Boundary Value Problems | Engineering Mathematics for Mechanical Engineering PDF Download

Introduction

A boundary value problem for a given differential equation consists of finding a solution of the given differential equation subject to a given set of boundary conditions. A boundary condition is a prescription some combinations of values of the unknown solution and its derivatives at more than one point. 

Let I = (a, b) ⊆ R be an interval. Let p, q, r : (a, b) → R be continuous functions.
Throughout this chapter we consider the linear second order equation given by 

Corresponding to ODE (5.1), there are four important kinds of (linear) boundary conditions. They are given by 

Remark 5.1 (On periodic boundary condition) If the coefficients of ODE (5.1) are periodic functions with period l = b − a and if φ is a solution of ODE (5.1) (note that this solution exists on R), then ψ defined by ψ(x) = φ(x + l) is also a solution. If φ satisfies the periodic boundary conditions, then ψ(a) = φ(a) and ψ ′ (a) = φ ′ (a). Since solutions to IVP are unique in the present case, it must be that ψ ≡ φ. In other words, φ is a periodic function of period l.
Boundary Value Problems do not behave as nicely as Initial value problems. For, there are BVPs for which solutions do not exist; and even if a solution exists there might be many more. Thus existence and uniqueness generally fail for BVPs. The following example illustrate all the three possibilities.

Adjoint forms, Lagrange identity

In mathematical physics there are many important boundary value problems corresponding to second order equations. In the studies of vibrations of a membrane, vibrations of a structure one has to solve a homogeneous boundary value problem for real frequencies (eigen values). As is wellknown in the case of symmetric matrices that there are only real eigen values and corresponding eigen vectors form a basis for the underlying vector space and thereby all symmetric matrices are diagonalisable. Self-adjoint problems can be thought of as corresponding ODE versions of symmetric matrices, and they play an important role in mathematical physics.

Let us consider the equation 

Integrating z[y] by parts from a to x, we have 

If we define the second order operator ∗ by 

then the equation (5.4) becomes 

The operator ^∗ is called the adjoint operator corresponding to the operator . It can be easily verified that adjoint of ^∗ is L itself. If and ^∗ are the same, then L is said to be self-adjoint .Thus, the necessary and sufficient condition for to be self-adjoint is thatwhich is satisfied if 

Thus if L is self-adjoint, we have
A general operator may not be self-adjoint but it can always be converted into a self-adjoint by suitably multiplying with a function.

Two-point boundary value problem

In this section we are going to set up the notations that we are going to use through out our discussion of BVPs. We consider the linear nonhomogeneous second order in the self-adjoint form described below:

Note: that the boundary conditions are in the most general form, and they include the first three conditions given at the beginning of our discussion on BVPs as special cases.

Let us introduce some nomenclature here
Assume hypothesis (HBVP). A nonhomogeneous boundary value problem consists of solving 

for given constants η₁ and η₂, and a given continuous function f on the interval [a, b]. 

The associated homogeneous boundary value problem is then given by

Let us list some properties of the solutions for BVP that are consequences of the linearity of the differential operator .

Fundamental solutions, Green’s functions

Fundamental solution of an ODE gives rise to a representation (integral) formula for solution of the nonhomogeneous equation. When we want to take care of boundary conditions, we impose boundary conditions on fundamental solutions and get Green’s functions. Thus Green’s functions give rise to a representation formula for solution of the nonhomogeneous BVP. The concepts of a fundamental solution as well as a Green’s function are defined in terms of the homogeneous BVPvassociated to the nonhomogeneous BVP.
Let Q denote the square Q := [a, b] × [a, b] in the xξ-plane. Let us partition Q by the line x = ξ and call the two resulting triangles Q₁ and Q₂. LetNote that the diagonal x = ξ belongs to both the triangles.  

Fundamental solution: A function γ(x, ξ) defined in Q is called a fundamental solution of the homogeneous differential equation [y] = 0 if it has the following properties:
(i) The function γ(x, ξ) is continuous in Q.
(ii) The first and second order partial derivatives w.r.t. variable x of the function γ(x, ξ) exist and continuous up to the boundary on Q₁ and Q₂.
(iii) Let ξ ∈ [a, b] be fixed. Then γ(x, ξ), considered as a function of x, satisfies [γ(., ξ)] = 0 at every point of the interval [a, b], except at ξ.(iv) The first derivate has a jump across the diagonal x = ξ, of magnitude 1/p, i.e.,

