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6.3 Partial Differential Equations • 263f := (x, t) →e (−(t+x)2) + piecewise(−t + x < 1 2 and t − x < 1 , 1, 0)2Now you can plot the solution.> plot3d( f, -8..8, 0..5, grid=[40,40] );Changing the Dependent Variable in a PDEThe following is the one-dimensional heat equation.> heat := diff(u(x,t),t) - k*diff(u(x,t), x,x) = 0;heat := ( ∂ ∂t∂2u(x, t)) − k ( u(x, t)) = 0∂x2 Try to find a solution of the form X(x)T (t) to this equation. Use theaptly named HINT option of pdsolve to suggest a course of action.> pdsolve( heat, u(x,t), HINT=X(x)*T(t));(u(x, t) = X(x) T(t)) &where[{ d dt T(t) = k _c 1 T(t), d 2dx 2 X(x) = _c 1 X(x)}]The result here is correct, but difficult to read.Alternatively, you can tell pdsolve to use separation of variables (asa product, ‘*‘) and then solve the resulting ODEs (using the ’build’option).> sol := pdsolve(heat, u(x,t), HINT=‘*‘, ’build’);
264 • Chapter 6: Examples from Calculussol := u(x, t) = e (√ _c 1 x) _C3 e (k _c 1 t) _C1 + _C3 e(k _c 1 t) _C2e (√ _c 1 x)Evaluate the solution at specific values for the constants.> S := eval( rhs(sol), {_C3=1, _C1=1, _C2=1, k=1, _c[1]=1} );S := e x e t + ete xYou can plot the solution.> plot3d( S, x=-5..5, t=0..5 );Checking the solution by evaluation with the original equation is agood idea.> eval( heat, u(x,t)=rhs(sol) );%1 _C3 k _c 1 e (k _c 1 t) _C1 + _C3 k _c 1 e(k _c 1 t) _C2%1− k (_c 1 %1 _C3 e (k _c 1 t) _C1 + _C3 e(k _c 1 t) _C2 _c 1) = 0%1%1 := e (√ _c 1 x)> simplify(%);0 = 0
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Learning Guidecommand the brillianc
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ii•Waterloo Maple Inc.57 Erb Stre
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iv • ContentsThe numer and denom
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vi • ContentsThe Parts of an Expr
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viii • Contents
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2 • Chapter 1: Introduction to Ma
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