दूसरे क्रम के टेंसरों के संबंध में अदिश (गणित), यूक्लिडियन सदिश और दूसरे क्रम के टेंसर के दिशात्मक व्युत्पन्न का सातत्य यांत्रिकी में अधिक उपयोग होता हैं। इन व्युत्पन्न का उपयोग अरेखीय लोच और प्लास्टिसिटी (भौतिकी) के सिद्धांतों में किया जाता है, विशेष रूप से संख्यात्मक अनुकरण के लिए एल्गोरिदम के डिजाइन में उपयोग किया जाता है।[1]
इस प्रकार दिशात्मक व्युत्पन्न इन व्युत्पन्नों को खोजने की व्यवस्थित विधि प्रदान करते है।[2]
सदिश और दूसरे क्रम के टेंसर के संबंध में व्युत्पन्न विभिन्न स्थितियों के लिए दिशात्मक व्युत्पन्न की परिभाषाएँ नीचे दी गई हैं। अतः यह माना जाता है कि कार्य पर्याप्त रूप से सुचारू होते हैं कि व्युत्पन्न लिया जा सकता है।
सदिशों के अदिश मान वाले कार्यों के व्युत्पन्न्स मान लीजिए कि f('v') सदिश 'v' का वास्तविक मान फलन है। फिर 'v' (या 'v' पर) के संबंध में f('v') का व्युत्पन्न 'सदिश' अपने डॉट उत्पाद के माध्यम से किसी भी वेक्टर यू के साथ परिभाषित किया गया है।
∂ f ∂ v ⋅ u = D f ( v ) [ u ] = [ d d α f ( v + α u ) ] α = 0 {\displaystyle {\frac {\partial f}{\partial \mathbf {v} }}\cdot \mathbf {u} =Df(\mathbf {v} )[\mathbf {u} ]=\left[{\frac {d}{d\alpha }}~f(\mathbf {v} +\alpha ~\mathbf {u} )\right]_{\alpha =0}}
सभी सदिश यू के लिए उपरोक्त डॉट उत्पाद अदिश उत्पन्न करता है और यदि यू इकाई सदिश होती है तब यू दिशा में वी पर 'एफ' का दिशात्मक व्युत्पन्न देता है।
गुण:
यदि f ( v ) = f 1 ( v ) + f 2 ( v ) {\displaystyle f(\mathbf {v} )=f_{1}(\mathbf {v} )+f_{2}(\mathbf {v} )} ∂ f ∂ v ⋅ u = ( ∂ f 1 ∂ v + ∂ f 2 ∂ v ) ⋅ u {\displaystyle {\frac {\partial f}{\partial \mathbf {v} }}\cdot \mathbf {u} =\left({\frac {\partial f_{1}}{\partial \mathbf {v} }}+{\frac {\partial f_{2}}{\partial \mathbf {v} }}\right)\cdot \mathbf {u} }
यदि f ( v ) = f 1 ( v ) f 2 ( v ) {\displaystyle f(\mathbf {v} )=f_{1}(\mathbf {v} )~f_{2}(\mathbf {v} )} ∂ f ∂ v ⋅ u = ( ∂ f 1 ∂ v ⋅ u ) f 2 ( v ) + f 1 ( v ) ( ∂ f 2 ∂ v ⋅ u ) {\displaystyle {\frac {\partial f}{\partial \mathbf {v} }}\cdot \mathbf {u} =\left({\frac {\partial f_{1}}{\partial \mathbf {v} }}\cdot \mathbf {u} \right)~f_{2}(\mathbf {v} )+f_{1}(\mathbf {v} )~\left({\frac {\partial f_{2}}{\partial \mathbf {v} }}\cdot \mathbf {u} \right)}
यदि f ( v ) = f 1 ( f 2 ( v ) ) {\displaystyle f(\mathbf {v} )=f_{1}(f_{2}(\mathbf {v} ))} ∂ f ∂ v ⋅ u = ∂ f 1 ∂ f 2 ∂ f 2 ∂ v ⋅ u {\displaystyle {\frac {\partial f}{\partial \mathbf {v} }}\cdot \mathbf {u} ={\frac {\partial f_{1}}{\partial f_{2}}}~{\frac {\partial f_{2}}{\partial \mathbf {v} }}\cdot \mathbf {u} } सदिशों के सदिश मूल्यवान कार्यों के व्युत्पन्न चूँकि f(v) सदिश v का सदिश मान फलन होता है। फिर v (या v पर) के संबंध में f(v) का व्युत्पन्न दूसरा क्रम टेन्सर है जो इसके डॉट उत्पाद के माध्यम से किसी भी सदिश यू के साथ परिभाषित किया गया है।
∂ f ∂ v ⋅ u = D f ( v ) [ u ] = [ d d α f ( v + α u ) ] α = 0 {\displaystyle {\frac {\partial \mathbf {f} }{\partial \mathbf {v} }}\cdot \mathbf {u} =D\mathbf {f} (\mathbf {v} )[\mathbf {u} ]=\left[{\frac {d}{d\alpha }}~\mathbf {f} (\mathbf {v} +\alpha ~\mathbf {u} )\right]_{\alpha =0}}
सभी सदिश यू के लिए। उपरोक्त डॉट उत्पाद सदिश उत्पन्न करता है, और यदि यू इकाई सदिश है, तो दिशात्मक यू में, v पर f का व्युत्पन्न देता है।
गुण:
यदि f ( v ) = f 1 ( v ) + f 2 ( v ) {\displaystyle \mathbf {f} (\mathbf {v} )=\mathbf {f} _{1}(\mathbf {v} )+\mathbf {f} _{2}(\mathbf {v} )} ∂ f ∂ v ⋅ u = ( ∂ f 1 ∂ v + ∂ f 2 ∂ v ) ⋅ u {\displaystyle {\frac {\partial \mathbf {f} }{\partial \mathbf {v} }}\cdot \mathbf {u} =\left({\frac {\partial \mathbf {f} _{1}}{\partial \mathbf {v} }}+{\frac {\partial \mathbf {f} _{2}}{\partial \mathbf {v} }}\right)\cdot \mathbf {u} }
यदि f ( v ) = f 1 ( v ) × f 2 ( v ) {\displaystyle \mathbf {f} (\mathbf {v} )=\mathbf {f} _{1}(\mathbf {v} )\times \mathbf {f} _{2}(\mathbf {v} )} ∂ f ∂ v ⋅ u = ( ∂ f 1 ∂ v ⋅ u ) × f 2 ( v ) + f 1 ( v ) × ( ∂ f 2 ∂ v ⋅ u ) {\displaystyle {\frac {\partial \mathbf {f} }{\partial \mathbf {v} }}\cdot \mathbf {u} =\left({\frac {\partial \mathbf {f} _{1}}{\partial \mathbf {v} }}\cdot \mathbf {u} \right)\times \mathbf {f} _{2}(\mathbf {v} )+\mathbf {f} _{1}(\mathbf {v} )\times \left({\frac {\partial \mathbf {f} _{2}}{\partial \mathbf {v} }}\cdot \mathbf {u} \right)}
यदि f ( v ) = f 1 ( f 2 ( v ) ) {\displaystyle \mathbf {f} (\mathbf {v} )=\mathbf {f} _{1}(\mathbf {f} _{2}(\mathbf {v} ))} ∂ f ∂ v ⋅ u = ∂ f 1 ∂ f 2 ⋅ ( ∂ f 2 ∂ v ⋅ u ) {\displaystyle {\frac {\partial \mathbf {f} }{\partial \mathbf {v} }}\cdot \mathbf {u} ={\frac {\partial \mathbf {f} _{1}}{\partial \mathbf {f} _{2}}}\cdot \left({\frac {\partial \mathbf {f} _{2}}{\partial \mathbf {v} }}\cdot \mathbf {u} \right)} दूसरे क्रम के टेंसरों के अदिश वैल्यू वाले कार्यों के व्युत्पन्न होने देना f ( S ) {\displaystyle f({\boldsymbol {S}})} S {\displaystyle {\boldsymbol {S}}} f ( S ) {\displaystyle f({\boldsymbol {S}})} S {\displaystyle {\boldsymbol {S}}} S {\displaystyle {\boldsymbol {S}}} T {\displaystyle {\boldsymbol {T}}}
∂ f ∂ S : T = D f ( S ) [ T ] = [ d d α f ( S + α T ) ] α = 0 {\displaystyle {\frac {\partial f}{\partial {\boldsymbol {S}}}}:{\boldsymbol {T}}=Df({\boldsymbol {S}})[{\boldsymbol {T}}]=\left[{\frac {d}{d\alpha }}~f({\boldsymbol {S}}+\alpha ~{\boldsymbol {T}})\right]_{\alpha =0}}
सभी दूसरे क्रम के टेंसरों के लिए
T {\displaystyle {\boldsymbol {T}}} .
