Syllabus for 18.305
- Review of Ordinary Differential Equations. ( Two lectures.)
- Review of complex analysis. ( Three Lecture.)
- Partial differential equations. (Six lectures.)
- WKB Method (Three lectures.)
- The Laplace method, the method of stationary phase, and the saddle point method. (Six lectures.)
- Boundary Layers. ( Four lectures.)
- Small Non-Linear Oscillations. ( Four lectures.)
Supplementary topic: Integrating factor
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Linear homogeneous second-order equation
Consider the linear, homogeneous, second order differential equation
$Y"+a(x)Y^{\prime}+b(x)Y=0.$ (1)
This equation is not readily solvablet if the coefficients $a(x)$ and $b(x)$ are not constants. In this section. we shall show that if one of its independent solutions is known, a second independent solution can be found.
We put
$Y_{2}(x)=Y_{1}(x)v(x),$ (2)
where $Y_{1}(x)$ is the first known solution and $Y_{2}(x)$ is the second solution. Substituting (2) into (1), we get
$(Y_{1}^{"}v+2Y_{1}^{\prime}v^{\prime}+Y_{1}v^{"})+a(x)(Y_{1}^{\prime}v+Y_{1}v^{\prime})+b(x)(Y_{1}v)=0.$
Since $Y_{1}(x)$ satisfies (1), the equation above can be reduced to
$(2Y_{1}^{\prime}v^{\prime}+Y_{1}v^{"})+a(x)(Y_{1}v^{\prime})=0,$
the other terms in the equation cancelling one another. Dividing the equation by $Y_{1}v^{\prime}$ and rearranging the terms in the equation, we get
$\dfrac{v"}{v^{\prime}}=-\dfrac{2Y_{1}^{\prime}}{Y_{1}}-a(x).$
Integrating, we get
$v^{\prime}(x)=\dfrac{\exp[-\int_{0}^{x}a(x^{^{\prime}})dx^{\prime}]}{Y_{1}^{2}(x)}$.
Integrating again, we get
$v(x)=\int_{0}^{x}dx^{\prime}\dfrac{\exp[-\int_{0}^{x^{\prime}}a(x^{"})dx^{"}]}{Y_{1}^{2}(x^{\prime})}.$
The second solution $Y_{2}(x)$ is therefore
$Y_{2}(x)=Y_{1}(x)\int_{0}^{x}dx^{\prime}\dfrac{\exp[-\int_{0}^{x^{\prime}}a(x^{"})dx^{"}]}{Y_{1}^{2}(x^{\prime})}.$ (3)
Thus we find the second solution $Y_{2}(x)\ $by introducing an integrating factor $Y_{1}(x),$which is a homogeneous solution.
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Linear inhomogeneous first-order equation
In this section we shall show that the first-order linear inhomogeneous differential equation
$y^{\prime}(x)+p(x)y(x)=q(x)$ (4)
is solvable.
First, the complementary solution $Y(x)$ of (4), satisfying
$Y^{\prime}(x)+p(x)Y(x)=0,$
or
$\dfrac{Y^{\prime}}{Y}=-p.$
is solved by
$Y(x)=\exp[-P(x)],$
where
$P(x)=\int_{0}^{x}p(x^{\prime})dx^{\prime}.$
As we shall see, the homogeneous solution $Y(x)$ is again the integrating factor for the inhomogeneous equation (4). We substitute
$y(x)=\exp[-P(x)]u(x)$ (5)
into (4) and get
$u^{\prime}(x)=\exp[P(x)]q(x).$
Integrating, we get
$u(x)=\int_{0}^{x}e^{P(x^{\prime})}q(x^{\prime})dx^{\prime}+c,$
where $c$ is a constant. Thus
$y(x)=e^{-P(x)}[\int_{0}^{x}e^{P(x^{\prime})}q(x^{\prime})dx^{\prime}+c].$
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Second-order linear inhomogenious differential equation
Finally, consider the second-order linear inhomogenious differential equation
$y"+a(x)y^{\prime}+b(x)y=f(x).$ (6)
This equation can be solved provided that $Y_{1}(x)$, one of its complementary solutions, is known.
To see this, we remind the reader that we have shown in Sec A how to find $Y_{2}$ of (6) once $Y_{1}$ is known. Then we can find the solution of (6) in the following way. We put
$y(x)=Y_{1}(x)v_{1}(x)+Y_{2}(2)v_{2}(x).$ (7)
From (7) we have
$y^{\prime}(x)=Y_{1}^{\prime}(x)v_{1}(x)+Y_{1}(x)v_{1}^{\prime}(x)+\ Y_{2}^{\prime}(2)v_{2}(x)+Y_{2}(2)v_{2}^{\prime}(x).$ (8)
We note that since (7) expresses a single funtion $y(x)$ by two functions $v_{1}(x)$ and $v_{2}(x),$ we have the freedom to impose a condition on the latter functions.
