The solution of Cauchy problem Let $a,b\;\epsilon\;\mathbb{R}$ be such that $a^2+b^2\neq\;0$. Then the Cauchy problem
$a\frac{\partial{u}}{\partial{x}}+b\frac{\partial{u}}{\partial{y}}=1$  ; $x,y\;\in\;\mathbb{R}$
$u(x,y)=x$ on $ax+by=1.$


*

*has more than one solution if either $a$ or $b$ is zero

*has no solution.

*has a unique solution.

*has infinitely many solution
My attempt:
Lagrange's auxiliary equations are $\frac{dx}{a}=\frac{dy}{b}=\frac{du}{1}$ . Solving these and using the conditions we get the solution involving $u,x,y$. As all the variables are presents in the solution so the P.D.E. has unique solution. But this method is too lengthy and laborious. As in the problem it is mention that it is a Cauchy problem , so I think it can be solved using Cauchy problem. Here it is a linear Cauchy problem. I know to solve a Cauchy problem. But without solving how we can determine that the problem has unique solution ?
Please help...
 A: As Herebrij and David pointed out; if one of the is zero, say $b=0$, then we have a unique solution.
Using Method of Characteristics, letting $x=x(t),y=y(t)$ and hence $u=u(t)$
$$\frac{du}{dt}=\frac{\partial u}{\partial x}\frac{dx}{dt}+\frac{\partial u}{\partial y}\frac{dy}{dt}$$
Comparing to $a\frac{\partial u}{\partial x}+b\frac{\partial u}{\partial x}=1$, we get
$$\frac{dx}{dt}=a,\frac{dy}{dt}=b, \frac{du}{dt}=1$$
$\frac{dx}{dt}=a\implies x=at\implies dt=\frac{dx}a$
$\frac{dy}{dt}=b\implies a\frac{dy}{dx}=b\implies ay-bx=y_0$
$\frac{du}{dt}=1\implies a\frac{du}{dx}=1\implies u(x,y)={x+F(y_0)\over a}={x+F(ay-bx)\over a}$
Now, $(\frac 1ax,0)$ lies on $ax+by=1$, so
$$u\left(\frac 1a x,0\right)=\frac 1a\left[\frac xa+F\left(-b\frac xa\right)\right]=\frac xa\implies F\left(-b\frac xa\right)=\frac xa\left[1-\frac 1a\right]$$
Letting $t=-b\frac xa$, we get
$$F(t)=\frac t{-b}\left[1-\frac 1a\right]$$
Substituting and simplifying, $u(x,y)=\frac{y(1-a)}{b}+x$
Which is unique for given $(a,b)$. Hence option 3.
A: Isn't $\partial u(x,y) /\partial x=1$ and $\partial u(x,y) /\partial y=0$ leaving just $a=1$ and therefore $b = 0$ and $x=1$ for the remaining equations? Therefore we only have infinitely many solutions because $y$ can vary over all the reals.
