3.1.6 Semi-inﬁnite domain

3.1.6.5 [238] nonhomogeneous BC
3.1.6.6 [239] I.C. not zero
3.1.6.7 [240] nonhomogeneous BC
3.1.6.8 [241] nonhomogeneous BC
3.1.6.9 [242] nonhomogeneous B.C.
3.1.6.10 [243] Unit triangle I.C.
3.1.6.11 [244] I.C. not at $$t=0$$
3.1.6.13 [246] Practice exam problem

3.1.6.1 [234] left end constant (general case)

problem number 234

Added July 6, 2019 Solve the heat equation for $$x>0,t>0$$ $u_t = k u_{xx}$ The boundary conditions are $$u(0,t)=A$$ and initial conditions $$u(x,0)=0$$

Mathematica

$\left \{\left \{u(x,t)\to \frac {x \text {Integrate}\left [\frac {A e^{-\frac {x^2}{4 k t-4 k K[2]}}}{(t-K[2])^{3/2}},\{K[2],0,t\},\text {Assumptions}\to \text {True}\right ]}{2 \sqrt {\pi } \sqrt {k}}\right \}\right \}$

Maple

$u \left (x , t\right ) = -A \erf \left (\frac {x}{2 \sqrt {k}\, \sqrt {t}}\right )+A$

Hand solution

Solving \begin {align} u_{t} & =ku_{xx\qquad }t>0,x>0\tag {1}\\ u\left ( 0,t\right ) & =A\nonumber \\ u\left ( x,0\right ) & =0\nonumber \end {align}

And $$u\left ( x,t\right ) <\infty$$ as $$x\rightarrow \infty$$. This means $$u\left ( x,t\right )$$ is bounded. This conditions is always needed to solve these problems.

Let $$U\left ( x,s\right )$$ be the Laplace transform of $$u\left ( x,t\right )$$. Deﬁned as $\mathcal {L}\left ( u,t\right ) =\int _{0}^{\infty }e^{-st}u\left ( x,t\right ) dt$ Applying Laplace transform to the original PDE (1) gives$sU\left ( x,s\right ) -u\left ( x,0\right ) =kU_{xx}\left ( x,s\right )$ But $$u\left ( x,0\right ) =0$$, therefore the above becomes$U_{xx}-\frac {s}{k}U=0$ The solution to this diﬀerential equation is$U\left ( x,s\right ) =c_{1}e^{\sqrt {\frac {s}{k}}x}+c_{2}e^{-\sqrt {\frac {s}{k}}x}$ Since $$u\left ( x,t\right )$$ is bounded in the limit as $$x\rightarrow \infty$$ and $$k>0$$, therefore it must be that $$c_{1}=0$$ to keep the solution bounded. The above simpliﬁes to\begin {equation} U\left ( x,s\right ) =c_{2}e^{-\sqrt {\frac {s}{k}}x}\tag {2} \end {equation} At $$x=0\,,u\left ( 0,t\right ) =A$$. Therefore $$U\left ( 0,s\right ) =\mathcal {L}\left ( u\left ( 0,t\right ) \right ) =\mathcal {L}\left ( A\right ) =\frac {1}{s}A$$. Hence at $$x=0$$ the above gives$\frac {1}{s}A=c_{2}$ Therefore (2) becomes\begin {equation} U\left ( x,s\right ) =\frac {A}{s}e^{-\sqrt {\frac {s}{k}}x}\tag {3} \end {equation} From tables, the inverse Laplace transform of the above is (since $$x>0,k>0$$)\begin {align*} u\left ( x,t\right ) & =A\operatorname {erfc}\left ( \frac {x}{2\sqrt {kt}}\right ) \\ & =A\left ( 1-\operatorname {erf}\left ( \frac {x}{2\sqrt {kt}}\right ) \right ) \end {align*}

