I'm struggling with the following question. I'm fairly new to PDEs as this is a question from an introductory course.

$$u_t + (x^2u)_x=0$$ $$u(x,0)=1$$

I know it's of the form linear transport equation with non-constant speed:

$$u_t+(v(x)u)_x=0$$ $$u(x,0)=u_0(x)$$

and thus I know the characteristic line $x(t)$ satisfies

$$\frac{dx}{dt}=v(x)=x^2$$ $$x(0)=x_0$$

and that the solution to the IVP is


However, I'm not sure how to find $x_0$, and then use this in the answer formula.


The text I'm using (and the above is based on) is Li Chen's lecture notes "Introduction to Partial Differential Equations" it states:

The Cauchy problem we consider in this part is \begin{equation}u_t+(v(x)u)_x=0, \text{ in } \mathbb{R}\times \mathbb{R}_+\end{equation} $$u(x,0)=u_0(x)$$ where $v(x)$ is a given Lipschitz continous function. A reformulation of the equation is $$u_t+v(x)u_x+v'(x)u=0.$$ If $v(x)$ is Lipschitz cont., then the characteristic line $x(t)$ satisfies $$\frac{dx}{dt}=v(x)=x^2$$ $$x(0)=x_0$$ With the help of this line, the equation is rewritten into $$\frac{du(x(t;x_0),t)}{dt}=-v'(x(t;x_0))u,$$ $$u(x(t;x_0),t)|_{t=0}=u_0(x_0).$$ By separation of variables in solving ODE, we have \begin{align*}\ln u(x(t;x_0),t)&=\ln u_0(x_0)+\int_0^t-v'(x(\tau;x_0))d\tau\\ &=\ln u_0(x_0)+\int^{x(t;x_0)}_{x_0}-\frac{v'(x)}{v(x)}dx\\ &=\ln u_0(x_0)-\ln v(x(t;x_0))+\ln v(x_0)\end{align*} where in the first equation the change of variable $x=x(\tau,x_0)$, $dx=x'd\tau=v(x)d\tau$ is used. So the problem has solution $$u(x(t;x_0),t)=u_0(x_0)\frac{v(x_0)}{v(x(t;x_0))}.$$ Therefore, finding the exact solution of the homogenous problem is reduced to the solvability of the characteristic line.

So in trying to solve my problem in this manner, I'm kinda confused with how I go from the PDE to the characteristic line (and $x(0)=x_0$), and how to interpret the $x_0$ value, since it is not given in my problem. Can anybody help me and walk me through it?

Thank you.


We have the pde $$ u_t+x^2u_x+2xu=0\\ u(x,0)=1 $$ which we examine along a characteristic curve $(t,x(t))$ which we require to satisfy the convenient conditions $$ x'(t)=x^2\\ \frac{d}{dt}u(t,x(t))=u_t+x'(t)u_x=-2x(t)u(t,x(t)) $$ So let's solve the odes. First we find $$ x(t)=\frac{1}{\frac{1}{x_0}-t} $$ then $$ u'(t,x(t))=-2x(t)u(t)=\frac{2}{t-\frac{1}{x_0}}u(t)\\ \implies u(t)=A(t-1/x_0)^2 $$ Now we impose that $u(0)=1$ to find $A=x_0^2$, yielding $$ u(t,x(t))=x_0^2(t-1/x_0)^2 $$ at which point we need to solve for $x_0$, which after all was just a dummy initial condition. We want our solution to be valid for any initial condition $x_0$. So we solve for $x_0$ in terms of $x$ to find $$ x_0=\frac{tx+1}{x} $$ which then yields the solution $$ u(t,x)=\frac{x^2}{(tx+1)^2}\frac{1}{x^2}=\frac{1}{(tx+1)^2} $$

edit: How $x'(t)=x^2$ is solved. Separate variables to find $$ \frac{dx}{x^2}=\mathrm dt\implies -\frac{1}{x}=t+c $$ for some $c$ which we will solve for in terms of initial condition $x(0)=x_0$, plug in $t=0$ and find $$ -\frac{1}{x_0}=c $$ yielding $$ -\frac{1}{x}=t-\frac{1}{x_0}\implies x=\frac{1}{\frac{1}{x_0}-t} $$ as we found before.

edit 2 explicitly solving other ode: $$ \frac{du}{u}=2(t-1/x_0)^{-1}\stackrel{\text{integrate both sides}}{\implies} \ln|u|=2\ln(t-1/x_0)+C=\ln(t-1/x_0)^2+C\\ \stackrel{\text{exponentiate both sides}}{\implies}u(x,t)=A(t-1/x_0)^2 $$


Let $u=f(t)g(x)$ then $u_t=f^{'}(t)g(x)$ and $(x^2u)_x=f(t)(2xg(x)+x^2g^{'}(x))$ and we have:

$\Large f^{'}(t)g(x)+f(t)(2xg(x)+x^2g^{'}(x))=0$

Dividing it by $u$:

$\Large {f^{'}(t)\over{f(t)}}+{{2xg(x)+x^2g^{'}(x)}\over g(x)}=0$

therefore there must be:

$\Large {f^{'}(t)\over{f(t)}}=k$ and $\Large{{2xg(x)+x^2g^{'}(x)}\over g(x)}=-k$

where $k$ is constant. The above two equations yield:

$\Large f(t)=Ae^{kt}\qquad\qquad\qquad g(x)={B{e^{k\over x}}\over{x^2}}$

finally we have:

$\Large u(x,t)=A{e^{kt+{k\over x}}\over{x^2}}$.

Since $A$ can vary with $k$ we write the general form using Fourier transform:

$\Large u(x,t)=\int_{-\infty}^{\infty}A(\omega){e^{(t+{1\over x})\omega}\over {x^2}}d\omega$

After imposing the condition of problem we have the final answer:

$\Huge u(x,t)={1\over{(xt+1)^2}}$

  • $\begingroup$ Thank you for your answer. Can you show it in terms of characteristics, since I was supposed to do it in this fashion? Thank you. $\endgroup$ – Morten Andersen Jan 5 '18 at 13:45
  • $\begingroup$ Don't mention it!....which characteristics and fashion you exactly mean? $\endgroup$ – Mostafa Ayaz Jan 5 '18 at 14:28
  • $\begingroup$ I have just made an edit in the question where I have cited the text I'm using. I have mentioned the part I'm confused about. $\endgroup$ – Morten Andersen Jan 5 '18 at 19:28
  • 2
    $\begingroup$ Sure!....My main reference is "Advanced Engineering Mathematics" written by Erwin Keryszig and is really useful specially for undergraduate courses. $\endgroup$ – Mostafa Ayaz Jan 6 '18 at 12:14
  • 1
    $\begingroup$ @qbert: You are welcome. $\endgroup$ – Hans Jan 7 '18 at 20:41

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