Heat Equation; Dirichlet BC

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We conside a general initial boundary value porlem for general linear heat equation with Dirichlet boundary conditions: Here, α > 0 and γ ≥ 0.

Alex:    1) Eq(4) needs LaTeX ,/p> 2) Eq(6) needs clean up

3) remove comments at right and write explanation explicitly over here!!!!

4) you need to explain where regularity conditions came from!!!

We further impose the conditions listed below that result from a later application of Green's theorem on the half-strip: \begin{align} \label{eq:heatgreencont} u, u_t, u_x, u_{xt}, u_{xx} &\in C(U) & &&\text{Continuity}\\ \label{eq:heatgreenpt0} u(x,t) \text{ and } u_x(x,t) &\xrightarrow[x \to \infty]{} 0 \text{ pointwise on } [0,T] & &&\text{Pointwise-convergence to zero}\\ \label{eq:heatgreenunibound} \abs{u(x,t)}, \:\: \abs{u_x(x,t)}& \leq c_0 &t\in[0,T], \: c_0 > 0,\: x \text{ large enough} &&\text{Uniform boundedness}\\ \label{eq:heatgreenregcondu} u, u_t, u_{xx} &\in L^1(R_T) & &&\text{Regularity condition on $u$}\\ \label{eq:heatgreenregcondf} f &\in L^1(R_T) & &&\text{Regularity condition on $f$} \end{align}

Alex:    1) explain wherre all these conditions came from; Eq(8) not clear written

2) explain L¹ Lebegue space and make reference

3) what is \abs ????

where U ⊂ ℝ² is some open set containing the infinite half-strip \( \displaystyle \quad \overline{R_T} , \) which is described in \ref{eq:heatrectangle}. \\ \\

Alex:    Describe R_T exlicitly---repetition is crucial.

Some immediate consequences of the continuity conditions \ref{eq:heatgreencont} are that

1. \( \displaystyle \quad \lim_{t\to 0} g(t) = g(0) = u(0,0) = u_0(0) = \lim_{x\to 0} u_0(x) ; \)
2. \( \displaystyle \quad \lim_{t\to 0} g'(t) = g'(0) = u_t(0,0) ; \)
3. \( \displaystyle \quad \lim_{x\to 0} u_0'(x) = u_0'(0) = u_x(0,0) . \)

Alex:    put semi-columns and period into LaTeX file

Local relation

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Our initial objective is to express the differential equation in divergence form, where r depends on the nonhomogenous part of the PDE rather than u. To obtain this form, we first multiply the heat equation by the reciprocal of the exponential in the Ehrenpreis integral formula \[ e^{\omega (k)t+ {\bf j}kx} u_t = \alpha e^{\omega (k)t+ {\bf j}kx} u_{xx} - \gamma e^{\omega (k)t+ {\bf j}kx} u + e^{\omega (k)t+ikx} f(x,t)\:. \] Alex:    make English!!!

where j is a unit imaginary vector on complex plane ℂ, so j² = −1.

and then make repeated use of the product rule. The term involving u_t on the left-hand side yields \begin{align*} e^{\omega (k)t+ikx} u_t &= \left(e^{\omega (k)t+ {\bf j}kx} u\right)_t - \omega (k)e^{\omega (k)t+ {\bf j}kx} u \end{align*} while the u_xx term on the right becomes \begin{align*} \alpha e^{\omega (k)t+ikx} u_{xx} &= \left(\alpha e^{\omega (k)t+ {\bf j}kx} u_x\right)_x - \alpha {\bf j}k \, e^{\omega (k)t+ {\bf j}kx}u_x \\ &= \left(\alpha e^{\omega (k)t+ {\bf j}kx} u_x\right)_x - \left(\alpha {\bf j}k \, e^{\omega (k)t+ {\bf j}kx}u\right)_x - \alpha k^2 e^{\omega (k)t+ {\bf j}kx}u\:. \end{align*} We substitute these terms back into the equation. \[ \left(e^{\omega (k)t+ {\bf j}kx} u\right)_t - \left(e^{\omega (k)t+ {\bf j}kx} \alpha (u_x- {\bf j}ku)\right)_x - e^{\omega (k)t+ {\bf j}kx} f(x,t) = \left[ \omega (k) - (\alpha k^2 + \gamma) \right] e^{\omega (k)t+ {\bf j}kx} u . \] The expression on the right implies that the dispersion relation: \begin{align} \label{eq:disprelation} \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \omega (k) &= \alpha k^2 + \gamma\:. \quad \quad \quad \quad &&\text{Dispersion relation} \end{align}

Alternatively, we could have deduced the dispersion relation by searching for a solution of the form u = e^{−ω (k) −jkx} $u = e^{-\omega (k)-ikx}$. However, the method we used additionally produces the local relation: \begin{align} \label{eq:localrelation} \quad \quad \left(e^{\omega (k)t+ikx} u\right)_t - \left(e^{\omega (k)t+ikx} \alpha (u_x-iku)\right)_x - e^{\omega (k)t+ikx} f(x,t) &= 0\:. &&\text{Local relation} \end{align} This equation is therefore represented in divergence form \ref{eq:divform} with \[ p(x,t,k) \:=\: e^{\omega (k)t+ikx} u\:, \quad \quad q(x,t,k) \:=\: -e^{\omega (k)t+ikx} \alpha (u_x-iku)\:, \quad \quad r(x,t,k) \:=\: e^{\omega (k)t+ikx} f(x,t)\:. \]

Global relation

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References

 

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