Intereting Posts

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For any set of formulas in propositional logic, there is an equivalent and independent set
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How may the method of characteristics be applied to solve a second order PDE? For instance, to solve the equation: $u_{tt}=u_{xx}-2u_t$.

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- On spectrum of periodic boundary value problem

Your equation is not suitable for displayng

the techniques called *Method of characteristics*.

Nevertheless, if you really need this equation, you

can construct its general solution. To do this, first

substitute $\;u=ve^{-t}$, then change variables

$\;\xi=2(t+x)\;$, $\;\eta=2(t-x)\;$,

reducing your equation to the simplest possible form $\;v_{\xi\eta}-v=0\;$.

General solution of the latter equation looks like

$$v(\xi,\eta)=\!\int\limits_0^{\xi}f(s)I_0\Bigl(2\sqrt{\eta(\xi-s)}\Bigr)\,ds

+\!\int\limits_0^{\eta}g(s)I_0\Bigl(2\sqrt{\xi(\eta-s)}\Bigr)\,ds

\tag{$\ast$}$$

for arbitrarily given functions $f,g\in C^1$ with notation $\;I_0\;$ standing for the *zero order modified Bessel function of the first kind*.

To obtain general solution $(\ast)$ apply the Laplace transform

$$

\hat{f}(p)\overset{\rm def}{=}\mathcal{L}[f(t)]=\int\limits_0^{\infty}f(t)e^{-pt}dt

$$

to the equation $v_{\xi\eta}-v=0$ *w.r.t.* any of the two variables,

say $\eta$, then integrate the equation $p\hat{v}_{\xi}-\hat{v}=v_{\xi}(\xi,0)

\overset{\rm not}{=}f(\xi)$ to find that

$$

\hat{v}(\xi,p)=\int\limits_0^{\xi}f(s)

\frac{e^{\frac{\xi-s}{p}}}{p}\,ds+ c(p)e^{\frac{\xi}{p}}.

$$

Since $\mathcal{L}\bigl[I_0\bigl(2\sqrt{\alpha t}\bigr)\bigr]=

\frac{1}{p}e^{\frac{\alpha}{p}}$,

the representation $(\ast)$ readily follows by

choosing $c(p)=\hat{g}(p)/p$.

Of course, general solution $(\ast)$ looks rather complicated, but for the last two and a half centuries, nothing better has yet been found to replace it.

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