गुण:
यदि f ( S ) = f 1 ( S ) + f 2 ( S ) {\displaystyle f({\boldsymbol {S}})=f_{1}({\boldsymbol {S}})+f_{2}({\boldsymbol {S}})} ∂ f ∂ S : T = ( ∂ f 1 ∂ S + ∂ f 2 ∂ S ) : T {\displaystyle {\frac {\partial f}{\partial {\boldsymbol {S}}}}:{\boldsymbol {T}}=\left({\frac {\partial f_{1}}{\partial {\boldsymbol {S}}}}+{\frac {\partial f_{2}}{\partial {\boldsymbol {S}}}}\right):{\boldsymbol {T}}}
यदि f ( S ) = f 1 ( S ) f 2 ( S ) {\displaystyle f({\boldsymbol {S}})=f_{1}({\boldsymbol {S}})~f_{2}({\boldsymbol {S}})} ∂ f ∂ S : T = ( ∂ f 1 ∂ S : T ) f 2 ( S ) + f 1 ( S ) ( ∂ f 2 ∂ S : T ) {\displaystyle {\frac {\partial f}{\partial {\boldsymbol {S}}}}:{\boldsymbol {T}}=\left({\frac {\partial f_{1}}{\partial {\boldsymbol {S}}}}:{\boldsymbol {T}}\right)~f_{2}({\boldsymbol {S}})+f_{1}({\boldsymbol {S}})~\left({\frac {\partial f_{2}}{\partial {\boldsymbol {S}}}}:{\boldsymbol {T}}\right)}
यदि f ( S ) = f 1 ( f 2 ( S ) ) {\displaystyle f({\boldsymbol {S}})=f_{1}(f_{2}({\boldsymbol {S}}))} ∂ f ∂ S : T = ∂ f 1 ∂ f 2 ( ∂ f 2 ∂ S : T ) {\displaystyle {\frac {\partial f}{\partial {\boldsymbol {S}}}}:{\boldsymbol {T}}={\frac {\partial f_{1}}{\partial f_{2}}}~\left({\frac {\partial f_{2}}{\partial {\boldsymbol {S}}}}:{\boldsymbol {T}}\right)}
===दूसरे क्रम के टेंसर === के टेन्सर मूल्यवान कार्यों के व्युत्पन्न
होने देना F ( S ) {\displaystyle {\boldsymbol {F}}({\boldsymbol {S}})} S {\displaystyle {\boldsymbol {S}}} F ( S ) {\displaystyle {\boldsymbol {F}}({\boldsymbol {S}})} S {\displaystyle {\boldsymbol {S}}} S {\displaystyle {\boldsymbol {S}}} T {\displaystyle {\boldsymbol {T}}}
∂ F ∂ S : T = D F ( S ) [ T ] = [ d d α F ( S + α T ) ] α = 0 {\displaystyle {\frac {\partial {\boldsymbol {F}}}{\partial {\boldsymbol {S}}}}:{\boldsymbol {T}}=D{\boldsymbol {F}}({\boldsymbol {S}})[{\boldsymbol {T}}]=\left[{\frac {d}{d\alpha }}~{\boldsymbol {F}}({\boldsymbol {S}}+\alpha ~{\boldsymbol {T}})\right]_{\alpha =0}}
सभी दूसरे क्रम के टेंसरों के लिए
T {\displaystyle {\boldsymbol {T}}} .
गुण:
यदि F ( S ) = F 1 ( S ) + F 2 ( S ) {\displaystyle {\boldsymbol {F}}({\boldsymbol {S}})={\boldsymbol {F}}_{1}({\boldsymbol {S}})+{\boldsymbol {F}}_{2}({\boldsymbol {S}})} ∂ F ∂ S : T = ( ∂ F 1 ∂ S + ∂ F 2 ∂ S ) : T {\displaystyle {\frac {\partial {\boldsymbol {F}}}{\partial {\boldsymbol {S}}}}:{\boldsymbol {T}}=\left({\frac {\partial {\boldsymbol {F}}_{1}}{\partial {\boldsymbol {S}}}}+{\frac {\partial {\boldsymbol {F}}_{2}}{\partial {\boldsymbol {S}}}}\right):{\boldsymbol {T}}}
यदि F ( S ) = F 1 ( S ) ⋅ F 2 ( S ) {\displaystyle {\boldsymbol {F}}({\boldsymbol {S}})={\boldsymbol {F}}_{1}({\boldsymbol {S}})\cdot {\boldsymbol {F}}_{2}({\boldsymbol {S}})} ∂ F ∂ S : T = ( ∂ F 1 ∂ S : T ) ⋅ F 2 ( S ) + F 1 ( S ) ⋅ ( ∂ F 2 ∂ S : T ) {\displaystyle {\frac {\partial {\boldsymbol {F}}}{\partial {\boldsymbol {S}}}}:{\boldsymbol {T}}=\left({\frac {\partial {\boldsymbol {F}}_{1}}{\partial {\boldsymbol {S}}}}:{\boldsymbol {T}}\right)\cdot {\boldsymbol {F}}_{2}({\boldsymbol {S}})+{\boldsymbol {F}}_{1}({\boldsymbol {S}})\cdot \left({\frac {\partial {\boldsymbol {F}}_{2}}{\partial {\boldsymbol {S}}}}:{\boldsymbol {T}}\right)}
यदि F ( S ) = F 1 ( F 2 ( S ) ) {\displaystyle {\boldsymbol {F}}({\boldsymbol {S}})={\boldsymbol {F}}_{1}({\boldsymbol {F}}_{2}({\boldsymbol {S}}))} ∂ F ∂ S : T = ∂ F 1 ∂ F 2 : ( ∂ F 2 ∂ S : T ) {\displaystyle {\frac {\partial {\boldsymbol {F}}}{\partial {\boldsymbol {S}}}}:{\boldsymbol {T}}={\frac {\partial {\boldsymbol {F}}_{1}}{\partial {\boldsymbol {F}}_{2}}}:\left({\frac {\partial {\boldsymbol {F}}_{2}}{\partial {\boldsymbol {S}}}}:{\boldsymbol {T}}\right)}
यदि f ( S ) = f 1 ( F 2 ( S ) ) {\displaystyle f({\boldsymbol {S}})=f_{1}({\boldsymbol {F}}_{2}({\boldsymbol {S}}))} ∂ f ∂ S : T = ∂ f 1 ∂ F 2 : ( ∂ F 2 ∂ S : T ) {\displaystyle {\frac {\partial f}{\partial {\boldsymbol {S}}}}:{\boldsymbol {T}}={\frac {\partial f_{1}}{\partial {\boldsymbol {F}}_{2}}}:\left({\frac {\partial {\boldsymbol {F}}_{2}}{\partial {\boldsymbol {S}}}}:{\boldsymbol {T}}\right)} ढाल, ∇ T {\displaystyle {\boldsymbol {\nabla }}{\boldsymbol {T}}} T ( x ) {\displaystyle {\boldsymbol {T}}(\mathbf {x} )}
∇ T ⋅ c = lim α → 0 d d α T ( x + α c ) {\displaystyle {\boldsymbol {\nabla }}{\boldsymbol {T}}\cdot \mathbf {c} =\lim _{\alpha \rightarrow 0}\quad {\cfrac {d}{d\alpha }}~{\boldsymbol {T}}(\mathbf {x} +\alpha \mathbf {c} )}
ऑर्डर n के टेंसर क्षेत्र का ग्रेडिएंट ऑर्डर n+1 का टेंसर क्षेत्र है।
कार्टेशियन निर्देशांक Note: the Einstein summation convention of summing on repeated indices is used below. यदि e 1 , e 2 , e 3 {\displaystyle \mathbf {e} _{1},\mathbf {e} _{2},\mathbf {e} _{3}} x 1 , x 2 , x 3 {\displaystyle x_{1},x_{2},x_{3}} T {\displaystyle {\boldsymbol {T}}}
∇ T = ∂ T ∂ x i ⊗ e i {\displaystyle {\boldsymbol {\nabla }}{\boldsymbol {T}}={\cfrac {\partial {\boldsymbol {T}}}{\partial x_{i}}}\otimes \mathbf {e} _{i}}
Proof
The vectors x and c can be written as x = x i e i {\displaystyle \mathbf {x} =x_{i}~\mathbf {e} _{i}} c = c i e i {\displaystyle \mathbf {c} =c_{i}~\mathbf {e} _{i}} y := x + αc . In that case the gradient is given by
∇ T ⋅ c = d d α T ( x 1 + α c 1 , x 2 + α c 2 , x 3 + α c 3 ) | α = 0 ≡ d d α T ( y 1 , y 2 , y 3 ) | α = 0 = [ ∂ T ∂ y 1 ∂ y 1 ∂ α + ∂ T ∂ y 2 ∂ y 2 ∂ α + ∂ T ∂ y 3 ∂ y 3 ∂ α ] α = 0 = [ ∂ T ∂ y 1 c 1 + ∂ T ∂ y 2 c 2 + ∂ T ∂ y 3 c 3 ] α = 0 = ∂ T ∂ x 1 c 1 + ∂ T ∂ x 2 c 2 + ∂ T ∂ x 3 c 3 ≡ ∂ T ∂ x i c i
चूंकि कार्टेसियन समन्वय प्रणाली में आधार सदिश भिन्न नहीं होते हैं, हमारे पास अदिश क्षेत्र के ग्रेडियेंट के लिए निम्नलिखित संबंध हैं ϕ {\displaystyle \phi } S {\displaystyle {\boldsymbol {S}}}
∇ ϕ = ∂ ϕ ∂ x i e i = ϕ , i e i ∇ v = ∂ ( v j e j ) ∂ x i ⊗ e i = ∂ v j ∂ x i e j ⊗ e i = v j , i e j ⊗ e i ∇ S = ∂ ( S j k e j ⊗ e k ) ∂ x i ⊗ e i = ∂ S j k ∂ x i e j ⊗ e k ⊗ e i = S j k , i e j ⊗ e k ⊗ e i {\displaystyle {\begin{aligned}{\boldsymbol {\nabla }}\phi &={\cfrac {\partial \phi }{\partial x_{i}}}~\mathbf {e} _{i}=\phi _{,i}~\mathbf {e} _{i}\\{\boldsymbol {\nabla }}\mathbf {v} &={\cfrac {\partial (v_{j}\mathbf {e} _{j})}{\partial x_{i}}}\otimes \mathbf {e} _{i}={\cfrac {\partial v_{j}}{\partial x_{i}}}~\mathbf {e} _{j}\otimes \mathbf {e} _{i}=v_{j,i}~\mathbf {e} _{j}\otimes \mathbf {e} _{i}\\{\boldsymbol {\nabla }}{\boldsymbol {S}}&={\cfrac {\partial (S_{jk}\mathbf {e} _{j}\otimes \mathbf {e} _{k})}{\partial x_{i}}}\otimes \mathbf {e} _{i}={\cfrac {\partial S_{jk}}{\partial x_{i}}}~\mathbf {e} _{j}\otimes \mathbf {e} _{k}\otimes \mathbf {e} _{i}=S_{jk,i}~\mathbf {e} _{j}\otimes \mathbf {e} _{k}\otimes \mathbf {e} _{i}\end{aligned}}}
वक्रीय निर्देशांक Note: the Einstein summation convention of summing on repeated indices is used below. यदि g 1 , g 2 , g 3 {\displaystyle \mathbf {g} ^{1},\mathbf {g} ^{2},\mathbf {g} ^{3}} वक्रीय निर्देशांक प्रणाली में सदिशों के आधार वाले सदिशों के सहप्रसरण और विपरीतप्रसरण होते हैं, जिन्हें बिंदुओं के निर्देशांक द्वारा निरूपित किया जाता है (ξ 1 , ξ 2 , ξ 3 {\displaystyle \xi ^{1},\xi ^{2},\xi ^{3}} T {\displaystyle {\boldsymbol {T}}} [3]
∇ T = ∂ T ∂ ξ i ⊗ g i {\displaystyle {\boldsymbol {\nabla }}{\boldsymbol {T}}={\frac {\partial {\boldsymbol {T}}}{\partial \xi ^{i}}}\otimes \mathbf {g} ^{i}}
इस परिभाषा से हमारे पास अदिश क्षेत्र के ग्रेडियेंट के लिए निम्नलिखित संबंध हैं
ϕ {\displaystyle \phi } , सदिश क्षेत्र v, और दूसरे क्रम का टेंसर क्षेत्र
S {\displaystyle {\boldsymbol {S}}} .
