The Heat Kernel to Solve an Initial Value ProblemHeat equation on the Whole LineInitial value problem for $y'...
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The Heat Kernel to Solve an Initial Value Problem
Heat equation on the Whole LineInitial value problem for $y' = alpha y - y^2$ with $y(0)=y_0 neq 0$ and $alpha > 0$DIfferential Equations Initial Value Problem Existence and UniquenessUse Duhamel’s principle and the heat kernel to solve the initial value problemUse the heat kernel “magic rule” to solve the initial value problemSolving the heat equation using the separation of variablesSolving $u_t=ku_{xx}$ for $ tge 0,-infty<x<infty$Solving the diffusion equation on the whole lineSolving the Heat/Diffusion Equation with Piecewise Continuous Initial ConditionUsing Duhamel's Principle and the Heat Kernel to Solve an IVP
$begingroup$
The Question I have is:
$$u_t(x, t) − ku_{xx}(x, t) = 0$$ $$∀x ∈ mathbb{R}, t > 0$$
subject to-
$u(x, 0) = x^2 − 3x − 1$ $∀x ∈ mathbb{R}.$
I started off with
$$u(x,t)=int_{mathbb{R}}e^frac{-(x-y)^2}{4kt}(y^2-3y-1)dy$$
and then I made $σ=y-x$ therefore I get
$$u(x,t)=frac{1}{sqrt{4 pi kt}}int_{mathbb{R}}e^frac{-σ^2}{4kt}((σ+x)^2-3(σ+x)-1)dσ$$
that simplifies to:
$$u(x,t)=frac{1}{sqrt{4 pi kt}}int_{mathbb{R}}e^frac{-σ^2}{4kt}(σ^2+2σx+x^2-3σ-3x-1)dσ$$
But from here I'm not sure how to integrate the answer and get any further. Any help would be greatly appreciated
ordinary-differential-equations heat-equation initial-value-problems
asked yesterday
kingking
405
$endgroup$
add a comment |
$begingroup$
The Question I have is:
$$u_t(x, t) − ku_{xx}(x, t) = 0$$ $$∀x ∈ mathbb{R}, t > 0$$
subject to-
$u(x, 0) = x^2 − 3x − 1$ $∀x ∈ mathbb{R}.$
I started off with
$$u(x,t)=int_{mathbb{R}}e^frac{-(x-y)^2}{4kt}(y^2-3y-1)dy$$
and then I made $σ=y-x$ therefore I get
$$u(x,t)=frac{1}{sqrt{4 pi kt}}int_{mathbb{R}}e^frac{-σ^2}{4kt}((σ+x)^2-3(σ+x)-1)dσ$$
that simplifies to:
$$u(x,t)=frac{1}{sqrt{4 pi kt}}int_{mathbb{R}}e^frac{-σ^2}{4kt}(σ^2+2σx+x^2-3σ-3x-1)dσ$$
But from here I'm not sure how to integrate the answer and get any further. Any help would be greatly appreciated
ordinary-differential-equations heat-equation initial-value-problems
asked yesterday
kingking
405
$endgroup$
add a comment |
$begingroup$
The Question I have is:
$$u_t(x, t) − ku_{xx}(x, t) = 0$$ $$∀x ∈ mathbb{R}, t > 0$$
subject to-
$u(x, 0) = x^2 − 3x − 1$ $∀x ∈ mathbb{R}.$
I started off with
$$u(x,t)=int_{mathbb{R}}e^frac{-(x-y)^2}{4kt}(y^2-3y-1)dy$$
and then I made $σ=y-x$ therefore I get
$$u(x,t)=frac{1}{sqrt{4 pi kt}}int_{mathbb{R}}e^frac{-σ^2}{4kt}((σ+x)^2-3(σ+x)-1)dσ$$
that simplifies to:
$$u(x,t)=frac{1}{sqrt{4 pi kt}}int_{mathbb{R}}e^frac{-σ^2}{4kt}(σ^2+2σx+x^2-3σ-3x-1)dσ$$
But from here I'm not sure how to integrate the answer and get any further. Any help would be greatly appreciated
