Find the solution to the initial condition $ u(x,0)=\sin x \ $

Question

Consider the following conservation law:

$$ u_t-xt^2 u_x=0 , \ -\infty <x <\infty, \ t> 0 \ $$

(a) Find its Characteristics lines.

(b) Find the solution to the initial condition $$ u(x,0)=\sin x \ $$

Answer:

(a)

$ u_t-xt^2u_x=0 $

The characteristic equation is

$ \frac{dt}{1}=\frac{dx}{-xt^2}=\frac{du}{0} \ $

Solving , we get

$$ u=constant=c , \\ \ln(x)-\ln(x_0) =-\frac{t^3}{3} \ $$

Now what to do ?

what would be characteristic line ?

Answer 1

Hint.

$$ u(x,t) = \Phi(xe^{\frac{t^3}{3}}) = \sin\left(xe^{\frac{t^3}{3}}\right) $$

and for $t=0\rightarrow u(x,0) = \sin(x)$

Answer 2

we have the Lagrange equations $$\frac {dt}{1}=\frac {dx}{-xt^2}$$ $$ \begin{cases} dz=0 \implies z=C_2 \\ -t^2 {dt}=\frac {dx}{x} \implies \ln(x)+\frac {t^3}3=C_1 \end{cases} $$ $$f(C_1)=C_2 \implies u(x,t)=f(\ln(x)+\frac {t^3}3)$$ $$ u(x,o)=\sin(x) \implies u(x,0)=f(\ln(x))=\sin(x) \implies f(x)=\sin(e^x)$$ Therefore $$u(x,t)=f(\ln(x)+\frac {t^3}3)$$ $$u(x,t)=\sin(\exp(\ln(x)+\frac {t^3}3)) $$ $$\boxed{u(x,t)=\sin(xe^{\frac {t^3}3})} $$