____________________________________________________________________________________
Taken from Maple pdsolve help pages, problem 4.
Solve for \(S \left ( x,y \right ) \) \begin {align*} S(x,y) \left ( \frac {\partial ^2 S}{\partial x \partial y} \right ) + \frac {\partial S}{\partial x} \frac {\partial S}{\partial y} &=1 \end {align*}
Mathematica ✗
ClearAll[s, x, y]; pde = s[x, y]*D[s[x, y], x, y] + D[s[x, y], x]*D[s[x, y], y] == 1; sol = AbsoluteTiming[TimeConstrained[DSolve[pde, s[x, y], {x, y}], 60*10]];
\[ \text {Failed} \]
Maple ✓
S:='S';x:='x'; y:='y'; pde := S(x,y)*diff(S(x,y),y,x) + diff(S(x,y),x)*diff(S(x,y),y) = 1; cpu_time := timelimit(60*10,CodeTools[Usage](assign('sol',pdsolve(pde,S(x,y),'build')),output='realtime'));
\[ S \left ( x,y \right ) ={\frac {\sqrt {2\,{\it \_c}_{{1}}x+{\it \_C1}}\sqrt {{\it \_C2}\,{{\it \_c}_{{1}}}^{2}+{\it \_c}_{{1}}y}}{{\it \_c}_{{1}}}} \]
____________________________________________________________________________________
Added December 20, 2018.
Solve for \(u(r,\theta )\)
\[ \frac {\partial ^2 u}{\partial r^2} + \frac {\partial ^2 u}{\partial \theta ^2} = 0 \]
With boundary conditions
\begin {align*} u(2,\theta )&=3 \sin (2 \theta )+1 \end {align*}
Mathematica ✗
ClearAll[u, r, theta]; pde = D[u[r, theta], {r, 2}] + D[u[r, theta], {theta, 2}] == 0; bc = u[2, theta] == 3*Sin[2*theta] + 1; sol = AbsoluteTiming[TimeConstrained[DSolve[{pde, bc}, u[r, theta], {r, theta}], 60*10]];
\[ \text {Failed} \]
Maple ✓
r:='r'; theta:='theta'; t:='t'; pde := diff(u(r, theta), r$2)+diff(u(r, theta), theta$2) = 0; bc := u(2, theta) = 3*sin(2*theta)+1; cpu_time := timelimit(60*10,CodeTools[Usage](assign('sol',pdsolve([pde, bc], u(r,theta),method = Fourier)),output='realtime'));
\[ u \left ( r,\theta \right ) =-3/2\,i{{\rm e}^{-2\,r+4+2\,i\theta }}+3/2\,i{{\rm e}^{2\,r-4-2\,i\theta }}+1 \]
____________________________________________________________________________________
Added December 20, 2018.
Solve for \(u(x,y)\)
\[ \frac {\partial ^2 u}{\partial x^2} + y \frac {\partial ^2 u}{\partial y^2} = 0 \]
With boundary conditions
\begin {align*} u(x,0)&=0 \\ \frac {\partial u}{\partial y}(x,0) &=x^2 \end {align*}
Mathematica ✓
ClearAll[u, x, y]; pde = D[u[x, y], {x, 2}] + y*D[u[x, y], {y, 2}] == 0; bc = {u[x, 0] == 0, Derivative[0, 1][u][x, 0] == x^2}; sol = AbsoluteTiming[TimeConstrained[DSolve[{pde, bc}, u[x, y], {x, y}], 60*10]];
\[ \left \{\left \{u(x,y)\to -y \left (y-x^2\right )\right \}\right \} \]
Maple ✓
x:='x'; y:='y'; u:='u'; pde := diff(u(x, y), x$2)+y*(diff(u(x, y), y$2)) = 0; bc:=u(x,0)=0, eval(diff(u(x,y),y),y=0)=x^2; cpu_time := timelimit(60*10,CodeTools[Usage](assign('sol',pdsolve([pde, bc], u(x, y))),output='realtime'));
\[ u \left ( x,y \right ) =y \left ( {x}^{2}-y \right ) \]
____________________________________________________________________________________
Added December 20, 2018.
