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Taken from Mathematica Symbolic PDE document
Solve for \(u(x,t)\) \[ \frac { \partial u}{\partial t} + \frac { \partial u}{\partial x} = 0 \]
Mathematica ✓
ClearAll[u, x, t]; pde = D[u[x, t], {t}] + D[u[x, t], {x}] == 0; sol = AbsoluteTiming[TimeConstrained[DSolve[pde, u[x, t], {x, t}], 60*10]];
\[ \left \{\left \{u(x,t)\to c_1(t-x)\right \}\right \} \]
Maple ✓
interface(showassumed=0); u:='u';x:='x';t:='t'; pde := diff(u(x, t), t) + diff(u(x, t),x) =0; cpu_time := timelimit(60*10,CodeTools[Usage](assign('sol',pdsolve(pde,u(x,t))),output='realtime'));
\[ u \left ( x,t \right ) ={\it \_F1} \left ( -x+t \right ) \]
Hand solution
\begin {equation} u_{t}+u_{x}=0\tag {1} \end {equation}
Let \(u\equiv u\left ( x\left ( t\right ) ,t\right ) \). Then \begin {equation} \frac {du}{dt}=\frac {\partial u}{\partial x}\frac {dx}{dt}+\frac {\partial u}{\partial t}\tag {2} \end {equation}
Comparing (1) to (2) then we see that
\begin {align} \frac {du}{dt} & =0\tag {3}\\ \frac {dx}{dt} & =1\tag {4} \end {align}
(3) says that \(u\) is constant. Since no initial conditions are given, let \(u=F\left ( x\left ( 0\right ) \right ) \) where \(F\) is arbitrary function. To find \(x\left ( 0\right ) \) we solve (4). The solution to (4) is \(x=x\left ( 0\right ) +t\). Hence \(x\left ( 0\right ) =x-t\). Therefore
\[ u\left ( x,t\right ) =F\left ( x-t\right ) \]
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Taken from Mathematica help pages
Solve for \(u(x,y)\) \[ 3 u_x + 5 u_y = x \]
Mathematica ✓
ClearAll[u, x, y]; sol = AbsoluteTiming[TimeConstrained[DSolve[3*D[u[x, y], x] + 5*D[u[x, y], y] == x, u[x, y], {x, y}], 60*10]];
\[ \left \{\left \{u(x,y)\to \frac {1}{6} \left (6 c_1\left (\frac {1}{3} (3 y-5 x)\right )+x^2\right )\right \}\right \} \]
Maple ✓
interface(showassumed=0); u:='u';x:='x';y:='y'; pde:=3*diff(u(x, y), x) + 5*diff(u(x, y), y) = x; cpu_time := timelimit(60*10,CodeTools[Usage](assign('sol',pdsolve(pde,u(x,y))),output='realtime'));
\[ u \left ( x,y \right ) =1/6\,{x}^{2}+{\it \_F1} \left ( -5/3\,x+y \right ) \]
Hand solution
Solve\begin {align} 3u_{x}+5u_{y} & =x\nonumber \\ u_{x}+\frac {5}{3}u_{y} & =\frac {x}{3}\tag {1} \end {align}
Solution
Let \(u=u\left ( y\left ( x\right ) ,x\right ) \). Then \begin {equation} \frac {du}{dx}=\frac {\partial u}{\partial y}\frac {dy}{dx}+\frac {\partial u}{\partial x}\tag {2} \end {equation} Comparing (1),(2) shows that \begin {align} \frac {du}{dx} & =\frac {x}{3}\tag {3}\\ \frac {dy}{dx} & =\frac {5}{3}\tag {4} \end {align}
Solving (3) gives\begin {align*} u & =\frac {x^{2}}{6}+C_{1}\\ C_{1} & =u-\frac {x^{2}}{6} \end {align*}
From (4)\begin {align*} y & =\frac {5}{3}x+C_{2}\\ C_{2} & =y-\frac {5}{3}x \end {align*}
Let \(C_{1}=F\left ( C_{2}\right ) \) where \(F\) is arbutrary function. This gives\begin {align*} u-\frac {x^{2}}{6} & =F\left ( y-\frac {5}{3}x\right ) \\ u\left ( x,y\right ) & =F\left ( y-\frac {5}{3}x\right ) +\frac {x^{2}}{6} \end {align*}
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Taken from Mathematica help pages
Solve for \(u(x,y)\) \[ x \frac { \partial u}{\partial x} + y \frac { \partial u}{\partial y} = -4 x y u(x,y) \] with initial value \(u(x,0)=e^{-x^2}\)
Mathematica ✓
ClearAll[u, x, y]; pde = x*D[u[x, y], y] + y*D[u[x, y], x] == -4*x*y*u[x, y]; ic = u[x, 0] == Exp[-x^2]; sol = AbsoluteTiming[TimeConstrained[DSolve[{pde, ic}, u[x, y], {x, y}], 60*10]];
\[ \left \{\left \{u(x,y)\to e^{-x^2-y^2}\right \}\right \} \]
Maple ✓
interface(showassumed=0); u:='u';x:='x';y:='y'; pde := x*diff(u(x, y), y) + y*diff(u(x, y), x) = -4*x*y*u(x, y); ic := u(x, 0) = exp(-x^2); cpu_time := timelimit(60*10,CodeTools[Usage](assign('sol',pdsolve([pde, ic], u(x, y))),output='realtime'));
\[ u \left ( x,y \right ) ={{\rm e}^{-{x}^{2}-{y}^{2}}} \]
Hand solution
Solve \[ xu_{y}+yu_{x}=-4xyu \] with \(u\left ( x,0\right ) =e^{-x^{2}}\).
