Solve the first order PDE

Question

$u u_x + (y + 1) u_y = u , x \in \mathbb{R}, y > 0$

$u(x, 0) = −3x , x \in \mathbb{R}.$

The question asks that we solve the initial value problem by solving a system of characteristic equations with initial conditions from the parametric description of the data curve Γ. I know that the characteristic equations are as follows:

$\frac{dx}{dτ}=u, \frac{dy}{dτ}=y+1 ,\frac{du}{dτ}=u$

I am unsure as to what to do with these from here, any help would be appreciated. Thanks

Answer 1

You should rather combine the equation, by solving for $d\tau$ to obtain:

$$\dfrac{dx}{u}=\dfrac{dy}{y+1}=\dfrac{du}{u}.$$

Use $$dx/u=du/u\implies dx = du \implies u = x + c_1$$

and $$dy/(y+1)=du/u \implies \ln u =\ln (y+1)+\ln c_2\implies c_2=\dfrac{u}{y+1}.$$

Now, we know that a constant $c_1$ can always be expressed as a function of another constant $c_2$.

Hence: $u=x+c_1=x+F(c_2)=x+F\left( \dfrac{u}{y+1}\right)$ $$\implies u=x+F\left( \dfrac{u}{y+1}\right)$$

Using: $u(x,0)=-3x$ implies

$$u(x,0)=x+F\left( \dfrac{u(x,0)}{0+1}\right)=-3x \implies F(-3x)=-4x$$ $$\implies F(-3x)=\dfrac{4}{3}(-3x) \implies F(z)=\dfrac{4}{3}z$$ $$\implies u=x+F\left(\dfrac{u}{y+1} \right)=x+\dfrac{4}{3}\dfrac{u}{y+1}$$

Solve for $u$ and you are done.

EDIT: To solve for $u$

$$u\left(1-\dfrac{4}{3}\dfrac{1}{y+1}\right)=x$$ $$u(x,y)=\dfrac{x}{1-\dfrac{4}{3}\dfrac{1}{y+1}}$$ $$u(x,y)=\dfrac{3x(y+1)}{3(y+1)-4}$$ $$u(x,y)=\dfrac{3x(y+1)}{3y-1}$$

Answer 2

The characteristic equations are $x=u+c$ and $\frac{u}{y+1}=d$. These are obtained by solving Lagrange's auxiliary equations. The general solution is of the form $$c=f(d)\qquad d=g(c)\qquad h(c,d)=0$$ Assuming the 2nd form we get $\frac{u}{y+1}=g(x-u)$. Given $u(x,0)=-3x$, we get $-3x=g(x+3x)\implies g(4x)=-3x\implies g(z)=-\frac{3}{4}z$. Thus $g(x-u)=-\frac{3}{4}(x-u)=\frac{3}{4}(u-x)$. Thus the complete solution is $$\frac{u}{y+1}=\frac{3}{4}(u-x)\implies u(x,y)=-\frac{3x(1+y)}{1-3y}$$