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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.

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    $\begingroup$ $u(s, τ ) = 12(f(s) + s) e^τ +12(f(s) − s) e^{−τ}$ is absurd. Probably there is a typo. $\endgroup$ – JJacquelin Sep 15 '18 at 6:51
  • $\begingroup$ Yep you are right. Fixed. @JJacquelin $\endgroup$ – stutsy Sep 15 '18 at 8:19
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Let us apply the method of characteristics.

  • $\frac{\text d t}{\text d \tau} = 1$, 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 $t \neq \text{argtanh}(-1/a)$.

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  • $\begingroup$ 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. $\endgroup$ – Steven May 1 '19 at 8:13
  • $\begingroup$ @Stuart-JamesBurney Done, answer edited. $\endgroup$ – Harry49 May 1 '19 at 10:49

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