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ME3255S2017/lecture_08/lecture_08.md

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Optimization First, let's find the minimum energy $\min(E_{LJ}(x))$ Brute force Golden Search Algorithm Parabolic Interpolation Using the solutions to minimization for the nonlinear spring constant Curve-fitting
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%plot --format svg

Optimization

Many problems involve finding a minimum or maximum based on given constraints. Engineers and scientists typically use energy balance equations to find the conditions of minimum energy, but this value may never go to 0 and the actual value of energy may not be of interest.

The Lennard-Jones potential is commonly used to model interatomic bonding.

$E_{LJ}(x)=4\epsilon \left(\left(\frac{\sigma}{x}\right)^{12}-\left(\frac{\sigma}{x}\right)^{6}\right)$

Considering a 1-D gold chain, we can calculate the bond length, $x_{b}$, with no force applied to the chain and even for tension, F. This will allow us to calculate the nonlinear spring constant of a 1-D gold chain.

TEM image of Gold chain

Computational Tools to Study and Predict the Long-Term Stability of Nanowires. By Martin E. Zoloff Michoff, Patricio Vélez, Sergio A. Dassie and Ezequiel P. M. Leiva

Model of Gold chain, from molecular dynamics simulation

Single atom gold chain mechanics

First, let's find the minimum energy $\min(E_{LJ}(x))$

Brute force

setdefaults
epsilon = 0.039; % kcal/mol
sigma = 2.934; % Angstrom
x=linspace(2.8,6,200); % bond length in Angstrom

Ex = lennard_jones(x,sigma,epsilon);

[Emin,imin]=min(Ex);

plot(x,Ex,x(imin),Emin,'o')

svg

x(imin-1)
x(imin)
x(imin+1)
ans =  3.2824
ans =  3.2985
ans =  3.3146

Golden Search Algorithm

We can't just look for a sign change for the problem (unless we can take a derivative) so we need a new approach to determine whether we have a maximum between the two bounds.

Rather than using the midpoint of initial bounds, here the problem is more difficult. We need to compare the values of 4 function evaluations. The golden search uses the golden ratio to determine two interior points.

golden ratio

Start with bounds of 2.5 and 6 Angstrom.

% define Au atomic potential
epsilon = 0.039; % kcal/mol
sigma = 2.934; % Angstrom
Au_x= @(x) lennard_jones(x,sigma,epsilon);

% calculate golden ratio
phi = 1/2+sqrt(5)/2;
% set initial limits
x_l=2.8; 
x_u=6; 

% Iteration #1
d=(phi-1)*(x_u-x_l);

x1=x_l+d; % define point 1
x2=x_u-d; % define point 2


% evaluate Au_x(x1) and Au_x(x2)

f1=Au_x(x1);
f2=Au_x(x2);
plot(x,Au_x(x),x_l,Au_x(x_l),'ro',x2,f2,'rs',x1,f1,'gs',x_u,Au_x(x_u),'go')
hold on;

if f2<f1
    plot([x_u,x1],[Au_x(x_u),f1],'k-')
    x_u=x1;
else 
    plot([x_l,x2],[Au_x(x_l),f2],'k-')
    x_l=x1;
end
hold off
%old_min = current_min;
current_min=min([f1,f2,Au_x(x_l),Au_x(x_u)])
%error_app=abs((current_min-old_min)/current_min)
current_min = -0.019959

svg

% Iteration #2
d=(phi-1)*(x_u-x_l);

x1=x_l+d; % define point 1
x2=x_u-d; % define point 2

% evaluate Au_x(x1) and Au_x(x2)

f1=Au_x(x1);
f2=Au_x(x2);
plot(x,Au_x(x),x_l,Au_x(x_l),'ro',x2,f2,'rs',x1,f1,'gs',x_u,Au_x(x_u),'go')
hold on;

if f2<f1
    plot([x_u,x1],[Au_x(x_u),f1],'k-')
    x_u=x1;
else 
    plot([x_l,x2],[Au_x(x_l),f2],'k-')
    x_l=x1;
end
hold off
old_min = current_min;
current_min=min([f1,f2,Au_x(x_l),Au_x(x_u)])
error_app=abs((current_min-old_min)/current_min)
current_min = -0.033707
error_app =  0.40787

svg

% Iteration #3
d=(phi-1)*(x_u-x_l);

x1=x_l+d; % define point 1
x2=x_u-d; % define point 2

% evaluate Au_x(x1) and Au_x(x2)

