I have asked this question two days ago. Now I re-derive the formulas and write new codes. But I still encounter the same error in using integral2.
clear
clc
% parameter settings
syms KI c r phi
KId = 0.000005;
G = 76.0;
nu = 0.31;
Vh = 2;
NA = 6.02214129*10^23;
b = 3.52*sqrt(6)/6;
A0 = G*b^2*(2-nu)/(8*pi*(1-nu));
lambda = 2;
m = 2.2;
kapa = 1.0;
beff = b;
delta2 = 1.0;
D = 1.9*10^(-14);
kB = 1.38064852*10^(-23);
R = 8.3144598;
E = 250;
Rc = 2*b;
for c0=0.0000005:0.0000005:0.0001
for T=1:1:300
for rou=b:b:10*b
for theta=0:0.01*pi:pi
h = (5*(1+nu)*D*Vh*cos(theta/2)*KI^2/(12*sqrt(2*pi)*kB*NA*T*KId))^(2/5);
B0 = G*Vh*c*h^2/(2*pi*(1-nu));
sigma1122 = ((cos(2*theta)*rou^2 + cos(2*phi)*r^2 - cos(theta - phi)*2*rou*r)*((sqrt(rou/r) + sqrt(r/rou))*cos((theta + phi)/2) - 2) + (sin(2*theta)*rou^2 - sin(2*phi)*r^2 - sin(theta - phi)*2*rou*r)*(sqrt(rou/r) - sqrt(r/rou))*sin((theta + phi)/2))*B0/((cos(2*theta)*rou^2 + cos(2*phi)*r^2 - cos(theta - phi)*2*rou*r)^2 + (sin(2*theta)*rou^2 - sin(2*phi)*r^2 - sin(theta - phi)*2*rou*r)^2); % 2*(psiHc+conj(psiHc));
sigma12 = KI*sin(theta)*cos(theta/2)/(2*sqrt(2*pi*rou));
c = c0*exp(((1+nu)*sigma1122/3-2*E*Vh*(c-c0)/(9*(1-nu)))*Vh/(R*T));
fun = #(r,phi) c*sin(2*phi)/r;
eq = pi*rou*A0*(log(8*rou*m/(lambda*b))-2)+kapa*G*beff^2*delta2-0.5*b*pi*rou^2*(sigma12-(G*Vh/(2*pi*(1-nu)*NA))*integral2(fun,0,Rc,0,2*pi));
solve(eq==0,KI);
end
end
end
end
The error information is:
Warning: The system is inconsistent. Solution does not exist.
In symengine (line 57)
In sym/privBinaryOp (line 903)
In / (line 297)
In #(r,phi)c*sin(2*phi)/r
In integral2Calc>integral2t/tensor (line 228)
In integral2Calc>integral2t (line 55)
In integral2Calc (line 9)
In integral2 (line 106)
In KIpredictor (line 44)
Error using integral2Calc>integral2t/tensor (line 231)
Input function must return 'double' or 'single' values. Found 'sym'.
Error in integral2Calc>integral2t (line 55)
[Qsub,esub] = tensor(thetaL,thetaR,phiB,phiT);
Error in integral2Calc (line 9)
> [q,errbnd] = integral2t(fun,xmin,xmax,ymin,ymax,optionstruct);
Error in integral2 (line 106)
> Q = integral2Calc(fun,xmin,xmax,yminfun,ymaxfun,opstruct);
Error in KIpredictor (line 44)
> eq =
pi*rou*A0*(log(8*rou*m/(lambda*b))-2)+kapa*G*beff^2*delta2-0.5*b*pi*rou^2*(sigma12-(G*Vh/(2*pi*(1-nu)*NA))*integral2(fun,0,Rc,0,2*pi));
I think the problem is due the complexity of the integrand. A detailed description of the integand is in the attached figure.
formula
Related
I'm trying to calculate a double integral but get an error.
