nx,Nx); dx(1,Nx) = -1;
26
dx = dx/hx;
iy = speye(Ny);
Dx = kron(iy,dx);
dy = spdiags(ones(Ny-1,1),1,Ny-1,Ny) ...
- spdiags(ones(Ny-1,1),0,Ny-1,Ny);
dy = dy/hy;
ix = speye(Nx);
Dy = kron(dy,ix);
% (squared) wave speeds
if (med)
al = zeros(Nx,Ny); al(:,My+1:Ny) = al0;
om = om0*ones(Nx,Ny-1); om(:,My) = 0;
else
al = zeros(Nx,Ny); al(:,My+1:Ny) = al0*(1 + rand(Nx,Ny-My))/2;
om = om0*(1 + rand(Nx,Ny-1))/2; om(:,My) = 0;
end
% second order differential operators
Lx = Dx’*spdiags(al(:),0,Nx*Ny,Nx*Ny)*Dx;
Ly = Dy’*spdiags(om(:),0,Nx*(Ny-1),Nx*(Ny-1))*Dy;
L = Lx + Ly;
% identities and zeros
I = speye(Nx*Ny);
I2 = kron(speye(2),I);
Z = sparse(Nx*Ny,Nx*Ny);
% non-reflecting boundary conditions through proper damping coefficient
c = zeros(Nx,Ny);
c(:,1) = sqrt(om(:,1))/hy;
c(:,My) = sqrt(om(:,My-1))/hy;
c(:,My+1) = sqrt(om(:,My+1))/hy;
c(:,Ny) = sqrt(om(:,Ny-1))/hy;
C = spdiags(c(:),0,Nx*Ny,Nx*Ny);
B = [Z,I;-L,-C];
% time stepping method
th = 1.0;
% starting state
u = zeros(Nx,Ny);
ut = zeros(Nx,Ny);
U = [u(:);ut(:)];
% association of serial excitations in left brain
r = randperm(Nx);
27
R = zeros(Nx,Nx);
nc = 5;
l = mod(Nx,nc);
for i=1:nc:(Nx-l)
for j=0:(nc-1)
for k=0:(nc-1)
R(r(i+j),r(i+k)) = 1;
end
end
end
if (l ~= 0)
i = i+nc;
for j=0:(l-1)
for k=0:(l-1)
R(r(i+j),r(i+k)) = 1;
end
end
end
% initialize activation measure
uact = sqrt((sum((sqrt(al(:))*(Dx*u(:))).^2) ...
+ sum((sqrt(om(:))*(Dy*u(:))).^2) ...
+ sum(ut(:).^2))*hx*hy);
% start time stepping
for k=1:Nt
% random excitation in right brain
urmax = max(abs(u(:,My+1:Ny)));
if (urmax < 0.01*u0)
u(round(Nx/2),round((My+Ny)/2)) = u0;
else
if (rand(1) < 0.005)
i = randi(Nx); j = randi(Ny-My-2)+My+1;
u(i,j) = u0;
end
end
% excitation from right brain into left brain
urmax = max(abs(u(:,My+1)));
ulmax = max(max(abs(u(:,1:My))));
if ((urmax > 0.005*u0) && (ulmax < 0.01*u0))
i = find(abs(u(:,My+1)) == urmax,1);
% if med, then only one impulse, otherwise assciated impulses
if (med)
u(i,My) = u0; ut(i,My) = -u0/hy;
else
r = find(R(:,i));
28
u(r,My) = u0; ut(r,My) = -u0/hy;
end
end
% if not med, excitation from left brain into right brain
Following are graphical representations of the results with med = false,
and with med = true,
31
Note that the mean energy level
1
|Ω|T
∫ T
0
∫Ω
[u2t +∇u>A2∇u]dxdt
is significantly lower in the model with meditation than in the model without.
To investigate the effect of random or uniform wave speeds, which presumably can be affected bymeditative practice, the eigenfunctions of the second order differential operator are shown as imagesfor the respective cases as follows. The following Matlab code used to produce these images.
% true if meditating
med = true;
% max wave speeds and max excitation
al0 = 1;
om0 = 1;
% there are Nx x Ny cells, Nt time steps and left-right interface at My
Nx = 51;
Ny = 51;
% dimensions of domain
xmin = 0; xmax = 1;
ymin = 0; ymax = 1;
% cell sizes
hx = (xmax - xmin)/Nx;
hy = (ymax - ymin)/Ny;
% first order differential operators
dx = spdiags(ones(Nx,1), 0,Nx,Nx) ...
- spdiags(ones(Nx,1),-1,Nx,Nx); dx(1,Nx) = -1;
dx = dx/hx;
iy = speye(Ny);
Dx = kron(iy,dx);
dy = spdiags(ones(Ny-1,1),1,Ny-1,Ny) ...
