1
$\begingroup$

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 $\phi$.

    1. 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.

$\endgroup$
2
$\begingroup$

$$xu_x+(1+y)u_y=x(1+y)+xu$$ 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 : $$u(x,y)=-\frac{1+y}{x}(x+1)+e^{x+1}\psi\left(-\frac{1+y+x}{x}\right)$$

$\endgroup$
0
$\begingroup$

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 $\psi$.

$\endgroup$
  • $\begingroup$ By solving (by hand) the ODE any solution must satisfy along any characteristic (they partition the plane). $\endgroup$ – Ayoub Feb 16 at 9:28

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.