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Introduction to Linear Algebra with Mathematica
Glossary
1D Heat Equation: Neumann BC
We consider the initial boundary value problem for the heat equation subject to the Neumann boundary conditions on half line:
\begin{align} \label{eqHeat.1}
u_t &= \alpha u_{xx} - \gamma\, u + f(x,t) , & x > 0, \quad 0 < t < \infty , \\ \label{eqHeat.2}
\frac{\partial u}{\partial x} (+0,t) &= -g(t) , & 0< t < \infty , \\ \label{eqHeat.3}
u(x,+0) &= u_0 (x) , & x> 0 ,
\\
u &\in 𝔏^2 (\mathbb{R}_{+} \times \mathbb{R}_{+}) , \qquad \mbox{regularity conditions},
\notag
\\
\int_0^{\infty} g(t)\,{\text d}t &= 0 , \qquad \mbox{compatibility condition}.
\notag
\end{align}
where subscripts indicate partial derivatives. For example, \( \displaystyle \quad u_t = \frac{\partial u}{\partial t} . \quad \)
The heat equation contains (phisical) real parameters, so α > 0 and γ ≥ 0.
All functions in the heat equation \eqref{eqHeat.1}, boundary condition \eqref{eqHeat.2}, and the initial condition \eqref{eqHeat.3} are assumed to be in the Hilbert space ℌ². Also all derivatives (ut, uxx) are assumed to be continuous in the open domain 0 < x < ∞, 0 < t < ∞
Example 1:
■
End of Example 1
To solve the given initial boundary value problem, we use the cosine Fourier transformation. So we multiply the differential equation \eqref{eqHeat.1} and the initial condition \eqref{eqHeat.3} by cosine (kx) and integrate with respect to x from 0 to ∞. This results in the equation
\[
\frac{{\text d} u^c}{{\text d}t} = \alpha \int_0^{\infty} \frac{\partial^2 u}{\partial x^2} \,\cos (kx) \,{\text d}x - \gamma \,u^c (k,t) + f^c (k,t) ,
\]
and the initial condition
\begin{equation} \label{eqHeat.4}
u^{c} (k, +0) = u^c_0 (k) = \int_0^{\infty} u_0 (x)\,\cos (kx)\, {\text d} x
\end{equation}
for the cosine Fourier transform of the unknown function
\[
u^{c} (k, t) = ℱ^s_{x \to k} \left[ u(x,t) \right] = \int_0^{\infty} u(x,t)\,\cos (kx)\,{\text d} x .
\]
The cosine Fourier transforms of the given function in Eq. \eqref{eqHeat.1} is
\[
f^{c} (k, t) = ℱ^s_{x \to k} \left[ f(x,t) \right] = \int_0^{\infty} f(x,t)\,\cos (kx)\,{\text d} x .
\]
Since the differential equation contains second derivative under integration, we integrate by parts twice and obtain
\begin{align*}
\int_0^{\infty} \frac{\partial^2 u}{\partial x^2} \,\cos (kx)\,{\text d}x &=
\left. \frac{\partial u}{\partial x}\,\cos (kx) \right\vert_{x=0}^{x\to +\infty} - \int_0^{\infty} \frac{\partial u}{\partial x} \,k\, \sin (kx)\,{\text d}x
\\
&= -u(x,t)\,k\,\sin (kx) \Large\vert_{x=0}^{x\to +\infty} - \int_0^{\infty} u(x,t)\, k^2 \cos (kx)\,{\text d}x
\\
&= k\, u(+0,t) - k^2 u^s (k,t) .
\end{align*}
This leads to the the ordinary differential equation for the sine Fourier transform of the unknown function:
\begin{equation} \label{eqHeat.5}
\frac{{\text d} u^s}{{\text d}t} = - \left( \alpha\,k^2 + \gamma \right) u^s (k,t) + k\alpha\, g (t) + f^s (k, t) .
\end{equation}
We ask Mathematica to solve the initial value problem \eqref{eqHeat.5}, \eqref{eqHeat.4}.
DSolve[{u'[t] + (a*k^2 + gamma)*u[t] == F[t], u[0] == u0}, u[t], t]
{{u[t] ->
E^(-gamma t -
a k^2 t) (u0 +
Inactive[Integrate][
E^(gamma K[1] + a k^2 K[1]) F[K[1]], {K[1], 0, t}])}}
\[
u^s (k, t) = e^{- (\alpha k^2 + \gamma )t} \left[ u_0^s (k) + \int_0^t e^{ (\alpha k^2 + \gamma ) \tau} \left( k\alpha\, g (\tau ) + f^s (k, \tau ) \right) {\text d} \tau \right] .
