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I'm just wondering if you can solve the first order PD wave-equation similar to the second order one by separation. For me however it turns out weird.

Having

$$\mathrm{\partial_t u(x,t) = -c\cdot \partial_x u(x,t)}$$

one might separate $\mathbf{u}$ into separate function of space (x) and time (t)

$$\mathrm{u_s = X(x)\,T(t)}$$

gaining the relation

$$\mathrm{\,\dfrac{\frac{d}{dt}T(t)}{T} = -c\, \dfrac{\frac{d}{dx}X(x)}{X}}$$

both sides have to equal a separation constant $\lambda$, yielding 2 ODE

$$\mathrm{\frac{d}{dt}T(t) = \lambda\cdot T}$$ $$\mathrm{ \lambda\cdot X = -c\,\frac{d}{dx}X(x)}$$

solving these classic ODE gives

$$\mathrm{u_s = X_0\,e^{-\lambda/c\,x}\,T_0\,e^{\lambda\,t}}$$

where $\mathrm{u(x,0) = X_0(x)}$ and $\mathrm{u(0,t) = T_0(t)}$ are the initial conditions

Other boundary conditions can be imposed but actually according to this file only one condition is necessary.

There the solution is plainly given as $$\mathrm{u(x,t) = X_0(x-c\,t)}$$

What is also the numeric result that I am getting.

But how to interpret the result from separating ? Especially what's the role of the separation constant $\lambda$ ?

Leon
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    Notice that if $u$ has to be bounded as $x\to\pm\infty$, then $\lambda$ must be purely imaginary. This gives the solution for a single Fourier mode, but now there aren't a countable set of these and you have to do a Fourier transform instead of a Fourier series to solve a general IVP – whpowell96 Feb 28 '24 at 23:11

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The method of separation of variables leads to a family of solutions but doesn't give all solutions. This is very clear in the present case : $$\frac{\partial u}{\partial t}+c \frac{\partial u}{\partial x}=0$$ In order to obtain all the solutions one need more advanced methods, especially the method of characteristics. The general solution of the PDE is : $$u(x,t)=F(x-ct)$$ where $F$ is an arbitrary function (any function, possibly piecewise).

You found $\quad u = X_0\,e^{-\lambda/c\,x}\,T_0\,e^{\lambda\,t}\quad$ which is correct. This solution is the particular case of exponential function with $\quad C=X_0T_0\quad$ and $\quad\chi=x-ct\quad$: $$F(\chi)=Ce^{-\frac{\lambda}{c}\chi}$$ One can see that whey are infinity many other solutions as many as they are infinity many arbitrary functions $F$.

One can interpret the result of "separating" as the selection among many functions $F$ only a kind of function (the exponential function in the present case). $\lambda$ is a parameter as well as $C$ which gives a family of solutions instead of a unique solution.

If a condition is specified for example $\quad u(x,0)=X_0(x)\quad$ this defines a unique function $\quad F(\chi)=X_0(\chi)\quad$ and the particular solution : $$u(x,t)=X_0(x-ct)$$ satisfies both the PDE and the condition.

If another condition is specified for example $\quad u(0,t)=T_0(t)\quad$ this defines another unique function $\quad F(\chi)=T_0(-\frac{1}{c}\chi)\quad$ and the particular solution : $$u(x,t)=T_0\left(-\frac{1}{c}(x-ct)\right)$$

If the two conditions $\quad u(x,0)=X_0(x)\quad$ and $\quad u(0,t)=T_0(t)\quad$ are specified together the function $F$ becomes a piecewise function.

JJacquelin
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