PDE25

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\Delta u=u_{{xx}}+u_{{yy}}=0\,


u(x,0)={\begin{cases}T_{0}&|x|<b\\0&|x|>b\end{cases}}\,
\lim _{{x^{2}+y^{2}\rightarrow \infty }}u(x,y)=0\,

-\infty <x<\infty ,\,\,\,\,\,0<y<\infty \,



Let u=XY\, so that

X''+\lambda ^{2}X=0,\,\,-\infty <x<\infty \,

Y''-\lambda ^{2}Y=0,\,\,0<y<\infty \,

X(x)={\begin{cases}{\frac  {1}{2}}A_{0}&\lambda =0\\c_{1}\cos \lambda x+c_{2}\sin \lambda x&\lambda >0\end{cases}}\,

Y(y)={\begin{cases}{\frac  {1}{2}}B_{0}&\lambda =0\\c_{3}e^{{\lambda x}}+c_{4}e^{{-\lambda x}}&\lambda >0\end{cases}}\,

The exponentials were chosen over the hyperbolic trig functions because one of the exponetials tends to zero as its argument tends to infinity.

To satisfy the limit condition, we must set c_{3}=A_{0}=B_{0}=0\,

The solution is

  • u(x,y)=\int _{0}^{\infty }\left[A_{\lambda }\cos \lambda x+B_{\lambda }\sin \lambda x\right]e^{{-\lambda x}}\,d\lambda \,

u(x,0)=\int _{0}^{\infty }A_{\lambda }\cos \lambda x+B_{\lambda }\sin \lambda x={\begin{cases}T_{0}&|x|<b\\0&|x|>b\end{cases}}\,

The Fourier coefficients are:

A_{\lambda }={\frac  {1}{\pi }}\int _{{-b}}^{b}T_{0}\cos \lambda x\,dx={\frac  {T_{0}}{\pi \lambda }}\sin \lambda x{\Bigg |}_{{-b}}^{b}={\frac  {2T_{0}\sin \lambda b}{\pi \lambda }}\,

B_{\lambda }={\frac  {1}{\pi }}\int _{{-b}}^{b}T_{0}\sin \lambda x\,dx={\frac  {-T_{0}}{\pi \lambda }}\cos \lambda x{\Bigg |}_{{-b}}^{b}=0\,

The final solution is:

u(x,y)={\frac  {2T_{0}}{\pi }}\int _{0}^{\infty }{\frac  {1}{\lambda }}\sin \lambda b\cos \lambda x\,d\lambda \,e^{{-\lambda y}}\,

Partial Differential Equations

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