Consider the initial value problem (IVP): $$u_t + x^3u_x = 0$$ $(x,t) ∈\mathbb{R} ×(0,∞)$, $u(x,0) = u_0(x)$, $x ∈R$, where $u_0 : \mathbb{R} → \mathbb{R}$ is a prescribed smooth and bounded function. Sketch the family of characteristic curves for IVP on the domain diagram. Obtain the solution $u : \mathbb{R} ×[0,∞) →\mathbb{R}$ to IVP.
This is a linear equation so we have the form $$a(x,t)u_t+b(x,t)u_x+ c(x,t)u=0$$ where $a(x,t)=x^3$ and $b(x,t)=1$ and $c(x,t)=0$.
Need to change coordinates from $(x,t)$ to $(x_0,s)$. We have: $$\frac{dx}{ds}=x^3, \, \, \, \, \, \, (2a)$$ $$\frac{dt}{ds}=1, \, \, \, \, \, \, (2b)$$
Then $$\frac{du}{ds}=\frac{dx}{ds}u_x+\frac{dt}{ds}u_t=x^3u_x+u_t$$ So $$\frac{du}{ds}=0, \, \, \, \, \, \, (3)$$
Solve $(2a)$ and $(2b)$ with condition $x(0)=x_0$ and $t(0)=0$ to get $$x^2= \frac1{2 (1/2x_0^2-s)}, \, \, \, \, t=s$$ respectively.
Solve $(3)$ with conditions $u(0)=f(x_0)$ which just gives $u=f(x_0)$.
I am stuck on this part:
When $$u_0(x) = e^{−x^2}$$ $x ∈\mathbb{R}$, sketch the solution to IVP on the $(x,u)$ plane for increasing values of $t > 0$. Describe the structure of the solution as $t →∞$.
Correct if I am wrong but we would have $$\exp \bigg(- \frac1{2t-1/s^2} \bigg)$$ but I don't know what this would look like...
the characteristic curve $C$ thru $x = a, t = 0$ is given by $$ \frac 1{a^2} - \frac 1{x^2} = 2t.$$ on the curve $C, u$ the value of $ u(x,t) = u_0(a).$
given $x, t$ you can solve for $a.$ we get $a = \frac 1{\sqrt{2t+ \frac1{x^2}}}.$ that is $$u(x,t) = u_0\left( \frac 1{\sqrt{2t+ \frac1{x^2}}}\right).$$