ordinary-differential-equations heat-equation initial-value-problems
asked yesterday
kingking
405
$endgroup$
The Question I have is:
$$u_t(x, t) − ku_{xx}(x, t) = 0$$ $$∀x ∈ mathbb{R}, t > 0$$
subject to-
$u(x, 0) = x^2 − 3x − 1$ $∀x ∈ mathbb{R}.$
I started off with
$$u(x,t)=int_{mathbb{R}}e^frac{-(x-y)^2}{4kt}(y^2-3y-1)dy$$
and then I made $σ=y-x$ therefore I get
$$u(x,t)=frac{1}{sqrt{4 pi kt}}int_{mathbb{R}}e^frac{-σ^2}{4kt}((σ+x)^2-3(σ+x)-1)dσ$$
that simplifies to:
$$u(x,t)=frac{1}{sqrt{4 pi kt}}int_{mathbb{R}}e^frac{-σ^2}{4kt}(σ^2+2σx+x^2-3σ-3x-1)dσ$$
But from here I'm not sure how to integrate the answer and get any further. Any help would be greatly appreciated
ordinary-differential-equations heat-equation initial-value-problems
ordinary-differential-equations heat-equation initial-value-problems
asked yesterday
kingking
405
asked yesterday
kingking
405
edited yesterday
king
asked yesterday
kingking
405
asked yesterday
kingking
405
asked yesterday
kingking
405
405
add a comment |
add a comment |
1 Answer
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$begingroup$
Substitute
$$u(x,t)=v(t)+x^2-3x-1$$
For $v(t)$ we get ODE problem
$$v'(t)=2k,quad v(0)=0.$$
$Rightarrow$ $v(t)=2kt$
Solution of Initial Value Problem is
$$u(x,t)=2kt+x^2-3x-1$$
answered yesterday
Aleksas DomarkasAleksas Domarkas
1,43116
$endgroup$
add a comment |
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1 Answer
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1 Answer
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$begingroup$
Substitute
$$u(x,t)=v(t)+x^2-3x-1$$
For $v(t)$ we get ODE problem
$$v'(t)=2k,quad v(0)=0.$$
$Rightarrow$ $v(t)=2kt$
Solution of Initial Value Problem is
$$u(x,t)=2kt+x^2-3x-1$$
answered yesterday
Aleksas DomarkasAleksas Domarkas
1,43116
$endgroup$
add a comment |
$begingroup$
Substitute
$$u(x,t)=v(t)+x^2-3x-1$$
For $v(t)$ we get ODE problem
$$v'(t)=2k,quad v(0)=0.$$
$Rightarrow$ $v(t)=2kt$
Solution of Initial Value Problem is
$$u(x,t)=2kt+x^2-3x-1$$
answered yesterday
Aleksas DomarkasAleksas Domarkas
1,43116
$endgroup$
add a comment |
$begingroup$
Substitute
$$u(x,t)=v(t)+x^2-3x-1$$
For $v(t)$ we get ODE problem
$$v'(t)=2k,quad v(0)=0.$$
$Rightarrow$ $v(t)=2kt$
Solution of Initial Value Problem is
$$u(x,t)=2kt+x^2-3x-1$$
answered yesterday
Aleksas DomarkasAleksas Domarkas
1,43116
$endgroup$
Substitute
$$u(x,t)=v(t)+x^2-3x-1$$
For $v(t)$ we get ODE problem
$$v'(t)=2k,quad v(0)=0.$$
$Rightarrow$ $v(t)=2kt$
Solution of Initial Value Problem is
$$u(x,t)=2kt+x^2-3x-1$$
answered yesterday
Aleksas DomarkasAleksas Domarkas
1,43116
answered yesterday
Aleksas DomarkasAleksas Domarkas
1,43116
answered yesterday
Aleksas DomarkasAleksas Domarkas
1,43116
answered yesterday
Aleksas DomarkasAleksas Domarkas
1,43116
1,43116
add a comment |
add a comment |
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