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3.1.6.2 [235] left end constant (special case)

problem number 235

Added July 6, 2019 Solve the heat equation for $$x>0,t>0$$ $u_t = k u_{xx}$ The boundary conditions are $$u(0,t)=A$$ and initial conditions $$u(x,0)=0$$, using \begin {align*} A &=60\\ k &=\frac {1}{10} \end {align*}

Mathematica

$\{\{u(x,t)\to \fbox {\text {Indeterminate}\text { if }x\leq 0}\}\}$ It fail if assumption $$x>0$$ is given. A bug

Maple

$u \left (x , t\right ) = -60 \erf \left (\frac {\sqrt {10}\, x}{2 \sqrt {t}}\right )+60$

Hand solution

Solving on semi-inﬁnite domain \begin {align*} u_{t} & =ku_{xx}\qquad t>0,x>0\\ u\left ( 0,t\right ) & =A\\ u\left ( x,0\right ) & =0 \end {align*}

With $$A=60,k=\frac {1}{10}$$

The general problem above was solved in 3.1.6.1 on page 1120 and the solution is$u\left ( x,t\right ) =A\left ( 1-\operatorname {erf}\left ( \frac {x}{2\sqrt {kt}}\right ) \right )$ Substituting the speciﬁc values given above into this solution gives$u\left ( x,t\right ) =60\left ( 1-\operatorname {erf}\left ( \frac {x}{2\sqrt {\frac {t}{10}}}\right ) \right )$ Animation is below

Source code used for the above

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3.1.6.3 [236] Logan p. 76. Left end general function of time (general case)

problem number 236

This is problem at page 76 from David J Logan text book.

Solve the heat equation for $$x>0,t>0$$ $u_t = k u_{xx}$ The boundary conditions are $$u(0,t)=f(t)$$ and initial conditions $$u(x,0)=0$$

Mathematica

$\left \{\left \{u(x,t)\to \frac {x \text {Integrate}\left [\frac {f(z) e^{-\frac {x^2}{4 k t-4 k z}}}{(t-z)^{3/2}},\{z,0,t\},\text {Assumptions}\to \text {True}\right ]}{2 \sqrt {\pi } \sqrt {k}}\right \}\right \}$

Maple

$u \left (x , t\right ) = \frac {x \left (\int _{0}^{t}\frac {{\mathrm e}^{\frac {x^{2}}{4 \left (-t +\zeta \right ) k}} f \left (\zeta \right )}{\left (t -\zeta \right )^{\frac {3}{2}}}d \zeta \right )}{2 \sqrt {\pi }\, \sqrt {k}}$

Hand solution

Solving on semi-inﬁnite domain \begin {align} u_{t} & =ku_{xx}\qquad t>0,x>0\tag {1}\\ u\left ( 0,t\right ) & =f\left ( t\right ) \nonumber \\ u\left ( x,0\right ) & =0\nonumber \end {align}

With $$k>0$$ and $$u\left ( x,t\right ) <\infty$$ as $$x\rightarrow \infty$$. This means $$u\left ( x,t\right )$$ is bounded. This conditions is always needed to solve these problems.

