Advanced Numerical solution of differential equations

In summary: AB-H^2$$$$A^2 + B^2 + 2AB > 4AB - 4H^2$$$$A^2 + B^2 - 2AB + 4H^2 > 0$$$$(A-B)^2 + 4H^2 > 0$$Since $H$ is a constant, this condition will always be satisfied as long as $A$ and $B$ are not equal. Therefore, the solutions of the system will always lie on ellipses as long as $A \neq B$.For part iii), you have correctly applied the explicit and symplectic Euler schemes to the system and obtained equations (6) and (
  • #1
ra_forever8
129
0
i) IF $\frac{dy}{dt} = - \frac{∂H}{∂z}, \frac{dz}{dt}= \frac{∂H}{∂y}$
where H is a function of $y$ and $z$, show that $H(y,z)$ is constant in time.
ii) Take a $H(y,z) =Ay^2 + 2Hyz + Bz^2$ where $A,B,H$ are constants and show that solutions of the system lie on ellipses.
iii) Apply the explicit Euler and the symplectic Euler schemes to the system in (ii) and check whether the area is preserved.

can anyone tell me how to start. totally confused.
=>

for i)
by using the chain rule, we observe that
$\frac{dH}{dt} = - \frac{∂H}{∂y} \frac{dy}{dt} + \frac{∂H}{∂z} \frac{dz}{dt}$ --------(1)
now we use the given equations $\frac{dy}{dt} = - \frac{∂H}{∂z}, \frac{dz}{dt}= \frac{∂H}{∂y}$
into (1), to obtain
$\frac{dH}{dt} = - \frac{∂H}{∂y} \frac{∂H}{∂z} + \frac{∂H}{∂z} \frac{∂H}{∂y}$
$\frac{dH}{dt}=0$
$H(t)= constant$
Therefore, $H(y,z)$ is constant in time.

ii)
From part i) $H(y,z)$ is constant. So,
$H(y,z) =Ay^{2 }+ 2Hyz + Bz^{2}= constant -------(2)$
are ellipses. but whether $(2)$ defines a family of ellipses or some other conic section depends on the values of $A,B,H$
Writing the function $H(y,z)$ defined in $(2)$ in vector-matrix form we have
\begin{equation}
H(y,z)= (y,z) \begin{bmatrix}
A & H\\
H & B\\
\end{bmatrix} \begin{pmatrix}
y\\
z\\
\end{pmatrix} -------------------(3)
\end{equation} Let the matrix \begin{equation}
M= \begin{bmatrix}
A & H\\
H & B\\
\end{bmatrix}
\end{equation}
occurring in (3), being symmetric may be diagonalised by some orthogonal matrix $O$, it will then take the form
\begin{equation}
O^{T} M O= \begin{bmatrix}
\lambda_{1} & 0\\
0 &\lambda_{2}\\
\end{bmatrix}
\end{equation}
$\lambda_{1}$ and $\lambda_{2}$ being the eigenvalues of $M$, satisfy its characteristics equation.
\begin{equation} \lambda^{2} -(A+B) \lambda + (AB-H^{2})=0 -------------------(4)
\end{equation}
and in the coordinate system $x_{1} - x_{2}$ which is the $y-z$ system transformed by $O$,
So $(3)$ takes the form
\begin{equation}
H(x_{1},x_{2})= (x_{1},x_{2}) \begin{bmatrix}
\lambda_{1} & 0\\
0 &\lambda_{2}\\
\end{bmatrix} \begin{pmatrix}
x_{1}\\
x_{2}\\
\end{pmatrix}
\end{equation}
\begin{equation}
= \lambda_{1} x^2_{1} + \lambda_{2} x^2_{2}
\end{equation}
The equation is
\begin{equation}
H(x_{1},x_{2})= \lambda_{1} x^2_{1} + \lambda_{2} x^2_{2}=constant
\end{equation}
represents an elliptical curve precisely when $\lambda_{1},\lambda_{2}$ are the same sign, in which case their sign must be shared with constant lest there be no curve whatsoever
The signs of $\lambda_{1},\lambda_{2}$ are governed by $(4)$ and via the quadratic formula, these signs are precisely when
$ det M = AB-H^{2}= \lambda_{1},\lambda_{2} >0$ ----------------(5)
when $(5)$ applies, the level of sets of $H(y,z)$ are ellipses but with their axes tilted to align themselves with the $x_{1}$ and $x_{2}$ axes in the transformed coordinates.