We shall choose it to be
$Y_{1}(x)v_{1}^{\prime}(x)+Y_{2}(2)v_{2}^{\prime}(x)=0.$ (9)
With (9), (8) becomes
$y^{\prime}(x)=Y_{1}^{\prime}(x)v_{1}(x)+Y_{2}^{\prime}(x)v_{2}(x).$ (10)
Differentiatng (10) we get,
$y"(x)=Y_{1}^{\prime}(x)v_{1}^{\prime}(x)+Y_{1}^{"}(x)v_{1}(x)+Y_{2}^{^{\prime}}(x)v_{2}^{\prime}(x)+Y_{2}^{"}(x)v_{2}(x).$ (11)
Substitute (7), (10) and (11) into (6), we get
$Y_{1}^{\prime}(x)v_{1}^{\prime}(x)+Y_{2}^{\prime}(x)v_{2}^{\prime}(x)=f(x),$ (12)
where we have made use of the fact that $Y_{1}(x)$ and $Y_{2}(x)$ are the complementay solutions of the differential equation.
Solving the simultaneus algebraic equations (9) and (12) for $v_{1}^{\prime}(x)$ and $v_{2}^{\prime}(x)$, with Kramer's rule
$v_{1}^{\prime}(x)=-\dfrac{f(x)Y_{2}(x)}{W(x)}$ (13)
and
$v_{2}^{\prime}(x)=\dfrac{f(x)Y_{1}(x)}{W(x)},$ (14)
where $W(x)$ is the Wronskian
$W(x)\equiv Y_{1}(x)Y_{2}^{\prime}(x)-Y_{1}^{\prime}(x)Y_{2}(x).$ (15)
From (13) and (14) we get
$v_{1}(x)=-\int_{0}^{x}\dfrac{f(x^{\prime})Y_{2}(x^{\prime})}{W(x^{\prime})}dx^{\prime}$ (16)
and
$v_{2}^{\prime}(x)=\int_{0}^{x}\dfrac{f(x^{\prime})Y_{2}(x^{\prime})}{W(x^{\prime})}dx^{\prime}.$ (17)
Thus a particular solution for (6) is (18)
$y(x)=-Y_{1}(x)\int_{0}^{x}\dfrac{f(x^{\prime})Y_{2}(x^{\prime})}{W(x^{\prime})}dx^{\prime}+Y_{2}(x)\int_{0}^{x}\dfrac{f(x^{\prime})Y_{1}(x^{\prime})}{W(x^{\prime})}dx^{\prime}.$ (19)
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The Wronskian
The Wronskian (15) can be expressed more simply. Let us re-write it as a determinant
$W(x)=\left\vert \begin{array} [Y_{1}(x) & Y_{2}(x)\\ Y_{1}^{\prime}(x) & Y_{2}^{\prime}(x) \end{array} \right\vert .$
Then
$W^{\prime}(x)=\left\vert \begin{array} [Y_{1}(x) & Y_{2}(x)\\ Y_{1}^{"}(x) & Y_{2}^{"}(x) \end{array} \right\vert =-a(x)\left\vert \begin{array} [Y_{1}(x) & Y_{2}(x)\\ Y_{1}^{\prime}(x) & Y_{2}^{\prime}(x) \end{array} \right\vert =-a(x)W(x).$
Therefore.
$W(x)=c\exp[-\int_{0}^{x}a(x^{\prime})dx^{\prime}].$ (20)
The constant $c$ above depends on the choice of $Y_{1}(x)$ and $Y_{2}(x)$. For example, if we choose
$Y_{1}(0)=1,$ $Y_{1}^{\prime}(0)=0,$ $Y_{2}(0)=0,$ $Y_{2}^{\prime}(0)=1,$
then
$c=1.$
With this choice, (20) is
$Y_{1}(x)Y_{2}^{\prime}(x)-Y_{1}^{\prime}(x)Y_{2}(x)=\exp[-\int_{0}^{x}a(x^{\prime})dx^{\prime}].$ (21)
With (21), (19) becomes
$y(x)=-Y_{1}(x)\int_{0}^{x}\exp[\int_{0}^{x^{\prime}}a(x^{"})dx^{^{"}}][Y_{1}(x)Y_{2}(x^{\prime})-Y_{2}(x)Y_{1}(x^{\prime})]f(x^{\prime})dx^{\prime}.$ (22)
Finally, we mention that viewing (21) as a differential equation for $Y_{2}(x)$ offers an alternative way to determine $Y_{2}(x) $from $Y_{1}(x)\ $ is given. This is because (21) is in the form of (4) if we identify $Y_{2}(x)$ as $y(x),$ $-Y_{1}^{\prime}(x)/Y_{1}(x)$ as $p(x),$ and $\exp[-\int_{0}^{x}a(x^{\prime})dx^{\prime}]/Y_{1}(x)$ as $q(x)$.