My code is:
clear
Gamma = rand([22,1]);
data_i = rand([1,10]);
Y_1 = 1;
fun = #(D_star, Y_0 ) fun1(D_star , Y_1 , Y_0 , Gamma , data_i);
p = integral2 ( fun , 0 , 1 , 0 , 1);
and the function fun1 is defined as
function [p] = fun1(D_star , Y_1 , Y_0 , Gamma , data_i)
gamma=Gamma(1:3);
beta_1=Gamma(4:5);
beta_0=Gamma(6:7);
alpha_D=Gamma(8);
alpha_1=Gamma(9);
alpha_0=Gamma(10);
sigma2_1=Gamma(11);
sigma2_0=Gamma(12);
Lambda=Gamma(13:17);
sigma2_M=Gamma(18:22);
Sigma2_temp = [1 ; sigma2_1 ; sigma2_0 ; sigma2_M];
Alpha = [alpha_D ; alpha_1 ; alpha_0 ; Lambda];
Sigma2 = Alpha * Alpha' + diag(Sigma2_temp);
Z = data_i(3:5);
X = data_i(3:4);
M = data_i(6:10);
Mu = [Z*gamma , X*beta_1 , X*beta_0 , zeros(1,5) ]';
YY = [D_star ; Y_1 ; Y_0 ; M' ];
p = mvnpdf(YY,Mu,Sigma2);
end
I don't think there is a problem in the definition of the functions because I can evaluate them (e.g. fun(1,1) give me an answer). The error message that I get is:
Error using vertcat
Dimensions of matrices being concatenated are not consistent.
Error in fun1 (line 23)
YY = [D_star ; Y_1 ; Y_0 ; M' ];
Error in #(D_star,Y_0)fun1(D_star,Y_1,Y_0,Gamma,data(n,:))
Error in integral2Calc>integral2t/tensor (line 228)
Z = FUN(X,Y); NFE = NFE + 1;
Error in integral2Calc>integral2t (line 55)
[Qsub,esub] = tensor(thetaL,thetaR,phiB,phiT);
Error in integral2Calc (line 9)
[q,errbnd] = integral2t(fun,xmin,xmax,ymin,ymax,optionstruct);
Error in integral2 (line 106)
Q = integral2Calc(fun,xmin,xmax,yminfun,ymaxfun,opstruct);
I tried also the command -quad2d- instead of -integral2- and got a similar error. Any ideas?
I found the problem. According to the documentation, when using -integral2-, the function inside the integral must get an array of both variables as an input and return an array as an output.
I get an error when taking the gradient of a symbolic function. Can anyone tell me why I get this error?
syms x1 x2 R
L0=0; %[m]
k1=8; %[N/m]
k2=4; %[N/m]
F1=5; %[N]
F2=10; %[N]
F = 0.5*k1*(sqrt(x1^2 + (L0-x2)^2) - L0)^2 + 0.5*k2*(sqrt(x1^2 + (L0+x2)^2)
- L0)^2 - F1*x1 - F2*x2;
f = matlabFunction(F);
R = 0.1;
GI = x1^2 + x2^2 - R^2;
gi = matlabFunction(GI);
epsilon=0.001;
xo=[0,0]';
k = 0;
r = (sqrt(5)-1) / 2;
rpe=0.01;
Merit_pe = #(x1,x2) f(x1,x2) + rpe*(max(0,gi(x1,x2)))^2;
g = gradient(Merit_pe, [x1 x2]);
Error:
Error using sym/max (line 97)
Input arguments must be convertible to
floating-point numbers.
Error in
Ex>#(x1,x2)f(x1,x2)+rpe*(max(0,gi(x1,x2)))^2
Error in sym>funchandle2ref (line 1249)
S = x(S{:});
Error in sym>tomupad (line 1154)
x = funchandle2ref(x);
Error in sym (line 163)
S.s = tomupad(x);
Error in sym/gradient (line 17)
args = privResolveArgs(sym(f));
Error in Ex (line 31)
g = gradient(Merit_pe, [x1 x2])
I think the max part is causing me trouble, but I still need to determine the gradient of this function. Any advise? (I could do it by hand I guess, but I'd rather not if I don't have to)
Directly taking the gradient of max(0,gi(x1,x2)) is not possible in matlab. Instead the function should be defined according to the following definition.
The function Merit_pe can then be defined as follows:
if gi(xo(1,1),xo(2,1)) > 0
Merit_pe = #(x1,x2) f(x1,x2) + rpe*(gi(x1,x2))^2;
else
Merit_pe = #(x1,x2) f(x1,x2);
end
The gradient can then be determined using:
g = gradient(Merit_pe, [x1 x2]);
as you can see in the last few lines below, I'm trying to store my function on multiple variables because it gets rather ugly. However, doing this yields the error shown below.
