2
$\begingroup$

I'm trying to learn how to find the general integral of a partial differential equation.

My module's paper usually asks after giving a pde, if there exists a singular integral, if so to find it and also asks the general integral.

Let's take the example:

$$pq=p+q$$

Where $p$ & $q$ are the partial derivatives of $z$ with respect to $x$ and $y$, respectively.

The complete integral has been found by considering a trial solution,

$$z=ax+by+c$$

which implies

$$z=ax+\frac{a}{a-1}y+c$$

The model answer also says there exists no singular integral for this.

The guide says to find the general integral by substituting $c=φ(a)$ in the complete integral found earlier

$$z=ax+\frac{a}{a-1}y + φ(a) \quad (A)$$

and then take the derivative of $(A)$ with respect to $a$ ,

$$0=x+\frac{(a-1)y - ay}{(a-1)^2} + φ'(a) \quad (B)$$

Then by $(A)$ and $(B)$, eliminate $a$ and obtain the general integral.

First of all, I don't know the relationship between singular integral, general integral and complete integral and what they mean.

Secondly, I don't see how I can eliminate $φ(a)$ and $φ'(a)$ from the two equations to get rid of $a$.

Any help is appreciated.

1 Answers 1

0

$z_xz_y=z_x+z_y$

$z_xz_y-z_x=z_y$

$z_x(z_y-1)=z_y$

$z_x=\dfrac{z_y}{z_y-1}$

$z_{xy}=-\dfrac{z_{yy}}{(z_y-1)^2}$

Let $u=z_y$ ,

Then $u_x=-\dfrac{u_y}{(u-1)^2}$

Follow the method in http://en.wikipedia.org/wiki/Method_of_characteristics#Example:

$\dfrac{dx}{dt}=1$ , letting $x(0)=0$ , we have $x=t$

$\dfrac{du}{dt}=0$ , letting $u(0)=u_0$ , we have $u=u_0$

$\dfrac{dy}{dt}=\dfrac{1}{(u-1)^2}=\dfrac{1}{(u_0-1)^2}$ , letting $y(0)=f(u_0)$ , we have $y(0)=f(u_0)+\dfrac{t}{(u_0-1)^2}=f(u)+\dfrac{x}{(u-1)^2}$ i.e. $u=F\left(y-\dfrac{x}{(u-1)^2}\right)$