Solution
Let \(u\equiv u\left ( x\left ( y\right ) ,y\right ) \). We’ve taken \(y\) as the independent variable for \(x\left ( y\right ) \) here, since the initial conditions has \(y\left ( 0\right ) \) in it. The PDE can be written as\begin {equation} u_{y}+\frac {y}{x}u_{x}=-4yu\tag {1} \end {equation} Then\begin {equation} \frac {du}{dy}=\frac {\partial u}{\partial x}\frac {dx}{dy}+\frac {\partial u}{\partial y}\tag {2} \end {equation} Comparing (1),(2) shows that \begin {align} \frac {du}{dy} & =-4yu\tag {3}\\ \frac {dx}{dy} & =\frac {y}{x}\tag {4} \end {align}
Solving (3) gives\begin {align} \ln \left \vert u\right \vert & =-\frac {4y^{2}}{2}+C_{1}\nonumber \\ u & =C_{1}e^{-2y^{2}}\tag {5} \end {align}
At \(y=0\), using initial conditions the above becomes\[ e^{-x\left ( 0\right ) ^{2}}=C_{1}\] (5) becomes\begin {align} u & =e^{-x\left ( 0\right ) ^{2}}e^{-2y^{2}}\nonumber \\ & =e^{-x\left ( 0\right ) ^{2}-2y^{2}}\tag {5A} \end {align}
All what is left is to find \(x\left ( 0\right ) \) to finish the solution. From (4)\begin {equation} \frac {x^{2}}{2}=\frac {y^{2}}{2}+C_{2}\tag {6} \end {equation} At \(y=0\)\[ \frac {x\left ( 0\right ) ^{2}}{2}=C_{2}\] Hence (6) becomes\begin {align*} \frac {x^{2}}{2} & =\frac {y^{2}}{2}+\frac {x\left ( 0\right ) ^{2}}{2}\\ x\left ( 0\right ) ^{2} & =x^{2}-y^{2} \end {align*}
Substituting the above in (5A) gives\begin {align*} u\left ( x\left ( y\right ) ,x\right ) & =e^{-\left ( x^{2}-y^{2}\right ) -2y^{2}}\\ & =e^{-x^{2}-y^{2}} \end {align*}
The following is a plot of the above solution showing the initial conditions are red line
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Taken from Mathematica help pages
Solve for \(u(x,t)\) \[ \frac { \partial u}{\partial t} + \frac { \partial u}{\partial x} = 0 \] with initial value \(u(x,0)=\sin x\) and boundary value \(u(0,t)=0\)
Mathematica ✓
ClearAll[u, x, t]; pde = D[u[x, t], t] + D[u[x, t], x] == 0; bc = u[0, t] == 0; ic = u[x, 0] == Sin[x]; sol = AbsoluteTiming[TimeConstrained[DSolve[{pde, ic, bc}, u[x, t], {x, t}], 60*10]];
\[ \{\{u(x,t)\to (\theta (t-x)-1) \sin (t-x)\}\} \]
Maple ✓
u:='u';x:='x';t:='t'; pde:=diff(u(x,t),t)+diff(u(x,t),x)=0; bc:=u(0,t)=0; ic:=u(x,0)=sin(x); cpu_time := timelimit(60*10,CodeTools[Usage](assign('sol',pdsolve([pde,ic,bc],u(x,t)) assuming x>0),output='realtime'));
\[ u \left ( x,t \right ) =-\sin \left ( -x+t \right ) {\it Heaviside} \left ( -t+x \right ) \]
Hand solution
Since initial and boundary conditions are given, the Laplace transform method will be used to solve this PDE. Let \(U\left ( x,s\right ) \) be the Laplace transform of \(u\left ( x,t\right ) \). Applying Laplace transform to the PDE gives\begin {align*} sU-u\left ( x,0\right ) +\frac {dU}{dx} & =0\\ \frac {dU}{dx}+sU & =\sin x \end {align*}
Integrating factor is \(\mu =e^{\int sdx}=e^{sx}\). Multiplying the above by \(\mu \) gives\[ \frac {d}{dx}\left ( Ue^{sx}\right ) =e^{sx}\sin x \] Integrating \begin {align*} Ue^{sx} & =\int e^{sx}\sin xdx+C\\ & =\frac {e^{sx}\left ( s\sin x-\cos x\right ) }{1+s^{2}}+C\\ U\left ( x,s\right ) & =\frac {s\sin x-\cos x}{1+s^{2}}+Ce^{-sx} \end {align*}
Applying boundary conditions \(U\left ( 0,s\right ) =0\) gives\begin {align*} 0 & =\frac {-1}{1+s^{2}}+C\\ C & =\frac {1}{1+s^{2}} \end {align*}
Hence \begin {align*} U\left ( x,s\right ) & =\frac {s\sin x-\cos x}{1+s^{2}}+\frac {e^{-sx}}{1+s^{2}}\\ & =\frac {s\sin x}{1+s^{2}}-\frac {\cos x}{1+s^{2}}+\frac {e^{-sx}}{1+s^{2}} \end {align*}
Applying inverse Laplace transform gives\begin {align*} u\left ( x,t\right ) & =\cos t\sin x-\cos x\sin t+\operatorname {Heaviside}\left ( t-x\right ) \sin \left ( t-x\right ) \\ & =-\sin \left ( t-x\right ) +\operatorname {Heaviside}\left ( t-x\right ) \sin \left ( t-x\right ) \\ & =\left ( \operatorname {Heaviside}\left ( t-x\right ) -1\right ) \sin \left ( t-x\right ) \end {align*}
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Taken from Mathematica help pages
Solve for \(u(x,t)\) \[ \frac { \partial u}{\partial t} + c \frac { \partial u}{\partial x} = 0 \] With initial conditions \(u(x,0)=e^{-x^2}\)
Mathematica ✓
ClearAll[u, x, t, c]; ic = u[x, 0] == Exp[-x^2]; pde = D[u[x, t], {t}] + c*D[u[x, t], {x}] == 0; sol = AbsoluteTiming[TimeConstrained[DSolve[{pde, ic}, u[x, t], {x, t}], 60*10]];
\[ \left \{\left \{u(x,t)\to e^{-(x-c t)^2}\right \}\right \} \]
Maple ✓
interface(showassumed=0); u:='u';x:='x';t:='t';c:='c'; pde := diff(u(x, t), t) + c* diff(u(x, t),x) =0; ic:=u(x,0)=exp(-x^2); cpu_time := timelimit(60*10,CodeTools[Usage](assign('sol',pdsolve([pde,ic],u(x,t))),output='realtime'));
\[ u \left ( x,t \right ) ={{\rm e}^{- \left ( tc-x \right ) ^{2}}} \]
Hand solution
Solve \begin {equation} u_{t}+cu_{x}=0\tag {1} \end {equation} with initial conditions \(u\left ( x,0\right ) =e^{-x^{2}}\).
Solution
Let \(u=u\left ( x\left ( t\right ) ,t\right ) \). Then \begin {equation} \frac {du}{dt}=\frac {\partial u}{\partial x}\frac {dx}{dt}+\frac {\partial u}{\partial t}\tag {2} \end {equation} Comparing (1),(2) shows that \begin {align} \frac {du}{dt} & =0\tag {3}\\ \frac {dx}{dt} & =c\tag {4} \end {align}
Solving (3) gives\begin {align} u & =u\left ( x\left ( 0\right ) \right ) \nonumber \\ & =e^{-x\left ( 0\right ) ^{2}}\tag {5} \end {align}
We need to find \(x\left ( 0\right ) \). From (4)\begin {align*} x & =ct+x\left ( 0\right ) \\ x\left ( 0\right ) & =x-ct \end {align*}
Then (5) becomes\[ u\left ( x\left ( t\right ) ,t\right ) =e^{-\left ( x-ct\right ) ^{2}}\]
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Added Nov 25, 2018.
