# 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 $\frac{d}{dt} f*y(0) = f(t)y(0) + \int_0^0 f(u) y^{\prime}(t-u) du = 0$.