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[求助]
求大神幫忙解決一下,用MATLAB求解動力學(xué)數(shù)據(jù)總是出錯~ 已有1人參與
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如題,本人是個Matlab方面的小白,臨時要用到,但自己糾結(jié)了好久還是沒弄會,只好來向諸位大師求助了,代碼及問題如下,多謝了 錯誤使用 odearguments (line 92) KINETICEQS 返回的矢量的長度為 5,但初始條件矢量的長度為 6。KINETICEQS 返回的矢量和初始條件矢量的元素數(shù)目必須相同。 出錯 ode45 (line 113) [neq, tspan, ntspan, next, t0, tfinal, tdir, y0, f0, odeArgs, odeFcn, ... 出錯 KineticsEst>ObjFunc4Fmincon (line 48) [t x] = ode45(@KineticEqs,tspan,x0,[],k); 出錯 fmincon (line 564) initVals.f = feval(funfcn{3},X,varargin{:}); 出錯 KineticsEst (line 19) [k,fval,flag] = fmincon(@ObjFunc4Fmincon,k0,[],[],[],[],lb,ub,[],[],x0,yexp); 原因: Failure in initial user-supplied objective function evaluation. FMINCON cannot continue. |
鐵蟲 (小有名氣)
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>> KineticsEst5 Local minimum possible. Constraints satisfied. fmincon stopped because the size of the current step is less than the default value of the step size tolerance and constraints are satisfied to within the default value of the constraint tolerance. <stopping criteria details> 使用函數(shù)fmincon()估計得到的參數(shù)值為: k1 = 0.0358 k2 = 0.0021 k3 = 5.6484 k4 = 22.7533 k5 = 14.7665 The sum of the squares is: 1.5e+00 Local minimum possible. lsqnonlin stopped because the final change in the sum of squares relative to its initial value is less than the default value of the function tolerance. <stopping criteria details> 使用函數(shù)lsqnonlin()估計得到的參數(shù)值為: k1 = 0.8360 k2 = 0.0128 k3 = 3.0166 k4 = 38.9765 k5 = 7.6220 Local minimum possible. lsqnonlin stopped because the size of the current step is less than the default value of the step size tolerance. <stopping criteria details> 以fmincon()的結(jié)果為初值,使用函數(shù)lsqnonlin()估計得到的參數(shù)值為: k1 = 0.0358 k2 = 0.0021 k3 = 5.6484 k4 = 22.7533 k5 = 14.7665 |

鐵蟲 (小有名氣)

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