∇ ϕ = ∂ ϕ ∂ ξ i g i ∇ v = ∂ ( v j g j ) ∂ ξ i ⊗ g i = ( ∂ v j ∂ ξ i + v k Γ i k j ) g j ⊗ g i = ( ∂ v j ∂ ξ i − v k Γ i j k ) g j ⊗ g i ∇ S = ∂ ( S j k g j ⊗ g k ) ∂ ξ i ⊗ g i = ( ∂ S j k ∂ ξ i − S l k Γ i j l − S j l Γ i k l ) g j ⊗ g k ⊗ g i {\displaystyle {\begin{aligned}{\boldsymbol {\nabla }}\phi &={\frac {\partial \phi }{\partial \xi ^{i}}}~\mathbf {g} ^{i}\\{\boldsymbol {\nabla }}\mathbf {v} &={\frac {\partial \left(v^{j}\mathbf {g} _{j}\right)}{\partial \xi ^{i}}}\otimes \mathbf {g} ^{i}=\left({\frac {\partial v^{j}}{\partial \xi ^{i}}}+v^{k}~\Gamma _{ik}^{j}\right)~\mathbf {g} _{j}\otimes \mathbf {g} ^{i}=\left({\frac {\partial v_{j}}{\partial \xi ^{i}}}-v_{k}~\Gamma _{ij}^{k}\right)~\mathbf {g} ^{j}\otimes \mathbf {g} ^{i}\\{\boldsymbol {\nabla }}{\boldsymbol {S}}&={\frac {\partial \left(S_{jk}~\mathbf {g} ^{j}\otimes \mathbf {g} ^{k}\right)}{\partial \xi ^{i}}}\otimes \mathbf {g} ^{i}=\left({\frac {\partial S_{jk}}{\partial \xi _{i}}}-S_{lk}~\Gamma _{ij}^{l}-S_{jl}~\Gamma _{ik}^{l}\right)~\mathbf {g} ^{j}\otimes \mathbf {g} ^{k}\otimes \mathbf {g} ^{i}\end{aligned}}}
जहां क्रिस्टोफेल प्रतीक है
Γ i j k {\displaystyle \Gamma _{ij}^{k}} का प्रयोग करके परिभाषित किया गया है
Γ i j k g k = ∂ g i ∂ ξ j ⟹ Γ i j k = ∂ g i ∂ ξ j ⋅ g k = − g i ⋅ ∂ g k ∂ ξ j {\displaystyle \Gamma _{ij}^{k}~\mathbf {g} _{k}={\frac {\partial \mathbf {g} _{i}}{\partial \xi ^{j}}}\quad \implies \quad \Gamma _{ij}^{k}={\frac {\partial \mathbf {g} _{i}}{\partial \xi ^{j}}}\cdot \mathbf {g} ^{k}=-\mathbf {g} _{i}\cdot {\frac {\partial \mathbf {g} ^{k}}{\partial \xi ^{j}}}}
बेलनाकार ध्रुवीय निर्देशांक बेलनाकार निर्देशांक में, ढाल द्वारा दिया जाता है
∇ ϕ = ∂ ϕ ∂ r e r + 1 r ∂ ϕ ∂ θ e θ + ∂ ϕ ∂ z e z ∇ v = ∂ v r ∂ r e r ⊗ e r + 1 r ( ∂ v r ∂ θ − v θ ) e r ⊗ e θ + ∂ v r ∂ z e r ⊗ e z + ∂ v θ ∂ r e θ ⊗ e r + 1 r ( ∂ v θ ∂ θ + v r ) e θ ⊗ e θ + ∂ v θ ∂ z e θ ⊗ e z + ∂ v z ∂ r e z ⊗ e r + 1 r ∂ v z ∂ θ e z ⊗ e θ + ∂ v z ∂ z e z ⊗
टेंसर क्षेत्र का विचलन टेंसर क्षेत्र का विचलन T ( x ) {\displaystyle {\boldsymbol {T}}(\mathbf {x} )}
( ∇ ⋅ T ) ⋅ c = ∇ ⋅ ( c ⋅ T T ) ; ∇ ⋅ v = tr ( ∇ v ) {\displaystyle ({\boldsymbol {\nabla }}\cdot {\boldsymbol {T}})\cdot \mathbf {c} ={\boldsymbol {\nabla }}\cdot \left(\mathbf {c} \cdot {\boldsymbol {T}}^{\textsf {T}}\right)~;\qquad {\boldsymbol {\nabla }}\cdot \mathbf {v} ={\text{tr}}({\boldsymbol {\nabla }}\mathbf {v} )} T {\displaystyle {\boldsymbol {T}}}
कार्टेशियन निर्देशांक Note: the Einstein summation convention of summing on repeated indices is used below. कार्तीय निर्देशांक प्रणाली में सदिश क्षेत्र v और दूसरे क्रम के टेंसर क्षेत्र के लिए हमारे पास निम्नलिखित संबंध हैं S {\displaystyle {\boldsymbol {S}}}
∇ ⋅ v = ∂ v i ∂ x i = v i , i ∇ ⋅ S = ∂ S i k ∂ x i e k = S i k , i e k {\displaystyle {\begin{aligned}{\boldsymbol {\nabla }}\cdot \mathbf {v} &={\frac {\partial v_{i}}{\partial x_{i}}}=v_{i,i}\\{\boldsymbol {\nabla }}\cdot {\boldsymbol {S}}&={\frac {\partial S_{ik}}{\partial x_{i}}}~\mathbf {e} _{k}=S_{ik,i}~\mathbf {e} _{k}\end{aligned}}} ∇ ⋅ S ≠ ∇ ⋅ S T . {\displaystyle {\boldsymbol {\nabla }}\cdot {\boldsymbol {S}}\neq {\boldsymbol {\nabla }}\cdot {\boldsymbol {S}}^{\textsf {T}}.} [4]
∇ ⋅ S = ∂ S k i ∂ x i e k = S k i , i e k {\displaystyle {\begin{aligned}{\boldsymbol {\nabla }}\cdot {\boldsymbol {S}}&={\cfrac {\partial S_{ki}}{\partial x_{i}}}~\mathbf {e} _{k}=S_{ki,i}~\mathbf {e} _{k}\end{aligned}}} ∇ ⋅ S {\displaystyle {\boldsymbol {\nabla }}\cdot {\boldsymbol {S}}} div S {\displaystyle \operatorname {div} {\boldsymbol {S}}}
अंतर इस बात से उपजा है कि क्या भेदभाव पंक्तियों या स्तंभों के संबंध में किया जाता है S {\displaystyle {\boldsymbol {S}}} S {\displaystyle \mathbf {S} } v {\displaystyle \mathbf {v} }
∇ ⋅ ( ∇ v ) = ∇ ⋅ ( v i , j e i ⊗ e j ) = v i , j i e i ⋅ e i ⊗ e j = ( ∇ ⋅ v ) , j e j = ∇ ( ∇ ⋅ v ) ∇ ⋅ [ ( ∇ v ) T ] = ∇ ⋅ ( v j , i e i ⊗ e j ) = v j , i i e i ⋅ e i ⊗ e j = ∇ 2 v j e j = ∇ 2 v {\displaystyle {\begin{aligned}{\boldsymbol {\nabla }}\cdot \left({\boldsymbol {\nabla }}\mathbf {v} \right)&={\boldsymbol {\nabla }}\cdot \left(v_{i,j}~\mathbf {e} _{i}\otimes \mathbf {e} _{j}\right)=v_{i,ji}~\mathbf {e} _{i}\cdot \mathbf {e} _{i}\otimes \mathbf {e} _{j}=\left({\boldsymbol {\nabla }}\cdot \mathbf {v} \right)_{,j}~\mathbf {e} _{j}={\boldsymbol {\nabla }}\left({\boldsymbol {\nabla }}\cdot \mathbf {v} \right)\\{\boldsymbol {\nabla }}\cdot \left[\left({\boldsymbol {\nabla }}\mathbf {v} \right)^{\textsf {T}}\right]&={\boldsymbol {\nabla }}\cdot \left(v_{j,i}~\mathbf {e} _{i}\otimes \mathbf {e} _{j}\right)=v_{j,ii}~\mathbf {e} _{i}\cdot \mathbf {e} _{i}\otimes \mathbf {e} _{j}={\boldsymbol {\nabla }}^{2}v_{j}~\mathbf {e} _{j}={\boldsymbol {\nabla }}^{2}\mathbf {v} \end{aligned}}} [4]
( ∇ ⋅ ) alt ( ∇ v ) = ( ∇ ⋅ ) alt ( v i , j e i ⊗ e j ) = v i , j j e i ⊗ e j ⋅ e j = ∇ 2 v i e i = ∇ 2 v {\displaystyle {\begin{aligned}\left({\boldsymbol {\nabla }}\cdot \right)_{\text{alt}}\left({\boldsymbol {\nabla }}\mathbf {v} \right)=\left({\boldsymbol {\nabla }}\cdot \right)_{\text{alt}}\left(v_{i,j}~\mathbf {e} _{i}\otimes \mathbf {e} _{j}\right)=v_{i,jj}~\mathbf {e} _{i}\otimes \mathbf {e} _{j}\cdot \mathbf {e} _{j}={\boldsymbol {\nabla }}^{2}v_{i}~\mathbf {e} _{i}={\boldsymbol {\nabla }}^{2}\mathbf {v} \end{aligned}}}
