Some PDE's like the heat or wave equation
ft(t,x) = fxx(t,x)
ftt(t,x) = fxx(t,x)
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can be solved similarly as ODE's.
xt (t) = A x(t)
xtt(t) = A x(t)
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Diagonalizing A
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on eigenspaces led to
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which are solved by
v(t)=exp(Lt) v
v(t)=v(0) cos(nt)+v'(0) sin(nt)/n
v(t)=v(0)+t v'(0)
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Diagonalizing T=D2
T sin(nx)=-n2sin(nx),
T cos(nx)=-n2cos(nx),
T 21/2 = 0.
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on eigenspaces leads to
f'(t)=-n2f(t)
f''(t)=-n2f(t)
f''(t)=0
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which are solved by
f(t)=exp(-n2 t) f(0)
f(t)=cos(nt) f(0)+sin(nt)/n f'(0)
f(t)=f(0)+t f'(0)
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For f written in the
Fourier basis
f(0,x) = a0 
+  an cos(n x)
+ bn sin(n x)
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we obtain a solution to the heat equation
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f(t,x) = a0
+ an exp(-n2 t) cos(n x)
+ bn exp(-n2 t) sin(n x)
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For f,f' written in the Fourier basis as
f(0,x) = a0
+ an cos(nx) + bn sin(nx)
f' (0,x) = a'0
+ a'n cos(nx) + b'n sin(nx)
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we obtain a solution to the wave equation
f(t,x) = (a0+a'0 t)
+(an cos(nt)+a'n sin(nt)/n) cos(nx)
+(bn cos(nt) + b'n sin(nt)/n) sin(nx)
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Fourier decomposition
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In this particular case, the Fourier coefficients bn belonging
to sin(nx) are zero because the function is an even function. Also
an is zero here if n is even.
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| f(x) |
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a1 cos(x) |
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a3 cos(3x) |
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a5 cos(5x) |
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a7 cos(7x) |
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a9 cos(9x) |
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Heat evolution
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The heat evolution is simple on each eigenfunction f=cos(nx) of D2.
Since f' = -n2 f for such functions, the decay is fast, if n
is large.
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Wave evolution
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The wave evolution is simple on each eigenfunction f=cos(nx) of D2.
Since f'' = -n2 f for such functions, the waves oscillate fast for
small wavelength.
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Using Symmetry
Remarks. In the case, when we are interested in the evolution on an interval like
[0, ], one can flip the graph at the y axes to obtain an
even function f(x)=f(-x) on [-,]
which has a pure cos series.
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For functions which are zero at 0 and ,
it makes also sense to continue on the left to an odd function f(x)=-f(x) which has a
sin Fourier series. This simplifies the formulas.
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2D heat and wave equation
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