Consider the initial value problem$\frac{d y}{d x}=f(x, y)$, y(x0) = y0where f is a twice continuously differentiable function on a rectangle containing the point (x0, y0). With the step-size h, let the first iterate of a second order scheme to approximate the solution of the above initial value problem be given byy1 = y0 + Pk1 + Qk2,where k1 = hf(x0, y0), k2 = hf(x0 + α0h, y0 + β0k1) and P, Q, α0, β0 ∈ &reals;.Which of the following statements are correct?

Question

Consider the initial value problem$\frac{d y}{d x}=f(x, y)$, y(x0) = y0where f is a twice continuously differentiable function on a rectangle containing the point (x0, y0). With the step-size h, let the first iterate of a second order scheme to approximate the solution of the above initial value problem be given byy1 = y0 + Pk1 + Qk2,where k1 = hf(x0, y0), k2 = hf(x0 + &alpha;0h, y0 + &beta;0k1) and P, Q, &alpha;0, &beta;0 &isin; &reals;.Which of the following statements are correct?

Prepp Team · Accepted Answer

Initial Value Problem and Numerical Scheme

We are given an initial value problem (IVP) of the form:

\[ \frac{d y}{d x}=f(x, y), \quad y(x_0) = y_0 \]

where $f(x, y)$ is a function that is twice continuously differentiable. We are considering a numerical scheme to approximate the solution at $x_1 = x_0 + h$ using a step-size $h$. The scheme is given by:

\[ y_1 = y_0 + Pk_1 + Qk_2 \]

where

\[ k_1 = hf(x_0, y_0) \] \[ k_2 = hf(x_0 + \alpha_0h, y_0 + \beta_0k_1) \]

and $P, Q, \alpha_0, \beta_0$ are real constants.

This type of scheme is a two-stage Runge-Kutta method. We want to find the conditions on $P, Q, \alpha_0, \beta_0$ for this scheme to be of second order.

Conditions for a Second Order Scheme

For a numerical method to be of second order, the numerical approximation $y_1$ must match the Taylor series expansion of the exact solution $y(x_0 + h)$ around $x_0$ up to terms of order $h^2$.

Taylor Expansion of the Exact Solution

The exact solution $y(x)$ at $x = x_0 + h$ can be expanded using the Taylor series around $x_0$:

\[ y(x_0 + h) = y(x_0) + h y'(x_0) + \frac{h^2}{2!} y''(x_0) + O(h^3) \]

From the IVP, $y'(x) = f(x, y(x))$, so $y'(x_0) = f(x_0, y_0)$.

Differentiating $y'(x) = f(x, y(x))$ with respect to $x$ using the chain rule gives $y''(x)$:

\[ y''(x) = \frac{d}{dx} f(x, y(x)) = \frac{\partial f}{\partial x}(x, y(x)) \frac{dx}{dx} + \frac{\partial f}{\partial y}(x, y(x)) \frac{dy}{dx} = f_x(x, y(x)) + f_y(x, y(x)) f(x, y(x)) \]

Evaluating at $x_0$, we get $y''(x_0) = f_x(x_0, y_0) + f_y(x_0, y_0) f(x_0, y_0)$.

Let $f_0 = f(x_0, y_0)$, $f_x^0 = f_x(x_0, y_0)$, and $f_y^0 = f_y(x_0, y_0)$. The Taylor expansion of the exact solution is:

\[ y(x_0 + h) = y_0 + h f_0 + \frac{h^2}{2} (f_x^0 + f_0 f_y^0) + O(h^3) \]

Expansion of the Numerical Approximation

The numerical approximation is $y_1 = y_0 + Pk_1 + Qk_2$. We have $k_1 = hf_0$. Now let's expand $k_2 = hf(x_0 + \alpha_0h, y_0 + \beta_0k_1)$.

Substitute $k_1 = hf_0$: $k_2 = hf(x_0 + \alpha_0h, y_0 + \beta_0 hf_0)$.