वक्रीय निर्देशांक Note: the Einstein summation convention of summing on repeated indices is used below. घुमावदार निर्देशांक में, सदिश क्षेत्र v और दूसरे क्रम के टेंसर क्षेत्र का विचलन S {\displaystyle {\boldsymbol {S}}}
∇ ⋅ v = ( ∂ v i ∂ ξ i + v k Γ i k i ) ∇ ⋅ S = ( ∂ S i k ∂ ξ i − S l k Γ i i l − S i l Γ i k l ) g k {\displaystyle {\begin{aligned}{\boldsymbol {\nabla }}\cdot \mathbf {v} &=\left({\cfrac {\partial v^{i}}{\partial \xi ^{i}}}+v^{k}~\Gamma _{ik}^{i}\right)\\{\boldsymbol {\nabla }}\cdot {\boldsymbol {S}}&=\left({\cfrac {\partial S_{ik}}{\partial \xi _{i}}}-S_{lk}~\Gamma _{ii}^{l}-S_{il}~\Gamma _{ik}^{l}\right)~\mathbf {g} ^{k}\end{aligned}}} ∇ ⋅ S = [ ∂ S i j ∂ q k − Γ k i l S l j − Γ k j l S i l ] g i k b j = [ ∂ S i j ∂ q i + Γ i l i S l j + Γ i l j S i l ] b j = [ ∂ S j i ∂ q i + Γ i l i S j l − Γ i j l S l i ] b j = [ ∂ S i j ∂ q k − Γ i k l S l j + Γ k l j S i l ] g i k b j {\displaystyle {\begin{aligned}{\boldsymbol {\nabla }}\cdot {\boldsymbol {S}}&=\left[{\cfrac {\partial S_{ij}}{\partial q^{k}}}-\Gamma _{ki}^{l}~S_{lj}-\Gamma _{kj}^{l}~S_{il}\right]~g^{ik}~\mathbf {b} ^{j}\\[8pt]&=\left[{\cfrac {\partial S^{ij}}{\partial q^{i}}}+\Gamma _{il}^{i}~S^{lj}+\Gamma _{il}^{j}~S^{il}\right]~\mathbf {b} _{j}\\[8pt]&=\left[{\cfrac {\partial S_{~j}^{i}}{\partial q^{i}}}+\Gamma _{il}^{i}~S_{~j}^{l}-\Gamma _{ij}^{l}~S_{~l}^{i}\right]~\mathbf {b} ^{j}\\[8pt]&=\left[{\cfrac {\partial S_{i}^{~j}}{\partial q^{k}}}-\Gamma _{ik}^{l}~S_{l}^{~j}+\Gamma _{kl}^{j}~S_{i}^{~l}\right]~g^{ik}~\mathbf {b} _{j}\end{aligned}}}
बेलनाकार ध्रुवीय निर्देशांक बेलनाकार निर्देशांक में
∇ ⋅ v = ∂ v r ∂ r + 1 r ( ∂ v θ ∂ θ + v r ) + ∂ v z ∂ z ∇ ⋅ S = ∂ S r r ∂ r e r + ∂ S r θ ∂ r e θ + ∂ S r z ∂ r e z + 1 r [ ∂ S θ r ∂ θ + ( S r r − S θ θ ) ] e r + 1 r [ ∂ S θ θ ∂ θ + ( S r θ + S θ r ) ] e θ + 1 r [ ∂ S θ z ∂ θ + S r z ] e z + ∂ S z r ∂ z e r + ∂ S z θ ∂ z e θ + ∂ S z z ∂ z e z {\displaystyle {\begin{aligned}{\boldsymbol {\nabla }}\cdot \mathbf {v} =\quad &{\frac {\partial v_{r}}{\partial r}}+{\frac {1}{r}}\left({\frac {\partial v_{\theta }}{\partial \theta }}+v_{r}\right)+{\frac {\partial v_{z}}{\partial z}}\\{\boldsymbol {\nabla }}\cdot {\boldsymbol {S}}=\quad &{\frac {\partial S_{rr}}{\partial r}}~\mathbf {e} _{r}+{\frac {\partial S_{r\theta }}{\partial r}}~\mathbf {e} _{\theta }+{\frac {\partial S_{rz}}{\partial r}}~\mathbf {e} _{z}\\{}+{}&{\frac {1}{r}}\left[{\frac {\partial S_{\theta r}}{\partial \theta }}+(S_{rr}-S_{\theta \theta })\right]~\mathbf {e} _{r}+{\frac {1}{r}}\left[{\frac {\partial S_{\theta \theta }}{\partial \theta }}+(S_{r\theta }+S_{\theta r})\right]~\mathbf {e} _{\theta }+{\frac {1}{r}}\left[{\frac {\partial S_{\theta z}}{\partial \theta }}+S_{rz}\right]~\mathbf {e} _{z}\\{}+{}&{\frac {\partial S_{zr}}{\partial z}}~\mathbf {e} _{r}+{\frac {\partial S_{z\theta }}{\partial z}}~\mathbf {e} _{\theta }+{\frac {\partial S_{zz}}{\partial z}}~\mathbf {e} _{z}\end{aligned}}}
टेंसर क्षेत्र का कर्ल ऑर्डर-एन > 1 टेन्सर क्षेत्र का कर्ल (गणित) । T ( x ) {\displaystyle {\boldsymbol {T}}(\mathbf {x} )}
( ∇ × T ) ⋅ c = ∇ × ( c ⋅ T ) ; ( ∇ × v ) ⋅ c = ∇ ⋅ ( v × c ) {\displaystyle ({\boldsymbol {\nabla }}\times {\boldsymbol {T}})\cdot \mathbf {c} ={\boldsymbol {\nabla }}\times (\mathbf {c} \cdot {\boldsymbol {T}})~;\qquad ({\boldsymbol {\nabla }}\times \mathbf {v} )\cdot \mathbf {c} ={\boldsymbol {\nabla }}\cdot (\mathbf {v} \times \mathbf {c} )}
प्रथम-क्रम टेंसर (सदिश) क्षेत्र का कर्ल सदिश क्षेत्र v और स्वेच्छ अचर सदिश c पर विचार करें। सूचकांक संकेतन में, क्रॉस उत्पाद किसके द्वारा दिया जाता है
v × c = ε i j k v j c k e i {\displaystyle \mathbf {v} \times \mathbf {c} =\varepsilon _{ijk}~v_{j}~c_{k}~\mathbf {e} _{i}} ε i j k {\displaystyle \varepsilon _{ijk}} ∇ ⋅ ( v × c ) = ε i j k v j , i c k = ( ε i j k v j , i e k ) ⋅ c = ( ∇ × v ) ⋅ c {\displaystyle {\boldsymbol {\nabla }}\cdot (\mathbf {v} \times \mathbf {c} )=\varepsilon _{ijk}~v_{j,i}~c_{k}=(\varepsilon _{ijk}~v_{j,i}~\mathbf {e} _{k})\cdot \mathbf {c} =({\boldsymbol {\nabla }}\times \mathbf {v} )\cdot \mathbf {c} } ∇ × v = ε i j k v j , i e k {\displaystyle {\boldsymbol {\nabla }}\times \mathbf {v} =\varepsilon _{ijk}~v_{j,i}~\mathbf {e} _{k}}
दूसरे क्रम के टेंसर क्षेत्र का कर्ल दूसरे क्रम के टेंसर के लिए S {\displaystyle {\boldsymbol {S}}}
c ⋅ S = c m S m j e j {\displaystyle \mathbf {c} \cdot {\boldsymbol {S}}=c_{m}~S_{mj}~\mathbf {e} _{j}} ∇ × ( c ⋅ S ) = ε i j k c m S m j , i e k = ( ε i j k S m j , i e k ⊗ e m ) ⋅ c = ( ∇ × S ) ⋅ c {\displaystyle {\boldsymbol {\nabla }}\times (\mathbf {c} \cdot {\boldsymbol {S}})=\varepsilon _{ijk}~c_{m}~S_{mj,i}~\mathbf {e} _{k}=(\varepsilon _{ijk}~S_{mj,i}~\mathbf {e} _{k}\otimes \mathbf {e} _{m})\cdot \mathbf {c} =({\boldsymbol {\nabla }}\times {\boldsymbol {S}})\cdot \mathbf {c} } ∇ × S = ε i j k S m j , i e k ⊗ e m {\displaystyle {\boldsymbol {\nabla }}\times {\boldsymbol {S}}=\varepsilon _{ijk}~S_{mj,i}~\mathbf {e} _{k}\otimes \mathbf {e} _{m}}
टेंसर क्षेत्र के कर्ल से संबंधित पहचान टेंसर क्षेत्र के कर्ल से जुड़ी सबसे अधिक उपयोग की जाने वाली पहचान, T {\displaystyle {\boldsymbol {T}}}
