Solving quasi-linear PDE, $x \frac{\partial{u}}{\partial{x}} - u\frac{\partial{u}}{\partial{t}} = t$ , $u(1,t) = t$, $-\infty < t < \infty$ [duplicate]

I am trying to solve the quasi-linear PDE $$x \frac{\partial{u}}{\partial{x}} - u\frac{\partial{u}}{\partial{t}} = t$$ , $$u(1,t) = t$$, $$-\infty < t < \infty$$ using method of characteristics.

$$\frac{\frac{dx}{dt}}{x} = \frac{\frac{dy}{dt}}{-u} = 0$$, implying $$(\frac{1}{x})\frac{dx}{dt} = -\frac{1}{u} \frac{dy}{dt}$$ implying $$\frac{1}{x}\frac{dx}{dt} + \frac{1}{u} \frac{dy}{dt} = 0$$ or $$\frac{d}{dt}(\ln|x|) + \frac{1}{u}\frac{dy}{dt} = 0$$.

Seems like I messed up somewhere but unable to find it out? There is no $$u$$ dependence in the RHS of the PDE, it is only $$t$$.

Follow up questions like to discuss -

After finding the solution to this PDE, I am trying to look at the maximal region where the solution is defined.

Iis the IVP wellposed? Are there some regions where the solution is not single-valued?

marked as duplicate by Dylan, Strants, clathratus, Cesareo, Lee David Chung LinMar 27 at 2:48

$$\dfrac{dx}{ds}=x$$ , letting $$x(0)=1$$ , we have $$x=e^s$$

$$\begin{cases}\dfrac{dt}{ds}=-u\\\dfrac{du}{ds}=t\end{cases}$$

$$\therefore\dfrac{d^2t}{ds^2}=-t$$

$$t=C_1\cos s+C_2\sin s$$

$$\therefore u=C_1\sin s-C_2\cos s$$

Hence $$\begin{cases}t=C_1\cos s+C_2\sin s\\u=C_1\sin s-C_2\cos s\end{cases}$$

$$t(0)=t_0$$ , $$u(0)=f(t_0)$$ :

$$\begin{cases}C_1=t_0\\C_2=-f(t_0)\end{cases}$$

$$\therefore\begin{cases}t=t_0\cos s-f(t_0)\sin s\\u=t_0\sin s+f(t_0)\cos s\end{cases}$$

$$\therefore\begin{cases}t_0=t\cos s+u\sin s=t\cos\ln x+u\sin\ln x\\f(t_0)=u\cos s-t\sin s=u\cos\ln x-t\sin\ln x\end{cases}$$

Hence $$u\cos\ln x-t\sin\ln x=f(t\cos\ln x+u\sin\ln x)$$

$$u(1,t)=t$$ :

$$f(t)=t$$

$$\therefore u\cos\ln x-t\sin\ln x=t\cos\ln x+u\sin\ln x$$

$$u(x,t)=\dfrac{t(\cos\ln x+\sin\ln x)}{\cos\ln x-\sin\ln x}$$

• Thanks! so trying the follow up questions - the solutions exist if the denominator is not zero that is $\cos(\ln(x)) - \sin(\ln(x)) \neq 0$. Is this correct? or there is some stronger condition? also I did the plot the solution here - desmos.com/calculator/ztufzpz9ry, so how do we know where the solution is single valued? I see that the solution is multivalued like for asingle $y$ there exists many $x$ values, finally how do i check the well posed ness of the IVP? – BAYMAX Mar 26 at 20:35