Solution of Differential Equation | Engineering Mathematics - Engineering Mathematics PDF Download

Initial and Boundary value problems

With an initial value problem we are given a differential equation together with the value of the unknown function and the appropriate number of its derivatives at the same point. Those values are called initial conditions. For example, for a second order differential equation the initial conditions are

• y(t₀) = y₀
• y'(t₀) = y₁

With a boundary value problem (BVP) we are given a differential equation together with values of the function and/or derivatives at different points (the boundary values). For a second order equation the boundary conditions commonly used are any one of the following types:

• y(x₀) = y₀, y(x₁) = y₁
• y'(x₀) = y₀, y'(x₁) = y₁
• y'(x₀) = y₀, y(x₁) = y₁
• y(x₀) = y₀, y'(x₁) = y₁

Throughout this chapter we will usually work with linear second order equations of the standard form

y'' + p(x) y' + q(x) y = g(x)

In earlier chapters a differential equation was called homogeneous if g(x) = 0 for all x. For a boundary value problem we extend that notion: the BVP is called homogeneous when both the differential equation is homogeneous and the boundary values are zero. For instance, for a second order BVP we call it homogeneous if

g(x) = 0, y(x₀) = 0, y(x₁) = 0

If any of these are nonzero we call the BVP non-homogeneous. It is important to remember that the adjective homogeneous (or non-homogeneous) refers to the combination of the differential equation and its boundary conditions, not only to the differential equation alone.

For linear ordinary differential equations (ODEs), an initial value problem typically has a unique solution under very mild regularity conditions on the coefficients. For boundary value problems the situation is more delicate: even for very nice linear equations a BVP may have no solution, a unique solution, or infinitely many solutions depending on the boundary conditions. The method of solution, however, is the same: solve the differential equation to obtain the general solution and then determine constants by applying the given boundary conditions.

Example 1

Solve the following BVP.

y'' + 4y = 0   y(0) = -2   y(π/4) = 1

Solution:

The general solution of the homogeneous differential equation is

y(x) = C₁ cos(2x) + C₂ sin(2x)

Apply the first boundary condition.

y(0) = C₁ = -2

Apply the second boundary condition.

y(π/4) = C₁ cos(π/2) + C₂ sin(π/2)

cos(π/2) = 0 and sin(π/2) = 1, therefore

1 = C₂

Thus the solution satisfying both boundary conditions is

y(x) = -2 cos(2x) + sin(2x)

Example 2

Solve the following BVP.

y'' + 4y = 0   y(0) = -2   y(2π) = -2

Solution:

The general solution remains

y(x) = C₁ cos(2x) + C₂ sin(2x)

Apply the first boundary condition.

y(0) = C₁ = -2

Apply the second boundary condition.

y(2π) = C₁ cos(4π) + C₂ sin(4π)

cos(4π) = 1 and sin(4π) = 0, therefore

-2 = C₁

The second condition gives the same equation for C₁ and imposes no condition on C₂.

Hence every function of the form

y(x) = -2 cos(2x) + C₂ sin(2x)

with arbitrary C₂ is a solution of the BVP. Therefore the BVP has infinitely many solutions.

Laplace Transform

Definition and basic properties

The Laplace transform converts an initial value problem for an ODE into an algebraic problem in the transform variable, which is often easier to solve. The usual procedure is:

• Take the Laplace transform of both sides of the differential equation.
• Use the transforms of derivatives to obtain an algebraic equation for the transform of the unknown function.
• Solve that algebraic equation for the transform.
• Find the inverse Laplace transform to recover the solution of the original problem.

Definition (Laplace transform): For a function f(t) defined for t ≥ 0 the Laplace transform F(s) is

F(s) = L{f}(s) = ∫₀^∞ e^-st f(t) dt for all s for which this integral converges.

Common examples and elementary transforms are given in standard tables.

Theorem (Linearity of the Laplace transform): If F(s) and G(s) are Laplace transforms of f and g valid for s > a, and α, β are real constants, then

L{α f + β g}(s) = α F(s) + β G(s) for s > a.

Inverse Laplace transform and uniqueness

Definition (Inverse Laplace transform): If G(s) is a given function and there exists g(t) such that

L{g}(s) = G(s) for s in an appropriate region, then g is called an inverse Laplace transform of G and we write

g(t) = L^-1{G}(t)

Theorem (Lerch's theorem - uniqueness): If f and g are continuous on [0, ∞) and their Laplace transforms agree on a half-line s > a, then f = g almost everywhere (in particular they are equal as continuous functions). This gives uniqueness of the inverse Laplace transform under standard hypotheses.