∇ × ( ∇ T ) = 0 {\displaystyle {\boldsymbol {\nabla }}\times ({\boldsymbol {\nabla }}{\boldsymbol {T}})={\boldsymbol {0}}} S {\displaystyle {\boldsymbol {S}}} ∇ × ( ∇ S ) = 0 ⟹ S m i , j − S m j , i = 0 {\displaystyle {\boldsymbol {\nabla }}\times ({\boldsymbol {\nabla }}{\boldsymbol {S}})={\boldsymbol {0}}\quad \implies \quad S_{mi,j}-S_{mj,i}=0}
== दूसरे क्रम के टेंसर == के निर्धारक का व्युत्पन्न
दूसरे क्रम के टेंसर के निर्धारक का व्युत्पन्न A {\displaystyle {\boldsymbol {A}}}
∂ ∂ A det ( A ) = det ( A ) [ A − 1 ] T . {\displaystyle {\frac {\partial }{\partial {\boldsymbol {A}}}}\det({\boldsymbol {A}})=\det({\boldsymbol {A}})~\left[{\boldsymbol {A}}^{-1}\right]^{\textsf {T}}~.} A {\displaystyle {\boldsymbol {A}}}
Proof
Let A {\displaystyle {\boldsymbol {A}}} f ( A ) = det ( A ) {\displaystyle f({\boldsymbol {A}})=\det({\boldsymbol {A}})}
∂ f ∂ A : T = d d α det ( A + α T ) | α = 0 = d d α det [ α A ( 1 α I + A − 1 ⋅ T ) ] | α = 0 = d d α [ α 3 det ( A ) det ( 1 α I + A − 1 ⋅ T ) ] | α = 0 . {\displaystyle {\begin{aligned}{\frac {\partial f}{\partial {\boldsymbol {A}}}}:{\boldsymbol {T}}&=\left.{\cfrac {d}{d\alpha }}\det({\boldsymbol {A}}+\alpha ~{\boldsymbol {T}})\right|_{\alpha =0}\\&=\left.{\cfrac {d}{d\alpha }}\det \left[\alpha ~{\boldsymbol {A}}\left({\cfrac {1}{\alpha }}~{\boldsymbol {\mathit {I}}}+{\boldsymbol {A}}^{-1}\cdot {\boldsymbol {T}}\right)\right]\right|_{\alpha =0}\\&=\left.{\cfrac {d}{d\alpha }}\left[\alpha ^{3}~\det({\boldsymbol {A}})~\det \left({\cfrac {1}{\alpha }}~{\boldsymbol {\mathit {I}}}+{\boldsymbol {A}}^{-1}\cdot {\boldsymbol {T}}\right)\right]\right|_{\alpha =0}.\end{aligned}}}
The determinant of a tensor can be expressed in the form of a characteristic equation in terms of the invariants I 1 , I 2 , I 3 {\displaystyle I_{1},I_{2},I_{3}}
det ( λ I + A ) = λ 3 + I 1 ( A ) λ 2 + I 2 ( A ) λ + I 3 ( A ) . {\displaystyle \det(\lambda ~{\boldsymbol {\mathit {I}}}+{\boldsymbol {A}})=\lambda ^{3}+I_{1}({\boldsymbol {A}})~\lambda ^{2}+I_{2}({\boldsymbol {A}})~\lambda +I_{3}({\boldsymbol {A}}).}
Using this expansion we can write
∂ f ∂ A : T = d d α [ α 3 det ( A ) ( 1 α 3 + I 1 ( A − 1 ⋅ T ) 1 α 2 + I 2 ( A − 1 ⋅ T ) 1 α + I 3 ( A − 1 ⋅ T ) ) ] | α = 0 = det ( A ) d d α [ 1 + I 1 ( A − 1 ⋅ T ) α + I 2 ( A − 1 ⋅ T ) α 2 + I 3 ( A − 1 ⋅ T ) α 3 ] | α = 0 = det ( A ) [ I 1 ( A − 1 ⋅ T ) + 2 I 2 ( A − 1 ⋅ T ) α + 3 I 3 ( A − 1 ⋅ T ) α 2 ] | α = 0 = det (
Recall that the invariant I 1 {\displaystyle I_{1}}
I 1 ( A ) = tr A . {\displaystyle I_{1}({\boldsymbol {A}})={\text{tr}}{\boldsymbol {A}}.}
Hence,
∂ f ∂ A : T = det ( A ) tr ( A − 1 ⋅ T ) = det ( A ) [ A − 1 ] T : T . {\displaystyle {\frac {\partial f}{\partial {\boldsymbol {A}}}}:{\boldsymbol {T}}=\det({\boldsymbol {A}})~{\text{tr}}\left({\boldsymbol {A}}^{-1}\cdot {\boldsymbol {T}}\right)=\det({\boldsymbol {A}})~\left[{\boldsymbol {A}}^{-1}\right]^{\textsf {T}}:{\boldsymbol {T}}.}
Invoking the arbitrariness of T {\displaystyle {\boldsymbol {T}}}
∂ f ∂ A = det ( A ) [ A − 1 ] T . {\displaystyle {\frac {\partial f}{\partial {\boldsymbol {A}}}}=\det({\boldsymbol {A}})~\left[{\boldsymbol {A}}^{-1}\right]^{\textsf {T}}~.}
== दूसरे क्रम के टेंसर == के आक्रमणकारियों के व्युत्पन्न
दूसरे क्रम के टेंसर के प्रमुख आविष्कार हैं
I 1 ( A ) = tr A I 2 ( A ) = 1 2 [ ( tr A ) 2 − tr A 2 ] I 3 ( A ) = det ( A ) {\displaystyle {\begin{aligned}I_{1}({\boldsymbol {A}})&={\text{tr}}{\boldsymbol {A}}\\I_{2}({\boldsymbol {A}})&={\frac {1}{2}}\left[({\text{tr}}{\boldsymbol {A}})^{2}-{\text{tr}}{{\boldsymbol {A}}^{2}}\right]\\I_{3}({\boldsymbol {A}})&=\det({\boldsymbol {A}})\end{aligned}}} A {\displaystyle {\boldsymbol {A}}} ∂ I 1 ∂ A = 1 ∂ I 2 ∂ A = I 1 1 − A T ∂ I 3 ∂ A = det ( A ) [ A − 1 ] T = I 2 1 − A T ( I 1 1 − A T ) = ( A 2 − I 1 A + I 2 1 ) T {\displaystyle {\begin{aligned}{\frac {\partial I_{1}}{\partial {\boldsymbol {A}}}}&={\boldsymbol {\mathit {1}}}\\[3pt]{\frac {\partial I_{2}}{\partial {\boldsymbol {A}}}}&=I_{1}~{\boldsymbol {\mathit {1}}}-{\boldsymbol {A}}^{\textsf {T}}\\[3pt]{\frac {\partial I_{3}}{\partial {\boldsymbol {A}}}}&=\det({\boldsymbol {A}})~\left[{\boldsymbol {A}}^{-1}\right]^{\textsf {T}}=I_{2}~{\boldsymbol {\mathit {1}}}-{\boldsymbol {A}}^{\textsf {T}}~\left(I_{1}~{\boldsymbol {\mathit {1}}}-{\boldsymbol {A}}^{\textsf {T}}\right)=\left({\boldsymbol {A}}^{2}-I_{1}~{\boldsymbol {A}}+I_{2}~{\boldsymbol {\mathit {1}}}\right)^{\textsf {T}}\end{aligned}}}
Proof
From the derivative of the determinant we know that
∂ I 3 ∂ A = det ( A ) [ A − 1 ] T . {\displaystyle {\frac {\partial I_{3}}{\partial {\boldsymbol {A}}}}=\det({\boldsymbol {A}})~\left[{\boldsymbol {A}}^{-1}\right]^{\textsf {T}}~.}
For the derivatives of the other two invariants, let us go back to the characteristic equation
det ( λ 1 + A ) = λ 3 + I 1 ( A ) λ 2 + I 2 ( A ) λ + I 3 ( A ) . {\displaystyle \det(\lambda ~{\boldsymbol {\mathit {1}}}+{\boldsymbol {A}})=\lambda ^{3}+I_{1}({\boldsymbol {A}})~\lambda ^{2}+I_{2}({\boldsymbol {A}})~\lambda +I_{3}({\boldsymbol {A}})~.}
Using the same approach as for the determinant of a tensor, we can show that