Let $$U\left ( x,s\right )$$ be the Laplace transform of $$u\left ( x,t\right )$$. Deﬁned as $\mathcal {L}\left ( u,t\right ) =\int _{0}^{\infty }e^{-st}u\left ( x,t\right ) dt$ Applying Laplace transform to the original PDE (1) gives$sU\left ( x,s\right ) -u\left ( x,0\right ) =kU_{xx}\left ( x,s\right )$ But $$u\left ( x,0\right ) =0$$, therefore the above becomes$U_{xx}-\frac {s}{k}U=0$ The solution to this diﬀerential equation is$U\left ( x,s\right ) =c_{1}e^{\sqrt {\frac {s}{k}}x}+c_{2}e^{-\sqrt {\frac {s}{k}}x}$ Since $$u\left ( x,t\right )$$ is bounded in the limit as $$x\rightarrow \infty$$ and $$k>0$$, therefore it must be that $$c_{1}=0$$ to keep the solution bounded. The above simpliﬁes to\begin {equation} U\left ( x,s\right ) =c_{2}e^{-\sqrt {\frac {s}{k}}x}\tag {2} \end {equation} At $$x=0\,,u\left ( 0,t\right ) =f\left ( t\right )$$. Therefore $$U\left ( 0,s\right ) =\mathcal {L}\left ( f\left ( t\right ) \right ) =F\left ( s\right )$$. Hence at $$x=0$$ the above gives$F\left ( s\right ) =c_{2}$ Therefore (2) becomes\begin {equation} U\left ( x,s\right ) =F\left ( s\right ) e^{-\sqrt {\frac {s}{k}}x}\tag {3} \end {equation} By convolution, the above becomes\begin {equation} u\left ( x,t\right ) =f\left ( t\right ) \circledast G\left ( x,t\right ) \tag {4} \end {equation} Where $$G\left ( x,t\right )$$ is the inverse transform of $$e^{-\sqrt {\frac {s}{k}}x}$$ which is $$\frac {xe^{\frac {-x^{2}}{4kt}}}{2\sqrt {k\pi }t^{\frac {3}{2}}}$$. Hence (4) becomes\begin {align*} u\left ( x,t\right ) & =f\left ( t\right ) \circledast \frac {xe^{\frac {-x^{2}}{4kt}}}{2\sqrt {k\pi }t^{\frac {3}{2}}}\\ & =\frac {x}{2\sqrt {k\pi }}\int _{0}^{t}\frac {f\left ( \tau \right ) }{\left ( t-\tau \right ) ^{\frac {3}{2}}}e^{\frac {-x^{2}}{4k\left ( t-\tau \right ) }}d\tau \end {align*}

For $$k=1$$$u\left ( x,t\right ) =\frac {x}{2\sqrt {\pi }}\int _{0}^{t}\frac {f\left ( \tau \right ) }{\left ( t-\tau \right ) ^{\frac {3}{2}}}e^{\frac {-x^{2}}{4\left ( t-\tau \right ) }}d\tau$

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3.1.6.4 [237] Left end function of time (special case)

problem number 237

Added July 7, 2019 Solve the heat equation for $$x>0,t>0$$ $u_t = k u_{xx}$ The boundary conditions are $$u(0,t)=\sin (t)$$ and initial conditions $$u(x,0)=0$$ using $$k=\frac {1}{10}$$

Mathematica

$\left \{\left \{u(x,t)\to \sqrt {\frac {5}{2 \pi }} x \text {Integrate}\left [\frac {\sin (K[2]) e^{-\frac {5 x^2}{2 (t-K[2])}}}{(t-K[2])^{3/2}},\{K[2],0,t\},\text {Assumptions}\to \text {True}\right ]\right \}\right \}$

Maple

$u \left (x , t\right ) = -\frac {\sqrt {10}\, x \left (\int _{0}^{t}-\frac {{\mathrm e}^{-\frac {5 x^{2}}{2 \zeta }} \sin \left (t -\zeta \right )}{\zeta ^{\frac {3}{2}}}d \zeta \right )}{2 \sqrt {\pi }}$

Hand solution

Solving

\begin {align} u_{t} & =ku_{xx}\qquad t>0,x>0\tag {1}\\ u\left ( 0,t\right ) & =f\left ( t\right ) \nonumber \\ u\left ( x,0\right ) & =0\nonumber \end {align}

Using $$k=\frac {1}{10}$$ and $$f\left ( t\right ) =\sin \left ( t\right )$$.

