5
The inverse problem of calculus of variations
It is often the case that the engineer starts from a differential equation with
certain boundary conditions, which is difficult to solve. Executing the inverse
of the Euler-Lagrange process and obtaining the variational formulation of
the boundary value problem may also be advantageous.
It is not necessarily easy, or may not even be possible to reconstruct the
variational problem from a differential equation. For differential equations,
partial or ordinary, containing a linear, self-adjoint, positive operator, the
task may be accomplished. Such an operator exhibits
(Au, v)=(u, Av),
where the parenthesis expression denotes a scalar product in the function
space of the solution of the differential equation. Positive definiteness of the
operator means
(Au, u) ≥ 0,
with zero attained only for the trivial (u = 0) solution. Let us consider the
differential equation of
Au = f,
where the operator obeys the above conditions and f is a known function. If
the differential equation has a solution, it corresponds to the minimum value
of the functional
I(u)=(Au, u)+2(u, f).
This may be proven by simply applying the appropriate Euler-Lagrange equa-
tion to this functional.
57
58 Applied calculus of variations for engineers
5.1 The variational form of Poisson’s equation
We demonstrate the inverse process through the example of Poisson’s equa-
tion, a topic of much interest for engineers:
Δu(x)=
∂
2
u
∂x
2
+
∂
2
u
∂y
2
= f(x, y).
Here the left-hand side is the well-known Laplace operator. We impose Dirich-
let type boundary conditions on the boundary of the domain of interest.
u(x, y)=0;(x, y) ∈ ∂D,
where D is the domain of solution and ∂D is its boundary. According to the
above proposition, we need to compute
(Au, u)=
D
u(
∂
2
u
∂x
2
+
∂
2
u
∂y
2
)dxdy.
Applying some vector calculus results in
(Au, u)=
∂D
(u
∂u
∂y
dx − u
∂u
∂x
dy)+
D
(
∂u
∂x
)
2
+(
∂u
∂y
)
2
dxdy.
Due to the boundary conditions, the first term vanishes and we obtain
(Au, u)=
D
(
∂u
∂x
)
2
+(
∂u
∂y
)
2
dxdy.
The right-hand side term of the differential equation is processed as
(u, f)=
D
uf(x, y)dxdy.
The variational formulation of Poisson’s equation finally is
D
((
∂u
∂x
)
2
+(
∂u
∂y
)
2
+2uf)dxdy =
D
F (x, y, u, u
x
,u
y
)dxdy = extremum.
To prove this, we will apply the Euler-Lagrange equation developed in Section
3.3. The terms for this particular case are:
∂F
∂u
=2f,
∂
∂x
∂F
∂u
x
=
∂
∂x
2u
x
=2
∂
2
u
∂x
2
,
and
∂
∂y
∂F
∂u
y
=
∂
∂y
2u
y
=2
∂
2
u
∂y
2
.
The inverse problem of calculus of variations 59
The resulting equation of
2f − 2
∂
2
u
∂x
2
− 2
∂
2
u
∂y
2
=0
is clearly equivalent with Poisson’s equation.
5.2 The variational form of eigenvalue problems
Eigenvalue problems of various kinds may also be formulated as variational
problems [12]. We consider the equation of the form
Δu(x) − λu(x)=0, (5.1)
where the unknown function u(x)definedondomainD is the eigensolution
and λ is the eigenvalue. The boundary condition is imposed as
u(x, y)=0,
on the perimeter ∂D of the domain D. The corresponding variational problem
is of the form
I =
D
((
∂u
∂x
)
2
+(
∂u
∂y
)
2
)dxdy = extremum, (5.2)
under the condition of
g(x, y)=
D
u
2
(x, y)dxdy =1.
This relation is proven as follows. Following the Lagrange solution of con-
strained variational problems introduced in Section 2.2, we can write
h(x, y)=u(x, y)+λg(x, y),
and
I =
D
((
∂u
∂x
)
2
+(
∂u
∂y
)
2
+ λu
2
(x, y))dxdy.
Note that the λ is still only in the role of the Lagrange multiplier, although
its name hints at its final meaning as well. Introducing
U(x, y)=u(x, y)+η(x, y)
the variational form becomes
I()=
D
((
∂u
∂x
+
∂η
∂x
)
2
+(
∂u
∂y
+
∂η
∂y
)
2
+ λ(u + η)
2
)dxdy.
60 Applied calculus of variations for engineers
The extremum is reached when
dI()
d
|
=0
=0,
which gives rise to the equation
2
D
(
∂u
∂x
∂η
∂x
+
∂u
∂y
∂η
∂y
+ λuη)dxdy =0. (5.3)
Green’s identity in its original three-dimensional form was exploited on sev-
eral occasions earlier; here we apply it for the special vector field
η∇u
in a two-dimensional domain. The result is
D
(∇η ·∇u)dA =
∂D
η(∇u · n)ds −
D
η∇
2
udA.
Since the tangent of the circumference is in the direction of
dx i
+ dy j,
the unit normal may be computed as
n
=
dy i
− dx j
dx
2
+ dy
2
.
Finally utilizing the arc length formula of
ds =
dx
2
+ dy
2
,
the line integral over the circumference of the domain becomes
∂D
η(
∂u
∂x
dy −
∂u
∂y
dx).
Applying the above for the first two terms of Equation (5.3) results in
D
(
∂u
∂x
∂η
∂x
+
∂u
∂y
∂η
∂y
)dxdy =
∂D
(η(
∂u
∂x
+
∂u
∂y
) · n
)ds −
D
(
∂
2
u
∂x
2
+
∂
2
u
∂y
2
)ηdxdy =
∂D
∂u
∂x
ηdy −
∂u
∂y
ηdx −
D
Δuηdxdy.
The integral over the boundary vanishes due to the assumption on η,and
substituting the remainder part into Equation (5.3) we obtain
−2
D
(Δu − λu)ηdxdy =0.
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