A fix for this would be to manually substitute k and kp, but this is precisely what I'm trying to avoid. Any help would be appreciated. thanks!
clc
clear
hbar = 1.055e-34;
mo = 9.1095e-31;
q = 1.602e-19;
kb=1.38065e-23;
T=298;
Ef = -0.1*q; % -100meV in units Joules
V0 = 1*q;
L = 1e-9;
k = #(E) (2*mo*E/hbar.^2)^.5;
kp = #(E) (2*mo*(V0-E)/hbar.^2)^.5;
fun = #(E) (exp(-2*kp*L) .* ((16*k.^2 .* kp.^2) ./ ((k.^2 + kp.^2).^2))) .* exp(-E./(kb*T));
Q = integral(fun,0,inf);
Error below
Undefined operator '*' for input arguments of type 'function_handle'.
Error in #(E)(exp(-2*kp*L).*((16*k.^2.*kp.^2)./((k.^2+kp.^2).^2))).*exp(-E./(kb*T))
Error in integralCalc/iterateScalarValued (line 314)
fx = FUN(t);
Error in integralCalc/vadapt (line 132)
[q,errbnd] = iterateScalarValued(u,tinterval,pathlen);
Error in integralCalc (line 83)
[q,errbnd] = vadapt(#AToInfInvTransform,interval);
Error in integral (line 88)
Q = integralCalc(fun,a,b,opstruct);
Error in PS3_2 (line 17)
Q = integral(fun,0,inf);
Use this here
k = #(E) (2.*mo.*E./hbar.^2).^.5;
kp = #(E) (2.*mo.*(V0-E)./hbar.^2).^.5;
fun = #(E) (exp(-2*kp(E)*L).*((16*k(E).^2.*kp(E).^2)./((k(E).^2+kp(E).^2).^2))).*exp(-E./(kb*T));
Q = integral(fun,0,inf);
I think you need to pass the argument E, then are you sure kb*T is correct? Maybe kp(E)*T? Then, you forgot the . is the sqrt for k and kp, or if it isn't a sqrt then the dot is on the wrong side of the power symbol.
I want to calculate the part-expectation of log-normal distribution via:
m = 1;
v = 2;
mu = log((m^2)/sqrt(v+m^2));
sigma = sqrt(log(v/(m^2)+1));
syms x;
d = x*lognpdf(x,mu,sigma);
int(d, x, 0, 10);
However, MATLAB says:
Error using symfun>validateArgNames (line 211) Second input must be a
scalar or vector of unique symbolic variables.
Error in symfun (line 45)
y.vars = validateArgNames(inputs);
Error in sym/subsasgn (line 771)
C = symfun(B,[inds{:}]);
Error in lognpdf (line 36) x(x <= 0) = Inf;
Error in untitled (line 7) d = x*lognpdf(x,mu,sigma);
I even tried to just calculate the integral of the pdf by:
m = 1;
v = 2;
mu = log((m^2)/sqrt(v+m^2));
sigma = sqrt(log(v/(m^2)+1));
syms x;
d = lognpdf(x,mu,sigma);
int(d, x, 0, 10);
But there are still errors, and MATLAB says:
Error using symfun>validateArgNames (line 211) Second input must be a
scalar or vector of unique symbolic variables.
Error in symfun (line 45)
y.vars = validateArgNames(inputs);
Error in sym/subsasgn (line 771)
C = symfun(B,[inds{:}]);
Error in lognpdf (line 36) x(x <= 0) = Inf;
Error in untitled (line 7) d = lognpdf(x,mu,sigma);
I really don't know what happened. Should the integral of the pdf be the cdf?
Similar to an answer several months ago, the Statistics Toolbox doesn't support the Symbolic Toolbox currently.
Therefore, you can proceed by hard coding the PDF itself and integrating it:
d = exp(-(log(x)-mu)^2/(2*sigma^2))/(x*sigma*sqrt(2*pi));
int(d, x, 0, 10);
Or you can use the logncdf function, which may be cleaner.
I got 2 questions here.
1. If I have y = a*x^2 + 5. What function can make it into y = a.*x.^2 +5. As you seen, dot was inserted.
It's easy, but kinda diffcult to describe, but please have patience with me. Thank you so much.
First, let me make a very simple example of my problem.
If I want to calucate Y = F(x=1)+ 2.^2, and I know F(x=1) = a+b,(a and b are syms). This means Y = a + b + 4. Problem is here,
Matlab give me error if I write down as.
F = function( .... ); <==== output of function is F(X=1), and = F(x=1) = a+b
Y = integral2( F + 2.^2, .. , .. ,..)
However, if I just copy the output of F as
Y = integral2( a+b + 2.^2, .. , .. ,..)