The general solution was solved in problem 3.1.6.3 on page 1132 and the solution was found to be$u\left ( x,t\right ) =\frac {x}{2\sqrt {k\pi }}\int _{0}^{t}\frac {f\left ( \tau \right ) }{\left ( t-\tau \right ) ^{\frac {3}{2}}}e^{\frac {-x^{2}}{4k\left ( t-\tau \right ) }}d\tau$ Replacing the given values above, the solution becomes$u\left ( x,t\right ) =\sqrt {\frac {5}{2}}\frac {x}{\sqrt {\pi }}\int _{0}^{t}\frac {\sin \left ( \tau \right ) }{\left ( t-\tau \right ) ^{\frac {3}{2}}}e^{\frac {-5x^{2}}{2\left ( t-\tau \right ) }}d\tau$ We could also use the following form of the solution$u\left ( x,t\right ) =\sqrt {\frac {5}{2}}\frac {x}{\sqrt {\pi }}\int _{0}^{t}\frac {\sin \left ( t-\tau \right ) }{\tau ^{\frac {3}{2}}}e^{\frac {-5x^{2}}{2\tau }}d\tau$ Animation is below

Source code used for the above

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3.1.6.5 [238] nonhomogeneous BC

problem number 238

Solve the heat equation $\frac { \partial u}{\partial t} = k \frac { \partial ^2 u}{\partial x^2}$ For $$x>0$$ and $$t>0$$. The boundary conditions is $$u(0,t)=1$$ and And initial condition $$u(x,0)=0$$

Mathematica

$\left \{\left \{u(x,t)\to \frac {x \text {Integrate}\left [\frac {e^{-\frac {x^2}{4 k t-4 k K[2]}}}{(t-K[2])^{3/2}},\{K[2],0,t\},\text {Assumptions}\to \text {True}\right ]}{2 \sqrt {\pi } \sqrt {k}}\right \}\right \}$

Maple

$u \left (x , t\right ) = \mathrm {erfc}\left (\frac {x}{2 \sqrt {k t}}\right )$

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3.1.6.6 [239] I.C. not zero

problem number 239

Solve the heat equation for $$u(x,t)$$ $\frac { \partial u}{\partial t}= \frac {1}{4} \frac { \partial ^2 u}{\partial x^2}$ With initial condition $u(x,t_0)= 10;$ And boundary conditions $u(-x_0,t) = 0$ For $$x>|x_0|$$ and $$t>|t_0|$$.

Mathematica

$\left \{\left \{u(x,t)\to \begin {array}{cc} \{ & \begin {array}{cc} 10 \text {erf}\left (\frac {x+\text {x0}}{\sqrt {t-\text {t0}}}\right ) & x+\text {x0}>0 \\ \text {Indeterminate} & \text {True} \\\end {array} \\\end {array}\right \}\right \}$ due to IC/BC not zero

Maple

$u \left (x , t\right ) = 10 \erf \left (\frac {x +\mathit {x0}}{\sqrt {t -\mathit {t0}}}\right )$

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3.1.6.7 [240] nonhomogeneous BC

problem number 240

Solve the heat equation

$\frac { \partial u}{\partial t} = k \frac { \partial ^2 u}{\partial x^2}$ For $$x>0$$ and $$t>0$$. The boundary conditions is $$u(0,t)=\mu$$ and And initial condition $$u(x,0)=\lambda$$

Mathematica

$\left \{\left \{u(x,t)\to \frac {x \sqrt {k t} \text {Integrate}\left [\frac {\lambda e^{-\frac {x^2}{4 k t-4 k K[2]}}}{(t-K[2])^{3/2}},\{K[2],0,t\},\text {Assumptions}\to \text {True}\right ]+\sqrt {k} \text {Integrate}\left [\mu \left (e^{-\frac {(x-K[1])^2}{4 k t}}-e^{-\frac {(K[1]+x)^2}{4 k t}}\right ),\{K[1],0,\infty \},\text {Assumptions}\to \text {True}\right ]}{2 \sqrt {\pi } k \sqrt {t}}\right \}\right \}$