Laplace transforms of derivatives

Theorem (Laplace transform of a derivative): If f is continuous on [0, ∞), f' is piecewise continuous on every finite interval [0, k], and f(t) grows at most exponentially so that the Laplace transform exists for s > s₀, then

L{f'}(s) = s F(s) - f(0), where F(s) = L{f}(s).

Theorem (Laplace transform of a higher derivative): Under similar hypotheses for f, f', ..., f^(n-1),

L{f⁽ⁿ⁾}(s) = sⁿ F(s) - s^n-1 f(0) - s^n-2 f'(0) - ... - f^(n-1)(0).

Examples illustrating these formulae are given in the following placeholders.

Convolution

Definition (Convolution): If f and g are defined for t ≥ 0 the convolution f * g is

(f * g)(t) = ∫₀^t f(τ) g(t - τ) dτ for t ≥ 0.

Convolution theorem: If F(s) = L{f}(s) and G(s) = L{g}(s) then

L{f * g}(s) = F(s) G(s).

Additional properties and examples:

Worked examples showing convolution computations and using the convolution theorem are indicated below.

Find f such that

More results relating differentiation under the transform and convolution are available.

Example problems solved by convolution methods:

Unit impulses and the Dirac delta

Dirac's delta function δ(t - a) is a distribution that models an idealised point impulse at t = a. Informally it satisfies

• δ(t - a) = 0 for t ≠ a
• ∫_-∞^∞ δ(t - a) dt = 1
• Filtering (sifting) property: ∫₀^∞ f(t) δ(t - a) dt = f(a) when a ≥ 0 and f is continuous at a.

Filtering property (theorem): Let a ≥ 0 and f be integrable on [0, ∞) and continuous at a. Then

∫₀^∞ f(t) δ(t - a) dt = f(a).

The Laplace transform of a shifted delta is

L{δ(t - a)}(s) = e^-as.

Examples involving δ(t - a) in forcing functions and their Laplace transform solutions are given below.

Systems, polynomial coefficient equations and further theorems

Laplace transform solution of linear systems: Apply the Laplace transform to each equation of the system and solve the resulting algebraic system for the transforms of the unknown functions. Then invert to find the time-domain solutions.

Differential equations with polynomial coefficients: There are transform rules relating multiplication by t and differentiation with respect to s in the transform domain. For example, if F(s) is the Laplace transform of f(t) and F is sufficiently differentiable, then

L{t f(t)}(s) = -dF/ds, and more generally

L{tⁿ f(t)}(s) = (-1)ⁿ dⁿF/dsⁿ.

Useful examples employing these rules are indicated in the following placeholders.

Additional examples and applications:

The Wave equation

The one-dimensional wave equation models transverse vibrations of a string and is written as

u_tt(x,t) = c² u_xx(x,t), where c is the wave speed, t denotes time and x denotes the spatial coordinate.

The unknown function u(x,t) represents the displacement of the string at position x and time t. The equation is derived by applying Newton's second law to an infinitesimal element of the string and taking the limit as the element length → 0.

Dividing by the element length and taking the limit we obtain the wave equation:

(1)

For a string of length 1 with fixed ends the boundary conditions are

u(0,t) = 0, u(1,t) = 0

Initial conditions supply the initial displacement and velocity, for example

u(x,0) = f(x), u_t(x,0) = g(x)

For the infinite string (-∞ < x < ∞, t > 0) the classical solution is given by D'Alembert's formula:

This expresses the solution as the sum of two travelling waves determined by initial displacement and velocity.

The Diffusion (or heat) Equation

The diffusion or heat equation models the distribution of temperature (or concentration) in a rod. The one-dimensional heat equation is

We denote the temperature by u(x,t), which depends on the spatial coordinate x and time t.

In the notation of diffusion problems the source term r(x) often appears; for the homogeneous heat equation r(x) = 0.

Typical initial condition is

u(x,0) = f(x)

If the initial condition is a Dirac delta concentrated at ξ, representing a point heat source at x = ξ, then the homogeneous heat equation

has the fundamental solution (heat kernel) given by

This fundamental solution describes the evolution of an initial point source and is used to build solutions for general initial data by convolution with f(x).

Example: The Braselton chemical reaction system (two components)

Consider a reaction-diffusion system on the unit interval modelling two interacting chemical components u(x,t) and v(x,t). A prototypical system takes the form

with x ∈ [0,1] and positive diffusion coefficients D₁, D₂ and positive reaction parameters B, C.

Typical boundary conditions (given in the model statement) are

u(0,t) = u(1,t) = C, v(0,t) = v(1,t) = B/C

Well-posedness and solution methods use separation of variables, eigenfunction expansions, or transform methods depending on the linearity and coefficients. For nonlinear reaction terms one studies existence, uniqueness and long-time behaviour using PDE theory.