Solve for \(u(x,y)\)
\[ \frac {\partial u}{\partial t} = - \frac {\partial ^3 u}{\partial x^2} \]
With initial conditions
\begin {align*} u(x,0)&=f(x) \end {align*}
Mathematica ✗
ClearAll[u, x, t, f]; pde = D[u[x, t], t] == -D[u[x, t], {x, 3}]; ic = u[x, 0] == f[x]; sol = AbsoluteTiming[TimeConstrained[DSolve[{pde, ic}, u[x, t], {x, t}], 60*10]];
\[ \text {Failed} \]
Maple ✓
x:='x'; t:='t'; u:='u'; pde := diff(u(x, t), t)=- diff(u(x, t), x$3); ic:=u(x,0)=f(x); cpu_time := timelimit(60*10,CodeTools[Usage](assign('sol',pdsolve([pde, ic],u(x,t))),output='realtime'));
\[ u \left ( x,t \right ) =1/4\,{\frac {1}{{\pi }^{2}}\int _{-\infty }^{\infty }\!4/3\,{\frac {\pi \,f \left ( -\zeta \right ) }{\sqrt [3]{-t}}\sqrt {-{\frac {x+\zeta }{\sqrt [3]{-t}}}}\BesselK \left ( 1/3,-2/9\,{\frac {\sqrt {3} \left ( x+\zeta \right ) }{\sqrt [3]{-t}}\sqrt {-{\frac {x+\zeta }{\sqrt [3]{-t}}}}} \right ) }\,{\rm d}\zeta } \]
____________________________________________________________________________________
Added December 20, 2018.
Solve for \(u(x,y)\)
\[ \frac {\partial u^2}{\partial x y} = \sin (x) \sin (y) \]
With boundary conditions
\begin {align*} u(x,0)&=1+\cos (x) \\ \frac {\partial u}{\partial y}(0,y) &= -2 \sin y \end {align*}
Mathematica ✗
ClearAll[u, x, y]; pde = D[u[x, y], y, x] == Sin[x]*Sin[y]; bc = {u[x, 0] == 1 + Cos[x], Derivative[0, 1][u][0, y] == -2*Sin[y]}; sol = AbsoluteTiming[TimeConstrained[DSolve[{pde, bc}, u[x, y], x, y], 60*10]];
\[ \text {Failed} \]
Maple ✓
x:='x'; y:='y'; u:='u'; pde := diff(u(x, y), y,x)=sin(x)*sin(y); bc:=u(x,0)=1+cos(x),eval( diff(u(x,y),y),x=0)=-2*sin(y); cpu_time := timelimit(60*10,CodeTools[Usage](assign('sol',pdsolve([pde, bc],u(x,y))),output='realtime'));
\[ u \left ( x,y \right ) =1/2\,\cos \left ( x-y \right ) +1/2\,\cos \left ( x+y \right ) +\cos \left ( y \right ) \]
____________________________________________________________________________________
Added December 20, 2018.
Example 25, Taken from https://www.mapleprimes.com/posts/209970-Exact-Solutions-For-PDE-And-Boundary--Initial-Conditions-2018
Solve for \(w(x_1,x_2,x_3,t)\)
\[ \frac {\partial w}{\partial t} = \frac {\partial w^2}{\partial x_1^2} + \frac {\partial w^2}{\partial x_2^2} + \frac {\partial w^2}{\partial x_3^2} \]
With initial condition \(w(x_1,x_2,x_3,1) = e^a x_1^2 +x_2 x_3\)
Mathematica ✗
ClearAll[w, x1, x2, x3, t, a]; pde = D[w[x1, x2, x3, t], t] == D[w[x1, x2, x3, t], {x1, 2}] + D[w[x1, x2, x3, t], {x2, 2}] + D[w[x1, x2, x3, t], {x3, 2}]; ic = w[x1, x2, x3, 1] == Exp[a]*x1^2 + x2*x3; sol = AbsoluteTiming[TimeConstrained[DSolve[{pde, ic}, w[x1, x2, x3, t], {x1, x2, x3, t}], 60*10]];
\[ \text {Failed} \]
Maple ✓
w:='w';x1:='x1';x2:='x2';x3:='x3';t:='t';a:='a'; pde := diff(w(x1, x2, x3, t), t) = diff(w(x1, x2, x3, t), x1$2)+diff(w(x1, x2, x3, t), x2$2)+diff(w(x1, x2, x3, t), x3$2); ic:= w(x1, x2, x3, 1) = exp(a)*x1^2+x2*x3; cpu_time := timelimit(60*10,CodeTools[Usage](assign('sol',pdsolve([pde, ic],w(x1,x2,x3,t))),output='realtime'));
\[ w \left ( {\it x1},{\it x2},{\it x3},t \right ) = \left ( {{\it x1}}^{2}+2\,t-2 \right ) {{\rm e}^{a}}+{\it x2}\,{\it x3} \]
____________________________________________________________________________________
Added December 20, 2018.
