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Unable to find value of T1

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MINATI PATRA
MINATI PATRA 2019 年 9 月 29 日
コメント済み: Walter Roberson 2019 年 10 月 10 日
syms L m x xx s theta_b h Z y U T0 T1 T2 k0 b Tinf Tb sumsol
k=6;
sym('u0(x)');('u1(x)');('u2(x)');('u3(x)');('u4(x)');('u5(x)');('u6(x)');('u7(x)');('u8(x)');('u9(x)');('u10(x)');
('u11(x)');('u12(x)');('u13(x)');('u14(x)');('u15(x)');('u16(x)');('u17(x)');('u18(x)');('u19(x)');('u20(x)');
s='u0(x)'+h*'u1(x)'+h^2*'u2(x)'+h^3*'u3(x)'+h^4*'u4(x)'+h^5*'u5(x)'+h^6*'u6(x)'+h^7*'u7(x)'+h^8*'u8(x)'+h^9*'u9(x)'...
+h^10*'u10(x)'+h^11*'u11(x)'+h^12*'u12(x)'+h^13*'u13(x)'+h^14*'u14(x)'+h^15*'u15(x)'+h^16*'u16(x)'+h^17*'u17(x)'...
+h^18*'u18(x)'+h^19*'u19(x)'+h^20*'u20(x)';
Ak=(1/factorial(k))*subs(diff(1/s*(diff(s,x))^2,h,k),h,0);Bk=(1/factorial(k))*subs(diff(s^(1-Z),h,k),h,0);Ck=(1/factorial(k))*subs(diff(s^(-Z),h,k),h,0);
Linverse = int(1/xx*int(y*x,x,0,xx),xx,L,x);
L=0.05;Tinf=273+25;Tb=273+150;h=4;beta=-1.3;Z=-1.3;k0=148*300^-beta;b=0.005;
T1=-beta*Linverse(subs(Ak(0),sym('u0(x)'),T0))+2*h*L/k0/b*Linverse(subs(Bk(0)/x,sym('u0(x)'),T0)) -2*h*L*Tinf/k0/b*Linverse(subs(Ck(0)/x,sym('u0(x)'),T0))
  5 件のコメント
Walter Roberson
Walter Roberson 2019 年 9 月 30 日
I guarantee that the code you posted never worked.
Ak=(1/factorial(k))*subs(diff(1/s*(diff(s,x))^2,h,k),h,0);Bk=(1/factorial(k))*subs(diff(s^(1-Z),h,k),h,0);Ck=(1/factorial(k))*subs(diff(s^(-Z),h,k),h,0);
That is a symbolic expression.
Ak(0)
That attempts to index the symbolic expression at location 0. 0 is not a valid subscript for symbolic expressions.
Ak(0) and Bk(0) and Ck(0) only make sense if Ak, Bk, and Ck are either functions or symbolic functions. Converting them to symbolic functions of one variable is not difficult, but which variable?
Likewise, Linverse(symbolic expression) only makes sense if Linverse is a function or symbolic function. You can convert to symbolic function, but function of what variable?
sym('u0(x)');('u1(x)');('u2(x)');('u3(x)');('u4(x)');('u5(x)');('u6(x)');('u7(x)');('u8(x)');('u9(x)');('u10(x)');
That creates the symbolic expression u0(x) in sufficiently old versions of MATLAB (in more recent versions it is an error.) It then throws away that symbolic expression. Then it creates the character vector 'u1(x)' and throws away the character vector. The sym() call does not enclose the 'u1(x)'
syms u0(x) u1(x) u2(x) u3(x) u4(x) u5(x) u6(x) u7(x) u8(x) u9(x) u10(x) u11(x) u12(x) u13(x) u14(x) u15(x) u16(x) u17(x) u18(x) u19(x) u20(x)
would be meaningful MATLAB.
s='u0(x)'+h*'u1(x)'+h^2*'u2(x)'+h^3*'u3(x)'+h^4*'u4(x)'+h^5*'u5(x)'+h^6*'u6(x)'+h^7*'u7(x)'+h^8*'u8(x)'+h^9*'u9(x)'...
That works more accidentally in older versions. 'u0(x)' is functioning as a character vector there, not as a symbolic expression. If you had 'u0(x)' + 'h*u1(x)' then that would fail due to trying to add character vectors of different lengths.
s = u0(x)+h*u1(x)+h^2*u2(x)+h^3*u3(x)+h^4*u4(x)+h^5*u5(x)+h^6*u6(x)+h^7*u7(x)+h^8*u8(x)+h^9*u9(x)...

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回答 (1 件)

Walter Roberson
Walter Roberson 2019 年 9 月 30 日
The below can at least execute without error. It is not clear that it is using the proper variable as the parameters for Ak, Bk, Ck.
syms L m x xx s theta_b h Z y U T0 T1 T2 k0 b Tinf Tb sumsol
k=6;
syms u0(x) u1(x) u2(x) u3(x) u4(x) u5(x) u6(x) u7(x) u8(x) u9(x) u10(x) u11(x) u12(x) u13(x) u14(x) u15(x) u16(x) u17(x) u18(x) u19(x) u20(x)
s = u0(x)+h*u1(x)+h^2*u2(x)+h^3*u3(x)+h^4*u4(x)+h^5*u5(x)+h^6*u6(x)+h^7*u7(x)+h^8*u8(x)+h^9*u9(x)...
+h^10*u10(x)+h^11*u11(x)+h^12*u12(x)+h^13*u13(x)+h^14*u14(x)+h^15*u15(x)+h^16*u16(x)+h^17*u17(x)...
+h^18*u18(x)+h^19*u19(x)+h^20*u20(x);
Ak(x)=(1/factorial(k))*subs(diff(1/s*(diff(s,x))^2,h,k),h,0);
Bk(x)=(1/factorial(k))*subs(diff(s^(1-Z),h,k),h,0);
Ck(x)=(1/factorial(k))*subs(diff(s^(-Z),h,k),h,0);
Linverse(x) = int(1/xx*int(y*x,x,0,xx),xx,L,x);
L=0.05;
Tinf=273+25;
Tb=273+150;
h=4;
beta=-1.3;
Z=-1.3;
k0=148*300^-beta;
b=0.005;
T1_raw = -beta*Linverse(subs(Ak(0),u0(x),T0))+2*h*L/k0/b*Linverse(subs(Bk(0)/x,u0(x),T0)) -2*h*L*Tinf/k0/b*Linverse(subs(Ck(0)/x,u0(x),T0));
T1 = simplify(subs(T1_raw));
disp(T1)
  2 件のコメント
Walter Roberson
Walter Roberson 2019 年 10 月 10 日
It appears you used a different login name before, so it does not know you are the same person who asked the Question. I cannot Accept it myself, as I cannot Accept my own answers .

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