College Math Teaching

November 3, 2011

Finding a Particular solution: the Convolution Method

Background for students
Remember that when one is trying to solve a non-homogeneous differential equation, say:
y^{\prime \prime} +3y^{\prime} +2y = cos(t) one finds the general solution to y^{\prime \prime} +3y^{\prime} +2y = 0 (which is called the homogeneous solution; in this case it is c_1 e^{-2t} + c_2 e^{-t} and then finds some solution to y^{\prime \prime} +3y^{\prime} +2y = cos(t) . This solution, called a particular solution, will not have an arbitrary constant. Hence that solution cannot meet an arbitrary initial condition.

But adding the homogenous solution to the particular solution yields a general solution with arbitrary constants which can be solved for to meet a given initial condition.

So how does one obtain a particular solution?

Students almost always learn the so-called “method of undetermined coefficients”; this is used when the driving function is a sine, cosine, e^{at} , a polynomial, or some sum and product of such things. Basically, one assumes that the particular solution has a certain form than then substitutes into the differential equation and then determines the coefficients. For example, in our example, one might try y_p = Acos(t) + Bsin(t) and then substitute into the differential equation to solve for A and B . One could also try a complex form; that is, try y_p = Ae^{it} and then determines A and then uses the real part of the solution.

A second method for finding particular solution is to use variation of parameters. Here is how that goes: one obtains two linearly independent homogeneous solutions y_1, y_2 and then seeks a particular solution of the form y_p = v_1y_1 + v_2y_2 where v_1 = -\int \frac{f(t)y_2}{W} dt and v_2 = \int \frac{f(t)y_1}{W} dt where W is the determinant of the Wronskian matrix. This method can solve differential equations like y^{\prime \prime} + y = tan(t) and sometimes is easier to use when the driving function is messy.
But sometimes it can lead to messy, non transparent solutions when “undetermined coefficients” is much easier; for example, try solving y^{\prime \prime} + 4y = cos(5t) with variation of parameters. Then try to do it with undetermined coefficients; though the answers are the same, one method yields a far “cleaner” answer.

There is a third way that gives a particular solution that meets a specific initial condition. Though this method can yield a not-so-easy-to-do-by-hand integral and can sometimes lead to what I might call an answer in obscured form, the answer is in the form of a definite integral that can be evaluated by numerical integration techniques (if one wants, say, the graph of a solution).

This method is the Convolution Method. Many texts introduce convolutions in the Laplace transform section but there is no need to wait until then.

What is a convolution?
We can define the convolution of two functions f and g to be:
f*g = \int_0^t g(u)f(t-u)du . Needless to say, f and g need to meet appropriate “integrability” conditions; this is usually not a problem in a differential equations course.

Example: if f = e^t, g=cos(t) , then f*g = \frac{1}{2}(e^t - cos(t) + sin(t)) . Notice that the dummy variable gets “integrated out” and the variable t remains.

There are many properties of convolutions that I won’t get into here; one interesting one is that f*g = g*f ; proving this is an interesting exercise in change of variable techniques in integration.

The Convolution Method
If y(t) is a homogenous solution to a second order linear differential equation that meets initial conditions: y(0)=0, y^{\prime}(0) =1 and f is the forcing function, then y_p = f*y is the particular solution that meets y_p(0)=0, y_p^{\prime}(0) =0

How might we use this method and why is it true? We’ll answer the “how” question first.

Suppose we want to solve y^{\prime \prime} + y = tan(t) . The homogeneous solution is y_h = c_1 cos(t) + c_2 sin(t) and it is easy to see that we need c_1 = 0, c_2 = 1 to meet the y_h(0)=0, y^{\prime}_h(0) =1 condition. So a particular solution is sin(t)*tan(t) = tan(t)*sin(t)= \int_0^t tan(u)sin(t-u)du = \int_0^t tan(u)(sin(t)cos(u)-cos(t)sin(u))du = sin(t)\int_0^t sin(u)du - cos(t)\int_0^t \frac{sin^2(u)}{cos(u)}du = sin(t)(1-cos(t)) -cos(t)ln|sec(t) + tan(t)| + sin(t)cos(t) = sin(t) -cos(t)ln|sec(t)+tan(t)|

This particular solution meets y_p(0)=0, y_p^{\prime}(0) = 0 .

Why does this work?
This is where “differentiation under the integral sign” comes into play. So we write f*y = \int_0^t f(u)y(t-u)du .
Then (f*y)^{\prime} = ?

Look at the convolution integral as g(x,z) = \int_0^x f(u)y(z-u)du . Now think of x(t) = t, z(t) = t . Then from calculus III: \frac{d}{dt} g(x,z) = g_x \frac{dx}{dt} + g_z \frac{dz}{dt} . Of course, \frac{dx}{dt}=\frac{dz}{dt}=1 .
g_x= f(x)y(z-x) by the Fundamental Theorem of calculus and g_z = \int_0^x f(u) y^{\prime}(z-u) du by differentiation under the integral sign.

So we let x = t, z = t and we see \frac{d}{dt} (f*y) = f(t)y(0) + \int_0^t f(u) y^{\prime}(t-u) du which equals \int_0^t f(u) y^{\prime}(t-u) du because y(0) = 0 . Now by the same reasoning \frac{d^2}{dt^2} (f*y) = f(t)y^{\prime}(0) + \int_0^t f(u) y^{\prime \prime}(t-u) du = f(t)+ \int_0^t f(u) y^{\prime \prime}(t-u) du because y^{\prime}(0) = 1 .
Now substitute into the differential equation y^{\prime \prime} + ay^{\prime} + by = f(t) and use the linear property of integrals to obtain f(t) + \int_0^t f(u) (y^{\prime \prime}(t-u) + ay^{\prime}(t-u) + by(t-u))du = f(t) + \int_0^t f(u) (0)du = f(t)

It is easy to see that (f*y)(0) = 0. Now check \frac{d}{dt} f*y(0) = f(t)y(0) + \int_0^0 f(u) y^{\prime}(t-u) du = 0 .

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7 Comments »

  1. “But adding the particular solution to the particular solution yields a general solution with arbitrary constants which can be solved for to meet a given initial condition. ”

    I believe you mean adding the “… homogeneous solution to the particular solution …”

    Good article, though.

    Comment by Josh — September 28, 2012 @ 7:59 pm

  2. […] Of course, the convolution method has utility beyond the Laplace transform setting. […]

    Pingback by Why I teach multiple methods for the inverse Laplace Transform. | College Math Teaching — November 12, 2013 @ 8:33 pm

  3. […] I’ve discussed a type of convolution integral with regards to solving differential equations […]

    Pingback by The convolution integral: do some examples in Calculus III or not? | College Math Teaching — August 31, 2014 @ 11:32 pm

  4. Elequently and acurate, thank you very much
    Matia Yoav, P.hD

    Comment by yoav matia — June 23, 2015 @ 11:26 am

  5. sorry for the erroneous spelling check:
    “Eloquently and accurate, thank you very much”

    Comment by yoav matia — June 23, 2015 @ 11:29 am

  6. I believe sin(t)*tan(t) = tan(t)*sin(t) is wrong as it should be sin(t)*tan(t) = -tan(t)*sin(t)

    Comment by Diego — November 28, 2015 @ 9:27 pm


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