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 EulerLagrange 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, selfadjoint, 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 EulerLagrange 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 lefthand side is the wellknown 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 righthand 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 EulerLagrange 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 threedimensional form was exploited on sev
eral occasions earlier; here we apply it for the special vector ﬁeld
η∇u
in a twodimensional 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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