MHB Help understanding the adjoint equation

[nacho-man]

So I have attached a screenshot of my lecturer's notes,
I can't quite understand the logical reasoning behind the link he's made between the two equations.

Previously he'd given us the fact that to check if an equation was adjoint,
you check if $ p_1' = p_0'' + p_2 $

for
$L = p_0u'' + p_1u' + p2u = r$

I am not sure if he has used this previous rule to here, but certainly I am a little dumbfounded.
I would appreciate any help!

Thanks

 

[Euge]

So I have attached a screenshot of my lecturer's notes,
I can't quite understand the logical reasoning behind the link he's made between the two equations.

Previously he'd given us the fact that to check if an equation was adjoint,
you check if $ p_1' = p_0'' + p_2 $

for
$L = p_0u'' + p_1u' + p2u = r$

I am not sure if he has used this previous rule to here, but certainly I am a little dumbfounded.
I would appreciate any help!

Thanks

Hi nacho,

Your professor may have checked that $L$ is adjoint, but that does not determine $M$. To have an adjoint, there must be an inner product defined. What was the inner product that he used?

 

[nacho-man]

Hi nacho,

Your professor may have checked that $L$ is adjoint, but that does not determine $M$. To have an adjoint, there must be an inner product defined. What was the inner product that he used?

Sorry, I am not entirely sure what you mean by inner product! :(

I hope my question was not vague, but I essentially want to know how the
first line of the equation expands to the second line in the screenshot I provided in the original post.

Thanks!

 

[Euge]

Ok, now I understand your question.

First, let's expand $(vp_1)'$. By the product rule, we have

$\displaystyle (vp_1)' = v' p_1 + v (p_1)'$.

Now for $(vp_0)''$. Like with $(vp_1)'$,

$\displaystyle (vp_0)' = v' p_0 + v (p_0)'$.

Differentiate both sides to get

$\displaystyle (vp_0)'' = (v' p_0)' + (v(p_0)')'$.

By the product rule we obtain

$\displaystyle (v' p_0)' = v'' p_0 + v' (p_0)'$

and

$\displaystyle (v (p_0)')' = v' (p_0)' + v (p_0)''$.

Adding the two results,

$\displaystyle (vp_0)'' = v'' p_0 + 2 v' (p_0)' + v (p_0)''$.

Now we can express

$M[v] = (v'' p_0 + 2v' (p_0)' + v (p_0)'') - (v' p_1 + v (p_1)') + vp_2$

$\displaystyle = p_0 v'' + [-p_1 + 2(p_0)'] v' + [p_2 - p_1 + (p_0)''] v$.