Problem 12.2.2 from Richard Haberman applied partial differential equations book, 5th edition
Solve for \(u(x,t)\) \[ \frac { \partial \omega }{\partial t} -3 \frac { \partial \omega }{\partial x} = 0 \] With initial conditions \(\omega (x,0)=\cos x\).
See my HW 12, Math 322, UW Madison.
Mathematica ✓
ClearAll[x, t, w]; pde = D[w[x, t], t] - 3*D[w[x, t], x] == 0; ic = w[x, 0] == Cos[x]; sol = AbsoluteTiming[TimeConstrained[DSolve[{pde, ic}, w[x, t], {x, t}], 60*10]];
\[ \{\{w(x,t)\to \cos (3 t+x)\}\} \]
Maple ✓
x:='x'; t:='t'; w:='w'; pde:=diff(w(x,t),t)-3*diff(w(x,t),x)=0; ic:=w(x,0)=cos(x); cpu_time := timelimit(60*10,CodeTools[Usage](assign('sol',pdsolve([pde,ic],w(x,t))),output='realtime'));
\[ w \left ( x,t \right ) =\cos \left ( x+3\,t \right ) \]
Hand solution
Solve \begin {equation} w_{t}-3w_{x}=0\tag {1} \end {equation} With I.C. \(w\left ( x,0\right ) =\cos x\)
Solution
Let \(w=w\left ( x\left ( t\right ) ,t\right ) \). Then \begin {equation} \frac {dw}{dt}=\frac {\partial w}{\partial x}\frac {dx}{dt}+\frac {\partial w}{\partial t}\tag {2} \end {equation} Comparing (1),(2) shows that \begin {align} \frac {dw}{dt} & =0\tag {3}\\ \frac {dx}{dt} & =-3\tag {4} \end {align}
Solving (3) gives\begin {align} w & =w\left ( x\left ( 0\right ) \right ) \nonumber \\ & =\cos \left ( x\left ( 0\right ) \right ) \tag {5} \end {align}
We need to find \(x\left ( 0\right ) \). From (4)\begin {align*} x & =-3t+x\left ( 0\right ) \\ x\left ( 0\right ) & =x+3t \end {align*}
Hence (5) becomes
\[ w\left ( x\left ( t\right ) ,t\right ) =\cos \left ( x+3t\right ) \]
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Added Nov 25, 2018.
Problem 12.2.4 from Richard Haberman applied partial differential equations book, 5th edition
Solve for \(u(x,t)\) \[ \omega _t +c \omega _x = 0 \] With \(c>0\). For \(x>0,t>0\) if \(\omega (x,0)=f(x)\) and \(\omega (0,t)=h(t)\).
See my HW 12, Math 322, UW Madison.
Mathematica ✗
ClearAll[x, t, w, f, h]; pde = D[w[x, t], t] + c*D[w[x, t], x] == 0; ic = w[x, 0] == f[x]; bc = w[0, t] == h[t]; sol = AbsoluteTiming[TimeConstrained[DSolve[{pde, ic, bc}, w[x, t], {x, t}, Assumptions -> c > 0 && x > 0 && t > 0], 60*10]];
\[ \text {Failed} \]
Maple ✓
x:='x'; t:='t'; w:='w';f:='f';h:='h';c:='c'; pde:=diff(w(x,t),t)+c*diff(w(x,t),x)=0; ic:=w(x,0)=f(x); bc:=w(0,t)=h(t); cpu_time := timelimit(60*10,CodeTools[Usage](assign('sol',pdsolve([pde,ic,bc],w(x,t)) assuming t>0,x>0,c>0),output='realtime'));
\[ w \left ( x,t \right ) ={\frac {1}{c} \left ( {\it Heaviside} \left ( {\frac {tc-x}{c}} \right ) h \left ( {\frac {tc-x}{c}} \right ) c-{\it invlaplace} \left ( {{\rm e}^{-{\frac {sx}{c}}}}\int ^{0}\!f \left ( {\it \_a} \right ) {{\rm e}^{{\frac {s{\it \_a}}{c}}}}{d{\it \_a}},s,t \right ) +{\it invlaplace} \left ( {{\rm e}^{-{\frac {sx}{c}}}}\int \!f \left ( x \right ) {{\rm e}^{{\frac {sx}{c}}}}\,{\rm d}x,s,t \right ) \right ) } \] Solution contains unresolved invlaplace calls
Hand solution
\begin {equation} \frac {\partial w}{\partial t}+c\frac {\partial w}{\partial x}=0\tag {1} \end {equation} Let \[ w\equiv w\left ( x\left ( t\right ) ,t\right ) \] Hence \begin {equation} \frac {dw}{dt}=\frac {\partial w}{\partial t}+\frac {\partial w}{\partial x}\frac {dx}{dt}\tag {2} \end {equation} Comparing given (1) and (2), we see that if we let \(\frac {dx}{dt}=c\) in (2), then we obtain (1). Hence we conclude that \(\frac {dw}{dt}=0\). Therefore, \(w\left ( x\left ( t\right ) ,t\right ) \) is constant. At \(t=0\), we are given that \begin {equation} w\left ( x\left ( t\right ) ,t\right ) =f\left ( x\left ( 0\right ) \right ) \qquad t=0\tag {3} \end {equation} We just now need to determine \(x\left ( 0\right ) \). This is found from \(\frac {dx}{dt}=c\), which has the solution \(x\left ( t\right ) =x\left ( 0\right ) +ct\,.\) Hence \(x\left ( 0\right ) =x\left ( t\right ) -ct\). Therefore (3) becomes\[ w\left ( x,t\right ) =f\left ( x-ct\right ) \] This is valid for \(x>ct.\) We now start all over again, and look at Let \[ w\equiv w\left ( x,t\left ( x\right ) \right ) \] Hence \begin {equation} \frac {dw}{dx}=\frac {\partial w}{\partial x}+\frac {\partial w}{\partial t}\frac {dt}{dx}\tag {4} \end {equation} Comparing (4) and (1), we see that if we let \(\frac {dt}{dx}=\frac {1}{c}\) in (4), then we obtain (1). Hence we conclude that \(\frac {dw}{dx}=0\). Therefore, \(w\left ( x,t\left ( x\right ) \right ) \) is constant. At \(x=0\), we are given that \begin {equation} w\left ( x,t\left ( x\right ) \right ) =h\left ( t\left ( 0\right ) \right ) \qquad x=0\tag {5} \end {equation} We just now need to determine \(t\left ( 0\right ) \). This is found from \(\frac {dt}{dx}=\frac {1}{c}\), which has the solution \(t\left ( x\right ) =t\left ( 0\right ) +\frac {1}{c}x\,.\) Hence \(t\left ( 0\right ) =t\left ( x\right ) -\frac {1}{c}x\). Therefore (5) becomes\[ w\left ( x,t\right ) =h\left ( t-\frac {1}{c}x\right ) \] Valid for \(t>\frac {x}{c}\) or \(x<ct\). Therefore, the solution is\[ w\left ( x,t\right ) =\left \{ \begin {array} [c]{ccc}f\left ( x-ct\right ) & & x>ct\\ h\left ( t-\frac {1}{c}x\right ) & & x<ct \end {array} \right . \]
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Added Nov 25, 2018.
