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Applications of the Laplace transform in solving integral equations. Laws for convolution. Abel’s integral equation. The tautochrone problem. Integro-differential equations. Conversion of linear differential equations into integral equations.
Theorem 1. Let
where a
x
b and f is assumed to be integrable on [a, b]. Then the function F(x) is continuous
and F'(x) = f(x) at each point where f(x) is continuous.
Theorem 2. Let L[F(t)] = f(s). Then
Proof. Let
Then by Theorem 1 we have
and
3) G"(t) = F(t).
Since G(0) = G'(0) = 0, we get
4) L[G"(t)] = s2 L[G(t)] - sG(0) - G'(0) = s2 L[G(t)]
Since from 3), G"(t) = F(t), we have
5) L[G"(t)] = f(s)
From 4) and 5) we obtain
6) s2 L[G(t)] = f(s)
Consequently,
or
QED
This result can also be written as
and can be generalized in the following theorem
Theorem 3. Let L[F(t)] = f(s). Then
Theorem 4. Let L[F(t)] = f(s). Then
Proof. By the convolution theorem
Thus
By Theorem 2 above
Thus
QED
This result can be written as
and can be generalized in the following theorem
Theorem 5. Let L[F(t)] = f(s). Then
Laws for convolution
Theorem 6. Given functions F(t), G(t) and H(t), convolution obeys the following laws:
F*G = G*H Commutative law
F*(G*H) = (F*G)*H Associative law
F*(G + H) = F*G + F*H Left distributive law
(F + G)*H = F*H + G*H Right distributive law
Def. Integral equation. An integral equation is an equation in which an unknown function occurs under an integral sign. It has the general form
where F(t) and K(u, t) are known functions, a and b are either given constants or functions of t, and Y(t) is an unknown function to be determined. A function Y(t) may or may not exist that satisfies the equation.
The function K(u, t) is called the kernel or nucleus of the equation. If a and b are constants, the equation is called a Fredholm integral equation. If a is a constant and b = t, it is called a Volterra integral equation.
Def. Integral equation of convolution type. An integral equation of type
is said to be of convolution type. It can be written as
Y(t) = F(t) + K(t)*Y(t)
If we take the Laplace transform of both sides we find, assuming L[F(t)] = f(s) and L[K(t)] = k(s) both exist, that
y(s) = f(s) + k(s) y(s)
or
The required solution can then be found by inversion.
Example. Solve the integral equation
Solution. The equation can be written
Y(t) = t2 + Y(t)*sin t
Taking the Laplace transform and using the convolution theorem, letting y = L[Y], we get
Solving for y we get
Inverting
Abel’s integral equation. The tautochrone problem. An important integral equation of convolution type is Abel’s integral equation
where Y(u) is the unknown function to be determined, G(t) is given, and α is a constant such that 0 < α < 1. The equation is associated with a problem known as the tautochrome problem in which it is desired to find the shape of a frictionless wire lying in a vertical plane such that a bead placed on the wire slides to the lowest point in the same time regardless of where the bead is placed initially. Solution of the problem reveals the shape to be that of a cycloid. See Murray R. Spiegel. Laplace Transforms. (Schaum) for details.
Def. Integro-differential equation. An integro-differential equation is an integral equation in which various derivatives of the unknown function Y(t) can also be present. For example,
is an integro-differential equation. The solution of such equations subject to given initial conditions can often be obtained by Laplace transform methods.
Example. Solve the equation
where Y(0) = 2.
Solution. The equation can be written
Y'(t) + 5 cos 2t * Y(t) = 10
Taking the Laplace transform and letting y = L[Y], we get
which can be reduced by the method of partial fractions to
Inverting we get
Conversion of linear differential equations into integral equations. It is possible to convert a linear differential equation into an integral equation. See the following example.
Example. Convert the differential equation
Y"(t) - 3Y'(t) + 2Y(t) = 4 sin t
where Y(0) = 1, Y'(0) = -2, into an integral equation.
Solution. We give two methods.
Method 1. Let Y"(t) = V(t). Since by the definition of an integral
we obtain, after evaluating the constant c,
Similarly,
Now using Theorem 4 above we obtain
Thus the differential equation becomes
or
Method 2. We first integrate both sides of the differential equation
Y"(t) - 3Y'(t) + 2Y(t) = 4 sin t
where Y(0) = 1, Y'(0) = -2.
We thus obtain
Substituting in Y'(0) = -2 and Y(0) = 1, we get
Integrating again from 0 to t as before, and using Theorem 4, we obtain
or
References
Murray R. Spiegel. Laplace Transforms. (Schaum)
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