Exact Differential Equations
Exact Differential Equations
Review of Integration Theory
- Let $f(x)$ be a function of $x$ defined on an interval $I: a \leq x \leq b$. Let $I$ be divided into $n$ subintervals and call $\Delta x_k$ the width of the $k$ th subinterval. Then if
exists as the number of subintervals increases in such a manner that the largest subinterval approaches zero, we say that
\[\tag{9.12} \lim_{n \to \infty} \sum\limits_{k=1}^n f(x_k) \Delta x_k = \int_{a}^{b} f(x)dx.\]This limit is called the Riemann intergral of $f(x)$ over $I$. If this limit does not exist, we say $f(x)$ is not Riemann integrable over $I$.
- If $f(x)$ is a continuous funtion of $x$ on an interval $I: a \leq x \leq b,$ and
then by the fundamental theorem of the calculus
\[\tag{9.14} F'(x) = f(x), a < x < b,\]or equivalently
\[\tag{9.15} \frac{d}{dx}\int_{x_0}^x f(u)du = f(x).\]- Let $P(x,y)$ and $Q(x,y)$ be functions of two independent variables $x,y$ both function being defined on a common domain $D$. If we wish to perform the following integration, the first with respect to $x$ ( $y$ constant), and the second with respect to $y$ ( $x$ constant)
we must be sure that $(x_0,y_0)$ is a point of $D$ and that the rectangle dtermined by the line segments joining the points $(x_0,y_0)$, $(x,y_0)$, $(x_0,y_0)$, and $(x_0,y)$ lies entirely in $D$. Since domains can be of various shapes, we must exclude domains where this rectangle either crosses regions outside of the domain, or domains with holes inside the rectangle.
Definition 9.19 A region is called a simply connected region if every simple closed curve lying entirely in the region encloses only points of the region.
To summarize: We can perform the integrations called for in (9.16) only if:
a) The common domain of definition of the functions is a simply connected region $R$.
b) The point $(x_0,y_0)$ is in $R$
c) The rectangle determined by the lines joining the points $(x_0,y_0)$, $(x,y_0)$, $(x_0,y_0)$, and $(x_0,y)$ lies entirely in $D$
Definition of an Exact Differential and of an Exact Differential Equation
Definition 9.23 A differential expression
\[\tag{9.24} P(x,y)dx + Q(x,y)dy\]is called an exact differential if it is the total differential of a function $f(x,y)$, i.e., if
\[\tag{9.241} P(x,y) = \frac{\partial}{\partial x}f(x,y) \quad \text{and} \quad Q(x,y) = \frac{\partial}{\partial y}f(x,y).\]Definition 9.27 The differential equation
\[\tag{9.28} P(x,y)dx + Q(x,y)dy = 0\]is called exact if there exists a function $f(x,y)$ such that its partial derivative with respect to $x$ is $P(x,y)$ and its partial derivative with respect to $y$ is $Q(x,y)$. In symbolic notation, the definition says that (9.28) is an exact differential equation if there exists a function $f(x,y)$ such that:
\[\tag{9.29} \frac{\partial}{\partial x}f(x,y) = P(x,y), \frac{\partial}{\partial y}f(x,y) = Q(x,y)\]A 1-parameter family of solutions of the exact differential equation (9.28) is then
\[\tag{9.291} f(x,y) = c\]Necessary and Sufficient Condition for Exactness and Method of Solving an Exact Differential Equation
Theorem 9.3 A necessary and sufficient condition that the differential equation
\[\tag{9.31} P(x,y)dx + Q(x,y)dx = 0\]be exact is that
\[\tag{9.32} \frac{\partial}{\partial y}P(x,y) = \frac{\partial}{\partial x}Q(x,y)\]where the functions defined by P(x,y) and Q(x,y), the partial derivatives in (9.32) and $\partial P(x,y)/\partial x, \partial Q(x,y)/\partial y$ exist and are continuous in a simply connected region R.
