PDE27

From Example Problems
Jump to: navigation, search
u_{{tt}}-c^{2}u_{{xx}}=2t=f(x,t)\,


u(x,0)=x^{2}=g(x)\,
u_{t}(x,0)=1=h(x)\,




In general, combining Duhammel's principle and D'Alembert's formula, the solution is:

u(x,t)={\frac  {1}{2}}\left[g(x+ct)+g(x-ct)\right]+{\frac  {1}{2c}}\int _{{x-ct}}^{{x+ct}}h(\xi )d\xi +{\frac  {1}{2c}}\int _{0}^{t}\int _{{x-c(t-s)}}^{{x+c(t-s)}}f(\xi ,s)\,d\xi \,ds\,.

In this particular case,

u(x,t)={\frac  {1}{2}}\left[(x-ct)^{2}+(x+ct)^{2}\right]+{\frac  {1}{2c}}\int _{{x-ct}}^{{x+ct}}d\xi +{\frac  {1}{2c}}\int _{0}^{t}\int _{{x-c(t-s)}}^{{x+c(t-s)}}2s\,d\xi \,ds\,

={\frac  {1}{2}}\left[2x^{2}+2c^{2}t^{2}\right]+{\frac  {1}{2c}}\left[(x+ct)-(x-ct)\right]+{\frac  {1}{c}}\int _{0}^{t}s\left[x+c(t-s)-x+c(t-s)\right]ds\,

=x^{2}+c^{2}t^{2}+{\frac  {1}{2c}}2ct+{\frac  {1}{c}}\int _{0}^{t}2sc(t-s)ds\,

And after simplifying, the soution is

u(x,t)=x^{2}+c^{2}t^{2}+t+{\frac  {1}{3}}t^{3}\,


Main Page : Partial Differential Equations : Laplace's Equation