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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?

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

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Follow the method in http://en.wikipedia.org/wiki/Method_of_characteristics#Example:

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

$u(1,t)=t$ :

$f(t)=t$

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

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

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  • $\begingroup$ 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? $\endgroup$ – BAYMAX Mar 26 at 20:35

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