Problem 12.2.5 (a) from Richard Haberman applied partial differential equations book, 5th edition
Solve for \(u(x,t)\) \[ \frac { \partial \omega }{\partial t} +c \frac { \partial \omega }{\partial x} = e^{2 x} \]
With \(\omega (x,0)=f(x)\).
See my HW 12, Math 322, UW Madison.
Mathematica ✓
ClearAll[x, t, w, f, h]; pde = D[w[x, t], t] + c*D[w[x, t], x] == Exp[2*x]; ic = w[x, 0] == f[x]; sol = AbsoluteTiming[TimeConstrained[Simplify[DSolve[{pde, ic}, w[x, t], {x, t}, Assumptions -> c > 0 && x > 0 && t > 0]], 60*10]];
\[ \left \{\left \{w(x,t)\to f(x-c t)+\frac {e^{2 x} \left (1-e^{-2 c t}\right )}{2 c}\right \}\right \} \]
Maple ✓
x:='x'; t:='t'; w:='w';c:='c'; pde:=diff(w(x,t),t)+c*diff(w(x,t),x)=exp(2*x); ic:=w(x,0)=f(x); cpu_time := timelimit(60*10,CodeTools[Usage](assign('sol',pdsolve([pde,ic],w(x,t)) assuming t>0,x>0,c>0),output='realtime'));
\[ w \left ( x,t \right ) =1/2\,{\frac {2\,f \left ( -tc+x \right ) c-{{\rm e}^{-2\,tc+2\,x}}+{{\rm e}^{2\,x}}}{c}} \]
Hand solution
Using the method of characteristics, the systems of characteristic lines are (from the PDE itself)\begin {align} \frac {dt}{ds} & =1\tag {1}\\ \frac {dx}{ds} & =c\tag {2}\\ \frac {du}{ds} & =e^{2x}\tag {3} \end {align}
With initial conditions at \(s=0\)\[ t\left ( 0\right ) =t_{1},x\left ( 0\right ) =t_{2},u\left ( 0\right ) =t_{3}\] And \(u\left ( x,0\right ) =f\left ( x\right ) \) becomes \begin {equation} t_{3}=f\left ( t_{2}\right ) ,t_{1}=0\tag {4} \end {equation} Equation (1) gives\begin {align} t & =s+t_{1}\nonumber \\ & =s\tag {5} \end {align}
Equation (2) gives\begin {equation} x=cs+t_{2}\tag {6} \end {equation} From (5,6) solving for \(t_{2}\) gives\begin {align} t_{2} & =x-cs\nonumber \\ & =x-ct\tag {7} \end {align}
Equation (3) gives\begin {align*} du & =e^{2x}ds\\ & =e^{2\left ( cs+t_{2}\right ) }ds \end {align*}
Integrating \[ u=\frac {e^{2\left ( cs+t_{2}\right ) }}{2c}+t_{3}\] Using (7,4,5) in the above gives the solution\begin {align*} u\left ( x,t\right ) & =\frac {e^{2\left ( ct+\left ( x-ct\right ) \right ) }}{2c}+f\left ( x-ct\right ) \\ & =\frac {1}{2c}e^{2x}+f\left ( x-ct\right ) \end {align*}
My solution is not the same as CAS, but it was verified OK using Maple pdetest.
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Added Nov 25, 2018.
Problem 12.2.5 (d) from Richard Haberman applied partial differential equations book, 5th edition
Solve for \(u(x,t)\) \[ \frac { \partial \omega }{\partial t} +3 t \frac { \partial \omega }{\partial x} = \omega (x,t) \]
with \(\omega (x,0)=f(x)\).
See my HW 12, Math 322, UW Madison.
