PDE chain rule

Bronn

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Suppose that f is a differentiable function of a single variable and F(x,y) is defined by F (x, y) = f (x2 − y).





a) Show that F satisfies the partial differential equation





∂F/∂x + 2x•∂F/∂y = 0

b) Given that F(0,y) = sin y for
all y, find a formula for F(x,y).






for a) I take LHS and subbed F for f (x2 y).
/∂x(f (x2 y)) + 2x•∂/∂y (f (x2 y))

which becomes

= 2x• f ' (x2 y) - 2x• f ' (x2 − y) = 0

so LHS = RHS

for b) I'm a bit confused

if F( 0, y ) = sin(y) thenF( 0, y) =
f (0y) = sin(y)

now the answer in the book is F(x,y) = sin ( y - x2 ) which is true F( 0, y ) = sin(y) but isn't the input
of f dependant on x2 − y, can you just change that?


thanks
 
Last edited:
Suppose that f is a differentiable function of a single variable and F(x,y) is defined by F (x, y) = f (x2 − y).





a) Show that F satisfies the partial differential equation





∂F/∂x + 2x•∂F/∂y = 0

b) Given that F(0,y) = sin y for
all y, find a formula for F(x,y).






for a) I take LHS and subbed F for f (x2 y).
/∂x(f (x2 y)) + 2x•∂/∂y (f (x2 y))

which becomes

= 2x• f ' (x2 y) - 2x• f ' (x2 − y) = 0

so LHS = RHS

for b) I'm a bit confused

if F( 0, y ) = sin(y) thenF( 0, y) =
f (0y) = sin(y)

now the answer in the book is F(x,y) = sin ( y - x2 ) which is true F( 0, y ) = sin(y) but isn't the input
of f dependant on x2 − y, can you just change that?


thanks
How about using the fact that sin is an odd function. Can you tell us what that means?
 
How about using the fact that sin is an odd function. Can you tell us what that means?

well it means f(-x) = -f(x) if i remeber correctly

but wouldn't that mean F(0,y) = f(0-y)= f(-y)=sin (-y) = -sin (y) ?


edit: i think i got it

F(x,y) = f(x2-y)

F(0,y) = f(02-y)= -sin(02 - y) = sin(y)

so F(x,y) = - sin (x 2 - y) = sin (y - x2 )

that look ok?
 
Last edited:
well it means f(-x) = -f(x) if i remeber correctly

but wouldn't that mean F(0,y) = f(0-y)= f(-y)=sin (-y) = -sin (y) ?
Yes! And what would sin (y-x2) equal?
Note 1: WHENEVER you want to change the way something looks like, you multiply by 1.
Note 2: There are many representatives for 1. Like 100%, 7/7, (-1)(-1), (2/3)(3/2), ....
 
well it means f(-x) = -f(x) if i remeber correctly

but wouldn't that mean F(0,y) = f(0-y)= f(-y)=sin (-y) = -sin (y) ?


edit: i think i got it

F(x,y) = f(x2-y)

F(0,y) = f(02-y)= -sin(02 - y) = sin(y)

so F(x,y) = - sin (x 2 - y) = sin (y - x2 )

that look ok?
Just as I was writing my post you got the hint. Good for you. You see, it really is better to give hints rather than solutions. Good job! And remember forever that sin is an odd function and that a-b = -(b-a)!
 
yes, thanks again. I was so stuck in seeing that the input HAS to be x2-y and didn't think you could represent it a different way. duh!
 
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