Maple

$u \left (x , t\right ) = \mu +\left (\lambda -\mu \right ) \mathrm {erfc}\left (\frac {x}{2 \sqrt {k t}}\right )$

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3.1.6.8 [241] nonhomogeneous BC

problem number 241

From Mathematica DSolve help pages. Solve the heat equation for $$u(x,t)$$ on half the line $$x>0$$ and $$t>0$$ $\frac { \partial u}{\partial t}= \frac { \partial ^2 u}{\partial x^2}$ With initial condition $u(x,0)= \cos x$ And boundary conditions $u(0,t)= 1$

Mathematica

$\left \{\left \{u(x,t)\to \begin {array}{cc} \{ & \begin {array}{cc} \frac {\text {Integrate}\left [\left (e^{-\frac {(x-K[1])^2}{4 t}}-e^{-\frac {(x+K[1])^2}{4 t}}\right ) \cos (K[1]),\{K[1],0,\infty \},\text {Assumptions}\to \text {True}\right ]+\sqrt {t} x \text {Integrate}\left [\frac {e^{-\frac {x^2}{4 (t-K[2])}}}{(t-K[2])^{3/2}},\{K[2],0,t\},\text {Assumptions}\to \text {True}\right ]}{2 \sqrt {\pi } \sqrt {t}} & x>0 \\ \text {Indeterminate} & \text {True} \\\end {array} \\\end {array}\right \}\right \}$

Maple

$u \left (x , t\right ) = -\frac {\erf \left (\frac {2 i t -x}{2 \sqrt {t}}\right ) {\mathrm e}^{-i x -t}}{2}+\frac {\erf \left (\frac {2 i t +x}{2 \sqrt {t}}\right ) {\mathrm e}^{i x -t}}{2}-\erf \left (\frac {x}{2 \sqrt {t}}\right )+1$

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3.1.6.9 [242] nonhomogeneous B.C.

problem number 242

Solve the heat equation for $$u(x,t)$$ on half the line $$x>0$$ and $$t>0$$ $u_t = k u_{xx}$ With initial condition $u(x,0)=0$ And boundary conditions $$u(0,t)=t$$. Solution is bounded at inﬁnity.

Mathematica

$\left \{\left \{u(x,t)\to \frac {x \text {Integrate}\left [\frac {K[2] e^{-\frac {x^2}{4 k t-4 k K[2]}}}{(t-K[2])^{3/2}},\{K[2],0,t\},\text {Assumptions}\to \text {True}\right ]}{2 \sqrt {\pi } \sqrt {k}}\right \}\right \}$

Maple

$u \left (x , t\right ) = -\frac {\sqrt {k}\, \sqrt {t}\, x \,{\mathrm e}^{-\frac {x^{2}}{4 k t}}+\sqrt {\pi }\, \left (\erf \left (\frac {x}{2 \sqrt {k}\, \sqrt {t}}\right )-1\right ) \left (k t +\frac {x^{2}}{2}\right )}{\sqrt {\pi }\, k}$

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3.1.6.10 [243] Unit triangle I.C.

problem number 243

From Mathematica DSolve help pages. Solve the heat equation for $$u(x,t)$$ on half the line $$x>0$$ and $$t>0$$ $u_t= u_{xx}$ With initial condition $u(x,0)= \text {UnitTriagle[x-3]}$ And boundary conditions $\frac { \partial u}{\partial x}(0,t)= 0$

Mathematica

$\left \{\left \{u(x,t)\to \begin {array}{cc} \{ & \begin {array}{cc} \Lambda (3-x) & t=0 \\ \frac {2 \int _0^{\infty }\frac {4 e^{-t K[1]^2} \cos (3 K[1]) \cos (x K[1]) \sin ^2\left (\frac {K[1]}{2}\right )}{K[1]^2}dK[1]}{\pi } & t>0 \\ \text {Indeterminate} & \text {True} \\\end {array} \\\end {array}\right \}\right \}$