Mathematica ✓
ClearAll[x, t, w, f]; pde = D[w[x, t], t] + 3*t*D[w[x, t], x] == w[x, t]; ic = w[x, 0] == f[x]; sol = AbsoluteTiming[TimeConstrained[Simplify[DSolve[{pde, ic}, w[x, t], {x, t}]], 60*10]];
\[ \left \{\left \{w(x,t)\to e^{-\sqrt {t^2}} f\left (x-\frac {3 t^2}{2}\right )\right \},\left \{w(x,t)\to e^{\sqrt {t^2}} f\left (x-\frac {3 t^2}{2}\right )\right \}\right \} \]
Maple ✓
x:='x'; t:='t'; w:='w';c:='c'; pde:=diff(w(x,t),t)+3*t*diff(w(x,t),x)=w(x,t); ic:=w(x,0)=f(x); cpu_time := timelimit(60*10,CodeTools[Usage](assign('sol',pdsolve([pde,ic],w(x,t))),output='realtime'));
\[ w \left ( x,t \right ) =f \left ( -3/2\,{t}^{2}+x \right ) {{\rm e}^{t}} \]
Hand solution
Solve \begin {equation} \frac {\partial w}{\partial t}+3t\frac {\partial w}{\partial x}=w\left ( x,t\right ) \tag {1} \end {equation}
With initial conditions \(w\left ( x,0\right ) =f\left ( x\right ) \)
Solution
Let \(w\equiv w\left ( x\left ( t\right ) ,t\right ) \) then \begin {equation} \frac {dw}{dt}=\frac {\partial w}{\partial x}\frac {dx}{dt}+\frac {\partial w}{\partial t}\tag {2} \end {equation}
Comparing (1,2) shows that
\begin {align} \frac {dw}{dt} & =w\tag {3}\\ \frac {dx}{dt} & =3t\tag {4} \end {align}
Solving (3) gives
\begin {equation} w=Ce^{t}\nonumber \end {equation}
From initial conditions at \(t=0\), the above becomes \(f\left ( x\left ( 0\right ) \right ) =C\). Hence the above becomes
\begin {equation} w\left ( x,t\right ) =f\left ( x\left ( 0\right ) \right ) e^{t}\tag {5} \end {equation}
From (4)
\begin {align*} x & =\frac {3}{2}t^{2}+x\left ( 0\right ) \\ x\left ( 0\right ) & =x-\frac {3}{2}t^{2} \end {align*}
Substituting the above in (5) gives
\[ w\left ( x\left ( t\right ) ,t\right ) =f\left ( x-\frac {3}{2}t^{2}\right ) e^{t}\]
Alternative solution
Using the method of characteristics, the systems of characteristic lines are (from the PDE itself)\begin {align} \frac {dt}{ds} & =1\tag {1}\\ \frac {dx}{ds} & =3t\tag {2}\\ \frac {dw}{ds} & =w\tag {3} \end {align}
With initial conditions at \(s=0\)\[ t\left ( 0\right ) =t_{1},x\left ( 0\right ) =t_{2},w\left ( 0\right ) =t_{3}\] And \(w\left ( x,0\right ) =f\left ( x\right ) \) becomes \begin {equation} t_{3}=f\left ( t_{2}\right ) ,t_{1}=0\tag {4} \end {equation} Equation (1) gives\begin {align} t & =s+t_{1}\nonumber \\ & =s\tag {5} \end {align}
Equation (2) gives, after replacing \(t\) by \(s\) from (5)\begin {align} \frac {dx}{ds} & =3s\nonumber \\ x & =\frac {3}{2}s^{2}+t_{2}\tag {6} \end {align}
Solving for \(t_{2}\) gives\begin {equation} t_{2}=x-\frac {3}{2}s^{2}\tag {7} \end {equation} Equation (3) gives\begin {align*} \ln w & =s+t_{3}\\ w & =t_{3}e^{s}\\ & =f\left ( t_{2}\right ) e^{s} \end {align*}
Using (7,5) in the above gives the solution\[ w\left ( x,t\right ) =f\left ( x-\frac {3}{2}t^{2}\right ) e^{t}\]
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Taken from Mathematica help pages
Solve for \(u(x,y)\) \[ 2 \frac { \partial u}{\partial x} + 5 \frac { \partial u}{\partial y} = u^2(x,y) + 1 \]
Mathematica ✓
ClearAll[u, x, y]; pde = 2*D[u[x, y], x] + 5*D[u[x, y], y] == u[x, y]^2 + 1; sol = AbsoluteTiming[TimeConstrained[Simplify[DSolve[pde, u[x, y], {x, y}]], 60*10]];
\[ \left \{\left \{u(x,y)\to \tan \left (c_1\left (y-\frac {5 x}{2}\right )+\frac {x}{2}\right )\right \}\right \} \]
Maple ✓
interface(showassumed=0); u:='u';x:='x';y:='y'; pde := 2* diff(u(x, y), x) + 5*diff(u(x, y), y) = u(x, y)^2 + 1; cpu_time := timelimit(60*10,CodeTools[Usage](assign('sol',pdsolve(pde, u(x, y))),output='realtime'));
\[ u \left ( x,y \right ) =\tan \left ( x/2+1/2\,{\it \_F1} \left ( -5/2\,x+y \right ) \right ) \]
Hand solution
Solve for \(u\left ( x,y\right ) \) in \(2u_{x}+5u_{y}=u^{2}+1.\) Using the Lagrange-charpit method, the characteristic equations are\[ \frac {dx}{2}=\frac {dy}{5}=\frac {du}{u^{2}+1}\] From the first pair of equation we obtain\begin {align*} 5dx & =2dy\\ 5x & =2y+C_{1}\\ C_{1} & =5x-2y \end {align*}
Now we can pick the pair \(\frac {dy}{5}=\frac {du}{u^{2}+1}\) or \(\frac {dx}{2}=\frac {du}{u^{2}+1}\) to solve for \(u\). It does not matter which. Using\[ \frac {dx}{2}=\frac {du}{u^{2}+1}\] Integrating gives\begin {align*} \frac {1}{2}x & =\arctan \left ( u\right ) +C_{2}\\ C_{2} & =\frac {1}{2}x-\arctan \left ( u\right ) \end {align*}
\(C_{1}\) and \(C_{2}\) are always related by \(C_{2}=F\left ( C_{1}\right ) \) where \(F\) is arbitrary function. Hence\begin {align*} \frac {1}{2}x-\arctan \left ( u\right ) & =F\left ( 5x-2y\right ) \\ \arctan \left ( u\right ) & =\frac {1}{2}x-F\left ( 5x-2y\right ) \\ u\left ( x,y\right ) & =\tan \left ( \frac {1}{2}x-F\left ( 5x-2y\right ) \right ) \end {align*}
