How to solve coupled linear 1st order PDE It is fairly straight forward to solve linear 1st order P

Marianna Stone

Marianna Stone

Answered question

2022-05-21

How to solve coupled linear 1st order PDE
It is fairly straight forward to solve linear 1st order PDEs by the method of characteristics. For example, if
t f + a x f = b f
we have that d f d t = b f on the characteristic curve of d x d t = a. From this we deduce that f ( t , x ) = g ( C ) e b t where x = a t + C.
Now, how does this work when f is multidimensional. Can I solve equations on the following form by characteristics, or by any other means?
t f i ( t , x ) + j A i j x f j ( t , x ) = j B i j f j ( t , x )
where the components of A and B might be dependent on x and t.
In particular, I am trying to solve the following,
{ t f + c t x g = ( a + 1 t ) f t g + c t x f = ( b + 1 t ) g
where f and g are functions of x and t , where t > t 0 > 0, c 0. Any help is highly appreciated.

Answer & Explanation

Haleigh Vega

Haleigh Vega

Beginner2022-05-22Added 13 answers

{ t f + c t x g = ( a + 1 t ) f   . . . . . . ( 1 ) t g + c t x f = ( b + 1 t ) g   . . . . . . ( 2 )
From ( 1 ),
t f + c t x g = ( a + 1 t ) f
c t x g = t f ( a + 1 t ) f
x g = t c t f ( a t c + 1 c ) f   . . . . . . ( 3 )
x t g = t c t t f 1 c t f ( a t c + 1 c ) t f a c f
x t g = t c t t f ( a t c + 2 c ) t f a c f   . . . . . . ( 4 )
From ( 2 ),
t g + c t x f = ( b + 1 t ) g
x t g + c t x x f = ( b + 1 t ) x g   . . . . . . ( 5 )
Put ( 3 ) and ( 4 ) into ( 5 ),
( 5 )
t c t t f ( a t c + 2 c ) t f a c f + c t x x f = ( b + 1 t ) ( t c t f ( a t c + 1 c ) f )
t c t t f ( a t c + 2 c ) t f a c f + c t x x f = ( b t c + 1 c ) t f + ( a b t c + a + b c + 1 c t ) f
c t x x f = t c t t f + ( ( a + b ) t c + 3 c ) t f + ( a b t c + 2 a + b c + 1 c t ) f
x x f = t 2 c 2 t t f + ( a + b ) t 2 + 3 t c 2 t f + a b t 2 + ( 2 a + b ) t + 1 c 2 f

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