If the problem does not depend on time and the material is isotropic, we get the boundary value problem for steady state heat conduction.
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PDE:
−
1
C
v
ρ
∇
∙
(
κ
∙
∇
T
)
=
Q
in
Ω
BCs:
T
=
T
¯
(
x
)
on
Γ
T
and
∂
T
∂
n
=
g
(
x
)
on
Γ
q
{\displaystyle {\begin{aligned}&&{\mathsf {The~boundary~value~problem~for~steady~heat~conduction}}\\&&\\&{\text{PDE:}}~~~&~~~-{\frac {1}{C_{v}~\rho }}{\boldsymbol {\nabla }}\bullet ({\boldsymbol {\kappa }}\bullet {\boldsymbol {\nabla T)}}=Q~~{\text{in}}~~\Omega \quad \\&{\text{BCs:}}~~~&~~~T={\overline {T}}(\mathbf {x} )~~{\text{on}}~~\Gamma _{T}~~{\text{and}}~~{\frac {\partial T}{\partial n}}=g(\mathbf {x} )~~{\text{on}}~~\Gamma _{q}\quad \\\end{aligned}}}
If the material is homogeneous the density, heat capacity, and the
thermal conductivity are constant. Define the thermal diffusivity
as
k
:=
κ
C
v
ρ
{\displaystyle k:={\frac {\kappa }{C_{v}~\rho }}}
Then, the boundary value problem becomes
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PDE:
−
k
∇
2
T
=
Q
in
Ω
BCs:
T
=
T
¯
(
x
)
on
Γ
T
and
∂
T
∂
n
=
g
(
x
)
on
Γ
q
{\displaystyle {\begin{aligned}&&{\mathsf {Poisson's~equation}}\\&&\\&{\text{PDE:}}~~~&~~~-k\nabla ^{2}T=Q~~{\text{in}}~~\Omega \quad \\&{\text{BCs:}}~~~&~~~T={\overline {T}}(\mathbf {x} )~~{\text{on}}~~\Gamma _{T}~~{\text{and}}~~{\frac {\partial T}{\partial n}}=g(\mathbf {x} )~~{\text{on}}~~\Gamma _{q}\quad \\\end{aligned}}}
where
∇
2
T
{\displaystyle \nabla ^{2}T}
Laplacian
∇
2
T
:=
∇
∙
∇
T
{\displaystyle \nabla ^{2}T:={\boldsymbol {\nabla }}\bullet {\boldsymbol {\nabla T}}}
Finally, if there is no internal source of heat, the value of
Q
{\displaystyle Q}
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u
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PDE:
∇
2
T
=
0
in
Ω
BCs:
T
=
T
¯
(
x
)
on
Γ
T
and
∂
T
∂
n
=
g
(
x
)
on
Γ
q
{\displaystyle {\begin{aligned}&&{\mathsf {Laplace's~equation}}\\&&\\&{\text{PDE:}}~~~&~~~\nabla ^{2}T=0~~{\text{in}}~~\Omega \quad \\&{\text{BCs:}}~~~&~~~T={\overline {T}}(\mathbf {x} )~~{\text{on}}~~\Gamma _{T}~~{\text{and}}~~{\frac {\partial T}{\partial n}}=g(\mathbf {x} )~~{\text{on}}~~\Gamma _{q}\quad \\\end{aligned}}}
The thin elastic membrane problem is another similar problem. See
Figure 1 for the geometry of the membrane.
The membrane is thin and elastic. It is initially planar and occupies
the 2D domain
Ω
{\displaystyle \Omega }
Γ
u
{\displaystyle \Gamma _{u}}
f
{\displaystyle \mathbf {f} }
x
{\displaystyle \mathbf {x} }
u
(
x
)
{\displaystyle \mathbf {u} (\mathbf {x} )}
The goal is to find the displacement
u
(
x
)
{\displaystyle \mathbf {u} (\mathbf {x} )}
Figure 1. The membrane problem.
It turns out that the equations for this problem are the same as those
for the heat conduction problem - with the following changes:
The time derivatives vanish.
The balance of energy is replaced by the balance of forces.
The constitutive equation is replaced by a relation that states that the vertical force depends on the displacement gradient (
∇
u
{\displaystyle {\boldsymbol {\nabla }}\mathbf {u} }
If the membrane if inhomogeneous, the boundary value problem is:
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m
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m
b
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n
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f
o
r
m
a
t
i
o
n
PDE:
−
∇
∙
(
E
∇
u
)
=
Q
in
Ω
BCs:
u
=
u
¯
(
x
)
on
Γ
u
and
∂
u
∂
n
=
g
(
x
)
on
Γ
t
{\displaystyle {\begin{aligned}&&{\mathsf {The~boundary~value~problem~for~membrane~deformation}}\\&&\\&{\text{PDE:}}~~~&~~~-{\boldsymbol {\nabla }}\bullet (E~{\boldsymbol {\nabla \mathbf {u} )}}=Q~~{\text{in}}~~\Omega \quad \\&{\text{BCs:}}~~~&~~~\mathbf {u} ={\bar {\mathbf {u} }}(\mathbf {x} )~~{\text{on}}~~\Gamma _{u}~~{\text{and}}~~{\frac {\partial u}{\partial n}}=g(\mathbf {x} )~~{\text{on}}~~\Gamma _{t}\quad \\\end{aligned}}}
For a homogeneous membrane, we get
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s
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′
s
E
q
u
a
t
i
o
n
PDE:
−
E
∇
2
u
=
Q
in
Ω
BCs:
u
=
u
¯
(
x
)
on
Γ
u
and
∂
u
∂
n
=
g
(
x
)
on
Γ
t
{\displaystyle {\begin{aligned}&&{\mathsf {Poisson's~Equation}}\\&&\\&{\text{PDE:}}~~~&~~~-E\nabla ^{2}\mathbf {u} =Q~~{\text{in}}~~\Omega \quad \\&{\text{BCs:}}~~~&~~~\mathbf {u} ={\bar {\mathbf {u} }}(\mathbf {x} )~~{\text{on}}~~\Gamma _{u}~~{\text{and}}~~{\frac {\partial u}{\partial n}}=g(\mathbf {x} )~~{\text{on}}~~\Gamma _{t}\quad \\\end{aligned}}}
Note that the membrane problem can be formulated in terms of a problem
of minimization of potential energy.
