Converting IVP to Volterra equations

In mathematical analysis, converting an Initial Value Problem (IVP) for a differential equation into an equivalent Volterra Integral Equation is a standard approach. This transformation is crucial in numerical methods, as integral equations can sometimes be easier to handle computationally.

Mathematical Background

Initial Value Problem (IVP)

An IVP typically involves solving a differential equation along with initial conditions. For example:

y(x)=f(x,y(x)),y(x0)=y0y'(x) = f(x, y(x)), \quad y(x_0) = y_0

where:

  • y(x)y(x) is the unknown function to be solved,
  • f(x,y(x))f(x, y(x)) is a given function,
  • x0x_0 is the initial point, and
  • y0y_0 is the initial value.

Volterra Integral Equation

A Volterra integral equation of the second kind has the general form:

y(x)=f(x)+x0xK(x,t)y(t)dty(x) = f(x) + \int_{x_0}^x K(x, t) y(t) \, dt

where:

  • f(x)f(x) is a given function,
  • K(x,t)K(x, t) is the kernel,
  • y(t)y(t) is the unknown function.

Procedure for Conversion

Step 1: Rewrite the Differential Equation

Start with the first-order IVP:

y(x)=f(x,y(x)),y(x0)=y0y'(x) = f(x, y(x)), \quad y(x_0) = y_0

Step 2: Integrate Both Sides

Integrate the differential equation from x0x_0 to xx:

x0xy(t)dt=x0xf(t,y(t))dt\int_{x_0}^x y'(t) \, dt = \int_{x_0}^x f(t, y(t)) \, dt

The left-hand side becomes:

y(x)y(x0)y(x) – y(x_0)

Thus:

y(x)=y0+x0xf(t,y(t))dty(x) = y_0 + \int_{x_0}^x f(t, y(t)) \, dt

Step 3: Express as a Volterra Integral Equation

The result is a Volterra integral equation of the second kind:

y(x)=y0+x0xf(t,y(t))dty(x) = y_0 + \int_{x_0}^x f(t, y(t)) \, dt

Here:

  • f(t,y(t))f(t, y(t)) plays the role of the kernel K(x,t)K(x, t),
  • y0y_0 is the initial value.

Example 1: Linear IVP

Problem:

Solve the IVP:

y(x)=2y(x)+4x,y(0)=1y'(x) = -2y(x) + 4x, \quad y(0) = 1

Conversion:

  1. Integrate both sides:

    y(x)=y(0)+0x[2y(t)+4t]dty(x) = y(0) + \int_0^x [-2y(t) + 4t] \, dt
  2. Substitute y(0)=1y(0) = 1:

    y(x)=1+0x[2y(t)+4t]dty(x) = 1 + \int_0^x [-2y(t) + 4t] \, dt
  3. Split the integral:

    y(x)=120xy(t)dt+0x4tdty(x) = 1 – 2 \int_0^x y(t) \, dt + \int_0^x 4t \, dt
  4. Evaluate the second term:

    0x4tdt=[2t2]0x=2x2\int_0^x 4t \, dt = [2t^2]_0^x = 2x^2

Thus, the Volterra equation is:

y(x)=120xy(t)dt+2x2y(x) = 1 – 2 \int_0^x y(t) \, dt + 2x^2

Example 2: Nonlinear IVP

Problem:

Solve the IVP:

y(x)=y2(x)+ex,y(0)=0y'(x) = y^2(x) + e^x, \quad y(0) = 0

Conversion:

  1. Integrate both sides:

    y(x)=y(0)+0x[y2(t)+et]dty(x) = y(0) + \int_0^x [y^2(t) + e^t] \, dt
  2. Substitute y(0)=0y(0) = 0:

    y(x)=0xy2(t)dt+0xetdty(x) = \int_0^x y^2(t) \, dt + \int_0^x e^t \, dt
  3. Simplify:

    y(x)=0xy2(t)dt+[et]0x=0xy2(t)dt+ex1y(x) = \int_0^x y^2(t) \, dt + [e^t]_0^x = \int_0^x y^2(t) \, dt + e^x – 1

Thus, the Volterra equation is:

y(x)=0xy2(t)dt+ex1y(x) = \int_0^x y^2(t) \, dt + e^x – 1

Numerical and Computational Considerations

  • Picard Iteration:
    Iterative methods like Picard iteration can be applied to solve the resulting Volterra integral equation numerically. Start with an initial guess y0(x)y_0(x) and refine it iteratively:

    yn+1(x)=y0+x0xf(t,yn(t))dty_{n+1}(x) = y_0 + \int_{x_0}^x f(t, y_n(t)) \, dt
  • Trapezoidal Rule:
    For numerical integration, apply the trapezoidal rule or other quadrature methods:

    x0xf(t,y(t))dti=0nwif(ti,y(ti))\int_{x_0}^x f(t, y(t)) \, dt \approx \sum_{i=0}^{n} w_i f(t_i, y(t_i))

    where wiw_i are weights.

References

  1. Books

    • Tricomi, F.G. Integral Equations. Dover Publications, 1985.
    • Kanwal, Ram P. Linear Integral Equations: Theory and Technique. Birkhäuser, 1996.
    • Kreyszig, Erwin. Advanced Engineering Mathematics. Wiley, 2011.
  2. Articles

    • “Integral Equation Methods in Differential Equations,” Mathematics of Computation, 1972.
    • “Numerical Techniques for Solving Volterra Equations,” SIAM Journal on Numerical Analysis, 1985.
  3. Online Resources

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