Math292 Textbook Duhamel's Principle

Section 2.1 Duhamel's Principle

Straight to the point....

Duhamel's Principle gives us a particular solution $Y (t)$ to a nonhomogeneous ODE $p (t) y^{″} + q (t) y^{'} + r (t) y = g (t) .$

Given a function $S (t; s)$ (variable $t$ and parameter $s$ ) that satisfies

\begin{aligned} p (t) S^{″} (t; s) + q (t) S^{'} (t; s) + r (t) S (t; s) & = 0, \\ S (s; s) & = 0, \\ S^{'} (s; s) & = 1, \end{aligned}

a particular solution to $p (t) y^{″} + q (t) y^{'} + r (t) y = g (t)$ is

\begin{aligned} Y (t) & = \int_{t_{0}}^{t} S (t; s) \frac{g (s)}{p (s)} d s . \end{aligned}

One thing to note here is that $Y (t_{0}) = Y^{'} (t_{0}) = 0 .$

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Duhamel's Principle also gives us a particular solution to a nonhomogeneous ODE of the form

p (t) y^{″} + q (t) y^{'} + r (t) y = g (t) .

However, it's often a lot easier to obtain this particular solution through Duhamel's principle than it is to obtain it using Variation of Parameters.

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Formally, say that we have a function

S (t; s)

(variable

t

and parameter

s

) that solves the following system:

\begin{aligned} p (t) S^{″} (t; s) + q (t) S^{'} (t; s) + r (t) S (t; s) & = 0, \\ S (s; s) & = 0, \\ S^{'} (s; s) & = 1. \end{aligned}

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Question 2.1.1. What does $S (t; s)$ or $S^{'} (t; s)$ mean?

The notation $S(t ; s)$ means that we treat $S$ not as a two-variable function, but as a one-variable function in $t$ with a parameter $s\text{.}$

Following this, we use $S'(t;s)$ as shorthand for

\begin{align*} S'(t;s) \amp = \frac{d}{dt} S(t;s). \end{align*}

For example, we might set $S(t ; s) = t^s\text{,}$ for some constant $s\text{.}$ Then, $S'(t;s) = st^{s-1}$ and $S''(t;s) = s(s-1)t^{s-2}\text{.}$

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Then, the particular solution of

p (t) y^{″} + q (t) y^{'} + r (t) y = g (t)

that Duhamel's principle gives us is

\begin{aligned} Y (t) & = \int_{t_{0}}^{t} S (t; s) \frac{g (s)}{p (s)} d s . \end{aligned}

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This is all a lot to absorb. The first thing to notice is that

S (t; s)

is a solution of the homogeneous system

p (t) y^{″} + q (t) y^{'} + r (t) y = 0 .

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Let's take

{y_{1}, y_{2}}

to be a set of fundamental solutions of the homogeneous ODE. Then, since

S (t; s)

is a solution of the homogeneous ODE,

\begin{aligned} S (t; s) & = c_{1} y_{1} (t) + c_{2} y_{2} (t) . \end{aligned}

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This was the info extracted from

p (t) S^{″} (t; s) + q (t) S^{'} (t; s) + r (t) S (t; s) = 0 .

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Looking now at the other conditions,

S (s; s) = 0

and

S^{'} (s; s) = 1,

notice that these two give us the following two equations.

\begin{aligned} c_{1} y_{1} (s) + c_{2} y_{2} (s) & = 0, \\ c_{1} y_{1}^{'} (s) + c_{2} y_{2}^{'} (s) & = 1. \end{aligned}

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Here,

y_{1} (s)

and

y_{2} (s)

are both knowns and the only unknowns in the above are

c_{1}, c_{2} .

This is just begging us to write the system above with a matrix. Let's do that:

\begin{aligned} [\begin{array}{c} y_{1} (s) & y_{2} (s) \\ y_{1}^{'} (s) & y_{2}^{'} (s) \end{array}] [\begin{array}{c} c_{1} \\ c_{2} \end{array}] & = [\begin{array}{c} 0 \\ 1 \end{array}] . \end{aligned}

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Solving this isn't too bad:

\begin{aligned} [\begin{array}{c} c_{1} \\ c_{2} \end{array}] & = {[\begin{array}{c} y_{1} (s) & y_{2} (s) \\ y_{1}^{'} (s) & y_{2}^{'} (s) \end{array}]}^{- 1} [\begin{array}{c} 0 \\ 1 \end{array}] \\ = \frac{1}{y_{1} (s) y_{2}^{'} (s) - y_{1}^{'} (s) y_{2} (s)} [\begin{array}{c} y_{2}^{'} (s) & - y_{2} (s) \\ - y_{1}^{'} (s) & y_{1} (s) \end{array}] [\begin{array}{c} 0 \\ 1 \end{array}] \\ = \frac{1}{y_{1} (s) y_{2}^{'} (s) - y_{1}^{'} (s) y_{2} (s)} [\begin{array}{c} - y_{2} (s) \\ y_{1} (s) \end{array}] . \end{aligned}

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A lot of things look pretty familiar here. The fraction above has a denominator equal to

y_{1} (s) y_{2}^{'} (s) - y_{1}^{'} (s) y_{2} (s),

which is exactly the Wronskian

W [y_{1}, y_{2}] (s) .

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Another thing to note is that, from the solution of

c_{1}, c_{2}

above, it's clear that both of these depend on

s .

So, let's write

c_{1} = c_{1} (s)

and

c_{2} = c_{2} (s) .

This way, we now have

\begin{aligned} S (t; s) & = c_{1} (s) y_{1} (t) + c_{2} (s) y_{2} (t), \end{aligned}

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where

\begin{aligned} c_{1} (s) = - \frac{y_{2} (s)}{W [y_{1}, y_{2}] (s)} & and c_{2} (s) = \frac{y_{1} (s)}{W [y_{1}, y_{2}] (s)} . \end{aligned}

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The familiarity of the above should become much more pronounced when we plug this into

Y (t) = \int_{t_{0}}^{t} S (t; s) g (s) / p (s) d s .

\begin{aligned} Y (t) & = \int_{t_{0}}^{t} [c_{1} (s) y_{1} (t) + c_{2} (s) y_{2} (t)] \frac{g (s)}{p (s)} d s \\ = \int_{t_{0}}^{t} \frac{y_{1} (s) y_{2} (t) - y_{1} (t) y_{2} (s)}{W [y_{1}, y_{2}] (s)} \cdot \frac{g (s)}{p (s)} d s . \end{aligned}

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Hey! This is Variation of Parameters exactly. So, Duhamel's principle is like a “re-skin” of Variation of Parameters. However, as we'll soon see with some examples, it can be a lot easier to work with than variation of parameters.

Section 2.1 Duhamel's Principle

Straight to the point....

Question 2.1.1. What does S(t;s) or S′(t;s) mean?

Question 2.1.1. What does $S (t; s)$ or $S^{'} (t; s)$ mean?