# Method of characteristics PDE $u_t +uu_x = x$

In this question you are asked to solve the initial value problem \begin{aligned} u_t +uu_x &= x,\\ u(x, 0) &= f(x) \end{aligned} using the method of characteristics. Using the parameters $s$ and $τ$, show that the solution can be expressed as

$$u(s, τ ) = \frac{1}{2}(f(s) + s) e^τ +\frac{1}{2} (f(s) − s) e ^{−τ}$$

I tried using Method of characteristics but cannot get the answer.

$\frac{dt}{ds} = 1$, $\frac{dx}{ds} =u$ and $\frac{du}{ds} =x$.

This gives $\frac{dx}{du} =x/u$ which does not give the solution as above. How do i get the above solution. Thanks.

• $u(s, τ ) = 12(f(s) + s) e^τ +12(f(s) − s) e^{−τ}$ is absurd. Probably there is a typo. Sep 15, 2018 at 6:51
• Yep you are right. Fixed. @JJacquelin Sep 15, 2018 at 8:19

• $$\frac{\text d t}{\text d \tau} = 1$$, letting $$t(0)=0$$, we know $$t=\tau$$.
• $$\frac{\text d x}{\text d \tau} = u$$ and $$\frac{\text d u}{\text d \tau} = x$$, so that $$\frac{\text d^2 x}{\text d \tau^2} = \frac{\text d}{\text d \tau}\frac{\text d x}{\text d \tau} = \frac{\text d u}{\text d \tau}$$, i.e., $$\frac{\text d^2 x}{\text d \tau^2} = x$$. Letting $$(x(0),u(0)) = (s,f(s))$$, we know \begin{aligned} x(\tau) &= \frac{1}{2}(f(s)+s) e^\tau - \frac{1}{2}(f(s)-s) e^{-\tau}\\ &= f(s) \sinh\tau + s\cosh\tau \, , \end{aligned} and \begin{aligned} u(x(\tau),\tau) &= \frac{1}{2}(f(s)+s) e^\tau + \frac{1}{2}(f(s) - s) e^{-\tau} \\ &= f(s) \cosh\tau + s\sinh\tau \, . \end{aligned}
To obtain $$u(x, t)$$, one eliminates $$s$$ by using the expression of $$x(\tau)$$. Here, it is hard to express $$s$$ in terms of $$x(\tau)$$ and $$\tau$$ for general initial data $$f$$, but it is possible in some particular cases. For instance, if $$f(s) = a s$$, then $$u(x,t)= x \frac{a+\tanh t}{1 + a\tanh t}$$ as long as $$t \neq \text{argtanh}(-1/a)$$.
• Could you explain how you arrived at $\frac{d^2 x}{d\tau^2}=x$? I shall put a bounty on this question if you wish. May 1, 2019 at 8:13