14.1 – Wärmeleitgleichung mit vorgegebener Lösung

Zeigen Sie, dass die Funktion

U\left( {x,t} \right) = {\left( {4\pi t} \right)^{-\frac{d}{2}}}{c^{-\frac{{{{\left| x \right|}^2}}}{{4t}}}}

die Wärmeleitgleichung löst.

Lösung

{u_t}-\Delta u = 0

u\left( {0,x} \right) = {u_0}\left( x \right)

Wir betrachten zunächst das Cauchy-Problem:

\Omega = {\mathbb{R}^d}, d.h. u \in C\left( {{\mathbb{R}^+} \times {\mathbb{R}^d}} \right)

Lösungsdarstellung:

u\left( {t,x} \right) = \int\limits_{{\mathbb{R}^d}} {{u_0}\left( y \right)U\left( {x-y,t} \right)dy}

U\left( {x,t} \right) = {\left( {4\pi t} \right)^{-\frac{d}{2}}}{e^{-\frac{{{{\left| x \right|}^2}}}{{4t}}}}

Zu zeigen:

{U_t}-\Delta U = 0

\frac{\partial }{{\partial {x_i}}}U = {\left( {4\pi t} \right)^{-\frac{d}{2}}}{e^{-\frac{{{{\left| x \right|}^2}}}{{4t}}}}\frac{{-2{x_i}}}{{4t}}

\frac{{{\partial ^2}}}{{\partial x_i^2}}U = {\left( {4\pi t} \right)^{-\frac{d}{2}}}{e^{-\frac{{{{\left| x \right|}^2}}}{{4t}}}}\frac{{4x_i^2}}{{16{t^2}}}-{\left( {4\pi t} \right)^{-\frac{d}{2}}}{e^{-\frac{{{{\left| x \right|}^2}}}{{4t}}}}\frac{1}{{2t}}

{\left| x \right|^2} = x_1^2+ \ldots {\text{+x}}_d^2

\frac{1}{{2t}}+\frac{1}{{2t}}+ \ldots {\text{+}}\frac{1}{{2t}} = \frac{d}{{2t}}

\Delta U = {\left( {4\pi t} \right)^{-\frac{d}{2}}}{e^{-\frac{{{{\left| x \right|}^2}}}{{4t}}}}\left[ {\frac{{{{\left| x \right|}^2}}}{{4t}}-\frac{d}{{2t}}} \right]

\frac{\partial }{{\partial t}}U = {\left( {4\pi } \right)^{-\frac{d}{2}}}{t^{-\frac{d}{2}-1}}\left[ {-\frac{d}{2}} \right]{e^{-\frac{{{{\left| x \right|}^2}}}{{4t}}}}+{\left( {4\pi t} \right)^{-\frac{d}{2}}}{e^{-\frac{{{{\left| x \right|}^2}}}{{4t}}}}\frac{{-{{\left| x \right|}^2}}}{{4{t^2}}}\left( {-1} \right)

= -\frac{d}{{2t}}{\left( {4\pi t} \right)^{-\frac{d}{2}}}{e^{-\frac{{{{\left| x \right|}^2}}}{{4t}}}}+{\left( {4\pi t} \right)^{-\frac{d}{2}}}{e^{-\frac{{{{\left| x \right|}^2}}}{{4t}}}}\frac{{{{\left| x \right|}^2}}}{{4{t^2}}}=\Delta u

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