Differential equation problems

[eljose]

let be the differential equation in the form...

$$L[y]=a_{0}y+a_{1}y´+a_{2}y´´-f(x)=0$$

then you could use your theory of differential equation to solve and get the solution y=y(x), my first question is if knowing the value of y1=Y(x) and y2=g(x) being y1 and y2 solutions of the differential equation, could we obtain the values of $$a_{0},a_{2},a_{1}$$ i think you could use the Wronskian W(x) and the solution to this come from setting W(x)=0 adn from this you get the values of the a,s..

the other question is given the differential operator..

$$L[G(x,s)]=\delta(x-s)$$ with G(x-s) known how could obtain the a,s? (this is the inverse to the Green operator theory)..

and the last question...given the differential operator..

$$L[y]=a_{0}+a_{1}D^{1}+a_{2}D^{2}=0$$

where the functions a,s are function of x a=A(x) for every a then how could you obtain the adjoint operator?..thanks.

 

[jvicens]

Hello, thank you for your questions. Let me address each of them individually.

1. If we know the values of y1=Y(x) and y2=g(x) being solutions of the differential equation, we can use the Wronskian W(x) to find the values of a0, a1, and a2. The Wronskian is defined as W(x) = y1y2' - y1'y2, where y1' and y2' are the derivatives of y1 and y2 with respect to x, respectively. If W(x) = 0, then a0, a1, and a2 can be determined by solving the resulting system of equations. This is known as the method of variation of parameters. Alternatively, if we know the initial conditions for the differential equation, we can use those to solve for a0, a1, and a2.

2. To find the values of a0, a1, and a2 in the differential operator L[G(x,s)] = δ(x-s), we can use the inverse Green operator theory. This involves solving the integral equation L[G(x,s)] = δ(x-s) for G(x,s), which can then be used to find the values of a0, a1, and a2. This method is commonly used in solving boundary value problems.

3. To obtain the adjoint operator for the differential operator L[y] = a0 + a1D^1 + a2D^2 = 0, we can use the definition of the adjoint operator, which is given by L^*[y] = (-1)^n(D^n)^*a0 + (-1)^(n-1)(D^(n-1))^*a1 + ... + (-1)(D^1)^*a(n-1) + a(n)^*, where a0, a1, ..., a(n) are the coefficients in L[y] and (D^k)^* denotes the adjoint of the kth derivative operator. This can be solved using the method of undetermined coefficients or other techniques for solving linear differential equations.

I hope this helps to answer your questions. Please let me know if you have any further inquiries or need clarification on any of the methods mentioned. Thank you.

 

[tacman]

Firstly, to answer the first question, knowing the values of y1 and y2 alone is not enough to determine the values of a0, a1, and a2. We would need additional information or conditions (such as initial conditions or boundary conditions) to solve for these coefficients. The Wronskian can be used to determine if the solutions y1 and y2 are linearly independent, but it does not provide information about the values of the coefficients.

For the second question, the Green operator theory can be used to find the inverse of the differential operator L. This involves using the Green's function G(x,s) to solve for the coefficients a0, a1, and a2. The Green's function is defined as the solution to the equation L[G(x,s)]=δ(x-s), where δ(x-s) is the Dirac delta function. Once the Green's function is known, the coefficients can be obtained by solving the integral equation a0G(x,s) + a1G'(x,s) + a2G''(x,s) = δ(x-s).

Finally, for the last question, to obtain the adjoint operator for L[y], we would need to use the concept of adjoint operators in functional analysis. The adjoint operator of L[y] is defined as the operator L*[y] that satisfies the equation ∫L[y]v dx = ∫yL*[v] dx, where v is an arbitrary function and the integrals are taken over the domain of the differential operator. The adjoint operator can be found by using integration by parts and solving for the coefficients a0, a1, and a2.