#### 4.1.6 Semi-inﬁnite domain

4.1.6.5 [239] nonhomogeneous BC
4.1.6.6 [240] I.C. not zero
4.1.6.7 [241] nonhomogeneous BC
4.1.6.8 [242] nonhomogeneous BC
4.1.6.9 [243] nonhomogeneous B.C.
4.1.6.10 [244] Unit triangle I.C.
4.1.6.11 [245] I.C. not at $$t=0$$
4.1.6.13 [247] Practice exam problem

##### 4.1.6.1 [235] left end constant (general 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 $$u(x,0)=0$$

Mathematica

$\left \{\left \{u(x,t)\to \frac {A x E_{\frac {1}{2}}\left (\frac {x^2}{4 k t}\right )}{2 \sqrt {\pi } \sqrt {k t}}\right \}\right \}$

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 $$U\left (0,s\right ) =\mathcal {L}\left (u\left (0,t\right ) \right ) =\mathcal {L}\relax (A) =\frac {1}{s}A$$. 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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##### 4.1.6.2 [236] left end constant (special case)

problem number 236

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 4.1.6.1 on page 1130 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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##### 4.1.6.3 [237] Logan p. 76. Left end general function of time (general case)

problem number 237

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\relax (t) \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 $$x=0\,,u\left (0,t\right ) =f\relax (t)$$. Therefore $$U\left ( 0,s\right ) =\mathcal {L}\left (f\relax (t) \right ) =F\relax (s)$$. Hence at $$x=0$$ the above gives$F\relax (s) =c_{2}$ Therefore (2) becomes\begin {equation} U\left (x,s\right ) =F\relax (s) e^{-\sqrt {\frac {s}{k}}x}\tag {3} \end {equation} By convolution, the above becomes\begin {equation} u\left (x,t\right ) =f\relax (t) \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\relax (t) \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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##### 4.1.6.4 [238] Left end function of time (special case)

problem number 238

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\relax (t) \nonumber \\ u\left (x,0\right ) & =0\nonumber \end {align}

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

The general solution was solved in problem 4.1.6.3 on page 1142 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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##### 4.1.6.5 [239] nonhomogeneous BC

problem number 239

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

$\left \{\left \{u(x,t)\to \frac {x E_{\frac {1}{2}}\left (\frac {x^2}{4 k t}\right )}{2 \sqrt {\pi } \sqrt {k t}}\right \}\right \}$

Maple

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

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##### 4.1.6.6 [240] I.C. not zero

problem number 240

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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##### 4.1.6.7 [241] nonhomogeneous BC

problem number 241

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

$\left \{\left \{u(x,t)\to \mu \text {erf}\left (\frac {x}{2 \sqrt {k t}}\right )+\frac {\lambda x E_{\frac {1}{2}}\left (\frac {x^2}{4 k t}\right )}{2 \sqrt {\pi } \sqrt {k t}}\right \}\right \}$

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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##### 4.1.6.8 [242] nonhomogeneous BC

problem number 242

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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##### 4.1.6.9 [243] nonhomogeneous B.C.

problem number 243

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

$\left \{\left \{u(x,t)\to \frac {\left (2 k t+x^2\right ) \operatorname {Erfc}\left (\frac {x}{2 \sqrt {k t}}\right )-\frac {2 x \sqrt {k t} e^{-\frac {x^2}{4 k t}}}{\sqrt {\pi }}}{2 k}\right \}\right \}$

Maple

$u \left (x , t\right ) = -\frac {2 \sqrt {k}\, t x \,{\mathrm e}^{-\frac {x^{2}}{4 k t}}+\sqrt {\pi }\, \left (\erf \left (\frac {x}{2 \sqrt {k}\, \sqrt {t}}\right )-1\right ) \left (2 k \,t^{\frac {3}{2}}+\sqrt {t}\, x^{2}\right )}{2 \sqrt {\pi }\, k \sqrt {t}}$

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##### 4.1.6.10 [244] Unit triangle I.C.

problem number 244

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

$\left \{\left \{u(x,t)\to \begin {array}{cc} \{ & \begin {array}{cc} \frac {1}{2} \left (\frac {\text {erf}\left (\frac {| x-4| }{2 \sqrt {t}}\right ) (x-4)^2}{| 4-x| }+(x+2) \text {erf}\left (\frac {x+2}{2 \sqrt {t}}\right )-2 (x+3) \text {erf}\left (\frac {x+3}{2 \sqrt {t}}\right )+(x+4) \text {erf}\left (\frac {x+4}{2 \sqrt {t}}\right )-\frac {2 (x-3)^2 \text {erf}\left (\frac {| x-3| }{2 \sqrt {t}}\right )}{| 3-x| }+\frac {(x-2)^2 \text {erf}\left (\frac {| x-2| }{2 \sqrt {t}}\right )}{| 2-x| }+\frac {2 \left (e^{-\frac {(x-4)^2}{4 t}}-2 e^{-\frac {(x-3)^2}{4 t}}+e^{-\frac {(x-2)^2}{4 t}}+e^{-\frac {(x+2)^2}{4 t}}-2 e^{-\frac {(x+3)^2}{4 t}}+e^{-\frac {(x+4)^2}{4 t}}\right ) \sqrt {t}}{\sqrt {\pi }}\right ) & x>0 \\ \text {Indeterminate} & \text {True} \\\end {array} \\\end {array}\right \}\right \}$

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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##### 4.1.6.11 [245] I.C. not at $$t=0$$

problem number 245

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

$\left \{\left \{u(x,t)\to \begin {array}{cc} \{ & \begin {array}{cc} \frac {5 \left (e^{-\frac {x^2}{t-\text {t0}+1}} \left (\text {erf}\left (\frac {x+(-t+\text {t0}-1) \text {x0}}{\sqrt {t^2-2 \text {t0} t+t+(\text {t0}-1) \text {t0}}}\right )+1\right )+e^{-\frac {(x-2 \text {x0})^2}{t-\text {t0}+1}} \operatorname {Erfc}\left (\frac {x+(t-\text {t0}-1) \text {x0}}{\sqrt {(t-\text {t0}) (t-\text {t0}+1)}}\right )\right )}{\sqrt {t-\text {t0}+1}} & x>\text {x0} \\ \text {Indeterminate} & \text {True} \\\end {array} \\\end {array}\right \}\right \}$

Maple

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

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##### 4.1.6.12 [246] Diﬀusion with advection

problem number 246

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

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

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##### 4.1.6.13 [247] Practice exam problem

problem number 247

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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