Find the solution of the Cauchy problem for the pde
$a\dfrac{\partial u }{\partial x}+b\dfrac{\partial u}{\partial y}=1$
with the initial condition $u(x,y)=x$ on $ax+by=1$.
By Lagrange's Equations
$\dfrac{dx}{a}=\dfrac{dy}{b}=\dfrac{dz}{1}$
If I solve the above I get ;$ay=bx+C$ and $y=bz+c_1,x=az+c_2$
But I am failing to use the boundary conditions given .
Please show some way to use them.
$$\frac{dx}{a}=\frac{dy}{b}=\frac{du}{1}$$ Only two characteristic curves are sufficient because this is a set of two characteristic equations (they are two sign =).
For example, we chose those characteristic curves :
From $\frac{dx}{a}=\frac{dy}{b}$ a first characteristic curve is : $ay-bx=c_1$
From $\frac{dx}{a}=\frac{du}{1}$ a second characteristic curve is : $u-\frac{x}{a}=c_2$
An implicit form of general solution is : $$\Phi\left(ay-bx\:,\:u-\frac{x}{a}\right)=0$$ where $\Phi$ is any differentiable function of two variables.
Solving the implicit equation for the second variable leads to the explicit equation :
$$u-\frac{x}{a}=F(ay-bx)$$ where $F$ is any differentiable function.
The general solution of the PDE is : $$u(x,y)=\frac{x}{a}+F(ay-bx)$$
This general solution can be obtained easily thanks to various methods. The above calculus is one example of method among several other equivalent methods. This not the difficult part of the task.
In fact, the difficult part of the task is to determine the function $F$ so that the initial condition be satisfied.
The condition is $u(x,y)=x$ on $ax+by=1$ that is $y=\frac{1-ax}{b}$
$u=x=\frac{x}{a}+F(a\frac{1-ax}{b}-bx) = \frac{x}{a} +F(\frac{a-(a^2+b^2)x}{b})$
$F(\frac{a-(a^2+b^2)x}{b})= x-\frac{x}{a}$ $$F(\frac{a-(a^2+b^2)x}{b})= \frac{a-1}{a}x$$
Let $X=\frac{a-(a^2+b^2)x}{b} \quad\to\quad x=\frac{a-bX}{a^2+b^2}$
$F(X)= \frac{a-1}{a}\frac{a-bX}{a^2+b^2} $ $$F(X)= \frac{a-1}{a^2+b^2} + \frac{b(1-a)}{a(a^2+b^2)}X$$
$$F(ay-bx)=\frac{a-1}{a^2+b^2} + \frac{b(1-a)}{a(a^2+b^2)}(ay-bx) $$
$u=\frac{x}{a}+F(ay-bx)= \frac{x}{a}+ \frac{a-1}{a^2+b^2} + \frac{b(1-a)}{a(a^2+b^2)}(ay-bx)$
Finally, the result is :
$$u(x,y)=\frac{a-1}{a^2+b^2} + \frac{b(1-a)}{a^2+b^2}y +\frac{a+b^2}{a^2+b^2}x$$
You can bing back this function $u(x,y)$ into the PDE and the initial condition and check that all agrees.