∂ ∂ A det ( λ 1 + A ) = det ( λ 1 + A ) [ ( λ 1 + A ) − 1 ] T . {\displaystyle {\frac {\partial }{\partial {\boldsymbol {A}}}}\det(\lambda ~{\boldsymbol {\mathit {1}}}+{\boldsymbol {A}})=\det(\lambda ~{\boldsymbol {\mathit {1}}}+{\boldsymbol {A}})~\left[(\lambda ~{\boldsymbol {\mathit {1}}}+{\boldsymbol {A}})^{-1}\right]^{\textsf {T}}~.}
Now the left hand side can be expanded as
∂ ∂ A det ( λ 1 + A ) = ∂ ∂ A [ λ 3 + I 1 ( A ) λ 2 + I 2 ( A ) λ + I 3 ( A ) ] = ∂ I 1 ∂ A λ 2 + ∂ I 2 ∂ A λ + ∂ I 3 ∂ A . {\displaystyle {\begin{aligned}{\frac {\partial }{\partial {\boldsymbol {A}}}}\det(\lambda ~{\boldsymbol {\mathit {1}}}+{\boldsymbol {A}})&={\frac {\partial }{\partial {\boldsymbol {A}}}}\left[\lambda ^{3}+I_{1}({\boldsymbol {A}})~\lambda ^{2}+I_{2}({\boldsymbol {A}})~\lambda +I_{3}({\boldsymbol {A}})\right]\\&={\frac {\partial I_{1}}{\partial {\boldsymbol {A}}}}~\lambda ^{2}+{\frac {\partial I_{2}}{\partial {\boldsymbol {A}}}}~\lambda +{\frac {\partial I_{3}}{\partial {\boldsymbol {A}}}}~.\end{aligned}}}
Hence
∂ I 1 ∂ A λ 2 + ∂ I 2 ∂ A λ + ∂ I 3 ∂ A = det ( λ 1 + A ) [ ( λ 1 + A ) − 1 ] T {\displaystyle {\frac {\partial I_{1}}{\partial {\boldsymbol {A}}}}~\lambda ^{2}+{\frac {\partial I_{2}}{\partial {\boldsymbol {A}}}}~\lambda +{\frac {\partial I_{3}}{\partial {\boldsymbol {A}}}}=\det(\lambda ~{\boldsymbol {\mathit {1}}}+{\boldsymbol {A}})~\left[(\lambda ~{\boldsymbol {\mathit {1}}}+{\boldsymbol {A}})^{-1}\right]^{\textsf {T}}}
or,
( λ 1 + A ) T ⋅ [ ∂ I 1 ∂ A λ 2 + ∂ I 2 ∂ A λ + ∂ I 3 ∂ A ] = det ( λ 1 + A ) 1 . {\displaystyle (\lambda ~{\boldsymbol {\mathit {1}}}+{\boldsymbol {A}})^{\textsf {T}}\cdot \left[{\frac {\partial I_{1}}{\partial {\boldsymbol {A}}}}~\lambda ^{2}+{\frac {\partial I_{2}}{\partial {\boldsymbol {A}}}}~\lambda +{\frac {\partial I_{3}}{\partial {\boldsymbol {A}}}}\right]=\det(\lambda ~{\boldsymbol {\mathit {1}}}+{\boldsymbol {A}})~{\boldsymbol {\mathit {1}}}~.}
Expanding the right hand side and separating terms on the left hand side gives
( λ 1 + A T ) ⋅ [ ∂ I 1 ∂ A λ 2 + ∂ I 2 ∂ A λ + ∂ I 3 ∂ A ] = [ λ 3 + I 1 λ 2 + I 2 λ + I 3 ] 1 {\displaystyle \left(\lambda ~{\boldsymbol {\mathit {1}}}+{\boldsymbol {A}}^{\textsf {T}}\right)\cdot \left[{\frac {\partial I_{1}}{\partial {\boldsymbol {A}}}}~\lambda ^{2}+{\frac {\partial I_{2}}{\partial {\boldsymbol {A}}}}~\lambda +{\frac {\partial I_{3}}{\partial {\boldsymbol {A}}}}\right]=\left[\lambda ^{3}+I_{1}~\lambda ^{2}+I_{2}~\lambda +I_{3}\right]{\boldsymbol {\mathit {1}}}}
or,
[ ∂ I 1 ∂ A λ 3 + ∂ I 2 ∂ A λ 2 + ∂ I 3 ∂ A λ ] 1 + A T ⋅ ∂ I 1 ∂ A λ 2 + A T ⋅ ∂ I 2 ∂ A λ + A T ⋅ ∂ I 3 ∂ A = [ λ 3 + I 1 λ 2 + I 2 λ + I 3 ] 1 . {\displaystyle {\begin{aligned}\left[{\frac {\partial I_{1}}{\partial {\boldsymbol {A}}}}~\lambda ^{3}\right.&\left.+{\frac {\partial I_{2}}{\partial {\boldsymbol {A}}}}~\lambda ^{2}+{\frac {\partial I_{3}}{\partial {\boldsymbol {A}}}}~\lambda \right]{\boldsymbol {\mathit {1}}}+{\boldsymbol {A}}^{\textsf {T}}\cdot {\frac {\partial I_{1}}{\partial {\boldsymbol {A}}}}~\lambda ^{2}+{\boldsymbol {A}}^{\textsf {T}}\cdot {\frac {\partial I_{2}}{\partial {\boldsymbol {A}}}}~\lambda +{\boldsymbol {A}}^{\textsf {T}}\cdot {\frac {\partial I_{3}}{\partial {\boldsymbol {A}}}}\\&=\left[\lambda ^{3}+I_{1}~\lambda ^{2}+I_{2}~\lambda +I_{3}\right]{\boldsymbol {\mathit {1}}}~.\end{aligned}}}
If we define I 0 := 1 {\displaystyle I_{0}:=1} I 4 := 0 {\displaystyle I_{4}:=0}
[ ∂ I 1 ∂ A λ 3 + ∂ I 2 ∂ A λ 2 + ∂ I 3 ∂ A λ + ∂ I 4 ∂ A ] 1 + A T ⋅ ∂ I 0 ∂ A λ 3 + A T ⋅ ∂ I 1 ∂ A λ 2 + A T ⋅ ∂ I 2 ∂ A λ + A T ⋅ ∂ I 3 ∂ A = [ I 0 λ 3 + I 1 λ 2 + I 2 λ + I 3 ] 1 . {\displaystyle {\begin{aligned}\left[{\frac {\partial I_{1}}{\partial {\boldsymbol {A}}}}~\lambda ^{3}\right.&\left.+{\frac {\partial I_{2}}{\partial {\boldsymbol {A}}}}~\lambda ^{2}+{\frac {\partial I_{3}}{\partial {\boldsymbol {A}}}}~\lambda +{\frac {\partial I_{4}}{\partial {\boldsymbol {A}}}}\right]{\boldsymbol {\mathit {1}}}+{\boldsymbol {A}}^{\textsf {T}}\cdot {\frac {\partial I_{0}}{\partial {\boldsymbol {A}}}}~\lambda ^{3}+{\boldsymbol {A}}^{\textsf {T}}\cdot {\frac {\partial I_{1}}{\partial {\boldsymbol {A}}}}~\lambda ^{2}+{\boldsymbol {A}}^{\textsf {T}}\cdot {\frac {\partial I_{2}}{\partial {\boldsymbol {A}}}}~\lambda +{\boldsymbol {A}}^{\textsf {T}}\cdot {\frac {\partial I_{3}}{\partial {\boldsymbol {A}}}}\\&=\left[I_{0}~\lambda ^{3}+I_{1}~\lambda ^{2}+I_{2}~\lambda +I_{3}\right]{\boldsymbol {\mathit {1}}}~.\end{aligned}}}
Collecting terms containing various powers of λ, we get
λ 3 ( I 0 1 − ∂ I 1 ∂ A 1 − A T ⋅ ∂ I 0 ∂ A ) + λ 2 ( I 1 1 − ∂ I 2 ∂ A 1 − A T ⋅ ∂ I 1 ∂ A ) + λ ( I 2 1 − ∂ I 3 ∂ A 1 − A T ⋅ ∂ I 2 ∂ A ) + ( I 3 1 − ∂ I 4 ∂ A 1 − A T ⋅ ∂ I 3 ∂ A ) = 0 . {\displaystyle {\begin{aligned}\lambda ^{3}&\left(I_{0}~{\boldsymbol {\mathit {1}}}-{\frac {\partial I_{1}}{\partial {\boldsymbol {A}}}}~{\boldsymbol {\mathit {1}}}-{\boldsymbol {A}}^{\textsf {T}}\cdot {\frac {\partial I_{0}}{\partial {\boldsymbol {A}}}}\right)+\lambda ^{2}\left(I_{1}~{\boldsymbol {\mathit {1}}}-{\frac {\partial I_{2}}{\partial {\boldsymbol {A}}}}~{\boldsymbol {\mathit {1}}}-{\boldsymbol {A}}^{\textsf {T}}\cdot {\frac {\partial I_{1}}{\partial {\boldsymbol {A}}}}\right)+\\&\qquad \qquad \lambda \left(I_{2}~{\boldsymbol {\mathit {1}}}-{\frac {\partial I_{3}}{\partial {\boldsymbol {A}}}}~{\boldsymbol {\mathit {1}}}-{\boldsymbol {A}}^{\textsf {T}}\cdot {\frac {\partial I_{2}}{\partial {\boldsymbol {A}}}}\right)+\left(I_{3}~{\boldsymbol {\mathit {1}}}-{\frac {\partial I_{4}}{\partial {\boldsymbol {A}}}}~{\boldsymbol {\mathit {1}}}-{\boldsymbol {A}}^{\textsf {T}}\cdot {\frac {\partial I_{3}}{\partial {\boldsymbol {A}}}}\right)=0~.\end{aligned}}}
