General solution of a PDE- Lagrange or Characteristics method

Question

I am trying to find the general solution of the PDE:

$$xu_x + (1+y)u_y= x(1+y)+xu$$

1. If the initial condition is $$u(x,6x-1)=\phi(x)$$ then what is the necessary condition for $\phi$ that guarantees the existence of a solution? How can one solve the problem for the appropriate functions $\phi$.

   2. The same question as 1 if we change the initial condition to $$u(-1,y)=\psi(y)$$

What are the differences between 1 and 2?

I apologize in advance for asking in this way , because I am totally new to this subject and I am trying to learn some ideas. Thanks an advance.

Answer 1

$$xu_x+(1+y)u_y=x(1+y)+xu$$ Charpit-Lagrange equations : $$\frac{dx}{x}=\frac{dy}{1+y}=\frac{du}{x(1+y)+xu}=ds$$ A first characteristic equation comes from solving $\frac{dx}{x}=\frac{dy}{1+y}$ : $$\frac{1+y}{x}=c_1$$ A second characteristic equation comes from solving $\frac{dx}{x}=\frac{du}{x(1+y)+xu}=\frac{du}{x(c_1x)+xu}$

This is a first order linear ODE : $\frac{du}{dx}-u=c_1x$ $$ue^{-x}+c_1(x+1)e^{-x}=c_2$$ $c_1$ and $c_2$ are arbitrary related which leads to the general solution on the form of implicit equation : $$ue^{-x}+\frac{1+y}{x}(x+1)e^{-x}=F\left(\frac{1+y}{x}\right)$$ $F$ is an arbitrary function (to be determined according to boundary conditions). $$u(x,y)=-\frac{1+y}{x}(x+1)+e^xF\left(\frac{1+y}{x}\right)$$ FIRST CASE of boundary condition :

$u(x,6x-1)=\phi(x)=-\frac{1+(6x-1)}{x}(x+1)+e^xF\left(\frac{1+(6x-1)}{x}\right)$

$\phi(x)=-6(x+1)+e^xF\left(6\right)$

This implies that the function $\phi(x)$ must have a particular form : $$\phi(x)=-6(x+1)+e^xC\quad\text{where}\quad C=\text{constant}$$ In general the functions $\phi(x)$ have not this very particular form and as a consequence the function $F$ cannot be determined. The problem has no solution fitting to the boundary condition. If by luck the function $\phi(x)$ has the above particular form, the solution is $\quad u(x,y)=-\frac{1+y}{x}(x+1)+e^xC$

SECONDCASE of boundary condition :

$u(-1,y)=\psi(y)=-\frac{1+y}{-1}(-1+1)+e^{-1}F\left(\frac{1+y}{-1}\right)=e^{-1}F(-1-y)$

Let $X=-1-y\quad;\quad y=-X-1$

$\psi(-X-1)=e^{-1}F(X)$

The function $F(X)$ is determined : $$F(X)=e\:\psi(-X-1)$$ Now we can put it into the general solution where $X=\frac{1+y}{x}$ : $$u(x,y)=-\frac{1+y}{x}(x+1)+e^x e\:\psi\left(-\frac{1+y}{x}-1\right)$$ In this case, the problem has a well determined solution fitting to the boundary condition : $$u(x,y)=-\frac{1+y}{x}(x+1)+e^{x+1}\psi\left(-\frac{1+y+x}{x}\right)$$

Answer 2

The idea of the method of characteristics is to find curves on the plane (characteristics) where solutions to the original PDEs satisfy a simple ODE. Let's see how thiw works here :

Let $\gamma=(x(s);y(s))$ be a parametrised curve. Then, along $\gamma$ any solution the original PDE satisfies

$$\begin{array}{} xu_x+(1+y)u_y&=&x(1+y)+xu\\ x'u_x+y'u_y&=&du \end{array}$$

Therefore if $x'=x$ and $y'=1+y$ then $u$ satisfies a nice ODE along $\gamma$, namely : $$\frac{du}{ds}=x(s)(1+y(s))+x(s)u(s).$$

Now, it's easy to see that such $\gamma$ are just half-lines starting from $(0,-1)$ (draw them !).

But $y=6x-1$ is a line going through $(0,-1)$!

So, $u$ and therefore $\phi$ must satisfy the following ODE (where $s\in\mathbb{R}$):

$$\begin{array}{} \phi'(s)&=&s(6s-1+1)+s\phi(s)\\ &=&6s^2+s\phi(s) \end{array} $$

The point is that $y=6x-1$ isn't transverse to the characteristic curves it meets. So there are further constraints on the kind of initial conditions we can consider.

2) By contrast, the line $x=-1$ is transverse to every characteristic curves it meets (draw it !), so there's no constraint on $\psi$.