Maple

$u \left (x , t\right ) = \frac {t \,{\mathrm e}^{-\frac {\left (x -4\right )^{2}}{4 t}}-2 t \,{\mathrm e}^{-\frac {\left (x -3\right )^{2}}{4 t}}+t \,{\mathrm e}^{-\frac {\left (x -2\right )^{2}}{4 t}}+t \,{\mathrm e}^{-\frac {\left (x +2\right )^{2}}{4 t}}-2 t \,{\mathrm e}^{-\frac {\left (x +3\right )^{2}}{4 t}}+t \,{\mathrm e}^{-\frac {\left (x +4\right )^{2}}{4 t}}+\frac {\left (\left (x -4\right ) \erf \left (\frac {x -4}{2 \sqrt {t}}\right )+\left (-2 x +6\right ) \erf \left (\frac {x -3}{2 \sqrt {t}}\right )+\left (x -2\right ) \erf \left (\frac {x -2}{2 \sqrt {t}}\right )+\left (x +2\right ) \erf \left (\frac {x +2}{2 \sqrt {t}}\right )+\left (-2 x -6\right ) \erf \left (\frac {x +3}{2 \sqrt {t}}\right )+\left (x +4\right ) \erf \left (\frac {x +4}{2 \sqrt {t}}\right )\right ) \sqrt {\pi }\, \sqrt {t}}{2}}{\sqrt {\pi }\, \sqrt {t}}$

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3.1.6.11 [244] I.C. not at $$t=0$$

problem number 244

Solve for $$u(x,t)$$ for $$t>0,x>0$$ $u_t= \frac {1}{4} u_{xx}$ With initial condition $u(x,t_0)= 10 e^{-x^2}$ And boundary conditions $\frac { \partial u}{\partial x}(x_0,t)= 0$

Mathematica

$\left \{\left \{u(x,t)\to \begin {array}{cc} \{ & \begin {array}{cc} 10 e^{-x^2} & t-\text {t0}=0 \\ \frac {2 \int _0^{\infty }\frac {5}{2} e^{-\frac {1}{4} K[1] (4 i \text {x0}+(t-\text {t0}+1) K[1])} \sqrt {\pi } \cos ((x-\text {x0}) K[1]) \left (\operatorname {Erfc}\left (\text {x0}-\frac {1}{2} i K[1]\right )+e^{2 i \text {x0} K[1]} \operatorname {Erfc}\left (\text {x0}+\frac {1}{2} i K[1]\right )\right )dK[1]}{\pi } & t-\text {t0}>0 \\ \text {Indeterminate} & \text {True} \\\end {array} \\\end {array}\right \}\right \}$

Maple

$u \left (x , t\right ) = -\frac {5 \left (-\erf \left (\frac {x +\left (-t +\mathit {t0} -1\right ) \mathit {x0}}{\sqrt {t -\mathit {t0} +1}\, \sqrt {t -\mathit {t0}}}\right )+\left (\erf \left (\frac {-\mathit {t0} \mathit {x0} +x +\left (t -1\right ) \mathit {x0}}{\sqrt {t -\mathit {t0} +1}\, \sqrt {t -\mathit {t0}}}\right )-1\right ) {\mathrm e}^{\frac {4 \left (-x +\mathit {x0} \right ) \mathit {x0}}{-t +\mathit {t0} -1}}-1\right ) {\mathrm e}^{\frac {x^{2}}{-t +\mathit {t0} -1}}}{\sqrt {t -\mathit {t0} +1}}$

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problem number 245

Solve for $$u(x,t)$$ in $u_t = u_{xx} - u_x$

For $$t>0,x>0$$. With boundary conditions $$u(0,t)=0$$ and intitial conditions $$u(x,0)=f(x)$$

