Please check ODE operator solutions

In summary, an ODE operator solution is a mathematical function or equation that satisfies a given ordinary differential equation (ODE). It is important to check ODE operator solutions to ensure their accuracy and validity. This can be done through methods such as substitution, verifying the solution, and plotting. Some common errors to watch out for include incorrect application of the operator and incorrect substitution. ODE operator solutions have various real-world applications, including in physics, engineering, and economics for modeling and predicting complex systems and phenomena.
  • #1
ognik
643
2
1) Given $\mathcal{L}u=0$ and $ g\mathcal{L}u$ is self-adjoint, show that for the adjoint operator $ \bar {\mathcal{L}}, \bar{\mathcal{L}}(gu)=0$

Is it enough to say that if self-adjoint, then $ \mathcal{L}= \bar {\mathcal{L}} $. I assume g represents a function of x (so no inner products with g) that transforms a non self-adjoint ODE into a self-adjoint ODE, therefore $ \bar{\mathcal{L}}(gu)=g\bar{\mathcal{L}}(u)=g{\mathcal{L}}(u)= 0 $?

2)For a 2nd order diff. operator $ \mathcal{L} $ that is self adjoint, show that $ <y_2 | \mathcal{L} y_1 > - <y_1 | \mathcal{L} y_2 > = \int_{a}^{b}[y_2 \mathcal{L}y_1 - y_1 \mathcal{L} y_2] \,dx$

My question is that the book has been using, for example, $ <y_2 | \mathcal{L} y_1 >= \int_{a}^{b}y_2^* \mathcal{L}y_1 dx $ so why no * (complex conjugates) in this problem? Is it a typo or am I about to learn something new?

Thanks
 
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  • #2
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1) Yes, it is enough to say that if $g\mathcal{L}u$ is self-adjoint, then $\mathcal{L}=\bar{\mathcal{L}}$. This is because for a self-adjoint operator, the adjoint operator is the same as the original operator. Therefore, $g\bar{\mathcal{L}}(u)=g\mathcal{L}(u)=0$.

2) It is not a typo; it is a different notation for the same thing. In the notation used in the book, $y_2^*$ represents the complex conjugate of $y_2$. In the notation used in the problem, $<y_2|\mathcal{L}y_1>$ and $<y_1|\mathcal{L}y_2>$ are already complex conjugates of each other, so there is no need for an additional $*$ symbol. Both notations are commonly used in mathematics, so it is just a matter of personal preference.
 

FAQ: Please check ODE operator solutions

What is an ODE operator solution?

An ODE operator solution refers to a mathematical function or equation that satisfies a given ordinary differential equation (ODE). It is a solution to the ODE that is dependent on the operator used in the equation.

Why is it important to check ODE operator solutions?

It is important to check ODE operator solutions to ensure the accuracy and validity of the solution. This helps to avoid any errors or mistakes in the solution and provides confidence in the results obtained.

How can I check ODE operator solutions?

There are several methods for checking ODE operator solutions, such as using substitution, verifying the solution satisfies the ODE, and plotting the solution to visually inspect its behavior.

What are some common errors to watch out for when checking ODE operator solutions?

Some common errors when checking ODE operator solutions include incorrect application of the operator, incorrect substitution, and forgetting to include all necessary terms in the solution.

Can ODE operator solutions be used in real-world applications?

Yes, ODE operator solutions have various real-world applications such as in physics, engineering, and economics. They are used to model and predict the behavior of complex systems and phenomena.

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