\]
Application of the inverse sine Fourier transform to the latter gives the required solution
\[
u(x,t) = \frac{2}{\pi} \int_0^{\infty} u^s (k, t)\,\sin (kx)\,{\text d} k = I_{u0} + I_g + I_f ,
\]
where
\begin{align*}
I_{u0} &= \frac{2}{\pi} \int_0^{\infty} \sin (kx)\,e^{- (\alpha k^2 + \gamma )t} \,{\text d} k \,\int_0^{\infty} {\text d}\xi u_0 (\xi )\,\sin (k\xi ) ,
\\
&= \frac{1}{2\sqrt{\pi\alpha t}}\, e^{-\gamma t} \,\int_0^{+\infty} {\text d}\xi u_0 (\xi ) \left[ e^{- (x-\xi )^2 /(4\alpha t)} - e^{- (x+\xi )^2 /(4\alpha t)} \right]
\\
I_{g} &= \frac{2}{\pi} \int_0^{\infty} \sin (kx)\,{\text d} k \,k\alpha \, \int_0^t e^{-\gamma (t- \tau )} \,e^{- \alpha k^2 (t-\tau )} \,g(\tau ) \,{\text d} \tau
\\
&= \frac{x}{2\sqrt{\pi\alpha} } \,\int_0^{t} e^{-\gamma (t- \tau )} \, e^{-x^2 /(4\alpha (t-\tau ))} \left( t- \tau \right)^{-3/2} g(\tau )\,{\text d}\tau ,
\\
I_{f} &= \frac{2}{\pi} \int_0^{+\infty} \sin (kx)\,e^{- (\alpha k^2 + \gamma )(t- \tau )} \,{\text d} k \,\int_0^t f^s (k, \tau )\,{\text d}\tau
\\
&= \frac{2}{\pi} \int_0^{\infty} \sin (kx)\,e^{- (\alpha k^2 + \gamma )(t- \tau )} \,{\text d} k \,\int_0^t {\text d}\tau \int_0^{\infty} f(\xi , \tau )\,\sin (k\xi )\,{\text d}\xi
\\
&= \frac{1}{2\sqrt{\pi\alpha}} \int_0^{\infty} f(\xi, \tau ) \,{\text d}\xi \int_0^t {\text d}\tau \left[ e^{(x-\xi )^2 /(4\alpha (t-\tau ))} - e^{(x+\xi )^2 /(4\alpha (t-\tau ))} \right] e^{-\gamma (t-\tau )} .
\end{align*}
Integrate[Sin[k*x]*Sin[k*xi]*Exp[-a*k^2], {k, 0, Infinity}]
ConditionalExpression[((E^(-((x - xi)^2/(4 a))) -
E^(-((x + xi)^2/(4 a)))) Sqrt[\[Pi]])/(4 Sqrt[a]), Re[a] > 0]
Integrate[Exp[-a*k^2]*k*Sin[k*x], {k, 0, Infinity}]
ConditionalExpression[(E^(-(x^2/(4 a))) Sqrt[\[Pi]] x)/(4 a^(3/2)),
Re[a] > 0]
Integrate[Exp[-a*k^2]*Sin[k*x]*Sin[k*xi], {k, 0, Infinity}]
ConditionalExpression[((E^(-((x - xi)^2/(4 a))) -
E^(-((x + xi)^2/(4 a)))) Sqrt[\[Pi]])/(4 Sqrt[a]), Re[a] > 0]
- Boyd, J.P. and Flyer, N., Compatibility conditions for time-dependent partial differential equations and the rate of convergence of Chebyshev and Fourier spectral methods, Computer Methods in Applied Mechanics and Engineering, 1999, Volume 175, Issues 3–4, Pages 281--309. https://doi.org/10.1016/S0045-7825(98)00358-2
- Chatziafratis, A., Boundary behaviour of the solution of the heat equation on the half line via the Fokas unified transform method, 2024, https://doi.org/10.48550/arXiv.2401.08331
- Chatziafratis, A., Fokas, A., Aifantis, E.C., Variations of heat equation on the half-line via the Fokas method, 2024, First published: 08 September 2024 https://doi.org/10.1002/mma.10303
- Chatziafratis, A., Mantzavinos, D., Boundary behavior for the heat equation on the half-line, 2022, https://doi.org/10.1002/mma.8245
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