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Taken from Mathematica Symbolic PDE document
Solve for \(u(x,y)\) \[ \frac { \partial u}{\partial x} + u(x,y) \frac { \partial u}{\partial y} = 0 \]
Mathematica ✓
ClearAll[u, x, y]; pde = D[u[x, y], {x}] + u[x, y]*D[u[x, y], {y}] == 0; sol = AbsoluteTiming[TimeConstrained[Simplify[DSolve[pde, u[x, y], {x, y}]], 60*10]];
\[ \text {Solve}\left [u(x,y)=c_1\left (x-\frac {y}{u(x,y)}\right ),u(x,y)\right ] \] Implicit solution
Maple ✓
interface(showassumed=0); u:='u';x:='x';y:='y'; pde := diff(u(x, y), x) + u(x,y)*diff(u(x, y),y) =0; cpu_time := timelimit(60*10,CodeTools[Usage](assign('sol',pdsolve(pde,u(x,y))),output='realtime')); sol:=DEtools:-remove_RootOf(sol);
\[ -y+xu \left ( x,y \right ) +{\it \_F1} \left ( u \left ( x,y \right ) \right ) =0 \]
Hand solution
Solve for \(u\left ( x,y\right ) \) in \(u_{x}+u\ u_{y}=0.\) Using the Lagrange-Charpit method, the characteristic equations are\[ \frac {dx}{1}=\frac {dy}{u}=\frac {du}{0}\] From the first pair of equation we obtain\[ u=\frac {dy}{dx}\] But \(du=0\) or \(u=C_{2}\). Hence the above becomes
\begin {align*} \frac {dy}{dx} & =C_{2}\\ y & =xC_{2}+C_{1}\\ C_{1} & =y-xC_{2} \end {align*}
Since \(C_{2}=F\left ( C_{1}\right ) \) where \(F\) is arbitrary function, then \[ u\left ( x,y\right ) =F\left ( y-ux\right ) \]
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Taken from Mathematica Symbolic PDE document
Solve for \(u(x,y)\) \[ u_x+ u(x,y) u_y = 0 \] With \(u(x,0)=\frac {1}{x+1}\)
Mathematica ✓
ClearAll[u, x, y]; pde = D[u[x, y], {x}] + u[x, y]*D[u[x, y], {y}] == 0; ic = u[x, 0] == 1/(x + 1); sol = AbsoluteTiming[TimeConstrained[Simplify[DSolve[{pde, ic}, u[x, y], {x, y}]], 60*10]];
\[ \left \{\left \{u(x,y)\to \frac {y+1}{x+1}\right \}\right \} \]
Maple ✓
interface(showassumed=0); u:='u';x:='x';y:='y'; pde := diff(u(x, y), x) + u(x,y)*diff(u(x, y),y) =0; ic:=u(x,0)=1/(x+1); cpu_time := timelimit(60*10,CodeTools[Usage](assign('sol',pdsolve([pde,ic],u(x,y))),output='realtime'));
\[ u \left ( x,y \right ) ={\frac {y+1}{x+1}} \]
Hand solution
Using the method of characteristics, the systems of characteristic lines are (from the PDE itself)\begin {align} \frac {dx}{ds} & =1\tag {1}\\ \frac {dy}{ds} & =u\tag {2}\\ \frac {du}{ds} & =0\tag {3} \end {align}
With initial conditions at \(s=0\)\[ x\left ( 0\right ) =t_{1},y\left ( 0\right ) =t_{2},u\left ( 0\right ) =t_{3}\] We are given that \(u\left ( x,0\right ) =\frac {1}{1+x}\). This initial condition translates to\begin {equation} t_{3}=\frac {1}{1+t_{1}},t_{2}=0\tag {4} \end {equation} Equation (1) gives\begin {equation} x=s+t_{1}\tag {5} \end {equation} Equation (2) gives\begin {align} y & =su+t_{2}\nonumber \\ & =su\tag {7} \end {align}
Equation (3) gives \[ u=t_{3}\] Hence the solution is \begin {align*} u & =t_{3}\\ & =\frac {1}{1+t_{1}}\\ & =\frac {1}{1+\left ( x-s\right ) }\\ & =\frac {1}{1+\left ( x-\frac {y}{u}\right ) } \end {align*}
Solving for \(u\) gives\begin {align*} u\left ( 1+\left ( x-\frac {y}{u}\right ) \right ) & =1\\ u+xu-y & =1\\ u\left ( 1+x\right ) & =1+y\\ u & =\frac {1+y}{1+x} \end {align*}
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Taken from Mathematica Symbolic PDE document
Solve for \(u(x,y)\) \[ x \frac { \partial u}{\partial x} + y \frac { \partial u}{\partial y} + \frac {1}{2} \left ( \left ( \frac { \partial u}{\partial x} \right )^2 + \left ( \frac { \partial u}{\partial y} \right )^2 \right )=0 \]
Mathematica ✓
ClearAll[u, x, y]; pde = u[x, y] == x*D[u[x, y], {x}] + y*D[u[x, y], {y}] + (1/2)*(D[u[x, y], {x}]^2 + D[u[x, y], {y}]^2); sol = AbsoluteTiming[TimeConstrained[Simplify[DSolve[pde, u[x, y], {x, y}]], 60*10]];
\[ \left \{\left \{u(x,y)\to c_1 x+c_2 y+\frac {1}{2} \left (c_1^2+c_2^2\right )\right \}\right \} \]
Maple ✓
interface(showassumed=0); u:='u';x:='x';y:='y'; pde := x*diff(u(x, y), x) + y*diff(u(x, y),y) + 1/2 * ( diff(u(x, y), x)^2 + diff(u(x, y), y)^2)=0; cpu_time := timelimit(60*10,CodeTools[Usage](assign('sol',pdsolve(pde,u(x,y),'build')),output='realtime'));
\[ u \left ( x,y \right ) =-1/2\,{x}^{2}-1/2\,x\sqrt {{x}^{2}+2\,{\it \_c}_{{1}}}-{\it \_c}_{{1}}\ln \left ( x+\sqrt {{x}^{2}+2\,{\it \_c}_{{1}}} \right ) +{\it \_C1}-1/2\,{y}^{2}-1/2\,y\sqrt {{y}^{2}-2\,{\it \_c}_{{1}}}+{\it \_c}_{{1}}\ln \left ( y+\sqrt {{y}^{2}-2\,{\it \_c}_{{1}}} \right ) +{\it \_C2} \]
Hand solution
