# Initial value problem for the PDE $u_t + x^3u_x = 0$

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...

• sorry, edited .. – snowman Nov 8 '15 at 16:03
• Is it really wave equation? – Empty Nov 8 '15 at 16:04
• Use Lagrange's auxiliary equation to solve – Empty Nov 8 '15 at 16:06
• The characteristics are the solutions to $x'=x^3,x(0)=x_0$ for each $x_0 \in \mathbb{R}$. The solution will be ("formally") constant along these curves. – Ian Nov 8 '15 at 16:11
• You made some mistakes in solving for the solution to the characteristic equation; you should find $x=(x_0^{-2} - 2t)^{-1/2}$. Then $u(t,x(t))=u_0(x_0)$. So to write $u(t,x)$ you need to solve $x=(x_0^{-2} - 2t )^{-1/2}$ for $x_0$ in terms of $t$ and $x$. (This algebra step amounts to following the characteristic curve backward in time from $(t,x)$ to $(0,x_0)$. When there are no singularities, it is equally correct to just do this directly, rather than going forward and then backward as I've written here.) – Ian Nov 8 '15 at 20:01

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 $$u(x,t) = u_0\left( \frac 1{\sqrt{2t+ \frac1{x^2}}}\right).$$

Hints

1) First, I recommend to take a look at this post.

2) Consider the following change of variables

$$\left\{ \matrix{ x = x(u,v) \hfill \cr t = t(u,v) \hfill \cr} \right.,\,\,\,\,\,\left\{ \matrix{ {{\partial x} \over {\partial u}} = {x^3} \hfill \cr {{\partial t} \over {\partial u}} = 1 \hfill \cr} \right.$$

and define

$$z(u,v)=u(x(u,v),t(u,v))$$

then the PDE will tell you that

$${{\partial z} \over {\partial u}} = 0$$

• If we have $a(x,y)u_x+b(x,y)u_y=f(x,y,u)$ then you have $dx/a=dy/b=du/f$ so in my case we have $dt/1=dx/x^3=du/0$. Is this not a problem... since my $f=0$ (RHS=$0$). – snowman Nov 8 '15 at 16:56
• @snowman I think you will confuse yourself trying to write it that way. You just need to know that if $u_t(t,x) + v(t,x) u_x(t,x) = f(t,x)$ then the characteristics satisfy the ODE $x'(t)=v(t,x(t))$ and $u(t,x(t))$ satisfies the ODE $\frac{du}{dt}(t,x(t)) = f(t,x(t))$. So because your $f$ is zero and your $v$ is $x^3$, your $u$ is constant along the solutions to $x'=x^3$. – Ian Nov 8 '15 at 17:30
• @Ian OK I have edited my post and tried to do a full solution but I am stuck on the last bit. Please have a look and tell me what to do!! – snowman Nov 8 '15 at 19:46