f1=Au_x(x1);
f2=Au_x(x2);
plot(x,Au_x(x),x_l,Au_x(x_l),'ro',x2,f2,'rs',x1,f1,'gs',x_u,Au_x(x_u),'go')
hold on;

if f2<f1
    plot([x_u,x1],[Au_x(x_u),f1],'k-')
    x_u=x1;
else 
    plot([x_l,x2],[Au_x(x_l),f2],'k-')
    x_l=x1;
end
hold off
old_min = current_min;
current_min=min([f1,f2,Au_x(x_l),Au_x(x_u)])
error_app=abs((current_min-old_min)/current_min)
current_min = -0.038904
error_app =  0.13359

svg

% Iteration #3
d=(phi-1)*(x_u-x_l);

x1=x_l+d; % define point 1
x2=x_u-d; % define point 2

% evaluate Au_x(x1) and Au_x(x2)

f1=Au_x(x1);
f2=Au_x(x2);
plot(x,Au_x(x),x_l,Au_x(x_l),'ro',x2,f2,'rs',x1,f1,'gs',x_u,Au_x(x_u),'go')
hold on;

if f2<f1
    plot([x_u,x1],[Au_x(x_u),f1],'k-')
    x_u=x1;
else 
    plot([x_l,x2],[Au_x(x_l),f2],'k-')
    x_l=x1;
end
hold off
old_min = current_min;
current_min=min([f1,f2,Au_x(x_l),Au_x(x_u)])
error_app=abs((current_min-old_min)/current_min)
current_min = -0.038904
error_app = 0

svg

Parabolic Interpolation

Near a minimum/maximum, the function resembles a parabola. With three data points, it is possible to fit a parabola to the function. So, given a lower and upper bound, we can choose the midpoint and fit a parabola to the 3 x,f(x) coordinates.

$ x_{4} =x_{2} - \frac{1}{2} \frac{ \left(x_{2} -x_{1} \right)^{2} \left[f \left(x_{2} \right)-f \left(x_{3} \right) \right]- \left(x_{2} -x_{3} \right)^{2} \left[f \left(x_{2} \right)-f \left(x_{1} \right) \right]}{ \left(x_{2} -x_{1} \right) \left[f \left(x_{2} \right)-f \left(x_{3} \right) \right]- \left(x_{2} -x_{3} \right) \left[f \left(x_{2} \right)-f \left(x_{1} \right) \right]}$

Where $x_{4}$ location of the maximum or minimum of the parabola.

% define Au atomic potential
epsilon = 0.039; % kcal/mol
sigma = 2.934; % Angstrom
Au_x= @(x) lennard_jones(x,sigma,epsilon);

% set initial limits
x_l=2.8; 
x_u=3.5; 

% Iteration #1
x1=x_l;
x2=mean([x_l,x_u]);
x3=x_u;

% evaluate Au_x(x1), Au_x(x2) and Au_x(x3)
 
f1=Au_x(x1);
f2=Au_x(x2);
f3=Au_x(x3);
p = polyfit([x1,x2,x3],[f1,f2,f3],2);
x_fit = linspace(x1,x3,20);
y_fit = polyval(p,x_fit);

plot(x,Au_x(x),x_fit,y_fit,[x1,x2,x3],[f1,f2,f3],'o')
hold on
if f2<f1 && f2<f3
    x4=x2-0.5*((x2-x1)^2*(f2-f3)-(x2-x3)^2*(f2-f1))/((x2-x1)...
    *(f2-f3)-(x2-x3)*(f2-f1));
    f4=Au_x(x4);

    if x4>x2
        plot(x4,f4,'*',[x1,x2],[f1,f2])
        x1=x2;
        f1=f2;
    else
        plot(x4,f4,'*',[x3,x2],[f3,f2])
        x3=x2;
        f3=f2;
    end
    x2=x4; f2=f4;
else
    error('no minimum in bracket')
end
hold off

svg

p = polyfit([x1,x2,x3],[f1,f2,f3],2);
x_fit = linspace(x1,x3,20);
y_fit = polyval(p,x_fit);

plot(x,Au_x(x),x_fit,y_fit,[x1,x2,x3],[f1,f2,f3],'o')
hold on
if f2<f1 && f2<f3
    x4=x2-0.5*((x2-x1)^2*(f2-f3)-(x2-x3)^2*(f2-f1))/((x2-x1)...
    *(f2-f3)-(x2-x3)*(f2-f1));
    f4=Au_x(x4);

    if x4>x2
        plot(x4,f4,'*',[x1,x2],[f1,f2])
        x1=x2;
        f1=f2;
    else
        plot(x4,f4,'*',[x3,x2],[f3,f2])
        x3=x2;
        f3=f2;
    end
    x2=x4; f2=f4;
else
    error('no minimum in bracket')
end
hold off

