Question

3. Consider another Volterra integral equation (a) Solve the integral equation (4) by using the Laplace transform. (b) Convert the integral equation (4) into an initial value problem, as in Problem 2. (c) Solve the initial value problem in part (b), and verify the solution is the same as the one in part (a)

Answer 1

(a) $\phi (t)-\int_{0}^{t}(t-\xi )\phi (\xi )d\xi =1$

Apply Laplace transform on both sides :

$L(\phi (t))-L(\int_{0}^{t}(t-\xi )\phi (\xi )d\xi) =L(1)$

Using the formula for Laplace transforms:

$(\phi (s))-\phi (s)/s^2=1/s$

=> $\phi (s)[1-1/s^2]=1/s$

=> $\phi (s)[(s^2-1)/s^2]=1/s$

=> $\phi (s)=s^2/s(s^2-1)$

=> $\phi (s)=s/(s^2-1)$

=> $\phi (s)=s/(s-1)(s+1)$ {Factorizing denominator}

=> $\phi (s)=\frac{1}{2(s-1)}-\frac{1}{2(s+1)}$ { Using partial fractions }

Taking inverse laplace transform:

$\phi (t)=\frac{e^t}{2}-\frac{e^{-t}}{2} = 1/2(e^t-e^{-t})$

(b) The given integral equation can also be written as :

$\phi (t)-t\int_{0}^{t}\phi (\xi )d\xi +\int_{0}^{t}\xi \phi (\xi )d\xi =1$

=> $\phi (t)=1+t\int_{0}^{t}\phi (\xi )d\xi -\int_{0}^{t}\xi \phi (\xi )d\xi$

Differentiating with respect to t :

$\phi' (t)=\int_{0}^{t}\phi (\xi )d\xi+t\phi (t) -t \phi (t)$

=> $\phi' (t)=\int_{0}^{t}\phi (\xi )d\xi$

Again differentiating with respect to t :

=> $\phi"(t)=\phi (t)$

=> $\phi"(t)-\phi (t)=0$

The initial value problem is therefore:

$\phi"(t)-\phi (t)=0; \phi (0)=1, \phi '(0)=0$

(c) The characteristic equation is given by :

r² -1 =0

=> r² = 1 => r =-1, +1

The general solution is given by :

$\phi (t)=Ae^t + Be^{-t}$

=> $\phi '(t)=Ae^t - Be^{-t}$

Putting initial conditions:

$\phi (0)=A + B=1$

$\phi' (0)=A - B=0$

Solving these equations , we get :

2A = 1 => A=1/2 and B = 1/2

Hence particular solution is given by :

$\phi (t)=1/2(e^t+e^{-t})$

This is same as we have got in part a)