Example 26, Taken from https://www.mapleprimes.com/posts/209970-Exact-Solutions-For-PDE-And-Boundary--Initial-Conditions-2018
Solve for \(w(x_1,x_2,x_3,t)\)
\[ \frac {\partial w}{\partial t} = \frac {\partial w^2}{\partial x_1 x_2} + \frac {\partial w^2}{\partial x_1 x_3} + \frac {\partial w^2}{\partial x_3^2} + \frac {\partial w^2}{\partial x_2 x_3} \]
With initial condition \(w(x_1,x_2,x_3,t_0) = e^{x_1} +x_2 -3 x_3\)
Mathematica ✗
ClearAll[w, x1, x2, x3, t, t0]; pde = D[w[x1, x2, x3, t], t] == D[w[x1, x2, x3, t], x1, x2] + D[w[x1, x2, x3, t], x1, x3] + D[w[x1, x2, x3, t], {x3, 2}] - D[w[x1, x2, x3, t], x2, x3]; ic = w[x1, x2, x3, t0] == Exp[x1] + x2 - 3*x3; sol = AbsoluteTiming[TimeConstrained[DSolve[{pde, ic}, w[x1, x2, x3, t], {x1, x2, x3, t}], 60*10]];
\[ \text {Failed} \]
Maple ✓
w:='w';x1:='x1';x2:='x2';x3:='x3';t:='t';t0:='t0'; pde := diff(w(x1, x2, x3, t), t)= diff(w(x1,x2,x3,t),x1,x2)+diff(w(x1,x2,x3,t),x1,x3)+diff(w(x1,x2,x3,t),x3$2)-diff(w(x1,x2,x3,t),x2,x3) ; ic:= w(x1, x2, x3, t0) = exp(x1)+x2-3*x3; cpu_time := timelimit(60*10,CodeTools[Usage](assign('sol',pdsolve([pde, ic],w(x1,x2,x3,t))),output='realtime'));
\[ w \left ( {\it x1},{\it x2},{\it x3},t \right ) ={{\rm e}^{{\it x1}}}+{\it x2}-3\,{\it x3} \]
____________________________________________________________________________________
Added December 20, 2018.
