# Solving an inhomogeneous Burgers' equation with the method of characteristics

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.

• You just need to solve two of the three equalities. Why not solve $$\frac{du}{dx} = \frac{1}{5}$$ and $$\frac{du}{dt} = u$$? – Mattos Feb 26 at 2:43
• I guess you could do this. But after you solve the two ODEs, what then? The concept is escaping me (especially the geometric interpretation). – user557493 Feb 26 at 3:47
• See here for what to do after solving the ODEs. – Mattos Feb 26 at 3:51

$$\frac{dt}{1}=\frac{dx}{5u}=\frac{du}{u}\qquad\text{is OK.}$$ 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 : $$u=e^t \left(u-\frac{x}{5}\right)^{13/14}$$

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:

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.