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Solve the IVP $$ \begin{cases} u_t + cu_x = 1 & c \in \mathbb{R} \\ u(x,0) = \sin x \end{cases}$$

To solve this, I have used characteristics as follows:

Note that $$\frac{\partial u}{\partial t} = u_x \frac{dx}{dt} + u_t$$ So this equation is equal to the one in the IVP if $\frac{\partial u}{\partial t} = 1$ and $\frac{dx}{dt} = c$

$$\frac{\partial u}{\partial t} = 1 \Rightarrow u(x,t) = t + A(x)$$ Using the initial condition, we find that $A(x) = \sin x$ so that $u(x,t) = t + \sin x$.

But this is not the correct solution and I have not used the fact that $\frac{dx}{dt} = c$ either.

Can someone explain where I am wrong and perhaps give a rigorous answer to the problem using my method so I can also see just how to solve the problem properly as I am currently doing it without really understanding anything...

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I would suggest that the most critical information here is $ dx/dt = c$ as it shows how constant values propagate. So using this fact we have:

$ \frac{dx}{dt} = c \rightarrow x = ct + x_0$

Which describes the characteristic line on which the information is propagated. Now let us go back to the solution of the first part: $ \frac{du}{dt} = 1 \rightarrow u = t+A$ to say that $A$ is dependent on $x$ is quite wrong over here since we presume $x$ is dependent on $t$ and it just creates a confusion (for me at least). Then, to find A we go to the initial condition:

$ u(x,t) = t+A \rightarrow u(x(0),0) = u(x_0,0) = \sin(x_0) = A$

If we want to go back to the original coordinates we then apply $ x_0 = x-ct$ and we get: $u(x,t) = t+\sin(x-ct)$

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  • $\begingroup$ How do you know that the $x_0$ found from the first ODE is the same as the one in the initial condition? That is where I am most stuck right now... $\endgroup$ – PhysicsMathsLove May 11 '18 at 16:56
  • $\begingroup$ Follow the math ;) ! The $x_0$ is in a sense just the initial value of $x$. As I see it, the method of characteristics describes the evolution of $x$ and $u$ as independent variables. Hence, what you obtain at the end is two variables where $x_0$ is simply the initial value of $x$ and $u$ follows $x$ on the line... I don't do a good job in explaining it but I think you may find sources online that do a better job in this regard. $\endgroup$ – Snifkes May 11 '18 at 17:07
  • $\begingroup$ Thank you - that is really helpful! $\endgroup$ – PhysicsMathsLove May 11 '18 at 19:05
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Note that

$$ du = u_x dx+u_t dt\\ 1 = c u_x+u_t $$

so

$$ \frac{du}{1} = \frac{dx}{c}=\frac{dt}{1} $$

then

$$ x-ct = C_1\\ u = t + C_2 $$

concluding that

$$ u = t + \phi(x-ct) $$

The contour conditions say

$$ u(0,x) = \sin x\Rightarrow \phi(x) = \sin(x) $$

and finally

$$ u(t,x) = t + \sin(x-ct) $$

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  • $\begingroup$ Thanks again LutzL. $\endgroup$ – Cesareo May 11 '18 at 18:40
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Answer to the question : "why is my method wrong?"

Your equation $\frac{\partial u}{\partial t} = u_x \frac{dx}{dt} + u_t$ is false.

It should be $\quad\frac{du}{dt} = u_x \frac{dx}{dt} + u_t$

because $\quad du=u_xdx+u_tdt$.

As a consequence, the continuation of your calculus fails : In your method of calculus $x$ is function of $t$ and $A(x)$ is not a constant with respect to $t$.

Since you already got several rigorous answers to the problem, there is no interest for me to edit again the solution.

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