Example 27, Taken from https://www.mapleprimes.com/posts/209970-Exact-Solutions-For-PDE-And-Boundary--Initial-Conditions-2018
Solve for \(w(x_1,x_2,x_3,t)\)
\[ \frac {\partial w^2}{\partial t^2} = \frac {\partial w^2}{\partial x_1 x_2} + \frac {\partial w^2}{\partial x_1 x_3} + \frac {\partial w^2}{\partial x_3^2} - \frac {\partial w^2}{\partial x_2 x_3} \]
With initial condition \begin {align*} w(x_1,x_2,x_3,t_0) &= x_1^3 x_2^2 + x_3 \\ \frac {\partial w}{\partial t}(x_1,x_2,x_3,t_0) &= -x_2 x_3 + x_1 \end {align*}
Mathematica ✗
ClearAll[w, x1, x2, x3, t, t0]; pde = D[w[x1, x2, x3, t], {t, 2}] == D[w[x1, x2, x3, t], x1, x2] + D[w[x1, x2, x3, t], x1, x3] + D[w[x1, x2, x3, t], {x3, 2}] - D[w[x1, x2, x3, t], x2, x3]; ic = {w[x1, x2, x3, t0] == x1^3*x2^2 + x3, Derivative[0, 0, 0, 1][w][x1, x2, x3, t0] == -(x2*x3) + x1}; sol = AbsoluteTiming[TimeConstrained[DSolve[{pde, ic}, w[x1, x2, x3, t], {x1, x2, x3, t}], 60*10]];
\[ \text {Failed} \]
Maple ✓
w:='w';x1:='x1';x2:='x2';x3:='x3';t:='t';t0:='t0'; pde := diff(w(x1, x2, x3, t), t$2)= diff(w(x1,x2,x3,t),x1,x2)+diff(w(x1,x2,x3,t),x1,x3)+diff(w(x1,x2,x3,t),x3$2)-diff(w(x1,x2,x3,t),x2,x3); ic:= w(x1, x2, x3, t0) = x1^3*x2^2+x3, eval( diff( w(x1,x2,x3,t),t),t=t0)=-x2*x3+x1; cpu_time := timelimit(60*10,CodeTools[Usage](assign('sol',pdsolve([pde, ic],w(x1,x2,x3,t))),output='realtime'));
\[ w \left ( {\it x1},{\it x2},{\it x3},t \right ) =1/2\,{{\it t0}}^{4}{\it x1}+1/6\, \left ( -12\,{\it x1}\,t-1 \right ) {{\it t0}}^{3}+1/6\, \left ( 18\,{\it x1}\,{t}^{2}+18\,{{\it x1}}^{2}{\it x2}+3\,t \right ) {{\it t0}}^{2}+1/6\, \left ( -36\,t{{\it x1}}^{2}{\it x2}+ \left ( -12\,{t}^{3}-6 \right ) {\it x1}-3\,{t}^{2}+6\,{\it x2}\,{\it x3} \right ) {\it t0}+{{\it x1}}^{3}{{\it x2}}^{2}+3\,{t}^{2}{{\it x1}}^{2}{\it x2}+1/6\, \left ( 3\,{t}^{4}+6\,t \right ) {\it x1}+1/6\,{t}^{3}-{\it x2}\,{\it x3}\,t+{\it x3} \]
____________________________________________________________________________________
Added January 2, 2018.
Solve for \(u(x,t)\) with \(x>0,t>0\)
\[ u_t = -\beta u_x + D u_{xx} \]
Assuming \(\beta >0,D>0\)
Mathematica ✗
ClearAll[u, x, t, beta, d]; pde = D[u[x, t], t] == beta*D[u[x, t], x] + d*D[u[x, t], {x, 2}]; sol = AbsoluteTiming[TimeConstrained[DSolve[pde, u[x, t], {x, t}, Assumptions -> {beta > 0, d > 0, x > 0, t > 0}], 60*10]];
\[ \text {Failed} \]
Maple ✓
u:='u';x:='x';t:='t';beta:='beta';d:='d'; pde:=diff(u(x,t),t)=-beta*diff(u(x,t),x)+d*diff(u(x,t),x$2); cpu_time := timelimit(60*10,CodeTools[Usage](assign('sol',pdsolve(pde,u(x,t),'build') assuming d>0,beta>0,x>0,t>0),output='realtime'));
\[ u \left ( x,t \right ) ={\frac {{\it \_C3}\,{\it \_C1}}{{{\rm e}^{{\it \_c}_{{1}}t}}}\sqrt {{{\rm e}^{{\frac {x\beta }{d}}}}}{{\rm e}^{1/2\,{\frac {x\sqrt {{\beta }^{2}-4\,d{\it \_c}_{{1}}}}{d}}}}}+{\frac {{\it \_C3}\,{\it \_C2}}{{{\rm e}^{{\it \_c}_{{1}}t}}}\sqrt {{{\rm e}^{{\frac {x\beta }{d}}}}}{{\rm e}^{-1/2\,{\frac {x\sqrt {{\beta }^{2}-4\,d{\it \_c}_{{1}}}}{d}}}}} \]