my question is related to the already answered question:

Solving cauchy hyperbolic second order pde

but in my case I want to solve:

$u_{tt} + a u + b u_t + d u_x = c^2 u_{xx}$

In the "book Partial Differential Equations of Mathematical Physics and Integral Equations" by Ronald B. Guenther and John W. Lee:


the solution is given on page 121 (equation 6-20). Unfortunately, I am not sure how to derive the second integral with the modified Bessel function of first order besseli(1,...) in this equation. Is it possible, that the constant k is missing in this integral? Furthermore, considering the book as correct, how to implement the solution into MATLAB? For the evaluation of the boundaries I have 0/0... If I use integration by parts I have to divide by $k$ (but $k=0$ for $a=b=d=0$).

Thank you very much. Best


Okay, I found the mistake, the book is slightly wrong. Now I still have the problem how to implement the expression

$ \int_{\;\beta}^\alpha I_1( 2\sqrt{k(\alpha-\eta)(\eta-\beta)} ) \frac{\displaystyle k(\alpha-\eta)}{\displaystyle \sqrt{k(\alpha-\eta)(\eta-\beta)}}\, \mathrm{d}\eta $


You can first convert the PDE to the form of $v_{\xi\eta}=kv$ , then the approach is similar to Method of charactersitics and second order PDE..

  • $\begingroup$ Yes, thats how it is done. Then you can apply the method of Riemann. So far it is clear, my question regards the equation (some steps after the transformation). $\endgroup$ – lapricorn1 Oct 12 '15 at 7:09

I implented it with finite differences in Matlab and it is working. Maybe it is helpful for someone one day:

R = @(alpha,beta,xi,eta) besseli(0,2*sqrt(k*(alpha - xi).*(eta - beta)));
R_diff = @(alpha,beta,xi,eta) (R(alpha,beta,xi,eta+dx) - R(alpha,beta,xi,eta-dx))/(2*dx);
int_2 = @(alpha,beta) 

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