Then, invoking the arbitrariness of λ, we have
I 0 1 − ∂ I 1 ∂ A 1 − A T ⋅ ∂ I 0 ∂ A = 0 I 1 1 − ∂ I 2 ∂ A 1 − I 2 1 − ∂ I 3 ∂ A 1 − A T ⋅ ∂ I 2 ∂ A = 0 I 3 1 − ∂ I 4 ∂ A 1 − A T ⋅ ∂ I 3 ∂ A = 0 . {\displaystyle {\begin{aligned}I_{0}~{\boldsymbol {\mathit {1}}}-{\frac {\partial I_{1}}{\partial {\boldsymbol {A}}}}~{\boldsymbol {\mathit {1}}}-{\boldsymbol {A}}^{\textsf {T}}\cdot {\frac {\partial I_{0}}{\partial {\boldsymbol {A}}}}&=0\\I_{1}~{\boldsymbol {\mathit {1}}}-{\frac {\partial I_{2}}{\partial {\boldsymbol {A}}}}~{\boldsymbol {\mathit {1}}}-I_{2}~{\boldsymbol {\mathit {1}}}-{\frac {\partial I_{3}}{\partial {\boldsymbol {A}}}}~{\boldsymbol {\mathit {1}}}-{\boldsymbol {A}}^{\textsf {T}}\cdot {\frac {\partial I_{2}}{\partial {\boldsymbol {A}}}}&=0\\I_{3}~{\boldsymbol {\mathit {1}}}-{\frac {\partial I_{4}}{\partial {\boldsymbol {A}}}}~{\boldsymbol {\mathit {1}}}-{\boldsymbol {A}}^{\textsf {T}}\cdot {\frac {\partial I_{3}}{\partial {\boldsymbol {A}}}}&=0~.\end{aligned}}}
This implies that
∂ I 1 ∂ A = 1 ∂ I 2 ∂ A = I 1 1 − A T ∂ I 3 ∂ A = I 2 1 − A T ( I 1 1 − A T ) = ( A 2 − I 1 A + I 2 1 ) T {\displaystyle {\begin{aligned}{\frac {\partial I_{1}}{\partial {\boldsymbol {A}}}}&={\boldsymbol {\mathit {1}}}\\{\frac {\partial I_{2}}{\partial {\boldsymbol {A}}}}&=I_{1}~{\boldsymbol {\mathit {1}}}-{\boldsymbol {A}}^{\textsf {T}}\\{\frac {\partial I_{3}}{\partial {\boldsymbol {A}}}}&=I_{2}~{\boldsymbol {\mathit {1}}}-{\boldsymbol {A}}^{\textsf {T}}~\left(I_{1}~{\boldsymbol {\mathit {1}}}-{\boldsymbol {A}}^{\textsf {T}}\right)=\left({\boldsymbol {A}}^{2}-I_{1}~{\boldsymbol {A}}+I_{2}~{\boldsymbol {\mathit {1}}}\right)^{\textsf {T}}\end{aligned}}}
== दूसरे क्रम की पहचान टेंसर == का व्युत्पन्न
होने देना 1 {\displaystyle {\boldsymbol {\mathit {1}}}} A {\displaystyle {\boldsymbol {A}}}
∂ 1 ∂ A : T = 0 : T = 0 {\displaystyle {\frac {\partial {\boldsymbol {\mathit {1}}}}{\partial {\boldsymbol {A}}}}:{\boldsymbol {T}}={\boldsymbol {\mathsf {0}}}:{\boldsymbol {T}}={\boldsymbol {\mathit {0}}}} 1 {\displaystyle {\boldsymbol {\mathit {1}}}} A {\displaystyle {\boldsymbol {A}}}
स्वयं के संबंध में दूसरे क्रम के टेंसर का व्युत्पन्न होने देना A {\displaystyle {\boldsymbol {A}}}
∂ A ∂ A : T = [ ∂ ∂ α ( A + α T ) ] α = 0 = T = I : T {\displaystyle {\frac {\partial {\boldsymbol {A}}}{\partial {\boldsymbol {A}}}}:{\boldsymbol {T}}=\left[{\frac {\partial }{\partial \alpha }}({\boldsymbol {A}}+\alpha ~{\boldsymbol {T}})\right]_{\alpha =0}={\boldsymbol {T}}={\boldsymbol {\mathsf {I}}}:{\boldsymbol {T}}} ∂ A ∂ A = I {\displaystyle {\frac {\partial {\boldsymbol {A}}}{\partial {\boldsymbol {A}}}}={\boldsymbol {\mathsf {I}}}} I {\displaystyle {\boldsymbol {\mathsf {I}}}} I = δ i k δ j l e i ⊗ e j ⊗ e k ⊗ e l {\displaystyle {\boldsymbol {\mathsf {I}}}=\delta _{ik}~\delta _{jl}~\mathbf {e} _{i}\otimes \mathbf {e} _{j}\otimes \mathbf {e} _{k}\otimes \mathbf {e} _{l}} ∂ A T ∂ A : T = I T : T = T T {\displaystyle {\frac {\partial {\boldsymbol {A}}^{\textsf {T}}}{\partial {\boldsymbol {A}}}}:{\boldsymbol {T}}={\boldsymbol {\mathsf {I}}}^{\textsf {T}}:{\boldsymbol {T}}={\boldsymbol {T}}^{\textsf {T}}} I T = δ j k δ i l e i ⊗ e j ⊗ e k ⊗ e l {\displaystyle {\boldsymbol {\mathsf {I}}}^{\textsf {T}}=\delta _{jk}~\delta _{il}~\mathbf {e} _{i}\otimes \mathbf {e} _{j}\otimes \mathbf {e} _{k}\otimes \mathbf {e} _{l}} A {\displaystyle {\boldsymbol {A}}} ∂ A ∂ A = I ( s ) = 1 2 ( I + I T ) {\displaystyle {\frac {\partial {\boldsymbol {A}}}{\partial {\boldsymbol {A}}}}={\boldsymbol {\mathsf {I}}}^{(s)}={\frac {1}{2}}~\left({\boldsymbol {\mathsf {I}}}+{\boldsymbol {\mathsf {I}}}^{\textsf {T}}\right)} I ( s ) = 1 2 ( δ i k δ j l + δ i l δ j k ) e i ⊗ e j ⊗ e k ⊗ e l {\displaystyle {\boldsymbol {\mathsf {I}}}^{(s)}={\frac {1}{2}}~(\delta _{ik}~\delta _{jl}+\delta _{il}~\delta _{jk})~\mathbf {e} _{i}\otimes \mathbf {e} _{j}\otimes \mathbf {e} _{k}\otimes \mathbf {e} _{l}}
== दूसरे क्रम के टेंसर == के व्युत्क्रम का व्युत्पन्न
होने देना A {\displaystyle {\boldsymbol {A}}} T {\displaystyle {\boldsymbol {T}}}