Mathematica

$\left \{\left \{u(x,t)\to e^{\frac {x}{2}-\frac {t}{4}} \left (\begin {array}{cc} \{ & \begin {array}{cc} \frac {\text {Integrate}\left [e^{-\frac {K[1]}{2}} \left (e^{-\frac {(x-K[1])^2}{4 t}}-e^{-\frac {(x+K[1])^2}{4 t}}\right ) f(K[1]),\{K[1],0,\infty \},\text {Assumptions}\to \text {True}\right ]}{2 \sqrt {\pi } \sqrt {t}} & x>0 \\ \text {Indeterminate} & \text {True} \\\end {array} \\\end {array}\right )\right \}\right \}$

Maple

$u \left (x , t\right ) = \left (\mathcal {L}^{-1}\left (\frac {\int \frac {\sqrt {{\mathrm e}^{\sqrt {4 s +1}\, x}}\, f (x )}{\sqrt {{\mathrm e}^{x}}}d x}{\sqrt {4 s +1}\, \sqrt {{\mathrm e}^{\sqrt {4 s +1}\, x}}}, s , t\right )-\mathcal {L}^{-1}\left (\frac {\left (\int \frac {f (x )}{\sqrt {{\mathrm e}^{x}}\, \sqrt {{\mathrm e}^{\sqrt {4 s +1}\, x}}}d x \right ) \sqrt {{\mathrm e}^{\sqrt {4 s +1}\, x}}}{\sqrt {4 s +1}}, s , t\right )+\mathcal {L}^{-1}\left (\frac {\int _{}^{0}\frac {f \left (\mathit {\_a} \right )}{\sqrt {{\mathrm e}^{\mathit {\_a}}}\, \sqrt {{\mathrm e}^{\sqrt {4 s +1}\, \mathit {\_a}}}}d\mathit {\_a}}{\sqrt {4 s +1}\, \sqrt {{\mathrm e}^{\sqrt {4 s +1}\, x}}}, s , t\right )-\mathcal {L}^{-1}\left (\frac {\int _{}^{0}\frac {\sqrt {{\mathrm e}^{\sqrt {4 s +1}\, \mathit {\_a}}}\, f \left (\mathit {\_a} \right )}{\sqrt {{\mathrm e}^{\mathit {\_a}}}}d\mathit {\_a}}{\sqrt {4 s +1}\, \sqrt {{\mathrm e}^{\sqrt {4 s +1}\, x}}}, s , t\right )\right ) {\mathrm e}^{\frac {x}{2}}$

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3.1.6.13 [246] Practice exam problem

problem number 246

From Math 5587 midterm I, Fall 2016, practice exam, problem 13.

Solve for $$u(x,t)$$ with IC $$u(x,0)=x^2+1$$ and BC $$u_t(0,t)=1$$ for $$x>0,t>0$$ \begin {align*} u_{t} = u_{xx} \end {align*}

Mathematica

$\left \{\left \{u(x,t)\to \begin {array}{cc} \{ & \begin {array}{cc} \frac {\text {Integrate}\left [\left (e^{-\frac {(x-K[1])^2}{4 t}}-e^{-\frac {(x+K[1])^2}{4 t}}\right ) \left (K[1]^2+1\right ),\{K[1],0,\infty \},\text {Assumptions}\to \text {True}\right ]+\sqrt {t} x \text {Integrate}\left [\frac {e^{-\frac {x^2}{4 (t-K[2])}}}{(t-K[2])^{3/2}},\{K[2],0,t\},\text {Assumptions}\to \text {True}\right ]}{2 \sqrt {\pi } \sqrt {t}} & x>0 \\ \text {Indeterminate} & \text {True} \\\end {array} \\\end {array}\right \}\right \}$

Maple

$u \left (x , t\right ) = x^{2}+2 t -2 \mathcal {L}^{-1}\left (\frac {{\mathrm e}^{-\sqrt {s}\, x}}{s^{2}}, s , t\right )+1$

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