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Using Duhamel's Principle and the Heat Kernel to Solve an IVP

Heat equation on the Whole LineDiffusion Equation on the Half LineDuhamel's principle in constructing heat kerneldirect relationship between diffusion and wave equationSolve the initial value problem for this inhomogeneous heat equation.Heat equation with a positive coefficientUse Duhamel’s principle and the heat kernel to solve the initial value problemUse the heat kernel “magic rule” to solve the initial value problemUnderstanding Duhamel's principle, PDEProperties of the 1-d Heat Kernel

1

1

$begingroup$

I need to solve this:

$u_t(x,t)-ku_{xx}(x,t)=xte^{-t^2}$

By using Duhamel's Principle and the Heat Kernel.

So far this is what I've done:

$u_t(x,t)-ku_{xx}(x,t)=xte^{-t^2}$

where u(x,0)=0

$u(x,t)= int h(x,t-sigma|sigma)dsigma$

is the solution to the IVP

where h solves

$h_t(x,t|sigma)-kh_xx(x,t|sigma)=0$ for all t>0

subject to $h(x,0|sigma)=f(sigma)$

where I believe $f(sigma)=ysigma e^{-sigma^2}$

and then using the Heat Kernel Method

$h(x,t|sigma)=frac{1}{sqrt{4 pi k t}}int ysigma e^{-sigma^2} $$e^frac{-(x-y)^2}{4kt}dy$

where I substitute in z=x-y and therfroe dy=-dz to get

$h(x,t|sigma)=frac{1}{sqrt{4 pi k t}}int (z+x)sigma e^{-sigma^2} $$e^frac{-z^2}{4kt}dz$

But I'm not really sure where to go from here? I'd be grateful if anyone could suggest anything

ordinary-differential-equations heat-equation initial-value-problems

share|cite|improve this question

asked yesterday

kingking

405

$endgroup$

add a comment | 

1

1

$begingroup$

I need to solve this:

$u_t(x,t)-ku_{xx}(x,t)=xte^{-t^2}$

By using Duhamel's Principle and the Heat Kernel.

So far this is what I've done:

$u_t(x,t)-ku_{xx}(x,t)=xte^{-t^2}$

where u(x,0)=0

$u(x,t)= int h(x,t-sigma|sigma)dsigma$

is the solution to the IVP

where h solves

$h_t(x,t|sigma)-kh_xx(x,t|sigma)=0$ for all t>0

subject to $h(x,0|sigma)=f(sigma)$

where I believe $f(sigma)=ysigma e^{-sigma^2}$

and then using the Heat Kernel Method

$h(x,t|sigma)=frac{1}{sqrt{4 pi k t}}int ysigma e^{-sigma^2} $$e^frac{-(x-y)^2}{4kt}dy$

where I substitute in z=x-y and therfroe dy=-dz to get

$h(x,t|sigma)=frac{1}{sqrt{4 pi k t}}int (z+x)sigma e^{-sigma^2} $$e^frac{-z^2}{4kt}dz$

But I'm not really sure where to go from here? I'd be grateful if anyone could suggest anything

ordinary-differential-equations heat-equation initial-value-problems

share|cite|improve this question

asked yesterday

kingking

405

$endgroup$

add a comment | 

$begingroup$

I need to solve this:

$u_t(x,t)-ku_{xx}(x,t)=xte^{-t^2}$

By using Duhamel's Principle and the Heat Kernel.

So far this is what I've done:

$u_t(x,t)-ku_{xx}(x,t)=xte^{-t^2}$

where u(x,0)=0

$u(x,t)= int h(x,t-sigma|sigma)dsigma$

is the solution to the IVP

where h solves

$h_t(x,t|sigma)-kh_xx(x,t|sigma)=0$ for all t>0

subject to $h(x,0|sigma)=f(sigma)$

where I believe $f(sigma)=ysigma e^{-sigma^2}$

and then using the Heat Kernel Method

$h(x,t|sigma)=frac{1}{sqrt{4 pi k t}}int ysigma e^{-sigma^2} $$e^frac{-(x-y)^2}{4kt}dy$

where I substitute in z=x-y and therfroe dy=-dz to get

$h(x,t|sigma)=frac{1}{sqrt{4 pi k t}}int (z+x)sigma e^{-sigma^2} $$e^frac{-z^2}{4kt}dz$

But I'm not really sure where to go from here? I'd be grateful if anyone could suggest anything

ordinary-differential-equations heat-equation initial-value-problems

share|cite|improve this question

asked yesterday

kingking

405

$endgroup$

share|cite|improve this question

asked yesterday

kingking

405

share|cite|improve this question

asked yesterday

kingking

405

asked yesterday

kingking

405

asked yesterday

kingking

405