∂ ∂ A ( A − 1 ) : T = − A − 1 ⋅ T ⋅ A − 1 {\displaystyle {\frac {\partial }{\partial {\boldsymbol {A}}}}\left({\boldsymbol {A}}^{-1}\right):{\boldsymbol {T}}=-{\boldsymbol {A}}^{-1}\cdot {\boldsymbol {T}}\cdot {\boldsymbol {A}}^{-1}} ∂ A i j − 1 ∂ A k l T k l = − A i k − 1 T k l A l j − 1 ⟹ ∂ A i j − 1 ∂ A k l = − A i k − 1 A l j − 1 {\displaystyle {\frac {\partial A_{ij}^{-1}}{\partial A_{kl}}}~T_{kl}=-A_{ik}^{-1}~T_{kl}~A_{lj}^{-1}\implies {\frac {\partial A_{ij}^{-1}}{\partial A_{kl}}}=-A_{ik}^{-1}~A_{lj}^{-1}} ∂ ∂ A ( A − T ) : T = − A − T ⋅ T T ⋅ A − T {\displaystyle {\frac {\partial }{\partial {\boldsymbol {A}}}}\left({\boldsymbol {A}}^{-{\textsf {T}}}\right):{\boldsymbol {T}}=-{\boldsymbol {A}}^{-{\textsf {T}}}\cdot {\boldsymbol {T}}^{\textsf {T}}\cdot {\boldsymbol {A}}^{-{\textsf {T}}}} ∂ A j i − 1 ∂ A k l T k l = − A j k − 1 T l k A l i − 1 ⟹ ∂ A j i − 1 ∂ A k l = − A l i − 1 A j k − 1 {\displaystyle {\frac {\partial A_{ji}^{-1}}{\partial A_{kl}}}~T_{kl}=-A_{jk}^{-1}~T_{lk}~A_{li}^{-1}\implies {\frac {\partial A_{ji}^{-1}}{\partial A_{kl}}}=-A_{li}^{-1}~A_{jk}^{-1}} A {\displaystyle {\boldsymbol {A}}} ∂ A i j − 1 ∂ A k l = − 1 2 ( A i k − 1 A j l − 1 + A i l − 1 A j k − 1 ) {\displaystyle {\frac {\partial A_{ij}^{-1}}{\partial A_{kl}}}=-{\cfrac {1}{2}}\left(A_{ik}^{-1}~A_{jl}^{-1}+A_{il}^{-1}~A_{jk}^{-1}\right)}
Proof
Recall that
∂ 1 ∂ A : T = 0 {\displaystyle {\frac {\partial {\boldsymbol {\mathit {1}}}}{\partial {\boldsymbol {A}}}}:{\boldsymbol {T}}={\boldsymbol {\mathit {0}}}}
Since A − 1 ⋅ A = 1 {\displaystyle {\boldsymbol {A}}^{-1}\cdot {\boldsymbol {A}}={\boldsymbol {\mathit {1}}}}
∂ ∂ A ( A − 1 ⋅ A ) : T = 0 {\displaystyle {\frac {\partial }{\partial {\boldsymbol {A}}}}\left({\boldsymbol {A}}^{-1}\cdot {\boldsymbol {A}}\right):{\boldsymbol {T}}={\boldsymbol {\mathit {0}}}}
Using the product rule for second order tensors
∂ ∂ S [ F 1 ( S ) ⋅ F 2 ( S ) ] : T = ( ∂ F 1 ∂ S : T ) ⋅ F 2 + F 1 ⋅ ( ∂ F 2 ∂ S : T ) {\displaystyle {\frac {\partial }{\partial {\boldsymbol {S}}}}[{\boldsymbol {F}}_{1}({\boldsymbol {S}})\cdot {\boldsymbol {F}}_{2}({\boldsymbol {S}})]:{\boldsymbol {T}}=\left({\frac {\partial {\boldsymbol {F}}_{1}}{\partial {\boldsymbol {S}}}}:{\boldsymbol {T}}\right)\cdot {\boldsymbol {F}}_{2}+{\boldsymbol {F}}_{1}\cdot \left({\frac {\partial {\boldsymbol {F}}_{2}}{\partial {\boldsymbol {S}}}}:{\boldsymbol {T}}\right)}
we get
∂ ∂ A ( A − 1 ⋅ A ) : T = ( ∂ A − 1 ∂ A : T ) ⋅ A + A − 1 ⋅ ( ∂ A ∂ A : T ) = 0 {\displaystyle {\frac {\partial }{\partial {\boldsymbol {A}}}}({\boldsymbol {A}}^{-1}\cdot {\boldsymbol {A}}):{\boldsymbol {T}}=\left({\frac {\partial {\boldsymbol {A}}^{-1}}{\partial {\boldsymbol {A}}}}:{\boldsymbol {T}}\right)\cdot {\boldsymbol {A}}+{\boldsymbol {A}}^{-1}\cdot \left({\frac {\partial {\boldsymbol {A}}}{\partial {\boldsymbol {A}}}}:{\boldsymbol {T}}\right)={\boldsymbol {\mathit {0}}}}
or,
( ∂ A − 1 ∂ A : T ) ⋅ A = − A − 1 ⋅ T {\displaystyle \left({\frac {\partial {\boldsymbol {A}}^{-1}}{\partial {\boldsymbol {A}}}}:{\boldsymbol {T}}\right)\cdot {\boldsymbol {A}}=-{\boldsymbol {A}}^{-1}\cdot {\boldsymbol {T}}}
Therefore,
∂ ∂ A ( A − 1 ) : T = − A − 1 ⋅ T ⋅ A − 1 {\displaystyle {\frac {\partial }{\partial {\boldsymbol {A}}}}\left({\boldsymbol {A}}^{-1}\right):{\boldsymbol {T}}=-{\boldsymbol {A}}^{-1}\cdot {\boldsymbol {T}}\cdot {\boldsymbol {A}}^{-1}}
भागों द्वारा एकीकरण कार्यक्षेत्र
Ω {\displaystyle \Omega } , इसकी सीमा
Γ {\displaystyle \Gamma } और जावक इकाई सामान्य
n {\displaystyle \mathbf {n} } सातत्य यांत्रिकी में टेंसर व्युत्पन्न से संबंधित अन्य महत्वपूर्ण ऑपरेशन भागों द्वारा एकीकरण है। भागों द्वारा एकीकरण के सूत्र को इस प्रकार लिखा जा सकता है
∫ Ω F ⊗ ∇ G d Ω = ∫ Γ n ⊗ ( F ⊗ G ) d Γ − ∫ Ω G ⊗ ∇ F d Ω {\displaystyle \int _{\Omega }{\boldsymbol {F}}\otimes {\boldsymbol {\nabla }}{\boldsymbol {G}}\,d\Omega =\int _{\Gamma }\mathbf {n} \otimes ({\boldsymbol {F}}\otimes {\boldsymbol {G}})\,d\Gamma -\int _{\Omega }{\boldsymbol {G}}\otimes {\boldsymbol {\nabla }}{\boldsymbol {F}}\,d\Omega } F {\displaystyle {\boldsymbol {F}}} G {\displaystyle {\boldsymbol {G}}} n {\displaystyle \mathbf {n} } ⊗ {\displaystyle \otimes } ∇ {\displaystyle {\boldsymbol {\nabla }}} F {\displaystyle {\boldsymbol {F}}} ∫ Ω ∇ G d Ω = ∫ Γ n ⊗ G d Γ . {\displaystyle \int _{\Omega }{\boldsymbol {\nabla }}{\boldsymbol {G}}\,d\Omega =\int _{\Gamma }\mathbf {n} \otimes {\boldsymbol {G}}\,d\Gamma \,.} ∫ Ω F i j k . . . . G l m n . . . , p d Ω = ∫ Γ n p F i j k . . . G l m n . . . d Γ − ∫ Ω G l m n . . . F i j k . . . , p d Ω . {\displaystyle \int _{\Omega }F_{ijk....}\,G_{lmn...,p}\,d\Omega =\int _{\Gamma }n_{p}\,F_{ijk...}\,G_{lmn...}\,d\Gamma -\int _{\Omega }G_{lmn...}\,F_{ijk...,p}\,d\Omega \,.} F {\displaystyle {\boldsymbol {F}}} G {\displaystyle {\boldsymbol {G}}} ∫ Ω F ⋅ ( ∇ ⋅ G ) d Ω = ∫ Γ n ⋅ ( G ⋅ F T ) d Γ − ∫ Ω ( ∇ F ) : G T d Ω . {\displaystyle \int _{\Omega }{\boldsymbol {F}}\cdot ({\boldsymbol {\nabla }}\cdot {\boldsymbol {G}})\,d\Omega =\int _{\Gamma }\mathbf {n} \cdot \left({\boldsymbol {G}}\cdot {\boldsymbol {F}}^{\textsf {T}}\right)\,d\Gamma -\int _{\Omega }({\boldsymbol {\nabla }}{\boldsymbol {F}}):{\boldsymbol {G}}^{\textsf {T}}\,d\Omega \,.} ∫ Ω F i j G p j , p d Ω = ∫ Γ n p F i j G p j d Γ − ∫ Ω G p j F i j , p d Ω . {\displaystyle \int _{\Omega }F_{ij}\,G_{pj,p}\,d\Omega =\int _{\Gamma }n_{p}\,F_{ij}\,G_{pj}\,d\Gamma -\int _{\Omega }G_{pj}\,F_{ij,p}\,d\Omega \,.}
यह भी देखें संदर्भ
↑ J. C. Simo and T. J. R. Hughes, 1998, Computational Inelasticity , Springer
↑ J. E. Marsden and T. J. R. Hughes, 2000, Mathematical Foundations of Elasticity , Dover.
↑ R. W. Ogden, 2000, Nonlinear Elastic Deformations , Dover.
↑ Jump up to: 4.04.1 Hjelmstad, Keith (2004). संरचनात्मक यांत्रिकी के मूल तत्व . Springer Science & Business Media. p. 45. ISBN 9780387233307