Solve $uu_x+u_y=-\frac{1}{2}u$ with characteristics

Question

I'm stuck trying to solve $uu_x+u_y=-\frac{1}{2}u$ satisfying $u(x,2x)=x^2$, this solution also needs to be in some parametrised form with $x(r,s), y(r,s),u(r,s)$.

So far I have this:

$\dfrac{dx}{ds}=u \longrightarrow x(r,s)=us+x_0$

$\dfrac{dy}{ds}=1 \longrightarrow y(r,s)=s+y_0$

$\dfrac{du}{ds}=-\dfrac{1}{2}u \longrightarrow u(r,s)=u_0e^{-s/2}$

If we have that $y(0)=0$ we can see that $y=s$, so we can replace each instance of $s$ with $y$. I know I will end up with an implicit function, but I don't know how to apply the $u(x,2x)=x^2$ condition from here. I was thinking about directly plugging in $u(x,2x)=u_0e^{-x}=x^2\Rightarrow u_0=x^2e^x$ and then after letting $x(0)=r$ (the initial curve) and some substitutions end up with $u=(2uy-x)^2e^{-x}e^{-y/2}$ but I don't think that's correct.

I feel like there's a step missing before I start doing that. I have only just been introduced to PDEs, so all of this is new, any help is welcome and I appreciate your patience with me.

As an aside, I know there is another question that is asking the exact same thing that I am, but they introduce some $\Phi$ and $F$ notations which I have never seen before and leaves me more confused than I initially was.

Answer 1

$$2uu_x+2u_y=-u$$ Charpit-Lagrange characteristic ODEs : $$\frac{dx}{2u}=\frac{dy}{2}=\frac{du}{-u}$$ First characteristic from $\frac{dx}{2u}=\frac{du}{-u}$ $$2u+x=c_1$$ Second characteristic from $\frac{dy}{2}=\frac{du}{-u}$ $$u\:e^{y/2}=c_2$$ General solution of the PDE on implicit form $c_2=F(c_1)$ : $$\boxed{u\:e^{y/2}=F\left(2u+x\right)}$$ $F$ is an arbitrary function (to be determined according to some boundary condition).

Condition : $u(x,2x)=x^2$ $$x^2\:e^{2x/2}=F\left(2x^2+x\right)$$ $X=2x^2+x\quad\implies\quad x=\frac14\left(-1+\sqrt{1+8X} \right)$

The sign minus of the square root is excluded (in puting $X=2x^2+x$ into the square root). $$F(X)=\frac{1}{16}\left(-1+\sqrt{1+8X} \right)^2 e^{-\frac14\left(-1+\sqrt{1+8X} \right)} $$

We put $F(X)$ into the above general solution where $X=2u+x$ :

$$\boxed{u\:e^{y/2}=\frac{1}{16}\left(-1+\sqrt{1+8(2u+x)} \right)^2 e^{-\frac14\left(-1+\sqrt{1+8(2u+x)} \right)}}$$ This is the solution on the form of implicit equation which satisfies both the PDE and the condition.

Answer 2

Let us apply the method of characteristics, with initial condition $u(x_0, 2x_0) = x_0^2$:

• $\frac{\text d}{\text d s} y = 1$, letting $y(0) = 2x_0$ we know $y = 2x_0 + s$.
• $\frac{\text d}{\text d s} u = -\frac12 u$, letting $u(0) = x_0^2$ we know $u = x_0^2 e^{-s/2}$
• $\frac{\text d}{\text d s} x = u$, letting $x(0) = x_0$ we know $x = x_0 - 2 x_0^2 (e^{-s/2} - 1)$

The latter is rewritten $$ x = x_0 - 2 u\, (1 - e^{y/2-x_0}) . $$ Resolution w.r.t. $x_0$ yields the expression $$ x_0 = W_n\left(-2u e^{y/2-2 u - x} \right) + 2 u + x $$ where $W_n$ is the analytic continuation of the product log function, to be substituted in $ue^{y/2} = x_0^2 e^{x_0}$ to get the final implicit expression.