svg

p = polyfit([x1,x2,x3],[f1,f2,f3],2);
x_fit = linspace(x1,x3,20);
y_fit = polyval(p,x_fit);

plot(x,Au_x(x),x_fit,y_fit,[x1,x2,x3],[f1,f2,f3],'o')
hold on
if f2<f1 && f2<f3
    x4=x2-0.5*((x2-x1)^2*(f2-f3)-(x2-x3)^2*(f2-f1))/((x2-x1)...
    *(f2-f3)-(x2-x3)*(f2-f1));
    f4=Au_x(x4);

    if x4>x2
        plot(x4,f4,'*',[x1,x2],[f1,f2])
        x1=x2;
        f1=f2;
    else
        plot(x4,f4,'*',[x3,x2],[f3,f2])
        x3=x2;
        f3=f2;
    end
    x2=x4; f2=f4;
else
    error('no minimum in bracket')
end
hold off

svg

Parabolic interpolation does not converge in many scenarios even though it it a bracketing method. Instead, functions like fminbnd in Matlab and Octave use a combination of the two (Golden Ratio and Parabolic)

Using the solutions to minimization for the nonlinear spring constant

Now, we have two routines to find minimums of a univariate function (Golden Ratio and Parabolic). Let's use these to solve for the minimum energy given a range of applied forces to the single atom gold chain

$E_{total}(\Delta x) = E_{LJ}(x_{min}+\Delta x) - F \cdot \Delta x$

epsilon = 0.039; % kcal/mol
epsilon = epsilon*6.9477e-21; % J/atom
epsilon = epsilon*1e18; % aJ/J
% final units for epsilon are aJ

sigma = 2.934; % Angstrom
sigma = sigma*0.10; % nm/Angstrom
x=linspace(2.8,6,200)*0.10; % bond length in um

Ex = lennard_jones(x,sigma,epsilon);

%[Emin,imin]=min(Ex);

[xmin,Emin] = fminbnd(@(x) lennard_jones(x,sigma,epsilon),0.28,0.6)

plot(x,Ex,xmin,Emin,'o')
ylabel('Lennard Jones Potential (aJ/atom)')
xlabel('bond length (nm)')

Etotal = @(dx,F) lennard_jones(xmin+dx,sigma,epsilon)-F.*dx;

xmin =  0.32933
Emin =   -2.7096e-04

svg

N=50;
dx = zeros(1,N); % [in nm]
F_applied=linspace(0,0.0022,N); % [in nN]
for i=1:N
    optmin=goldmin(@(dx) Etotal(dx,F_applied(i)),-0.001,0.035);
    dx(i)=optmin;
end

plot(dx,F_applied)
xlabel('dx (nm)')
ylabel('Force (nN)')
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svg

For this function, it is possible to take a derivative and compare the analytical result:

dx_full=linspace(0,0.06,50);
F= @(dx) 4*epsilon*6*(sigma^6./(xmin+dx).^7-2*sigma^12./(xmin+dx).^13)
plot(dx_full,F(dx_full),dx,F_applied)
F =

@(dx) 4 * epsilon * 6 * (sigma ^ 6 ./ (xmin + dx) .^ 7 - 2 * sigma ^ 12 ./ (xmin + dx) .^ 13)

svg

Curve-fitting

Another example is minimizing error in your approximation of a function. If you have data (now we have Force-displacement data) we can fit this to a function, such as:

$F(x) = K_{1}\Delta x + \frac{1}{2} K_{2}(\Delta x)^{2}$

function SSE = sse_of_parabola(K,xdata,ydata)
    % calculate the sum of squares error for a parabola given a function, func, and xdata and ydata
    % output is SSE=sum of squares error
    K1=K(1);
    K2=K(2);
    y_function = K1*xdata+1/2*K2*xdata.^2;
    SSE = sum((ydata-y_function).^2);
end
    
[K,SSE_min]=fminsearch(@(K) sse_of_parabola(K,dx,F_applied),[1,1]);
fprintf('\nThe nonlinear spring constants are K1=%1.2f nN/nm and K2=%1.2f nN/nm^2\n',K)
fprintf('The mininum sum of squares error = %1.2e',SSE_min)
The nonlinear spring constants are K1=0.16 nN/nm and K2=-5.98 nN/nm^2
The mininum sum of squares error = 7.35e-08

plot(dx,F_applied,'o',dx,K(1)*dx+1/2*K(2)*dx.^2)

svg