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I have the following equation:

\begin{align} xu_x + u_y &= 3u \\ u(x, 0) &= x\end{align}

I wanted to solve it with Lagrange's method:

$$ \text{d}t = \frac{\text{d}x}{x} = \text{d}y = \frac{\text{d}u}{3u}$$

Eq.1: $$ \frac{\text{d}x}{x} = \text{d}y \quad \implies \quad \text{ln}(x) = y + C$$ which gives the first surface $$ \phi_1 = xe^{-y} $$

and Eq.2: $$ \text{d}y = \frac{\text{d}u}{3u}$$ which gives the second surface:

$$\phi_2 = ue^{-3y}$$

so $$u(x, y) = F(ue^{-3y}, xe^{-y}) \implies u(x, y) = ue^{-3y} + f(xe^{-y}) $$ which gives: $$ u(x, y) = \frac{1}{1-e^{-3y}}\cdot f(xe^{-y}) $$

but I can't plug in the initial condition $ u(x, 0) = x $ due to division by zero... Help would be appreciated.

Vítězslav Štembera
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Edward Josef
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  • I see you're using the method of characteristic, but I'm not familiar with the conventions (notation, terminology) you're using, so I cannot follow the reasoning. There must a mistake somewhere, as I got a different result: $u(x,y) = e^{3y} \cdot u(x e^{-y}, 0)$. – Michał Miśkiewicz Jan 16 '22 at 11:26

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Your calculus is correct up to : $$u(x, y) = F(ue^{-3y}, xe^{-y}) \implies u(x, y) = ue^{-3y} + f(xe^{-y}) \quad\text{is not correct.}$$ The correct relationship is : $$F(ue^{-3y}, xe^{-y})=0 \implies ue^{-3y}=f\left( xe^{-y}\right)$$ $$u(x,y)=e^{3y}f\left( xe^{-y}\right)$$ Then is easy to find the particular solution which satisfies $u(x,0)=x$ : $$u(x,y)=x\,e^{2y}$$ Note : $F(ue^{-3y}, xe^{-y})=0$ is equivalent to $G(ue^{-3y}, xe^{-y})=$constant , with arbitrary related functions $F$ and $G$.

JJacquelin
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