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t>0,\,\,0<x<\infty \,

Since the boundary condition makes the solution equal to zero at x=0, the Fourier sine transform is most appropriate.

U(x,t)=\int _{0}^{\infty }u(y,t)\sin(xy)\,dy\,

Take the partial wrt to t and use the DE in the transform integral.

{\partial  \over \partial t}U(x,t)=\int _{0}^{\infty }u_{t}(y,t)\sin(xy)\,dy=k\int _{0}^{\infty }u_{{yy}}(y,t)\sin(xy)\,dy\,

Integrate the integral on the right twice by parts.

=k\left([u_{y}(y,t)\sin(xy)]_{0}^{\infty }-x\int _{0}^{\infty }u_{y}\cos(xy)\,dy\right)\,

The first term evaluated at 0 is 0 because of the sine function, and evaluated at infinity is 0 because we assume that there is no heat flow past an infinite boundary.

=-kx\left([u(y,t)\cos(xy)]_{0}^{\infty }+x\int _{0}^{\infty }u(y,t)\sin(xy)\,dy\right)\,

The boundary condition at 0 gives u(0,t)=0 and again we assume that there is no heat at infinity. Otherwise, there would be heat flow past that point.

=-kx^{2}\int _{0}^{\infty }u(y,t)\sin(xy)\,dy\,

This integral is the Fourier sine transform of u(y,t). This can be written


This is an easy ODE for U. The solution follows.

\int _{0}^{t}{U_{t}(x,\tau ) \over U(x,\tau )}\,d\tau =-kx^{2}\int _{0}^{t}\,d\tau \,

[\ln U(x,\tau )]_{{\tau =0}}^{t}=-kx^{2}t\,

\ln U(x,t)-\ln U(x,0)=-kx^{2}t\,


The solution is the inverse transform of this last equation.

U(x,0)\, is the Fourier transform of u(x,0)\,, so it can be written like this:

U(x,0)=\int _{0}^{\infty }u(y,0)\sin(xy)\,dy=\int _{0}^{\infty }f(y)\sin(xy)\,dy\,

So, the solution is

u(x,t)={2 \over \pi }\int _{0}^{\infty }\!\!\left(\int _{0}^{\infty }f(z)\sin(xz)\,dz\right)e^{{-ky^{2}t}}\sin(xy)\,dy\,

Partial Differential Equations

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