4.1.6 Semi-infinite domain

   4.1.6.1 [235] left end constant (general case)
   4.1.6.2 [236] left end constant (special case)
   4.1.6.3 [237] Logan p. 76. Left end general function of time (general case)
   4.1.6.4 [238] Left end function of time (special case)
   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 \(t=0\)
   4.1.6.12 [246] Diffusion with advection
   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 \(u(x,0)=0\)

pict
Figure 4.116:PDE specification

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*}

pict
Figure 4.117:PDE specification

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

pict
Figure 4.118:Source code

pict
Figure 4.119:Code for animation

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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 \(u(x,0)=0\)

pict
Figure 4.120:PDE specification

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 \(k=\frac {1}{10}\)

pict
Figure 4.121:PDE specification

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

pict
Figure 4.122:Source code

pict
Figure 4.123:Code for animation

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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 \(u(x,0)=0\)

pict
Figure 4.124:PDE specification

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

Added December 20, 2018.

From https://www.mapleprimes.com/posts/209970-Exact-Solutions-For-PDE-And-Boundary--Initial-Conditions-2018

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 \(t>|t_0|\).

pict
Figure 4.125:PDE specification

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 \(u(x,0)=\lambda \)

pict
Figure 4.126:PDE specification

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 \[ u(0,t)= 1 \]

pict
Figure 4.127:PDE specification

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.

pict
Figure 4.128:PDE specification

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 \[ \frac { \partial u}{\partial x}(0,t)= 0 \]

pict
Figure 4.129:PDE specification

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

Added December 20, 2018.

Taken from https://www.mapleprimes.com/posts/209970-Exact-Solutions-For-PDE-And-Boundary--Initial-Conditions-2018

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 \[ \frac { \partial u}{\partial x}(x_0,t)= 0 \]

pict
Figure 4.130:PDE specification

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

Added April 5, 2019.

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 \(u(x,0)=f(x)\)

pict
Figure 4.131:PDE specification

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

Added May 23, 2019.

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*}

pict
Figure 4.132:PDE specification

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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