Construction of Green’s functions

In this paragraph we are going to construct Green’s functions under the assumption that the homogeneous BVP has only trivial solution. Let (λ₁, λ₂) 6= (0, 0) be such that a₁λ₁ + a₂λ₂ = 0 and let φ₁ be solution of [y] = 0 satisfying φ₁(a) = λ₁ and φ ′ ₁ (a) = λ₂. Choose another solution φ₂ of [y] = 0 similarly. This way of choosing φ1 and φ₂ make sure that both are non-trivial solutions.
Note that φ₁ and φ₂ form a fundamental pair of solutions of [y] = 0, since we assumed that homogeneous BVP has only trivial solutions.By Lagrange’s identity (5.20), we get d dx p(φ ′ ₁φ₂ − φ₁φ ′ ₂ ) = 0.
This implies p(φ ′ 1φ₂ − φ₁φ ′ ₂ ) ≡ c, a constant and non-zero, (5.44)
as a consequence of (φ ′ ₁φ2 − φ₁φ ′ 2 ) being the wronskian corresponding to a fundamental pair of solutions.
Green’s function is then given by
Verifying that G(x, ξ) has the required properties for it to be a Green’s function is left as an exercise.

Generalised Green’s function

In Section 5.3 we constructed Green’s function for the homogeneous BVP when the latter problem had only trivial solutions. Now we investigate the case where the homogeneous BVP has non-trivial solutions.
Let us look at what happens in the case of linear system of equations of size k in k variables. Let A be a k × k matrix, and b ∈ R k . Suppose that Ax = 0 has non-trivial solutions. In this case we know that Ax = b does not have solution for every b ∈ R k . We also know that Ax = b has a solution if and only if b belongs to orthogonal complement of a certain subspace of R k , i.e., b must satisfy a compatibility condition. Note that, in this case, if Ax = b has one solution, then there are infinitely many solutions.
We are going to see that there exists an analogue of Green’s function, called Generalised Green’s function. We also prove that a situation similar to that of matricial analogy given above occurs here too; and the compatibility condition for the existence of a solution to nonhomogeneous BVP is a kind of Fredholm alternative.
Let us recall the notations Q₁ and Q₂.

Generalised Green’s function: Let φ₀ be a solution of [y] = 0 with homogeneous boundary conditions, such that . A function Γ(x, ξ) defined in Q is called a generalised Green’s function if Γ has the following properties:
(i) The function Γ(x, ξ) is continuous in Q.
(ii) The first and second order partial derivatives w.r.t. variable x of the function Γ(x, ξ) exist and continuous up to the boundary on Q₁ and Q₂.
(iii) Let ξ ∈ [a, b] be fixed. Then γ(x, ξ), considered as a function of x, satisfies 

at every point of the interval [a, b], except at ξ, satisfying the homogeneous boundary conditions U₁[y] = 0 and U₂[y] = 0.
(iv) The first derivate has a jump across the diagonal x = ξ, of magnitude 1/p, i.e., 

(v) The function Γ(x, ξ) satisfies the condition

The following result characterises the class of functions f, for which the nonhomogeneous equation [y] = f has a solution satisfying homogeneous boundary conditions.

Solved Numericals

Q1. We have $\frac{d}{d}\left(\frac{1}{3}\frac{d}{y}\right)+𝜆(3x2+1):$,y = 0, y(0) = 0, y(π) = 0 then the corresponding eigen value and eigen function are:
Solution:

Differentiating both sides we get

Then the ODE becomes

Auxillary equation is  

Solution is

Hence solution is

Q2. Consider the differential equation .If x = 0 at t = 0 and x = 1 at t = 1, the value of x at t = 2 is: 
Solution: 

• Let d/dx = D
• So, the differential equation becomes : (D² - 3D + 2)x = 0
• Now, as x cannot be zero, the function in the bracket will be zero.

Substituting the value in the solution: $𝑥=𝑐1𝑒𝐷1𝑡+𝑐2𝑒𝐷2𝑡$

$𝑥=𝑐1𝑒𝑡+𝑐2𝑒2𝑡$Applying the boundary condition: x = 0 at t = 0: we get : $𝑐1=-𝑐2$
Applying the second boundary condition: $𝑥=1,𝑡=1:$x = t, t = 1  It is obtained that : $𝑐1=\frac{1}{𝑒},𝑐2=\frac{1}{𝑒}$

 

Q3. For λ ∈ ℝ, consider the boundary value problem

Which of the following statement is true?
Solution: Given: Given D.E. is 

Concept used: We will put $𝑥=𝑒𝑧$x = e^zand then transform it in z
Putting $𝑥=𝑒𝑧$ we get x = e^z$𝑥2\frac{𝑑}{2}=\frac{𝑑}{2}-\frac{𝑑}{𝑦}$, we get

$\frac{𝑑}{𝑦}=\frac{𝑑}{𝑦}$

Now or D.E. reduced to

$(\frac{𝑑}{2}+\frac{𝑑}{𝑑}+\lambda )𝑦=0$
Now, using the notation $\frac{𝑑}{𝑑}=𝐷$  our equation reduced to

Auxillary equation
m² + m + λ = 0

Converting it into the function of x by putting

Now using initial values

So (i) and (ii) has a trivial solution for λ = 1/4$\lambda =1/4$
Except that it has non-trivial independent solution.