ODE and PDE, Basics II
January 28, 2021
1. Separable Differential Equations
Set \(x\) as the independent variable and \(y\) is a function of \(x\), we can generally write the first-order differential equation in the following form:
\[M(x, y)+N(x, y) \frac{d y}{d x}=0\tag{1}\]where \(M\) and \(N\) are two functions of \(x\) and \(y\). Let’s say, when \(M\) is a function of \(x\) and \(N\) is the function of \(y\) only, equation (1) becomes:
\[M(x)dx + N(y)dy = 0 \tag{2}\]Not quite easy actually, details here: $$ \begin{aligned} M(x)dx + N(y)dy &= 0 \\[2ex] \implies \frac{d H_1(x)}{dx}dx + \frac{dN(y)}{dy}dy &= 0 \\ \implies dH_1(x) + dH_2(y) &= 0 \\ \implies d(H_1(x) + H_2(y)) &= 0 \\ \implies H_1(x) + H_2(y) &= c \end{aligned} $$
This can be easily integrated to solution:
\[H_{1}(x)+H_{2}(y)=c \tag{3}\]where \(H_{1}(x) = \int M(x) dx\) and \(H_{2}(y) = \int N(y) dy\). The constant \(c\) is determined by the initial value.
An example, we want to solve the following first-order differential equation:
$$
\frac{d y}{d x}=\frac{x^{2}}{1-y^{2}}
$$
which is apparently separable.
Solution:
Rewrite it in the form of equation (2):
$$
-x^{2}dx+\left(1-y^{2}\right)dy=0
$$
So:
$$
H_{1}(x) = \int (-x^2) dx = -\frac{1}{3}x^3 + c_1 \\ H_{2}(y) = \int (1-y^2) dy = y-\frac{1}{3}y^3 + c_2
$$
We have:
$$
-\frac{1}{3}x^3 + y - \frac{1}{3}y^3 = c
$$
Constant is determined by some boudnary condistions
2. Existence and Uniqueness Theorem
For a first-order linear equation, as long as functions \(p\) and \(g\) are continuous on an open interval \(I, \alpha< t < \beta\) with \(t=t_0\) inside, there exists a unique function \(y=\phi(t)\) satisfies: $$y'+p(t)y=g(t)$$ for each \(t\) in \(I\), and also satisfies initial condition \(phi(t_0) = y_0\) when an arbitary initial value \(y_0\) is given.
For a first-order non-linear equation, as long as functions \(f\) and \(\partial f/\partial y\) are continuous in some open rectangle \( \alpha< t < \beta, \gamma < y < \delta\) with \( (t_0,y_0) \) inside, there exists a unique function \(y=\phi(t)\) in some interval \( t_{0}-h< t < t_{0}+h \)say, a small interval around initial point \(t=t_0\) containing \(\alpha < t < \beta\) satisfies: $$y' = f(t,y), \quad \text{with } \phi(t_0) = y_0$$
One note here: breaking of the conditions in the above two theorems does not mean there is no solutions, there could possibally be multiple solutionsSee B&D Book, Chap 2.4, Example 2, Page 53. .