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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?

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$$u_t+(1-2u)u_x=0$$ 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 $F(X,Y)$:

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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 : $$u=\frac{1}{4}+\frac{3}{4}H\left(x-(1-2u)t \right)$$

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  • $\begingroup$ 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
    Commented Dec 18, 2015 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
    Commented Dec 18, 2015 at 16:51
  • $\begingroup$ Yes, I know. That's why I used implicit differentiation. $\endgroup$
    – Richard
    Commented Dec 18, 2015 at 16:53
  • $\begingroup$ I differentiated both sides of the EQUATION w.r.t $x$ then w.r.t $t$. $\endgroup$
    – Richard
    Commented Dec 18, 2015 at 16:54
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    $\begingroup$ 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
    Commented Dec 18, 2015 at 19:10

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