Question

ODE为y'' + lam * y(x)=0，其中包含BC y'(0)-y(0)=0和y'(1)=0。

Maple dsolve仅提供简单的解决方案y(x)=0。人们怎么能要求Maple提供Mathematica默认的非平凡解决方案呢？

这就是我试过的

ode:= diff(y(x),x$2)+ lambda*y(x)=0;
bc:=D(y)(0) - y(0)=0, D(y)(1)=0;
dsolve({ode,bc},y(x));

在Windows 7上的Maple 2017.3上给出y(x)=0。

虽然Mathematica上的相同命令给出了

ClearAll[y,x,lam];
DSolve[{y''[x]+lam y[x]==0,y'[0]-y[0]==0,y'[1]==0},y[x],x]

Mathematica graphics

手动解决这个问题是正确的。

Maple是否具有单独的特征值边界值dsolver？需要单独包装吗？我试图通过键入assuming(lambda>0)来帮助Maple，但它没有什么区别。

我确信Maple可以做到这一点，因为我知道Maple在微分方程中非常强大，我只需要正确的选项或命令来完成它。

Answer 1

这不是自动生成的分段结果。

第二次调用solve涉及解决cos(lambda^(1/2))而不是简单的名称。（求解lambda会产生RootOf，并且处理任何替换似乎都很尴尬。）

但是这里有一些东西。（我希望有更好的方式......）

restart;

ans:=dsolve({diff(diff(y(x),x),x)+lambda*y(x) = 0,
             D(y)(0)=K, D(y)(1) = 0},y(x)):

ans:=collect(ans,K);  # for beauty's sake

                 /          1/2                1/2            1/2   \
                |sin(lambda    x)   cos(lambda   ) cos(lambda    x)|
  ans := y(x) = |---------------- + -------------------------------| K
                |         1/2                1/2           1/2     |
                \   lambda             lambda    sin(lambda   )    /

solve(eval(subs(x=0,y(0)=K,ans)),{K});

                {K = 0}

eval(ans, %);

               y(x) = 0

eqn:=solve(eval(subs(x=0,y(0)=K,ans)),{cos(lambda^(1/2))});

                        1/2          1/2           1/2
      eqn := {cos(lambda   ) = lambda    sin(lambda   )}

eval(ans,eqn);

         /          1/2                      \
         |sin(lambda    x)             1/2   |
  y(x) = |---------------- + cos(lambda    x)| K
         |         1/2                       |
         \   lambda                          /

[编辑：如果您对lambda的真实价值感兴趣，那么这里还有别的东西。]

restart;
de := diff(diff(y(x),x),x) + lambda*y(x) = 0:
bcs := D(y)(0) = K, D(y)(1) = 0:
ans := dsolve({de,bcs}, y(x)):
ans := collect(ans, K);

                    /   /      (1/2)  \
                    |sin\lambda      x/
      ans := y(x) = |------------------
                    |         (1/2)    
                    \   lambda         

              /      (1/2)\    /      (1/2)  \\  
           cos\lambda     / cos\lambda      x/|  
         + -----------------------------------| K
                    (1/2)    /      (1/2)\    |  
              lambda      sin\lambda     /    /  

# The following equation must hold in order that your
# original boundary condition  D(y)(0) = y(0) holds.

eqn := eval(subs(x=0, y(0)=K,ans));

                             /      (1/2)\     
                        K cos\lambda     /     
        eqn := K = ----------------------------
                         (1/2)    /      (1/2)\
                   lambda      sin\lambda     /

spec := [solve(eqn)];

         [                          
         [                          
         [                          
 spec := [{K = 0, lambda = lambda}, 
         [                          

    /                                              2\ ]
    |                      /             (1/2)    \ | ]
   <                       |        /  2\         |  >]
    |K = K, lambda = RootOf\tan(_Z) \_Z /      - 1/ | ]
    \                                               / ]

all_spec := map(allvalues, spec):

# Select only those elements of `all_spec` which satisfy `eqn`.
# (One of the float RootOfs returned by `allvalues` doesn't
# satisfy `eqn` above.

all_spec := select(u->is(simplify(fnormal(evalf[20](eval((rhs-lhs)(eqn),u))),
                                  zero)
                         = 0),
                   all_spec):

map(print, all_spec):

                {K = 0, lambda = lambda}

 /                                                            2\ 
 |                      /             (1/2)                  \ | 
<                       |        /  2\                       |  >
 |K = K, lambda = RootOf\tan(_Z) \_Z /      - 1, 0.8603335890/ | 
 \                                                             / 

 /                                                           2\ 
 |                      /             (1/2)                 \ | 
<                       |        /  2\                      |  >
 |K = K, lambda = RootOf\tan(_Z) \_Z /      - 1, 3.425618459/ | 
 \                                                            / 

# These below are the mentioned solutions, using the above RootOf's
# having so-called "float descriptors", which means they can
# be further approximated to arbitrary precision using `evalf`.

p_sols := {seq([eval(ans, K=eval(K,all_spec[i])),
                [lambda=eval(lambda,all_spec[i])]],
               i=1..nops(all_spec))}:

#map(print, p_sols):

# For fun, now approximate using default precision.

fp_sols := evalf(p_sols):

seq(print(eval(fp_sols[i][1], fp_sols[i][2])), i=1..nops(fp_sols)):

                       y(x) = 0.

        y(x) = (1.162339833 sin(0.8603335890 x)
           + 1.000000000 cos(0.8603335890 x)) K

        y(x) = (0.2919180907 sin(3.425618459 x)
           + 1.000000002 cos(3.425618459 x)) K

如何获得这个二阶边界值特征值ODE的非平凡解？

1 个答案: