求助：求解对称的微分方程组时结果不对称

代码如下：
feq1 = {D[u[t, x], t, t] + D[u[t, x], x, x, x, x] == 0};
feq2 = {D[v[t, x], t, t] + D[v[t, x], x, x, x, x] == 0};
bc1 = {v[t, 0] == 0, Derivative[0, 1][v][t, 0] == 0,
Derivative[0, 2][v][t, 1] == 0,
Derivative[0, 3][v][t, 1] == u[t, 1]};
bc2 = {u[t, 0] == 0, Derivative[0, 1][u][t, 0] == 0,
Derivative[0, 2][u][t, 1] == 0,
Derivative[0, 3][u][t, 1] == If[t < 1, t, 1]};
st1 = {u[0, x] == 0, Derivative[1, 0][v][0, x] == 0};
st2 = {v[0, x] == 0, Derivative[1, 0][u][0, x] == 0};
sol = NDSolve[{feq1, feq2, st1, st2, bc1, bc2}, {v, u}, {t, 0,
10}, {x, 0, 1}];
Plot[{Evaluate[v[t, 1] /. First[sol]],
Evaluate[u[t, 1] /. First[sol]]}, {t, 0, 10}, PlotRange -> {-1, 1}]
此时运行代码，速出的图像如下：

这是不正确的解。
现在如果调换边界条件bc1和bc2如下：
bc1 = {v[t, 0] == 0, Derivative[0, 1][v][t, 0] == 0,
Derivative[0, 2][v][t, 1] == 0,
Derivative[0, 3][v][t, 1] == If[t < 1, t, 1]};
bc2 = {u[t, 0] == 0, Derivative[0, 1][u][t, 0] == 0,
Derivative[0, 2][u][t, 1] == 0,
Derivative[0, 3][u][t, 1] == v[t, 1]};
则输出图像如下：

这是正确的解。
同样对称的方程为何仅仅调换边界条件的顺序就会得到如此不同的解？