Now it works!!!
Ok. Please allow me to talk about my code here. I am trying to find a double interation by using integral2. One part of my equcation(which is Y) is from another int output(which is F). Matlab will give ERROR for code below:
clear all;
a=4;
la1=1/(pi*500^2); la2= la1*5;
p1=25; p2=p1/25;
sgma2=10^(-11);
index=1;
g=2./a;
syms r u1 u2
powe= -2 ;
seta= 10^powe;
xNor = ( (u2./u1).^(a./2) + 1 ).^(2./a);
x = (xNor).^(0.5) * seta^(-1/a);
fun1 = r./(1+ r.^a );
out1 = int(fun1, x, Inf) ; %== This is my F in my example
q=pi.*(la1.*p1.^(2./a)+la2.*p2.^(2./a));
yi = #(u2,u1) exp(-u2.*(1+2.*...
( out1 )./... %=== out1 is the problem here.
( (( (u2./u1).^(a./2) + 1 ).^(2./a)).*seta.^(-2./a)))).*...
exp(-sgma2.*q.^(-a./2).* seta.*u2.^(a./2)./...
((( (u2./u1).^(a./2) + 1 ).^(2./a)).^(a./2)) );
maxF =#(u2) u2;
out2 = integral2(yi,0,Inf,0 ,maxF) % == this is Y in my previous example.
However, since I know the out1 = pi/4 - atan(10*(u2^2/u1^2 + 1)^(1/2))/2 (no dot,1/2, not 1./2). Instead of writing down out1, I will just type the equaction and add dot in the
yi = #(u2,u1) exp(-u2.*(1+2.*...
( pi./4 - atan(10.*(u2.^2./u1.^2 + 1).^(1./2))./2 )./... %=== not "out1"
( (( (u2./u1).^(a./2) + 1 ).^(2./a)).*seta.^(-2./a)))).*...
exp(-sgma2.*q.^(-a./2).* seta.*u2.^(a./2)./...
((( (u2./u1).^(a./2) + 1 ).^(2./a)).^(a./2)) );
Now the code is working!!!! The final output is = 0.9957.
Dear friends, I already spend a long time on this, but I still can not find out the problem. Could you please take a deeper look for me. Please copy the code to you matlab and test. Thank you so much.
Below is the error given by matlab, if I just use "out1" in yi = #(u2,u1) ......
Error using integralCalc/finalInputChecks (line 511)
Input function must return 'double' or 'single' values. Found 'sym'.
Error in integralCalc/iterateScalarValued (line 315)
finalInputChecks(x,fx);
Error in integralCalc/vadapt (line 133)
[q,errbnd] = iterateScalarValued(u,tinterval,pathlen);
Error in integralCalc (line 76)
[q,errbnd] = vadapt(#AtoBInvTransform,interval);
Error in
integral2Calc>#(xi,y1i,y2i)integralCalc(#(y)fun(xi*ones(size(y)),y),y1i,y2i,opstruct.integralOptions)
(line 18)
innerintegral = #(x)arrayfun(#(xi,y1i,y2i)integralCalc( ...
Error in
integral2Calc>#(x)arrayfun(#(xi,y1i,y2i)integralCalc(#(y)fun(xi*ones(size(y)),y),y1i,y2i,opstruct.integralOptions),x,ymin(x),ymax(x))
(line 18)
innerintegral = #(x)arrayfun(#(xi,y1i,y2i)integralCalc( ...
Error in integralCalc/iterateScalarValued (line 314)
fx = FUN(t);
Error in integralCalc/vadapt (line 133)
[q,errbnd] = iterateScalarValued(u,tinterval,pathlen);
Error in integralCalc (line 84)
[q,errbnd] = vadapt(#AToInfInvTransform,interval);
Error in integral2Calc>integral2i (line 21)
[q,errbnd] = integralCalc(innerintegral,xmin,xmax,opstruct.integralOptions);
Error in integral2Calc (line 8)
[q,errbnd] = integral2i(fun,xmin,xmax,ymin,ymax,optionstruct);
Error in integral2 (line 107)
Q = integral2Calc(fun,xmin,xmax,yminfun,ymaxfun,opstruct);
Error in ref7_equ11n2 (line 129)
out2 = integral2(yi,0,Inf,0 ,maxF)
The documentation of the function integral2 states:
integral2 - Numerically evaluate double integral
You won't be able to use integral2 with symbolic expressions, unless you convert them to function handles using matlabfunction.