Using implicit differentiation, verify that u = f ( x − t u ) satisfies ∂ u ( x , t ) ∂ t + u ( x , t ) ∂ u ( x , t ) ∂ x .Could someone explain how implicit differentiation works with a pde?

Question

Using implicit differentiation, verify that   u  =  f  (  x  −  t  u  ) satisfies             ∂      u      (      x      ,      t      )              ∂      t        +  u  (  x  ,  t  )            ∂      u      (      x      ,      t      )              ∂      x      .Could someone explain how implicit differentiation works with a pde?

geoforoiunpwd · Accepted Answer

Differentiating   u  =  f  (  x  −  t    u  ) first with respect to   t and then with respect to   x we get:      u    t    =      f    ′    (  x  −  t    u  )  (  −  u  −  t        u    t    )  ,      u    x    =      f    ′    (  x  −  t    u  )  (  1  −  t        u    x    )  .Multiplying the second equation by u and adding both equations we get      u    t    +  u        u    x    =  −  t        f    ′    (  x  −  t    u  )  (      u    t    +  u        u    x    )    ⟹    (  1  +  t        f    ′    )  (      u    t    +  u        u    x    )  =  0.If       f    ′    ≥  0 (i.e.   f is increasing) then   1  +  t        f    ′    &amp;gt;  1 for all   t  ≥  0 and       u    t    +  u        u    x    =  0 follows.If       f    ′    &amp;lt;  0 at some point, there is a       T          max       such that   1  +  t        f    ′    &amp;gt;  0 on   [  0  ,      T          max        ),

Using implicit differentiation, verify that u=f(x−tu) satisfies (del u(x,t))/(del t)+u(x,t)(del u(x,t))/(del x). Could someone explain how implicit differentiation works with a pde?

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