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...

  • $\begingroup$ sorry, edited .. $\endgroup$ – snowman Nov 8 '15 at 16:03
  • $\begingroup$ Is it really wave equation? $\endgroup$ – Empty Nov 8 '15 at 16:04
  • $\begingroup$ Use Lagrange's auxiliary equation to solve $\endgroup$ – Empty Nov 8 '15 at 16:06
  • $\begingroup$ 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. $\endgroup$ – Ian Nov 8 '15 at 16:11
  • $\begingroup$ 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.) $\endgroup$ – 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 $$u(x,t) = u_0\left( \frac 1{\sqrt{2t+ \frac1{x^2}}}\right).$$



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


then the PDE will tell you that

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

  • $\begingroup$ 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=$0$). $\endgroup$ – snowman Nov 8 '15 at 16:56
  • 1
    $\begingroup$ @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 $x'=x^3$. $\endgroup$ – Ian Nov 8 '15 at 17:30
  • $\begingroup$ @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!! $\endgroup$ – snowman Nov 8 '15 at 19:46

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.