Assuming the solution is \(u\left ( x,y\right ) =X\left ( x\right ) +Y\left ( y\right ) \). Substituting this into the PDE gives\begin {align*} xX^{\prime }+yY^{\prime }+\frac {1}{2}\left ( \left ( X^{\prime }\right ) ^{2}+\left ( Y^{\prime }\right ) ^{2}\right ) & =0\\ \frac {1}{2}\left ( X^{\prime }\right ) ^{2}+xX^{\prime } & =-\frac {1}{2}\left ( Y^{\prime }\right ) ^{2}-yY^{\prime } \end {align*}
The above is possible when each side is equal to same constant, say \(C_{1}\). This gives two ODE’s\begin {align} \frac {1}{2}\left ( X^{\prime }\right ) ^{2}+xX^{\prime } & =C_{1}\tag {1}\\ \frac {1}{2}\left ( Y^{\prime }\right ) ^{2}+yY^{\prime } & =-C_{1}\tag {2} \end {align}
ODE (1) becomes\begin {align*} \left ( X^{\prime }\right ) ^{2}+2xX^{\prime }-2C_{1} & =0\\ X^{\prime } & =\frac {-b}{2a}\pm \frac {1}{2a}\sqrt {b^{2}-4ac}\\ & =\frac {-2x}{2}\pm \frac {1}{2}\sqrt {4x^{2}+8C_{1}}\\ & =-x\pm \sqrt {x^{2}+2C_{1}} \end {align*}
For the case \(X^{\prime }=-x+\sqrt {x^{2}+2C_{1}}\), the solution is \begin {align*} X\left ( x\right ) & =\int -x+\sqrt {x^{2}+2C_{1}}dx+C_{2}\\ & =-\frac {x^{2}}{2}+\frac {x\sqrt {x^{2}+2C_{1}}}{2}+C_{1}\ln \left ( x+\sqrt {x^{2}+2C_{1}}\right ) +C_{2} \end {align*}
For the case \(X^{\prime }=-x-\sqrt {x^{2}+2C_{1}}\), the solution is\begin {align*} X\left ( x\right ) & =\int -x-\sqrt {x^{2}+2C_{1}}dx+C_{2}\\ & =-\frac {x^{2}}{2}-\frac {x\sqrt {x^{2}+2C_{1}}}{2}-C_{1}\ln \left ( x+\sqrt {x^{2}+2C_{1}}\right ) +C_{2} \end {align*}
Combining the above two solutions to one gives\begin {equation} X\left ( x\right ) =-\frac {x^{2}}{2}\pm \frac {x\sqrt {x^{2}+2C_{1}}}{2}\pm C_{1}\ln \left ( x+\sqrt {x^{2}+2C_{1}}\right ) +C_{2}\tag {3} \end {equation} ODE (2) becomes\begin {align*} \left ( Y^{\prime }\right ) ^{2}+2yY^{\prime }+2C_{1} & =0\\ Y^{\prime } & =\frac {-b}{2a}\pm \frac {1}{2a}\sqrt {b^{2}-4ac}\\ & =\frac {-2y}{2}\pm \frac {1}{2}\sqrt {4y^{2}-8C_{1}}\\ & =-y\pm \sqrt {y^{2}-2C_{1}} \end {align*}
For the case \(Y^{\prime }=-y+\sqrt {x^{2}-2C_{1}}\), the solution is \begin {align*} Y\left ( y\right ) & =\int -y+\sqrt {y^{2}-2C_{1}}dy+C_{2}\\ & =\frac {-y^{2}}{2}+\frac {y\sqrt {y^{2}-2C_{1}}}{2}-C_{1}\ln \left ( y+\sqrt {y^{2}-2C_{1}}\right ) +C_{3} \end {align*}
For the case \(Y^{\prime }=-y-\sqrt {y^{2}-2C_{1}}\), the solution is\begin {align*} Y\left ( y\right ) & =\int -y-\sqrt {y^{2}+2C_{1}}dy+C_{2}\\ & =-\frac {y^{2}}{2}-\frac {y\sqrt {x^{2}-2C_{1}}}{2}+C_{1}\ln \left ( y+\sqrt {y^{2}-2C_{1}}\right ) +C_{3} \end {align*}
Combining the above two solutions to one gives\begin {equation} Y\left ( x\right ) =-\frac {y^{2}}{2}\pm \frac {y\sqrt {y^{2}-2C_{1}}}{2}\pm C_{1}\ln \left ( y+\sqrt {y^{2}-2C_{1}}\right ) +C_{3}\tag {4} \end {equation} From (3,4) the final solution is\begin {align*} u\left ( x,y\right ) & =X\left ( x\right ) +Y\left ( x\right ) \\ & =\left ( -\frac {x^{2}}{2}\pm \frac {x\sqrt {x^{2}+2C_{1}}}{2}\pm C_{1}\ln \left ( x+\sqrt {x^{2}+2C_{1}}\right ) +C_{2}\right ) +\left ( -\frac {y^{2}}{2}\pm \frac {y\sqrt {y^{2}-2C_{1}}}{2}\pm C_{1}\ln \left ( y+\sqrt {y^{2}-2C_{1}}\right ) +C_{3}\right ) \\ & =-\frac {x^{2}}{2}\pm \frac {x}{2}\sqrt {x^{2}+2C_{1}}\pm C_{1}\ln \left ( x+\sqrt {x^{2}+2C_{1}}\right ) -\frac {y^{2}}{2}\pm \frac {y}{2}\sqrt {y^{2}-2C_{1}}\pm C_{1}\ln \left ( y+\sqrt {y^{2}-2C_{1}}\right ) +C_{4} \end {align*}
Where \(C_{4}=C_{2}+C_{3}\).
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Taken from Mathematica Symbolic PDE document
Solve for \(u(x,y)\) \[ x \frac { \partial u}{\partial x} + y \frac { \partial u}{\partial y} + \frac {1}{2} \left ( \left ( \frac { \partial u}{\partial x} \right )^2 + \left ( \frac { \partial u}{\partial y} \right )^2 \right ) = 0 \] With \(u(x,0)= \frac {1}{2} (1-x^2)\)
Mathematica ✓
ClearAll[u, x, y]; pde = u[x, y] == x*D[u[x, y], {x}] + y*D[u[x, y], {y}] + (1/2)*(D[u[x, y], {x}]^2 + D[u[x, y], {y}]^2); ic = u[x, 0] == (1*(1 - x^2))/2; sol = AbsoluteTiming[TimeConstrained[DSolve[{pde, ic}, u[x, y], {x, y}], 60*10]];
\[ \left \{\left \{u(x,y)\to \frac {1}{2} \left (-x^2-2 y+1\right )\right \}\right \} \]
Maple ✓
interface(showassumed=0); u:='u';x:='x';y:='y'; pde := x*diff(u(x, y), x) + y*diff(u(x, y),y) + 1/2 * ( diff(u(x, y), x)^2 + diff(u(x, y), y)^2)=0; ic:=u(x,0)=1/2*(1-x^2); cpu_time := timelimit(60*10,CodeTools[Usage](assign('sol',pdsolve([pde,ic],u(x,y))),output='realtime'));
\[ u \left ( x,y \right ) =-1/2\, \left ( x-y+1 \right ) \left ( x-y-1 \right ) \]
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Taken from Mathematica DSolve help pages