We use the Taylor expansion of $f(x, y)$ around $(x_0, y_0)$ for the term $f(x_0 + \alpha_0h, y_0 + \beta_0 hf_0)$. Let $\Delta x = \alpha_0h$ and $\Delta y = \beta_0 hf_0$. The expansion is:

\[ f(x_0 + \Delta x, y_0 + \Delta y) = f(x_0, y_0) + \Delta x f_x(x_0, y_0) + \Delta y f_y(x_0, y_0) + O((\Delta x)^2, (\Delta y)^2, \Delta x \Delta y) \] \[ f(x_0 + \alpha_0h, y_0 + \beta_0 hf_0) = f_0 + (\alpha_0h) f_x^0 + (\beta_0 hf_0) f_y^0 + O(h^2) \]

Substituting this back into the expression for $k_2$:

\[ k_2 = h \left[ f_0 + \alpha_0h f_x^0 + \beta_0 hf_0 f_y^0 + O(h^2) \right] \] \[ k_2 = hf_0 + \alpha_0h^2 f_x^0 + \beta_0 h^2 f_0 f_y^0 + O(h^3) \]

Now substitute $k_1$ and $k_2$ into the expression for $y_1$:

\[ y_1 = y_0 + P(hf_0) + Q(hf_0 + \alpha_0h^2 f_x^0 + \beta_0 h^2 f_0 f_y^0 + O(h^3)) \] \[ y_1 = y_0 + Phf_0 + Qhf_0 + Q\alpha_0h^2 f_x^0 + Q\beta_0 h^2 f_0 f_y^0 + O(h^3) \] \[ y_1 = y_0 + (P+Q)hf_0 + h^2 (Q\alpha_0 f_x^0 + Q\beta_0 f_0 f_y^0) + O(h^3) \]

Comparing Expansions to Find Conditions

For the numerical scheme to be second order, the coefficients of powers of $h$ in the expansion of $y_1$ must match those in the Taylor expansion of the exact solution up to order $h^2$. Comparing

\[ y(x_0 + h) = y_0 + h f_0 + \frac{h^2}{2} (f_x^0 + f_0 f_y^0) + O(h^3) \]

and

\[ y_1 = y_0 + (P+Q)hf_0 + h^2 (Q\alpha_0 f_x^0 + Q\beta_0 f_0 f_y^0) + O(h^3) \]

We equate the coefficients of $h$, $h^2 f_x^0$, and $h^2 f_0 f_y^0$:

Coefficient of $h$: $P+Q = 1$
Coefficient of $h^2 f_x^0$: $Q\alpha_0 = \frac{1}{2}$
Coefficient of $h^2 f_0 f_y^0$: $Q\beta_0 = \frac{1}{2}$

These three equations are the conditions for the scheme to be second order. Note that from the second and third conditions, if $Q \ne 0$, then $\alpha_0 = \beta_0$.

Checking the Given Options

We check each option to see if the provided values of $P, Q, \alpha_0, \beta_0$ satisfy the derived conditions:

Option	Given Values (α₀, β₀, P, Q)	Condition 1 (P+Q=1)	Condition 2 (Qα₀=1/2)	Condition 3 (Qβ₀=1/2)	Satisfies Conditions?
1	α₀=2, β₀=2, P=3/4, Q=1/4	$ \frac{3}{4} + \frac{1}{4} = 1 $ (Yes)	$ \frac{1}{4} \times 2 = \frac{1}{2} $ (Yes)	$ \frac{1}{4} \times 2 = \frac{1}{2} $ (Yes)	Yes
2	α₀=3, β₀=3, P=5/6, Q=1/6	$ \frac{5}{6} + \frac{1}{6} = 1 $ (Yes)	$ \frac{1}{6} \times 3 = \frac{3}{6} = \frac{1}{2} $ (Yes)	$ \frac{1}{6} \times 3 = \frac{3}{6} = \frac{1}{2} $ (Yes)	Yes
3	α₀=2, β₀=2, P=1/4, Q=3/4	$ \frac{1}{4} + \frac{3}{4} = 1 $ (Yes)	$ \frac{3}{4} \times 2 = \frac{6}{4} = \frac{3}{2} \ne \frac{1}{2} $ (No)	$ \frac{3}{4} \times 2 = \frac{6}{4} = \frac{3}{2} \ne \frac{1}{2} $ (No)	No
4	α₀=3, β₀=3, P=1/6, Q=5/6	$ \frac{1}{6} + \frac{5}{6} = 1 $ (Yes)	$ \frac{5}{6} \times 3 = \frac{15}{6} = \frac{5}{2} \ne \frac{1}{2} $ (No)	$ \frac{5}{6} \times 3 = \frac{15}{6} = \frac{5}{2} \ne \frac{1}{2} $ (No)	No

Based on the checks, Options 1 and 2 satisfy the conditions for the scheme to be a second order method.

Prepp Team · Answer

If &alpha;0 = 2, then &beta;0 = 2, $P=\frac{1}{4}, Q=\frac{3}{4}$

Prepp Team · Answer

If &beta;0 = 3, then &alpha;0 = 3, $P=\frac{1}{6}, Q=\frac{5}{6}$