Solving First-Order PDE: $u_x+2u_y+2u=0$

[Markov2]

Solve

$u_x+2u_y+2u=0,$ $x,y\in\mathbb R$ where $u(x,y)=F(x,y)$ in the curve $y=x.$

I don't know what does mean with the $y=x.$ Well I set up the following $\dfrac{dx}{1}=\dfrac{dy}{2}=\dfrac{du}{-2}
,$ is that correct? but I don't know what's next.

Thanks for the help!

 

[TheEmptySet]

Think about the idea behind the method of characteristics. We start by parameterizing the independant variables x and y.

[tex]x=x(s,t) \quad y=y(s,t) \implies u(x,y)=u(x(r,s),y(r,s))[/tex]

Now take the derivative with respect to s to get

[tex]\frac{du}{dx}\frac{dx}{ds}+\frac{du}{dy}\frac{dy}{ds}=\frac{d}{ds}u(x(r,s),y(r,s))[/tex]

Compare this with the PDE to get

[tex]\frac{dx}{ds}=1 \quad \frac{dy}{ds}=2 \quad \frac{du}{ds}=-2u[/tex]

Now we need some intial conditions to solve this system of ODE's

Since we know values of the line [tex]y=x[/tex] we should get

[tex]x(0,t)=t \quad y(0,t)=t \quad u(0,t)=f(t)[/tex]

Solving each of the ODE's gives

[tex]x=s+g(t) \implies t=0+g(t) \implies x=s+t[/tex]

[tex]y=2s+h(t) \implies t=2\cdot 0+h(t) \implies y=2s+t[/tex]

[tex]u=Ae^{-2s} \implies f(t)=A \implies u=f(t)e^{-2s}[/tex]Now we just solve for the characteristics to eliminate t and s. This gives

[tex]s=y-x \quad t=2x-y[/tex]

So the solution is

[tex]u(x,y)=f(2x-y)e^{2(x-y)}[/tex]

We can check that this satisfies the PDE

[tex]u_x=[2f'(2x-y)+2f(2x-y)]e^{2(x-y)}[/tex]

[tex]u_y=[-f'(2x-y)-2f(2x-y)]e^{2(x-y)}[/tex]

[tex]u_x+2u_y=-2f(2x-y)e^{2(x-y)}=-2u[/tex]