Consider the initial value problem \(\frac{d y}{d x}=f(x, y)\), y(x0) = y0 where 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 by y1 = y0 + Pk1 + Qk2, where k1 = hf(x0, y0), k2 = hf(x0 + α0h, y0 + β0k1) and P, Q, α0, β0 ∈ ℝ. Which of the following statements are correct?
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.
Important Questions from Initial Value Problem
Initial value problem, \(\rm x \frac{d y}{d x}=y\), y(0) = 0, x > 0
Consider the eigenvalue problem
((1 + x4)y')' + λy = 0, x ∈ (0, 1),
y(0) = 0, y(1) + 2y'(1) = 0.
Then which of the following statements are true?
Consider the following two initial value ODEs
(A) \(\frac{dx}{dt}=x^3,x(0)=1;\)
(B) \(\frac{dx}{dt}=x\sin x^2,x(0)=2.\)
Related to these ODEs, we make the following assertions.
I. The solution to (A) blows up in finite time.
II. The solution to (B) blows up in finite time.
Which of the following statements is true?
Let y0 > 0, z0 > 0 and α > 1.
Consider the following two differential equations:
\(\begin{aligned} &(*)\left\{\begin{array}{l} \frac{d y}{d t}=y^\alpha \quad \text { for } t>0, \\ y(0)=y_0 \end{array}\right. \\ &(* *)\left\{\begin{array}{l} \frac{d z}{d t}=-z^\alpha \quad \text { for } t>0, \\ z(0)=z_0 \end{array}\right. \end{aligned}\)
We say that the solution to a differential equation exists globally if it exists for all t > 0.Which of the following statements is true?
Let f ∶ ℝ2 → ℝ be a locally Lipschitz function. Consider the initial value problem
ẋ = f(t, x), x(t0) = x0
for (t0, x0) ∈ ℝ2. Suppose that J(t0, x0) represents the maximal interval of existence for the initial value problem. Which of the following statements is true?