(IS THIS ENOUGH TO ANSWER THE PART ii), did I missed anythings, ? do I need to calculate $\lambda_{1},\lambda_{2}$?)
for iii)
when explicit euler is applied to part i) equations, it gives:
$ y_{n+1} = y_{n} - h z_{n}$
$ z_{n+1} = z_{n} + h y_{n}$
now squaring both sides and rearranging both equations gives
$ y^{2}_{n+1}+z^{2}_{n+1} = (1+h^{2}) (y^{2}_{n} + z^{2}_{n})$--------------(6)

after that I really don't know, I guess I need to find area of ellipses by comparing (6) with $H(y,z) =Ay^2 + 2Hyz + Bz^2$ but how?

and same for the symmetric euler applied to part i)equations gives:
$ y_{n+1} = y_{n} - h z_{n}$
$ z_{n+1} = z_{n} + h y_{n+1}$
now squaring both sides and rearranging both equations gives
$ y^{2}_{n}+z^{2}_{n} = (1-h^{2}+h^{4}) (y^{2}_{n+1}- 2h^{3} y_{n+1} z_{n+1}+(1+h^{2}) z^{2}_{n})$--------------(7)
again
to find area of ellipses by comparing (7) with $H(y,z) =Ay^2 + 2Hyz + Bz^2$ , but this time
it gives
$A= 1-h^{2}+h^{4}, H= -h^{3}, B=1+h^{2}$
so,
$AB-H^{2}= (1-h^{2}+h^{4})(1+h^{2})-(-h^{3})$
$= 1-h^{2}+h^{4}+h^{2}-h^{4}+h^{6}+h^{6}$
$= 1+2h^{6}$

where $AB-H^{2}=\lambda_{1} \lambda_{2}$
and $a= \frac{1}{\sqrt{\lambda_{1}}} , b=\frac{1}{\sqrt{\lambda_{2}}}$

SO, AREA = $\frac{\pi}{\sqrt{(1+2h^{6})}}$ ( area of $ellipse = \pi a b)$
I am sure with my answer for sympetic euler and not sure for explicit euler to find area.
please someone help will be really grateful.

THANK YOU
 
Last edited:
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  • #2
for reading my answer.
For part ii), you have correctly shown that the solutions of the system lie on ellipses. To determine the specific shape and orientation of these ellipses, you need to find the values of $A$, $B$, and $H$ that satisfy the condition in equation (5).
To do this, you can use the characteristic equation (4) and the fact that the determinant of a symmetric matrix is equal to the product of its eigenvalues. This means that $AB-H^2 = \lambda_1 \lambda_2$, where $\lambda_1$ and $\lambda_2$ are the eigenvalues of the matrix $M$.
Solving the characteristic equation (4) for $\lambda_1$ and $\lambda_2$, we get:
$$\lambda_1 = \frac{1}{2}(A+B + \sqrt{(A+B)^2 - 4(AB-H^2)})$$
$$\lambda_2 = \frac{1}{2}(A+B - \sqrt{(A+B)^2 - 4(AB-H^2)})$$
Now, we need to consider the different cases based on the sign of the expression inside the square root. This will determine the shape and orientation of the ellipses.
If $(A+B)^2 - 4(AB-H^2) > 0$, then both eigenvalues will be positive, and the level sets of $H(y,z)$ will be ellipses with axes aligned with the $x_1$ and $x_2$ axes, as you have correctly stated.
If $(A+B)^2 - 4(AB-H^2) < 0$, then both eigenvalues will be negative, and the level sets of $H(y,z)$ will be hyperbolas.
If $(A+B)^2 - 4(AB-H^2) = 0$, then one eigenvalue will be positive and one will be negative, and the level sets of $H(y,z)$ will be parabolas.
Therefore, to ensure that the solutions lie on ellipses, we need to have $(A+B)^2 - 4(AB-H^2) > 0$. This condition can be rewritten as:
$$\frac{(A+B)^2}{4} > AB-H^2$$
$$\frac{A^2 + 2AB + B^2}{4} >
 

1. What is the purpose of advanced numerical solutions of differential equations?

Advanced numerical solutions of differential equations are used to approximate the solutions of complex mathematical models, such as those used in physics, engineering, and other scientific fields. These methods allow for the efficient and accurate computation of solutions to differential equations that may not have closed-form analytical solutions.

2. What are some common techniques used in advanced numerical solutions of differential equations?

Some common techniques used in advanced numerical solutions of differential equations include finite difference methods, finite element methods, and spectral methods. Each of these methods has its own strengths and weaknesses and may be more suitable for certain types of differential equations or boundary conditions.

3. How do advanced numerical solutions differ from traditional analytical solutions of differential equations?

Traditional analytical solutions of differential equations involve finding a closed-form solution that can be expressed in terms of known functions. Advanced numerical solutions, on the other hand, use computational techniques to approximate the solution and may not have a closed-form expression. These methods are often necessary for complex or nonlinear systems that do not have analytical solutions.

4. What are the benefits of using advanced numerical solutions of differential equations?

The use of advanced numerical solutions allows for the accurate and efficient computation of solutions to complex differential equations. They also allow for the analysis of systems that may not have analytical solutions, providing valuable insights into their behavior and dynamics. Additionally, these methods can handle a wide range of boundary conditions and can be easily adapted to different types of differential equations.

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Advanced numerical solutions of differential equations have numerous applications across various scientific fields. They are commonly used in computational fluid dynamics, structural analysis, and optimization problems in engineering. They also have applications in weather forecasting, population dynamics, and other areas of physics and biology where complex systems can be modeled using differential equations.

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