- spdiags(ones(Ny-1,1),0,Ny-1,Ny);
dy = dy/hy;
ix = speye(Nx);
Dy = kron(dy,ix);
% (squared) wave speeds
if (med)
al = al0*ones(Nx,Ny);
om = om0*ones(Nx,Ny-1);
else
al = al0*rand(Nx,Ny);
om = om0*rand(Nx,Ny-1);
end
32
% second order differential operators
Lx = Dx’*spdiags(al(:),0,Nx*Ny,Nx*Ny)*Dx;
Ly = Dy’*spdiags(om(:),0,Nx*(Ny-1),Nx*(Ny-1))*Dy;
L = Lx + Ly;
% eigenspace decomposition
[V,D] = eig(full(L));
% set up figure
h1 = figure(1);
close(h1);
h1 = figure(1);
set(h1,’Position’,[10 30 800 800]);
% display high energy states
for i=1:4
subplot(2,2,i)
imagesc(reshape(V(:,Nx*Ny-i+1),Nx,Ny))
axis image; axis off;
if (med)
title(sprintf([’High Energy Eigenfunction #%0.0f\n’, ...
’Uniform Wave Speeds’],Nx*Ny-i+1))
else
title(sprintf([’High Energy Eigenfunction #%0.0f\n’, ...
’Random Wave Speeds’],Nx*Ny-i+1))
end
end
% set up figure
h2 = figure(2);
close(h2);
h2 = figure(2);
set(h2,’Position’,[10 30 800 800]);
% display low energy states
for i=1:4
subplot(2,2,i)
imagesc(reshape(V(:,i),Nx,Ny))
axis image; axis off;
if (med)
title(sprintf([’Low Energy Eigenfunction #%0.0f\n’, ...
’Uniform Wave Speeds’],i))
else
title(sprintf([’Low Energy Eigenfunction #%0.0f\n’, ...
’Random Wave Speeds’],i))
end
end
33
For the case that α and ω are random throughout Ω, the lower energy eigenfunctions associated withthe 4 smallest eigenvalues are
34
and the higher energy eigenfunctions associated with the 4 largest eigenvalues are
35
For the case that α and ω are uniform throughout Ω, the lower energy eigenfunctions associated withthe 4 smallest eigenvalues are
36
and the higher energy eigenfunctions associated with the 4 largest eigenvalues are
Note that the forms of the low energy states are comparable when the wave speeds are either randomor uniform, although those associated with random wave speeds are noisier. Yet the forms of the
37
high energy states are conspicuously more localized for the random wave speeds in relation to broadlysupported one for the uniform wave speeds.
• Exercise 5: Comparison of Membrane Models
Task
Summarize the functionals used in the lecture to model membranes or strings and explain underwhat circumstances the minimizing or stationary state for one functional agrees with that of anotherfunctional.
Solution
For simplicity only clamped string models are considered. The observations have natural counterpartsfor membranes.
Let u ∈ R3 represent a point on a continuously smooth string in R3. Let the position u(t) along thestring be parameterized by t ∈ [0, T ]. Then the length of the string is
L = L(u) =
∫ T
0‖u′(t)‖dt. (1)
When the string is reparameterized according to an arclength variable
s(t) =
∫ t
0‖u′(τ)‖dτ, s′(t) = ‖u′(t)‖, t ∈ [0, T ], s(T ) = L
then it follows
u′(s) = u′(t)t′(s) = u′(t)/s′(t) =u′(t)
‖u′(t)‖and thus
‖u′(s)‖ = 1. (2)
The string of shortest length connecting two points
u(0) = u, u(T ) = u
must satisfy
0 =δL
δu(u;v) =
∫ T
0
u′(t) · v′(t)‖u′(t)‖
dt, ∀v ∈ C∞0 ([0, T ],R3)
or after integration by parts,
0 =
∫ T
0v(t) ·
(u′(t)
‖u′(t)‖
)′dt+
u′(t) · v(t)
‖u′(t)‖
∣∣∣∣t=Tt=0
=
∫ T
0v(t) ·
(u′(t)
‖u′(t)‖
)′dt, ∀v ∈ C∞0 ([0, T ],R3).
Thus u must satisfy the boundary value problem(u′(t)
‖u′(t)‖
)′= 0, u(0) = u, u(T ) = u
and is uniquely determined by
u′(s)∣∣s=s(t)
=u′(t)
‖u′(t)‖= c ∈ R3
38
or with the boundary conditions, c = (u− u)/L,
u(s) = (1− s/L)u + (s/L)u, s ∈ [0, L]. (3)
As expected, the string lies along the straight line connecting the endpoints.
Now let w ∈ R3 represent a point on the string in the presence of an external force per unit lengthf ∈ R3. Let the position w(s) along the string be parameterized by arclength s in the absence of theforce. Suppose w satisfies the boundary conditions as before,
w(0) = u, w(L) = u
but now, in contrast to (2), there holds in the presence of the force,
‖w′(s)‖ 6= 1, s ∈ [0, L].