Solve for \(u(x,y)\) \[ u(x,y)= x \frac { \partial u}{\partial x} + y \frac { \partial u}{\partial y} + \sin \left ( \frac { \partial u}{\partial x} + \frac {\partial u}{\partial y} \right ) \]
Mathematica ✓
ClearAll[u, x, y]; pde = u[x, y] == x*D[u[x, y], x] + y*D[u[x, y], y] + Sin[D[u[x, y], x] + D[u[x, y], y]]; sol = AbsoluteTiming[TimeConstrained[DSolve[pde, u[x, y], {x, y}], 60*10]];
\[ \left \{\left \{u(x,y)\to c_1 x+c_2 y+\sin \left (c_1+c_2\right )\right \}\right \} \]
Maple ✓
u:='u';x:='x';y:='y'; pde:= u(x,y)= x*diff(u(x,y),x)+y*diff(u(x,y),y)+sin(diff(u(x,y),x)+diff(u(x,y),y)); cpu_time := timelimit(60*10,CodeTools[Usage](assign('sol',pdsolve(pde,u(x,y))),output='realtime'));
\[ u \left ( x,y \right ) =x{\it \_c}_{{1}}+y{\it \_c}_{{2}}+\sin \left ( {\it \_c}_{{1}}+{\it \_c}_{{2}} \right ) \]
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Taken from Mathematica DSolve help pages
Solve for \(f(x,y)\) \begin {align*} \frac { \partial f}{\partial x} &= x y \cos (x y)+ \sin (x y) \\ \frac { \partial f}{\partial y} &= -e^{-y} +x^2 \cos (x y) \end {align*}
Mathematica ✓
ClearAll[f, x, y]; eq1 = D[f[x, y], x] == x*y*Cos[x*y] + Sin[x*y]; eq2 = D[f[x, y], y] == -E^(-y) + x^2*Cos[x*y]; sol = AbsoluteTiming[TimeConstrained[DSolve[{eq1, eq2}, f[x, y], {x, y}], 60*10]];
\[ \left \{\left \{f(x,y)\to c_1+x \sin (x y)+e^{-y}\right \}\right \} \]
Maple ✓
u:='u';x:='x';y:='y'; eq1:=diff(f(x,y),x)=x*y*cos(x*y)+sin(x*y); eq2:=diff(f(x,y),y)=-exp(-y)+x^2*cos(x*y); cpu_time := timelimit(60*10,CodeTools[Usage](assign('sol',pdsolve({eq1,eq2},f(x,y))),output='realtime'));
\[ \left \{ f \left ( x,y \right ) =x\sin \left ( yx \right ) +{{\rm e}^{-y}}+{\it \_C1} \right \} \]
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Taken from Maple pdsolve help pages
Solve for \(f(x,y)\) \begin {align*} x \frac { \partial f}{\partial y}- \frac { \partial f}{\partial x} &= \frac { f^2(x,y) g(x)}{h(y)} \end {align*}
Mathematica ✗
ClearAll[f, x, y, h, g]; pde = x*D[f[x, y], y] - D[f[x, y], x] == (f[x, y]^2*g[x])/h[y]; sol = AbsoluteTiming[TimeConstrained[DSolve[pde, f[x, y], {x, y}], 60*10]];
\[ \text {Failed} \]
Maple ✓
x:='x';y:='y';f:='f';g:='g';h:='h'; pde := x*diff(f(x,y),y)-diff(f(x,y),x)=f(x,y)^2*g(x)/h(y); cpu_time := timelimit(60*10,CodeTools[Usage](assign('sol',pdsolve(pde,f(x,y))),output='realtime'));
\[ f \left ( x,y \right ) = \left ( \int ^{x}\!{\frac {g \left ( {\it \_a} \right ) }{h \left ( -1/2\,{{\it \_a}}^{2}+1/2\,{x}^{2}+y \right ) }}{d{\it \_a}}+{\it \_F1} \left ( 1/2\,{x}^{2}+y \right ) \right ) ^{-1} \] Has unresolved integral in the answer
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Taken from Maple pdsolve help pages, probem 5
Solve for \(f(x,y,z)\) \begin {align*} f_x + (f_y)^2 &= f(x,y,z)+z \end {align*}
Mathematica ✓
ClearAll[f, x, y, z]; pde = D[f[x, y, z], x] + D[f[x, y, z], y]^2 == f[x, y, z] + z; sol = AbsoluteTiming[TimeConstrained[DSolve[pde, f[x, y, z], {x, y, z}], 60*10]];
\[ \left \{\left \{f(x,y,z)\to \frac {1}{4} \left (c_1(z){}^2 \text {ProductLog}\left (-\frac {e^{\frac {y}{c_1(z)}+\frac {c_2(z)}{c_1(z)}+x-1}}{c_1(z)}\right ){}^2+2 c_1(z){}^2 \text {ProductLog}\left (-\frac {e^{\frac {y}{c_1(z)}+\frac {c_2(z)}{c_1(z)}+x-1}}{c_1(z)}\right )-4 z\right )\right \}\right \} \]
Maple ✓
x:='x';y:='y';f:='f';z:='z'; pde := diff(f(x,y,z),x) + (diff(f(x,y,z),y))^2 = f(x,y,z)+z; cpu_time := timelimit(60*10,CodeTools[Usage](assign('sol',pdsolve(pde,f(x,y,z),'build')),output='realtime'));
\[ f \left ( x,y,z \right ) =-{\frac {{{\rm e}^{-x}}z{{\it \_C5}}^{2}+{{\rm e}^{x}}{{\it \_C3}}^{2}+{\it \_C3}\,y{\it \_C5}+z{\it \_C4}\,{\it \_C5}+{\it \_C1}\,{\it \_C5}}{{{\it \_C5}}^{2}{{\rm e}^{-x}}}} \]
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From example 3.5.4, page 212 nonlinear pde’s by Lokenath Debnath, 3rd edition.
Solve for \(u(x,y,z)\) \begin {align*} (y-z) u_x + (z-x) u_y + (x-y) u_z = 0 \end {align*}
Mathematica ✗
ClearAll[u, x, y, z]; pde = (y - z)*D[u[x, y, z], x] + (z - x)*D[u[x, y, z], y] + (x - y)*D[u[x, y, z], z] == 0; sol = AbsoluteTiming[TimeConstrained[DSolve[pde, u[x, y, z], {x, y, z}], 60*10]];
\[ \text {\$Aborted} \] Timed out
Maple ✓
x:='x';y:='y';u:='u';z:='z'; pde:=(y-z)*diff(u(x,y,z),x)+(z-x)*diff(u(x,y,z),y)+(x-y)*diff(u(x,y,z),z)=0; cpu_time := timelimit(60*10,CodeTools[Usage](assign('sol',pdsolve(pde,u(x,y,z),'build')),output='realtime'));
\[ u \left ( x,y,z \right ) ={{\rm e}^{1/2\,{\it \_C2}\,{x}^{2}}}{{\rm e}^{{\it \_C1}\,x}}{{\rm e}^{1/2\,{\it \_C2}\,{y}^{2}}}{{\rm e}^{{\it \_C1}\,y}}{\it \_C3}\,{\it \_C5}\,{\it \_C4}\,{{\rm e}^{1/2\,{\it \_C2}\,{z}^{2}}}{{\rm e}^{{\it \_C1}\,z}} \]