Proof of necessary condition To summarize an already brief proof, recall from Calculus 3 that the order of partial derivatives doesn’t matter, i.e. that $F_xy(x,y) == F_yx(x,y)$. Thus, if such a function $f(x,y)$ whose partial derivatives with respect to $x$ and $y$ are $P(x,y)$ and $Q(x,y)$, respectively, then the second partial derivatives of those functions with respect to $y$ and $x$ must also be equal.
Solution from proof of sufficient condition This proof takes several pages in the book and I won’t reproduce it here. It does give a method of finding the function which is a solution, and I’ll instead talk about how to find that function.
Following the proof in the book we find we can take either of two functions as solutions.
\[\tag{9.45} f(x,y) = \int_{x_0}^x P(x,y)dx + \int_{y_0}^yQ(x_0,y)dy = c\]or
\[\tag{9.47} f(x,y) = \int_{x_0}^x Q(x,y)dy + \int_{y_0}^yP(x,y_0)dy = c\]where $(x_0,y_0)$ is a point in $R$ and the rectangle determined by the lines joining the points $(x_0,y_0)$, $(x,y_0)$, $(x_0,y_0)$, and $(x_0,y)$ lies entirely in $R$.
We can pick which function to used based on which is easier to compute with, which may require some experimentation. One handy effect is picking $0$ for $x_0$ and taking the first function with $Q(x_0,y)$ causes all terms of $x$ in $Q$ to vanish, and similarly taking $y_0$ as $0$ when we use the second function.
Using this formulaic approach, we setup the integral, integrate, and have our solution.
Solution from Definition
An alternative approach, and one preferred by the authors (and by me), is to instead construct a solution from definition.
jmh: what follows is my own and has some weird notation I made up and maybe some shakey ideas. The book didn’t do this; it used an specific function, and that may be more useful to look at.
Given a differential equation of the form:
\[\tag{s.1} P(x,y)dx + Q(x,y)dy = 0\]first test for exactness. If it’s exact, we know by definition (9.27) there is a $f(x,y)$ where:
\[\tag{s.2} \frac{\partial}{\partial x} f(x,y) = P(x,y).\]Then, by integration, we have:
\[\tag{s.3} f(x,y) = P_{-x}(x,y) + R(y)\]jmh: * $P_{-1}$ here is my notation for the antiderivative of $P(x,y)$ * with respect to x without any constant of integration, since that’s what* $R(Y)$ is. Not sure what the convention is here. The book shows this by example rather than abstractly.
where $R(y)$ is the arbitrary constant of integration (in this case an unknown function of $y$ since we integrated with respect to $x$)
we can then differentiate with respect to y to obtain
\[\tag{s.4} \frac{\partial}{\partial y} f(x,y) = \frac{\partial}{\partial y}( P_{-x}(x,y) + R(y)) = P_{-x,y}(x,y) + R'(y)\]where $P_{-x,y}$ is the antiderivative of $P$ with respect to $x$ subsequently partially differentiated by $y$.
but by definition (9.27) we also have
\[\tag{s.5} \frac{\partial}{\partial y} f(x,y) = Q(x,y)\]Setting the right-most expression of (s.4) equal to the right-most expression of (s.5) we get:
\[\tag{s.6} P_{-x,y}(x,y) + R'(y) = Q(x,y)\]$P_{-x,y}$ and $Q(x,y)$ will be almost equivalent, with just some constant or function of $y$ left between them, let’s call that $W(y)$ - but be clear that $W(y)$ is a known function (because both $P_{-x,y}(x,y)$ and $Q(x,y)$ are known, so is their difference where $R’(y)$ is unknown still (as it’s the derivative of an unknown arbitrary constant of integration).
To find $R(y)$ integrate to get
\[\tag{s.7} R(Y) = \int R'(y) dy = \int W(y) dy\]Then substituting (s.7) into (s.3) we get
\[\tag{s.8} f(x,y) = P_{-x}(x,y) + \int W(y) dy\]which is the function we seek.