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Auxiliary condition: u is bounded.
t>0,\,\,x,y\in \Re ,\,\,\,z>0\,

Since x\, and y\, both range over the whole line, they should both be transformed.

Assume that a solution exists in the form

u(x,y,z)=\int \!\!\!\int _{\Re }e^{{i\lambda x+i\mu y}}U(\lambda ,\mu ,z)\,d\lambda d\mu \,

U(\lambda ,\mu ,z)=\left({\frac  {1}{2\pi }}\right)^{2}\int \!\!\!\int _{\Re }e^{{-i\lambda x-i\mu y}}u(x,y,z)\,dxdy\,

Take the second partial derivative with respect to z of both sides.

{\frac  {\partial ^{2}}{\partial z^{2}}}U(\lambda ,\mu ,z)=-\left({\frac  {1}{2\pi }}\right)^{2}\int \!\!\!\int _{\Re }e^{{-i\lambda x-i\mu y}}u_{{zz}}(x,y,z)\,dxdy=-\left({\frac  {1}{2\pi }}\right)^{2}\int \!\!\!\int _{\Re }e^{{-i\lambda x-i\mu y}}(u_{{xx}}+u_{{yy}})\,dxdy\,

=(\lambda ^{2}+\mu ^{2}){\frac  {1}{4\pi ^{2}}}\int \!\!\!\int _{\Re }e^{{-i\lambda x-i\mu y}}u(x,y,z)\,dxdy\,

=(\lambda ^{2}+\mu ^{2})U\,

The solution to the ODE {\frac  {\partial ^{2}}{\partial z^{2}}}U(\lambda ,\mu ,z)=(\lambda ^{2}+\mu ^{2})U is:

U(\lambda ,\mu ,z)=A(\lambda ,\mu )e^{{{\sqrt  {\lambda ^{2}+\mu ^{2}}}\,z}}+B(\lambda ,\mu )e^{{-{\sqrt  {\lambda ^{2}+\mu ^{2}}}\,z}}\,

To satisfy the auxiliary boundedness condition, we must set A(\lambda ,\mu )=0\,.

U(\lambda ,\mu ,z)=B(\lambda ,\mu )e^{{-{\sqrt  {\lambda ^{2}+\mu ^{2}}}\,z}}\,

So the solution is the inverse transform.

  • u(x,y,z)=\int \!\!\!\int _{\Re }e^{{i\lambda x+i\mu y}}B(\lambda ,\mu )e^{{-{\sqrt  {\lambda ^{2}+\mu ^{2}}}\,z}}\,d\lambda d\mu \,

Use the boundary condition now.

u(x,y,0)=\int \!\!\!\int _{\Re }e^{{i\lambda x+i\mu y}}B(\lambda ,\mu )\,d\lambda d\mu =f(x,y)\,

B(\lambda ,\mu )={\frac  {1}{4\pi ^{2}}}\int \!\!\!\int _{\Re }e^{{-i\lambda x-i\mu y}}f(x,y)\,dxdy\,

Finally the solution is

u(x,y,z)=\int \!\!\!\int _{\Re }e^{{i\lambda x+i\mu y}}{\frac  {1}{4\pi ^{2}}}\left(\int \!\!\!\int _{\Re }e^{{-i\lambda x-i\mu y}}f(x,y)\,dxdy\right)e^{{-{\sqrt  {\lambda ^{2}+\mu ^{2}}}\,z}}\,d\lambda d\mu \,

Partial Differential Equations

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