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

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

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

 

[blue_raver22]

,The adjoint equation is a mathematical concept that is used in many different fields, including physics, engineering, and mathematics. It is a mathematical tool that is used to solve certain types of problems, particularly those involving boundary value problems.

In simple terms, the adjoint equation is the dual equation to the original equation. It is derived by taking the original equation and flipping the sign of the highest order derivative. So for the equation $L = p_0u'' + p_1u' + p2u = r$, the adjoint equation would be $L^*[v] = -p_0v'' + p_1v' + p_2v = \phi$, where $v$ and $\phi$ are the adjoint variables.

The reason why the adjoint equation is important is because it allows us to solve boundary value problems by using the Green's function. The Green's function is a function that satisfies the adjoint equation, and it is used to solve the original equation by taking the inner product of the Green's function and the source term.

So to answer your question, the link between the two equations is that the adjoint equation is derived from the original equation and is used to solve boundary value problems. The rule that your lecturer gave you is a way to check if an equation is adjoint, but it is not the only way. It is important to understand the concept behind the adjoint equation and how it is used in solving problems. If you are still having trouble understanding, I would recommend talking to your lecturer or seeking additional resources to help clarify the concept for you.