There are a couple of questions I have which would help me understand the working (I have put the questions in bold).
Question: Find the solution to the partial differential equation
$$uu_x-u_y=2, \qquad y \gt 0$$
Subject to the initial condition $u(x,0)=x^2$ and determine where in the xy-plane this solution is valid.
My attempt:
Parameterise the initial data:
$$x = s, \qquad y = 0, \qquad u = x^2$$
Write the differential equations which determine the characteristics:
$$\frac{dx}{d\tau}=u, \qquad \frac{dy}{d\tau}=-1, \qquad \frac{du}{d\tau}=2.$$
1. Are the above called 'Characteristics', or are the 'solutions' to the differential equations called the characteristics of the PDE? For example. solving the third equation, $\bf{u = 2\tau+s^2}$.
Solving the ordinary differential equations, and eliminating $\tau$ I get the family of ground/base curves, in terms of the parameter, $s$:
$$x=y^2 – s^2y +s$$
2. Does this answer the part of the question 'determine where in the xy-plane this solution is valid.'?
Now to find the solution, since $y = -\tau, $ we just need to find an expression for $s$:
Using the quadratic formula for the equation corresponding to the family of base curves, I get:
$$s = \frac{1 \pm \sqrt{1-4y(x-y^2)}}{2y}$$
3. I don't know which sign I should choose for the problem and the initial conditions given?
Finally the solution to the partial differential equation is going to be:
$$u(x,y) = u = -2y+s^2.$$ – Where $s$ would be one of the two possible solutions above.
4. Is this correct?
5. Are there any resources which would assist my understanding of simple PDE's?
Best Answer
In fact, your result is : $$u(x,y) = -2y+\left(\frac{1 \pm \sqrt{1-4y(x-y^2)}}{2y}\right)^2.$$ These solutions both satisfy the PDE $uu_x-u_y=2$. But both doesn't agree with the condition $u(x,0)=x^2$.
So, you are right to ask about the sign of the square root.
If the sign is plus, obviously $u$ tends to infinity for $y$ tending to $0$, which contradicts the condition. This solution of the PDE must be rejected.
If the sign is minus, $u$ tends to $x^2$ for $y$ tending to $0$. This is easy to show thanks to limited series expansion. I let this for you.
Thus, the solution satisfying the PDE and the boundary condition is : $$u(x,y) = -2y+\left(\frac{1 - \sqrt{1-4y(x-y^2)}}{2y}\right)^2.$$