0
$\begingroup$

Consider the PDE: $$ u_t + (1-2u) u_x =0 \text{, where } x<0 \text{ and } t>0,$$ with initial and boundary data given by $$u(x,0) = \frac{1}{4} \text{ for } x<0, \text{ and } u(0,t) = 1, \text{ for } t>0.$$

I tried solving it by method of characteristics, but it merely gives a trivial solution, since the initial condition function $u(x,0)$ is simply a constant in this case. Any other suggestions of how to solve this PDE?

$\endgroup$
4
$\begingroup$

$$u_t+(1-2u)u_x=0$$ Applying the method of characteristics : $$\frac{dt}{1}=\frac{dx}{1-2u}=\frac{du}{0}$$ which gives $du=0$ hense the characteristics $(u)$ and $\left(x-(1-2u)t\right)$

Thus the general solution, on implicit form, is : $$F\left(u\:,\:x-(1-2u)t\right)=0$$

The function $F(X,Y)$ with $X=u$ and $Y=x-(1-2u)t$ is not a solution of the PDE. The solutions of the PDE are the functions $u(x,t)$ which are solutions of the equation $F\left(u\:,\:x-(1-2u)t\right)=0$

Verification, any derivable function $F(X,Y)$:

enter image description here

In addition : Look at the Boundary conditions : $$ $$\begin{cases} u(x,0)=\frac{1}{4} & \quad \text{for } x<0 \\ u(0,t)=1 & \quad \text{for } t>0 \\ \end{cases} The implicit equation $F\left(u\:,\:x-(1-2u)t \right)=0$ can be transformed to : $$u=f\left(x-(1-2u)t \right)$$ where $f$ is a function to be determined according to the boundary conditions. $$ \begin{cases} u(x,0)=\frac{1}{4}=f\left(x-(1-2u(x,0))0 \right)=f(x) & \quad x<0\\ u(0,t)=1=f\left(0-(1-2*1)t \right)=f(t) & \quad \text t>0 \\ \end{cases} $$ So, the function $f$ is defined by \begin{cases} f(\chi)=\frac{1}{4} & \quad \text{for } \chi<0 \\ f(\chi)=1 & \quad \text{for } \chi>0 \\ \end{cases} Hense $$f(\chi)=\frac{1}{4}+\frac{3}{4}H(\chi)$$ where $H(\chi)$ is the Heaviside step function. Finally, the solution expressed on implicit form is : $$u=\frac{1}{4}+\frac{3}{4}H\left(x-(1-2u)t \right)$$

$\endgroup$
  • $\begingroup$ Is this really a solution? Differentiating w.r.t $x$ and $t$ respectively and adding up gives $ [ (2u- \frac{5}{4})(x- (1-2u)t) + 2t (u-\frac{1}{4})(u-1) ][u_t + (1-2u) u_x] =0. $ Thus, the PDE is only satisified if we can prove that $(2u- \frac{5}{4})(x- (1-2u)t) + 2t (u-\frac{1}{4})(u-1) $ is non-zero. Moreover, it is not clear to me why the initial and boundary conditions are satisfied in your final implicit solution. Could you please explain to me??? $\endgroup$ – Richard Dec 18 '15 at 16:46
  • $\begingroup$ $(u-\frac{1}{4})(u-1)\left(x-(1-2u)t\right)$ is NOT a solution. $u$ is a solution. $(u-\frac{1}{4})(u-1)\left(x-(1-2u)t\right)=0$ not $=u$. $\endgroup$ – JJacquelin Dec 18 '15 at 16:51
  • $\begingroup$ Yes, I know. That's why I used implicit differentiation. $\endgroup$ – Richard Dec 18 '15 at 16:53
  • $\begingroup$ I differentiated both sides of the EQUATION w.r.t $x$ then w.r.t $t$. $\endgroup$ – Richard Dec 18 '15 at 16:54
  • 1
    $\begingroup$ The partial derivatives must be explicitly written. If this isn't done, the calculus cannot be completed. The detailed proof is now added to my first answer. Moreover, in order to avoid hard discussion about the particular notations I use in the choice of a convenient function $F(X,Y)$ fulfilling the boundary conditions, I remove this part from my answer. $\endgroup$ – JJacquelin Dec 18 '15 at 19:10

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.