The work per unit length performed by the external force is of the product, force × displacement,f(s) · [w(s)− u(s)], s ∈ [0, L], and the total work performed by the external force is
Je(w) =
∫ L
0f(s) · [w(s)− u(s)]ds.
The elastic force acting tangentially along the string in the direction w′(s)/‖w′(s)‖ is the tension T .The relative change in length of the string in response to a stress is the strain,
τ(s) =‖w′(s)‖ − ‖u′(s)‖
‖u′(s)‖= ‖w′(s)‖ − 1. (4)
Here, (2) has been applied. Let the tension be modelled to depend linearly on the strain,
T (s) = κτ(s)w′(s)
‖w′(s)‖, s ∈ [0, L]. (5)
Then the strain energy
Ji(w) =1
2
∫ L
0κ(‖w′(s)‖ − 1)2ds (6)
results from integrating the variational derivative,
δJi
δw(w;v) =
∫ L
0T (s) · v′(s)ds =
∫ L
0
[κ(‖w′(s)‖ − 1)
w′(s)
‖w′(s)‖
]· v′(s)ds (7)
with the tension in (5) × the displacement dv = v′(s)ds. Thus, the total energy combining theopposing work efforts is given by
J(w) = Ji(w)− Je(w) =1
2
∫ L
0κ(‖w′(s)‖ − 1)2ds−
∫ L
0f(s) · [w(s)− u(s)]ds
which should be minimized by the position w(s). The variational derivative of J is given by
δJ
δw(w;v) = −
∫ L
0v(s) ·
[κ(‖w′(s)‖ − 1)
w′(s)
‖w′(s)‖
]′+ f(s)
ds
39
and so the position of the string is determined by the solution to the boundary value problem,
−[κ(‖w′(s)‖ − 1)
w′(s)
‖w′(s)‖
]′= f(s), w(0) = u, w(0) = u.
If the tension is modelled to be (at least approximately) independent of strain,
T (s) = κw′(s)
‖w′(s)‖with ‖T (s)‖ = κ for s ∈ [0, L] (8)
then integrating the variational derivative
δJi
δw(w;v) =
∫ L
0T (s) · v′(s)ds =
∫ L
0
[κ
w′(s)
‖w′(s)‖
]· v′(s)ds
gives an elastic energy
Ji(w) =
∫ L
0κ(‖w′(s)‖ − 1)ds =
∫ L
0κτ(s)ds (9)
involving only the strain (4) itself. Comparing (9) with (1) shows that minimizing Ji minimizes thelength of the string. In this context, the total energy combining the opposing work efforts is given by
J(w) = Ji(w)− Je(w) =
∫ L
0κ(‖w′(s)‖ − 1)ds−
∫ L
0f(s) · [w(s)− u(s)]ds.
The variational derivative of J is given by
δJ
δw(w;v) = −
∫ L
0v(s) ·
[κ
w′(s)
‖w′(s)‖
]′+ f(s)
ds
and so the position of the string is determined by the solution to the boundary value problem,
−[κ
w′(s)
‖w′(s)‖
]′= f(s), w(0) = u, w(L) = u.
Consider the circumstances under which the strain model in (5) may be justifiably approximated bythe strain model in (8). Suppose first that the motion of the string is purely in the longitudinaldirection. Then
and for (5) and (8) to be comparable, it must be that
κ(√
1 + w′(s)2 − 1) ≈ κ or |w′(s)| ≈√
(κ/κ+ 1)2 − 1.
In the situation of small transverse displacements with |w′(s)| 1, κ must be sufficiently small inrelation to κ. When the string is very taut and |w′(s)| 1, then the form w(s) = (s, 0, w(s)) is notrealistic but rather
which reverts back to the first case considered, in which motion is in the longitudinal direction.
In the case of small transverse displacements with w(s) = (s, 0, w(s)), the strain may be approximatedas quadratic
τ(s) =√
1 + w′(s)2 − 1 ≈ 12w′(s)2.
Then elastic energy becomes
Ji(w) =1
2
∫ L
0κw′(s)2ds
and the position of the string is determined by the solution to the linear Poisson boundary valueproblem,
−κ∆w = f, w(0) = u, w(L) = u
where u = (0, 0, u) and u = (0, 0, u).
These formulations may be considered in the context of thermodynamics, in which the infinitesimalformulation of energy is given by a sum of terms of type with force × displacement,
dF = KdL − ΦdA. (10)
Here, dL and dA have qualities of changes in length and area respectively. Length is measured alongthe string and area corresponds to that between an infinitesimal piece of unloaded string and its loadedcounterpart. The terms K and Φ are force terms with appropriate units so that F corresponds toan energy. In light of foregoing considerations, it is evident that the delicate term is dL. With thetension model (8), dL in (10) may be associated with the variational derivative of the length in (1),
δL
δw(w;v) =
∫ L
0dL(w;v), dL(w;v) =
w′(s) · v′(s)‖w′(s)‖
ds. (11)
Then with K = κ, the elastic energy is given by (9). On the other hand, the following variant of dLfor (10) results from the tension model (5),