Show that if u ( x , t ) satisfies u t = k Δ u in a bounded region G, then for any L &gt; 0, u ( L x , L 2 t ) solves the same equation for x ∈ L − 1 G, where L − 1 G is the set of points L − 1 y for y ∈ G. Here k is a positive constant.I am considering L to be some form of a linear operator. In terms of sets, this statement makes sense, but I'm not quite too sure how to definitively show it.

Question

Show that if   u  (  x  ,  t  ) satisfies       u    t    =  k  Δ  u in a bounded region G, then for any   L  &amp;gt;  0,   u  (  L  x  ,      L    2    t  ) solves the same equation for   x  ∈      L          −      1        G, where       L          −      1        G is the set of points       L          −      1        y for   y  ∈  G. Here   k is a positive constant.I am considering   L to be some form of a linear operator. In terms of sets, this statement makes sense, but I&#039;m not quite too sure how to definitively show it.

Gretchen Bond · Accepted Answer

To be sure, you need to make difference between new and old variables, between new and old unkowns. Namely, let       u    t    (  x  ,  t  )  =  k      Δ    x    u  (  x  ,  t  ) for   x  ∈  G and   t  ∈  (  0  ,  T  ). Consider a function             u      ~        (  y  ,  s  )      =          d      e      f        u  (  L  y  ,      L    2    s  ). It is clear that                     u        ~              s    (  y  ,  s  )  =  k      Δ    y                      u        ~              s    (  y  ,  s  ) for   L  y  =  x  ∈  G and       L    2    s  =  t  ∈  (  0  ,  T  ), i.e., for   y  ∈      L          −      1        G and   s  ∈  (  0  ,      L          −      2        T  ).

Show that if u ( x , t ) satisfies u t </msub> = k <

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