Can this integral (of PDE solution) be solved?












1












$begingroup$


I study the heat equation from the book, Heat Conduction, Third Edition.



By using the method of separation of variables for the boundary condition of



$$BC1: frac{partial T}{partial x} Bigg|_{x=0}= 0$$
$$IC: T(x,t=0)=F(x)$$



the following solution is given,



$$T(x,t) = frac{1}{(4 pi alpha t)^{1/2}} int_{x'=0}^{infty} F(x')Bigglbraceexp left [ -frac{(x-x')^2}{4 alpha t}right ] + exp left [ -frac{(x+x')^2}{4 alpha t}right ] Biggrbrace dx'$$



Surprisingly, the section ends with this equation. I wonder if this integration can be generally solved to obtain a more tangible solution for the temperature.










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$endgroup$








  • 2




    $begingroup$
    In general, no. Even more so, you don’t gain anything from the evaluates form that you can’t see from the integral solution.
    $endgroup$
    – DaveNine
    Jan 26 at 0:10
















1












$begingroup$


I study the heat equation from the book, Heat Conduction, Third Edition.



By using the method of separation of variables for the boundary condition of



$$BC1: frac{partial T}{partial x} Bigg|_{x=0}= 0$$
$$IC: T(x,t=0)=F(x)$$



the following solution is given,



$$T(x,t) = frac{1}{(4 pi alpha t)^{1/2}} int_{x'=0}^{infty} F(x')Bigglbraceexp left [ -frac{(x-x')^2}{4 alpha t}right ] + exp left [ -frac{(x+x')^2}{4 alpha t}right ] Biggrbrace dx'$$



Surprisingly, the section ends with this equation. I wonder if this integration can be generally solved to obtain a more tangible solution for the temperature.










share|cite|improve this question









$endgroup$








  • 2




    $begingroup$
    In general, no. Even more so, you don’t gain anything from the evaluates form that you can’t see from the integral solution.
    $endgroup$
    – DaveNine
    Jan 26 at 0:10














1












1








1





$begingroup$


I study the heat equation from the book, Heat Conduction, Third Edition.



By using the method of separation of variables for the boundary condition of



$$BC1: frac{partial T}{partial x} Bigg|_{x=0}= 0$$
$$IC: T(x,t=0)=F(x)$$



the following solution is given,



$$T(x,t) = frac{1}{(4 pi alpha t)^{1/2}} int_{x'=0}^{infty} F(x')Bigglbraceexp left [ -frac{(x-x')^2}{4 alpha t}right ] + exp left [ -frac{(x+x')^2}{4 alpha t}right ] Biggrbrace dx'$$



Surprisingly, the section ends with this equation. I wonder if this integration can be generally solved to obtain a more tangible solution for the temperature.










share|cite|improve this question









$endgroup$




I study the heat equation from the book, Heat Conduction, Third Edition.



By using the method of separation of variables for the boundary condition of



$$BC1: frac{partial T}{partial x} Bigg|_{x=0}= 0$$
$$IC: T(x,t=0)=F(x)$$



the following solution is given,



$$T(x,t) = frac{1}{(4 pi alpha t)^{1/2}} int_{x'=0}^{infty} F(x')Bigglbraceexp left [ -frac{(x-x')^2}{4 alpha t}right ] + exp left [ -frac{(x+x')^2}{4 alpha t}right ] Biggrbrace dx'$$



Surprisingly, the section ends with this equation. I wonder if this integration can be generally solved to obtain a more tangible solution for the temperature.







calculus integration pde






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asked Jan 25 at 20:57









KimiaKimia

314




314








  • 2




    $begingroup$
    In general, no. Even more so, you don’t gain anything from the evaluates form that you can’t see from the integral solution.
    $endgroup$
    – DaveNine
    Jan 26 at 0:10














  • 2




    $begingroup$
    In general, no. Even more so, you don’t gain anything from the evaluates form that you can’t see from the integral solution.
    $endgroup$
    – DaveNine
    Jan 26 at 0:10








2




2




$begingroup$
In general, no. Even more so, you don’t gain anything from the evaluates form that you can’t see from the integral solution.
$endgroup$
– DaveNine
Jan 26 at 0:10




$begingroup$
In general, no. Even more so, you don’t gain anything from the evaluates form that you can’t see from the integral solution.
$endgroup$
– DaveNine
Jan 26 at 0:10










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