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Given the partial differential equation $$2\frac{\partial u}{\partial t} + 3\frac{\partial u}{\partial x} = 0$$ with auxiliary conditions $u = \sin(x)$ when $t = 0$, I tried to solve for $u(x, t)$ using the characteristic lines function. Here's my solution:

$$u(x, t) = f(bx - at) = f(3x - 2t)$$ Using the auxiliary conditions, $$\Longrightarrow \sin(x) = f(3x - 2(0)) = f(3x)$$ Substituting $w = 3x$, we have $$\sin{\left(\frac{w}{3}\right)} = f(w)$$ $$\therefore u(x, t) = \sin{\left(\frac{3x - 2t}{3}\right)} = \sin\left(x - \frac{2}{3}t\right)$$

[Edit] I differentiated $u(x, t)$ and found that $$2\frac{\partial u}{\partial t} + 3\frac{\partial u}{\partial x} \neq 0$$ But I did find that $$3\frac{\partial u}{\partial t} + 2\frac{\partial u}{\partial x} = 0$$ I would like to know where I went wrong.

Abdulrahman
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    Try $u(x,t)=f(2x-3t)$. – Gonçalo Aug 09 '23 at 23:57
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    To answer your question "I would like to know where I went wrong", you went wrong is saying the solution was $u = f(\color{red}{3}x - \color{blue}{2}t)$ when it is $u = f(\color{blue}{2} x - \color{red}{3} t)$. But without showing your working, we can't know where you went wrong in getting that particular solution. Finally, from Gonçalo's answer, you have $$u(x, 0) = f(2x) = \sin(x) \implies f(X) = \sin(X/2) \implies u(x,t) = \sin((2x - 3t)/2)$$ matching your result. – Matthew Cassell Aug 10 '23 at 06:25
  • Please can you explain why the solution is $f(2x - 3t)$ and not $f(3x - 2t)$? – Abdulrahman Aug 10 '23 at 08:19
  • @Gonçalo I see what you mean now. – Abdulrahman Aug 10 '23 at 08:20
  • You can just compute derivatives of your solution and substitute it in to the PDE to see what the values of $a, b$ should be $$u_{x} = bf', \quad u_{t} = -af' \implies 0 = 2 u_{t} + 3 u_{x} = (-2a + 3b) f' \implies 2a = 3b$$ Then let $a = 3$ and hence $b = 2$ (or let $a = 1 \implies b = 2/3$, they are equivalent statements). – Matthew Cassell Aug 13 '23 at 02:33

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