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I am trying to solve the PDE $$u_t+5uu_x=u,$$ subject the boundary condition $u(0,t)=e^{14t}.$

I first start by defining the set of characteristic equations, $$\frac{dt}{1}=\frac{dx}{5u}=\frac{du}{u}.$$ From here I obtain two ODEs, $$\frac{dx}{dt}=5u, \ \ \frac{du}{dx}=\frac{1}{5}.$$

From this we have, $$u=\frac{x}{5}+C_1, \ \ \ x=xt+C_2t+C_3.$$ But I am unsure of the following steps to proceed to a solution. Thank you kindly in advanced.

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  • $\begingroup$ You just need to solve two of the three equalities. Why not solve $$\frac{du}{dx} = \frac{1}{5}$$ and $$\frac{du}{dt} = u$$? $\endgroup$ – Mattos Feb 26 at 2:43
  • $\begingroup$ I guess you could do this. But after you solve the two ODEs, what then? The concept is escaping me (especially the geometric interpretation). $\endgroup$ – user557493 Feb 26 at 3:47
  • $\begingroup$ See here for what to do after solving the ODEs. $\endgroup$ – Mattos Feb 26 at 3:51
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$$\frac{dt}{1}=\frac{dx}{5u}=\frac{du}{u}\qquad\text{is OK.}$$ As you found it, a first characteristic equation comes from $\frac{du}{dx}=\frac{1}{5}$. $$u-\frac{x}{5}=C_1$$ For a second characteristic equation, you chose $\frac{dx}{dt}=5u$

This is not the simplest equation because it cannot be integrated directly. As already pointed out in comments, $\frac{dt}{1}=\frac{du}{u}$ is straightforward. It doesn't mater, any way leads to the same final result. So we will continue the way you chose. even if is is a bit more complicated.

$$\frac{dx}{dt}=5u=5(C_1+\frac{x}{5})=5C_1+x$$ This is a first order linear ODE easy to solve : $x=-5C_1+C_2e^t$

$xe^{-t}+5C_1e^{-t}=C_2$ $$xe^{-t}+5(u-\frac{x}{5})e^{-t}=C_2$$ $$5ue^{-t}=C_2$$

The general solution of the PDE is on the form of implicite equation $\Phi(C_1,C_2)=0$ or $C_2=F(C_1)$ or $C_1=G(C_2)$ where $F$ and $G$ are arbitrary functions (in fact one inverse from the other). $$5ue^{-t}=F\left(u-\frac{x}{5}\right)$$ $$u=\frac15 e^t F\left(u-\frac{x}{5}\right)$$ $F$ is an arbitrary function to be determined according yo the boundary condition.

$u(0,t)=e^{14t}=\frac15 e^t F\left(e^{14t}-\frac{0}{5}\right)$

$e^{14t}=\frac15 e^t F\left(e^{14t}\right)$

Let $X=e^{14t}$

$5e^{13t}= F\left(X\right)=5X^{13/14}$

Now the function $F(X)$ is determined. We put it into the above general solution where $X=u-\frac{x}{5}$ . The particular solution fitting to the boundary condition is : $$u=e^t \left(u-\frac{x}{5}\right)^{13/14}$$

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For this boundary-value problem, the method of characteristics gives

  • $\frac{\text d t}{\text d s} = 1$. Letting $t(0) = t_0$, we have $t=s+t_0$.
  • $\frac{\text d u}{\text d s} = u$. Letting $u(0) = e^{14\, t_0}$, we have $u = e^{14\, t_0}e^s = e^{13\, t_0}e^{t}$.
  • $\frac{\text d x}{\text d s} = 5 u$. Letting $x(0) = 0$, we have $x = 5\, e^{14\, t_0}(e^{s}-1) = 5 u - 5 e^{14\, t_0}$.

Therefore, we can write the following implicit equation $$ u = \left(u-\frac{x}{5}\right)^{13/14} e^{t} $$ which should be solved numerically/graphically. A sketch of the characteristic curves in the $x$-$t$ plane is given below:

char

We can observe that characteristic curves that carry different information intersect at the breaking time $t^* = \ln(14/13) \simeq 0.074$, which tells that the solution deduced from the method of characteristics becomes multi-valued within the domain of dependence, as shown in the figure below. Weak solutions must be considered.

plot

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